Constant-Expectation Counterexamples, Conditional Differences, Affine Stability and Fair Sums
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batch-20261014-advance-20261031-r07-martingale-foundations · 2026-10-14 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
1. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=1>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [1, -1], [-1, 1], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [1, -1], [-1, 1], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
2. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=2>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [2, -2], [-2, 2], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [2, -2], [-2, 2], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
3. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=3>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [3, -3], [-3, 3], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [3, -3], [-3, 3], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
4. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=4>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [4, -4], [-4, 4], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [4, -4], [-4, 4], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
5. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=5>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [5, -5], [-5, 5], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [5, -5], [-5, 5], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
6. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=6>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [6, -6], [-6, 6], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [6, -6], [-6, 6], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
7. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=7>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [7, -7], [-7, 7], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [7, -7], [-7, 7], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
8. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=8>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [8, -8], [-8, 8], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [8, -8], [-8, 8], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
9. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=9>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [9, -9], [-9, 9], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [9, -9], [-9, 9], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
10. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=10>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [10, -10], [-10, 10], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [10, -10], [-10, 10], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
11. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=11>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [11, -11], [-11, 11], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [11, -11], [-11, 11], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
12. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=12>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [12, -12], [-12, 12], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [12, -12], [-12, 12], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
13. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=13>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [13, -13], [-13, 13], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [13, -13], [-13, 13], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
14. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=14>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [14, -14], [-14, 14], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [14, -14], [-14, 14], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
15. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=15>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [15, -15], [-15, 15], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [15, -15], [-15, 15], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
16. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=16>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [16, -16], [-16, 16], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [16, -16], [-16, 16], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
17. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=17>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [17, -17], [-17, 17], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [17, -17], [-17, 17], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
18. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=18>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [18, -18], [-18, 18], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [18, -18], [-18, 18], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
19. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=19>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [19, -19], [-19, 19], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [19, -19], [-19, 19], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
20. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=20>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [20, -20], [-20, 20], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [20, -20], [-20, 20], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
21. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=21>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [21, -21], [-21, 21], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [21, -21], [-21, 21], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
22. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=22>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [22, -22], [-22, 22], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [22, -22], [-22, 22], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
23. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=23>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [23, -23], [-23, 23], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [23, -23], [-23, 23], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
24. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=24>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [24, -24], [-24, 24], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [24, -24], [-24, 24], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
25. Constant Total Expectations Can Fail Conditional Martingale Steps
Use two equiprobable outcomes. X0=0, X1=[a,−a], X2=[−a,a] with amplitude a=25>0. F0 is trivial and F1=F2 is the full information partition revealed by X1. Give (X0,X1,X2,expectations,martingale?,constant_expectations_do_not_imply_conditional_martingale_steps).
View answer and explanation
(0, [25, -25], [-25, 25], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps)
All three total expectations are zero, and the process is adapted. At time 1, however, the conditional expectation of X2 is X2 itself and differs from X1 on both outcomes. This is the specific counterexample, not a claim that every constant-expectation process fails. The exact result, retaining the requested order, is (0, [25, -25], [-25, 25], [0, 0, 0], No, constant_expectations_do_not_imply_conditional_martingale_steps).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-constant-expectation-counterexample · candidate
26. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[1, -1, 2, -2] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [1, -1], 0], [1, [2, -2], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [1, -1], 0], [1, [2, -2], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
27. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[2, -2, 3, -3] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [2, -2], 0], [1, [3, -3], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [2, -2], 0], [1, [3, -3], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
28. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[3, -3, 4, -4] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [3, -3], 0], [1, [4, -4], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [3, -3], 0], [1, [4, -4], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
29. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[4, -4, 5, -5] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [4, -4], 0], [1, [5, -5], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [4, -4], 0], [1, [5, -5], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
30. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[5, -5, 6, -6] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [5, -5], 0], [1, [6, -6], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [5, -5], 0], [1, [6, -6], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
31. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[6, -6, 7, -7] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [6, -6], 0], [1, [7, -7], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [6, -6], 0], [1, [7, -7], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
32. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[7, -7, 8, -8] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [7, -7], 0], [1, [8, -8], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [7, -7], 0], [1, [8, -8], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
33. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[8, -8, 9, -9] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [8, -8], 0], [1, [9, -9], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [8, -8], 0], [1, [9, -9], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
34. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[9, -9, 10, -10] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [9, -9], 0], [1, [10, -10], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [9, -9], 0], [1, [10, -10], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
35. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[10, -10, 11, -11] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [10, -10], 0], [1, [11, -11], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [10, -10], 0], [1, [11, -11], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
36. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[11, -11, 12, -12] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [11, -11], 0], [1, [12, -12], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [11, -11], 0], [1, [12, -12], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
37. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[12, -12, 13, -13] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [12, -12], 0], [1, [13, -13], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [12, -12], 0], [1, [13, -13], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
38. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[13, -13, 14, -14] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [13, -13], 0], [1, [14, -14], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [13, -13], 0], [1, [14, -14], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
39. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[14, -14, 15, -15] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [14, -14], 0], [1, [15, -15], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [14, -14], 0], [1, [15, -15], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
40. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[15, -15, 16, -16] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [15, -15], 0], [1, [16, -16], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [15, -15], 0], [1, [16, -16], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
41. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[16, -16, 17, -17] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [16, -16], 0], [1, [17, -17], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [16, -16], 0], [1, [17, -17], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
42. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[17, -17, 18, -18] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [17, -17], 0], [1, [18, -18], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [17, -17], 0], [1, [18, -18], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
43. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[18, -18, 19, -19] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [18, -18], 0], [1, [19, -19], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [18, -18], 0], [1, [19, -19], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
44. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[19, -19, 20, -20] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [19, -19], 0], [1, [20, -20], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [19, -19], 0], [1, [20, -20], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
45. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[20, -20, 21, -21] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [20, -20], 0], [1, [21, -21], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [20, -20], 0], [1, [21, -21], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
46. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[21, -21, 22, -22] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [21, -21], 0], [1, [22, -22], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [21, -21], 0], [1, [22, -22], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
47. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[22, -22, 23, -23] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [22, -22], 0], [1, [23, -23], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [22, -22], 0], [1, [23, -23], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
48. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[23, -23, 24, -24] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [23, -23], 0], [1, [24, -24], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [23, -23], 0], [1, [24, -24], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
49. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[24, -24, 25, -25] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [24, -24], 0], [1, [25, -25], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [24, -24], 0], [1, [25, -25], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
50. Zero Total Mean Is Not Enough: Check Each Information Block
Use four equiprobable outcomes with aligned increments=[25, -25, 26, -26] and current information block labels=[0, 0, 1, 1]. Give (conditional_rows,all_zero?,classification), with rows [block_label,increments_in_outcome_order,exact_block_mean] sorted by block label. Classification is conditional_mean_zero_difference or difference_fails_martingale_condition.
View answer and explanation
([[0, [25, -25], 0], [1, [26, -26], 0]], Yes, conditional_mean_zero_difference)
Average the increments within every nonempty information block. All block means must be zero; a globally cancelling mean alone does not establish a martingale difference. The exact result, retaining the requested order, is ([[0, [25, -25], 0], [1, [26, -26], 0]], Yes, conditional_mean_zero_difference).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-difference-mean · candidate
51. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=-12 and complete terminal values=[-13, -11, -12, -12]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-2, shift=3. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(-12, [-13, -11, -12, -12], -2, 3, 27, [29, 25, 27, 27], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-12, [-13, -11, -12, -12], -2, 3, 27, [29, 25, 27, 27], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
52. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=-11 and complete terminal values=[-12, -10, -11, -11]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-1, shift=4. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(-11, [-12, -10, -11, -11], -1, 4, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-11, [-12, -10, -11, -11], -1, 4, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
53. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=-10 and complete terminal values=[-11, -9, -10, -10]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=0, shift=5. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(-10, [-11, -9, -10, -10], 0, 5, 5, [5, 5, 5, 5], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-10, [-11, -9, -10, -10], 0, 5, 5, [5, 5, 5, 5], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
54. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=-9 and complete terminal values=[-10, -8, -9, -9]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=1, shift=6. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(-9, [-10, -8, -9, -9], 1, 6, -3, [-4, -2, -3, -3], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-9, [-10, -8, -9, -9], 1, 6, -3, [-4, -2, -3, -3], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
55. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=-8 and complete terminal values=[-9, -7, -8, -8]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=2, shift=7. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(-8, [-9, -7, -8, -8], 2, 7, -9, [-11, -7, -9, -9], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-8, [-9, -7, -8, -8], 2, 7, -9, [-11, -7, -9, -9], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
56. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=-7 and complete terminal values=[-8, -6, -7, -7]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-2, shift=8. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(-7, [-8, -6, -7, -7], -2, 8, 22, [24, 20, 22, 22], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-7, [-8, -6, -7, -7], -2, 8, 22, [24, 20, 22, 22], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
57. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=-6 and complete terminal values=[-7, -5, -6, -6]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-1, shift=9. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(-6, [-7, -5, -6, -6], -1, 9, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-6, [-7, -5, -6, -6], -1, 9, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
58. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=-5 and complete terminal values=[-6, -4, -5, -5]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=0, shift=10. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(-5, [-6, -4, -5, -5], 0, 10, 10, [10, 10, 10, 10], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-5, [-6, -4, -5, -5], 0, 10, 10, [10, 10, 10, 10], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
59. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=-4 and complete terminal values=[-5, -3, -4, -4]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=1, shift=11. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(-4, [-5, -3, -4, -4], 1, 11, 7, [6, 8, 7, 7], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-4, [-5, -3, -4, -4], 1, 11, 7, [6, 8, 7, 7], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
60. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=-3 and complete terminal values=[-4, -2, -3, -3]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=2, shift=12. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(-3, [-4, -2, -3, -3], 2, 12, 6, [4, 8, 6, 6], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-3, [-4, -2, -3, -3], 2, 12, 6, [4, 8, 6, 6], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
61. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=-2 and complete terminal values=[-3, -1, -2, -2]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-2, shift=13. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(-2, [-3, -1, -2, -2], -2, 13, 17, [19, 15, 17, 17], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-2, [-3, -1, -2, -2], -2, 13, 17, [19, 15, 17, 17], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
62. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=-1 and complete terminal values=[-2, 0, -1, -1]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-1, shift=14. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(-1, [-2, 0, -1, -1], -1, 14, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (-1, [-2, 0, -1, -1], -1, 14, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
63. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=0 and complete terminal values=[-1, 1, 0, 0]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=0, shift=15. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(0, [-1, 1, 0, 0], 0, 15, 15, [15, 15, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (0, [-1, 1, 0, 0], 0, 15, 15, [15, 15, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
64. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=1 and complete terminal values=[0, 2, 1, 1]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=1, shift=16. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(1, [0, 2, 1, 1], 1, 16, 17, [16, 18, 17, 17], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (1, [0, 2, 1, 1], 1, 16, 17, [16, 18, 17, 17], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
65. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=2 and complete terminal values=[1, 3, 2, 2]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=2, shift=17. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(2, [1, 3, 2, 2], 2, 17, 21, [19, 23, 21, 21], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (2, [1, 3, 2, 2], 2, 17, 21, [19, 23, 21, 21], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
66. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=3 and complete terminal values=[2, 4, 3, 3]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-2, shift=18. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(3, [2, 4, 3, 3], -2, 18, 12, [14, 10, 12, 12], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (3, [2, 4, 3, 3], -2, 18, 12, [14, 10, 12, 12], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
67. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=4 and complete terminal values=[3, 5, 4, 4]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-1, shift=19. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(4, [3, 5, 4, 4], -1, 19, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (4, [3, 5, 4, 4], -1, 19, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
68. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=5 and complete terminal values=[4, 6, 5, 5]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=0, shift=20. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(5, [4, 6, 5, 5], 0, 20, 20, [20, 20, 20, 20], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (5, [4, 6, 5, 5], 0, 20, 20, [20, 20, 20, 20], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
69. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=6 and complete terminal values=[5, 7, 6, 6]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=1, shift=21. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(6, [5, 7, 6, 6], 1, 21, 27, [26, 28, 27, 27], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (6, [5, 7, 6, 6], 1, 21, 27, [26, 28, 27, 27], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
70. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=7 and complete terminal values=[6, 8, 7, 7]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=2, shift=22. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(7, [6, 8, 7, 7], 2, 22, 36, [34, 38, 36, 36], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (7, [6, 8, 7, 7], 2, 22, 36, [34, 38, 36, 36], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
71. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=8 and complete terminal values=[7, 9, 8, 8]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-2, shift=23. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(8, [7, 9, 8, 8], -2, 23, 7, [9, 5, 7, 7], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (8, [7, 9, 8, 8], -2, 23, 7, [9, 5, 7, 7], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
72. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=9 and complete terminal values=[8, 10, 9, 9]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=-1, shift=24. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(9, [8, 10, 9, 9], -1, 24, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (9, [8, 10, 9, 9], -1, 24, 15, [16, 14, 15, 15], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
73. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=10 and complete terminal values=[9, 11, 10, 10]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=0, shift=25. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(10, [9, 11, 10, 10], 0, 25, 25, [25, 25, 25, 25], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (10, [9, 11, 10, 10], 0, 25, 25, [25, 25, 25, 25], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
74. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=11 and complete terminal values=[10, 12, 11, 11]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=1, shift=26. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(11, [10, 12, 11, 11], 1, 26, 37, [36, 38, 37, 37], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (11, [10, 12, 11, 11], 1, 26, 37, [36, 38, 37, 37], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
75. One Fixed Affine Transformation Preserves the Same-Filtration Martingale
Use the declared equiprobable one-step martingale with constant initial=12 and complete terminal values=[11, 13, 12, 12]. The initial information is trivial and the terminal information reveals these outcomes. Fixed scale=2, shift=27. Give (initial,terminal,scale,shift,transformed_initial,transformed_terminal,martingale?,affine_transform_preserves_same_filtration_martingale).
View answer and explanation
(12, [11, 13, 12, 12], 2, 27, 51, [49, 53, 51, 51], Yes, affine_transform_preserves_same_filtration_martingale)
First require the terminal mean to equal the initial value. Apply the same fixed scale and shift to both times; conditional expectation linearity preserves the martingale under the same filtration. Zero or negative scale is allowed. This does not permit time-dependent or random coefficients without further conditions. The exact result, retaining the requested order, is (12, [11, 13, 12, 12], 2, 27, 51, [49, 53, 51, 51], Yes, affine_transform_preserves_same_filtration_martingale).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-affine-stability · candidate
76. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=1 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(1, 1, [-1, 1], [1, 1], 2^1, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (1, 1, [-1, 1], [1, 1], 2^1, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
77. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=2 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(2, 1, [-2, 0, 2], [1, 2, 1], 2^2, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (2, 1, [-2, 0, 2], [1, 2, 1], 2^2, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
78. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=3 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(3, 1, [-3, -1, 1, 3], [1, 3, 3, 1], 2^3, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (3, 1, [-3, -1, 1, 3], [1, 3, 3, 1], 2^3, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
79. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=4 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(4, 1, [-4, -2, 0, 2, 4], [1, 4, 6, 4, 1], 2^4, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (4, 1, [-4, -2, 0, 2, 4], [1, 4, 6, 4, 1], 2^4, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
80. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=5 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(5, 1, [-5, -3, -1, 1, 3, 5], [1, 5, 10, 10, 5, 1], 2^5, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (5, 1, [-5, -3, -1, 1, 3, 5], [1, 5, 10, 10, 5, 1], 2^5, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
81. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=6 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(6, 1, [-6, -4, -2, 0, 2, 4, 6], [1, 6, 15, 20, 15, 6, 1], 2^6, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (6, 1, [-6, -4, -2, 0, 2, 4, 6], [1, 6, 15, 20, 15, 6, 1], 2^6, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
82. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=7 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(7, 1, [-7, -5, -3, -1, 1, 3, 5, 7], [1, 7, 21, 35, 35, 21, 7, 1], 2^7, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (7, 1, [-7, -5, -3, -1, 1, 3, 5, 7], [1, 7, 21, 35, 35, 21, 7, 1], 2^7, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
83. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=8 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(8, 1, [-8, -6, -4, -2, 0, 2, 4, 6, 8], [1, 8, 28, 56, 70, 56, 28, 8, 1], 2^8, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (8, 1, [-8, -6, -4, -2, 0, 2, 4, 6, 8], [1, 8, 28, 56, 70, 56, 28, 8, 1], 2^8, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
84. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=9 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(9, 1, [-9, -7, -5, -3, -1, 1, 3, 5, 7, 9], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1], 2^9, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (9, 1, [-9, -7, -5, -3, -1, 1, 3, 5, 7, 9], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1], 2^9, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
85. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=10 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(10, 1, [-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10], [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1], 2^10, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (10, 1, [-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10], [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1], 2^10, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
86. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=11 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(11, 1, [-11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11], [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1], 2^11, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (11, 1, [-11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11], [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1], 2^11, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
87. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=12 independent equiprobable signs, each increment ±scale with scale=1>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(12, 1, [-12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12], [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1], 2^12, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (12, 1, [-12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12], [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1], 2^12, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
88. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=1 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(1, 2, [-2, 2], [1, 1], 2^1, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (1, 2, [-2, 2], [1, 1], 2^1, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
89. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=2 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(2, 2, [-4, 0, 4], [1, 2, 1], 2^2, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (2, 2, [-4, 0, 4], [1, 2, 1], 2^2, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
90. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=3 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(3, 2, [-6, -2, 2, 6], [1, 3, 3, 1], 2^3, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (3, 2, [-6, -2, 2, 6], [1, 3, 3, 1], 2^3, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
91. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=4 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(4, 2, [-8, -4, 0, 4, 8], [1, 4, 6, 4, 1], 2^4, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (4, 2, [-8, -4, 0, 4, 8], [1, 4, 6, 4, 1], 2^4, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
92. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=5 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(5, 2, [-10, -6, -2, 2, 6, 10], [1, 5, 10, 10, 5, 1], 2^5, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (5, 2, [-10, -6, -2, 2, 6, 10], [1, 5, 10, 10, 5, 1], 2^5, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
93. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=6 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(6, 2, [-12, -8, -4, 0, 4, 8, 12], [1, 6, 15, 20, 15, 6, 1], 2^6, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (6, 2, [-12, -8, -4, 0, 4, 8, 12], [1, 6, 15, 20, 15, 6, 1], 2^6, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
94. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=7 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(7, 2, [-14, -10, -6, -2, 2, 6, 10, 14], [1, 7, 21, 35, 35, 21, 7, 1], 2^7, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (7, 2, [-14, -10, -6, -2, 2, 6, 10, 14], [1, 7, 21, 35, 35, 21, 7, 1], 2^7, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
95. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=8 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(8, 2, [-16, -12, -8, -4, 0, 4, 8, 12, 16], [1, 8, 28, 56, 70, 56, 28, 8, 1], 2^8, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (8, 2, [-16, -12, -8, -4, 0, 4, 8, 12, 16], [1, 8, 28, 56, 70, 56, 28, 8, 1], 2^8, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
96. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=9 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(9, 2, [-18, -14, -10, -6, -2, 2, 6, 10, 14, 18], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1], 2^9, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (9, 2, [-18, -14, -10, -6, -2, 2, 6, 10, 14, 18], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1], 2^9, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
97. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=10 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(10, 2, [-20, -16, -12, -8, -4, 0, 4, 8, 12, 16, 20], [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1], 2^10, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (10, 2, [-20, -16, -12, -8, -4, 0, 4, 8, 12, 16, 20], [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1], 2^10, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
98. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=11 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(11, 2, [-22, -18, -14, -10, -6, -2, 2, 6, 10, 14, 18, 22], [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1], 2^11, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (11, 2, [-22, -18, -14, -10, -6, -2, 2, 6, 10, 14, 18, 22], [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1], 2^11, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
99. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=12 independent equiprobable signs, each increment ±scale with scale=2>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(12, 2, [-24, -20, -16, -12, -8, -4, 0, 4, 8, 12, 16, 20, 24], [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1], 2^12, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (12, 2, [-24, -20, -16, -12, -8, -4, 0, 4, 8, 12, 16, 20, 24], [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1], 2^12, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
100. A Declared Fair Independent-Increment Model Supplies a Specific Martingale
Use steps=1 independent equiprobable signs, each increment ±scale with scale=3>0, on the full finite sign-product space with its natural revealed-sign filtration. Give (steps,scale,support,multiplicities,total_paths,martingale?,independent_fair_signed_increments_have_conditional_mean_zero). List support by increasing number of positive signs; retain total_paths as exact 2^steps text.
View answer and explanation
(1, 3, [-3, 3], [1, 1], 2^1, Yes, independent_fair_signed_increments_have_conditional_mean_zero)
The terminal value with k positive signs is scale*(2*k−steps), with binomial multiplicity choose(steps,k). Every next increment has conditional mean zero in this explicitly independent fair model. These premises do not follow from an arbitrary martingale, nor do stopping, convergence or Markov theorems follow from this finite calculation. The exact result, retaining the requested order, is (1, 3, [-3, 3], [1, 1], 2^1, Yes, independent_fair_signed_increments_have_conditional_mean_zero).
Reviewed step-by-step derivation
Finite martingale · qf-mod31-martdeep-fair-partial-sum · candidate
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