Goal and scope
Given the support S={(0,0),(0,1),(1,0),(1,1),(2,1)} with masses p(0,0)=1/6, p(0,1)=1/3, p(1,0)=1/4, p(1,1)=1/12, p(2,1)=1/6, verify that (S,p) is a two-dimensional finite joint distribution; from this one joint law compute the marginals of X and Y, the rectangle event P(X∈{1,2}, Y=1) and the single-pair mass P(X=2, Y=0); finally construct a different joint law with the same marginals, showing that two marginals do not in general determine the joint distribution.
Stays inside the MKO: a finite joint law whose support consists of distinct integer pairs with exact rational masses (assumption a1), treated as a mathematical model, not as data fitting or a continuous density (assumption a2). The guide validates one concrete table and proves the product-law lemma in general; it does not treat independence, covariance or conditional expectation (later nodes of the MKO), and does not change the canonical MKO or its English edition.
Main derivation
S01 · definition
Before
\text{table }(S,p)After
\text{(i) distinct pairs, (ii) } p(s)\ge 0 \text{ exact, (iii) } \sum_{s\in S}p(s)=1Conditions: none
Rule: mko_joint_def · Uses: G1, G2
Leaf basis: mko_joint_def + def_unfold
By the MKO, a table is a two-dimensional finite joint distribution only if it meets three requirements at once: (i) the listed pairs are distinct; (ii) every mass is a nonnegative exact number; (iii) the masses sum to exactly 1, not approximately 1. Each requirement is checked below, and no marginal or event is computed before all three pass.
S02 · implies
Before
(0,0),\ (0,1),\ (1,0),\ (1,1),\ (2,1)
After
\text{five pairwise distinct pairs}Conditions: none
Rule: cartesian_product_def · Uses: S01, G1, cartesian_product_def
Leaf basis: cartesian_product_def + case_analysis
Two ordered pairs are equal only if both coordinates agree. Group by first coordinate: two pairs with different first coordinates differ; with first coordinate 0, (0,0) and (0,1) differ in the second coordinate; with first coordinate 1, (1,0) and (1,1) differ in the second coordinate; first coordinate 2 has only (2,1). So all C(5,2)=10 comparisons give 'not equal', and requirement (i) holds.
S03 · implies
Before
p=\tfrac16,\ \tfrac13,\ \tfrac14,\ \tfrac1{12},\ \tfrac16After
p(s)>0\ \text{for every } s\in SConditions: none
Rule: order_basics · Uses: S01, G1
Leaf basis: order_basics + frac_rules
Each mass is a quotient of two positive integers, so it is a positive exact rational and in particular nonnegative. Requirement (ii) holds, and exact rationals also meet MKO assumption a1.
S04 · equivalent
Before
\sum_{s\in S}p(s)=\tfrac16+\tfrac13+\tfrac14+\tfrac1{12}+\tfrac16After
\sum_{s\in S}p(s)=1Conditions: none
Rule: frac_rules · Uses: S01, G1
Leaf basis: frac_rules + numeric_evaluation
Compute the total with exact fraction arithmetic (details in S04.1 and S04.2): the result is exactly 1, so requirement (iii) holds. Rounded decimals may not be used instead.
S05 · implies
Before
\text{(i), (ii), (iii) checked}After
(S,p)\ \text{is a finite joint law}Conditions: none
Rule: and_intro_elim · Uses: S01, S02, S03, S04
Leaf basis: mko_joint_def + and_intro_elim
The three requirements hold by S02, S03 and S04; combining them, by the definition in S01, (S,p) is a two-dimensional finite joint distribution. From now on any event probability may be aggregated from this one table.
S06 · definition
Before
C\subseteq\mathbb{Z}^2After
P((X,Y)\in C)=\sum_{s\in S\cap C}p(s)Conditions: (S,p)\ \text{valid}
Rule: joint_event_sum · Uses: S05, G2, joint_event_sum
Leaf basis: joint_event_sum + prob_axioms
A valid joint law determines every event probability: add the masses of the support pairs inside the event; pairs outside S contribute 0. This is additivity over disjoint single-pair events. For example C=ℤ² gives the total 1 of S04, in line with P(Ω)=1.
S07 · implies
Before
p_X(x)=\sum_{y:(x,y)\in S}p(x,y)After
p_X(0)=\tfrac12,\ p_X(1)=\tfrac13,\ p_X(2)=\tfrac16
Conditions: none
Rule: marginal_def · Uses: S06, marginal_def, G1
Leaf basis: marginal_def + joint_event_sum + frac_rules
The marginal of X is the probability of the event {X=x}={x}×ℤ; by S06 add the masses of the support pairs in that event. First list the pairs for each x (S07.1), then add (S07.2).
S08 · equivalent
Before
\sum_{x}p_X(x)=\sum_{x}\sum_{y:(x,y)\in S}p(x,y)After
\sum_{x}p_X(x)=\sum_{s\in S}p(s)=1Conditions: \text{the fibres }S\cap(\{x\}\times\mathbb{Z})\text{ partition }S
Rule: finite_sum_regroup · Uses: S04, S07.1, S07.2, finite_sum_regroup
Leaf basis: finite_sum_regroup + numeric_evaluation
The three sets of S07.1 partition S, so by regrouping the finite sum, adding the three marginal values gives the total mass of the support, which is 1. This holds for every valid finite joint law, not only this table; the numeric check here is 1/2+1/3+1/6=1.
S09 · implies
Before
p_Y(y)=\sum_{x:(x,y)\in S}p(x,y)After
p_Y(0)=\tfrac{5}{12},\ p_Y(1)=\tfrac{7}{12}Conditions: none
Rule: marginal_def · Uses: S06, marginal_def, G1
Leaf basis: marginal_def + joint_event_sum + frac_rules
Group by second coordinate: the pairs with Y=0 are (0,0) and (1,0), with total 2/12+3/12=5/12; the pairs with Y=1 are (0,1), (1,1) and (2,1), with total 4/12+1/12+2/12=7/12. They add to 1, consistent with regrouping as in S08.
S10 · implies
Before
P(X\in A,\ Y\in B)=P((X,Y)\in A\times B)
After
P(X\in A,\ Y\in B)=\tfrac14
Conditions: none
Rule: joint_event_sum · Uses: S06, G3
Leaf basis: joint_event_sum + cartesian_product_def
A rectangle event requires both coordinates to fall in their sets at the same time, so it is the event that (X,Y) belongs to A×B. By S06, first find S∩(A×B) (S10.1), then add the masses (S10.2).
S11 · implies
Before
P(X\in A)\,P(Y\in B)=\left(\tfrac13+\tfrac16\right)\tfrac{7}{12}After
\tfrac{7}{24}\neq\tfrac{6}{24}=P(X\in A,\ Y\in B)Conditions: none
Rule: numeric_evaluation · Uses: S07, S09, S10
Leaf basis: numeric_evaluation + frac_rules + prob_axioms
P(X∈A)=p_X(1)+p_X(2)=1/3+1/6=1/2 (the events {X=1} and {X=2} are disjoint) and P(Y∈B)=p_Y(1)=7/12, with product 7/24; but S10 gives the rectangle probability 1/4=6/24. So for this joint law the rectangle probability is not the product of the marginal probabilities and must be aggregated from the joint table itself.
S12 · implies
Before
P(X=2,\ Y=0)
After
0\neq\tfrac{5}{72}=p_X(2)\,p_Y(0)Conditions: none
Rule: joint_event_sum · Uses: S06, S07, S09, G1
Leaf basis: joint_event_sum + finite_sum_rules + frac_rules
The only pair of S with first coordinate 2 is (2,1), so (2,0)∉S; the event {(2,0)} contains no support pair and by S06 its probability is the empty sum 0. On the other hand p_X(2)p_Y(0)=(1/6)(5/12)=5/72>0. A single-pair mass must be read from the joint law, not obtained by multiplying marginals.
S13 · theorem_application
Before
p_X\ \text{on}\ \{0,1,2\},\ p_Y\ \text{on}\ \{0,1\}After
Q(x,y)=p_X(x)\,p_Y(y)\ \text{is a finite joint law with marginals}\ p_X,\ p_YConditions: p_X\ge 0,\ \sum_x p_X(x)=1; p_Y\ge 0,\ \sum_y p_Y(y)=1
Rule: L_product_law · Uses: S07, S08, S09
Leaf basis: L_product_law + numeric_evaluation
Apply Lemma L_product_law with S_X={0,1,2} and S_Y={0,1}; its hypotheses come from S07, S08 and S09. The six masses of Q are Q(0,0)=5/24, Q(0,1)=7/24, Q(1,0)=5/36, Q(1,1)=7/36, Q(2,0)=5/72 and Q(2,1)=7/72. The lemma itself is proved in general in S13.1 to S13.3.
S14 · implies
Before
J(2,0)=0,\ Q(2,0)=\tfrac{5}{72}After
J\neq Q\ \text{while}\ J,\ Q\ \text{have the same marginals}Conditions: none
Rule: function_def · Uses: S07, S09, S12, S13
Leaf basis: function_def + exists_intro + quantifier_negation
Regard both joint laws as mass functions on ℤ² (zero off the support). The original table J has mass 0 at (2,0) (S12), while Q has mass 5/72 there; the two functions differ at a point, so J≠Q. Yet both have the marginals p_X and p_Y (J by S07 and S09, Q by S13). One counterexample refutes the universal claim that two marginals determine the joint distribution.
Substeps and lemmas
S04.1 · equivalent
Before
\tfrac16+\tfrac13+\tfrac14+\tfrac1{12}+\tfrac16After
\tfrac{2}{12}+\tfrac{4}{12}+\tfrac{3}{12}+\tfrac{1}{12}+\tfrac{2}{12}Conditions: k\neq 0
Rule: frac_rules · Uses: G1
Leaf basis: frac_rules
Use a/b=(ak)/(bk) for k≠0 to give every fraction the denominator 12: k=2 for 1/6, k=4 for 1/3, k=3 for 1/4 and k=1 for 1/12. No term changes its value.
S04.2 · equivalent
Before
\tfrac{2}{12}+\tfrac{4}{12}+\tfrac{3}{12}+\tfrac{1}{12}+\tfrac{2}{12}After
\tfrac{12}{12}=1Conditions: 12\neq 0
Rule: frac_rules · Uses: S04.1
Leaf basis: frac_rules + numeric_evaluation
Add over the common denominator, a/c+b/c=(a+b)/c: the numerators give 2+4+3+1+2=12, so the total is 12/12=1.
S07.1 · equivalent
Before
S\cap(\{x\}\times\mathbb{Z}),\ x=0,1,2After
\{(0,0),(0,1)\},\ \{(1,0),(1,1)\},\ \{(2,1)\}Conditions: none
Rule: set_operations_def · Uses: G1, set_operations_def
Leaf basis: set_operations_def + set_extensionality
Check the first coordinate of each of the five pairs of S and put it into the matching set. Each pair lies in exactly one set; the three sets are pairwise disjoint and their union is S.
S07.2 · equivalent
Before
p(0,0)+p(0,1),\ p(1,0)+p(1,1),\ p(2,1)
After
\tfrac12,\ \tfrac13,\ \tfrac16
Conditions: none
Rule: frac_rules · Uses: S07.1, G1
Leaf basis: frac_rules + numeric_evaluation
1/6+1/3=1/6+2/6=3/6=1/2; 1/4+1/12=3/12+1/12=4/12=1/3; p(2,1)=1/6.
S10.1 · equivalent
Before
S\cap(A\times B)
After
\{(1,1),\ (2,1)\}Conditions: none
Rule: cartesian_product_def · Uses: G1, G3, cartesian_product_def
Leaf basis: cartesian_product_def + set_operations_def + set_extensionality
Check each pair: (0,0) and (0,1) have x=0∉A; (1,0) has y=0∉B; (1,1) has 1∈A and 1∈B; (2,1) has 2∈A and 1∈B. So the intersection is exactly {(1,1),(2,1)}.
S10.2 · equivalent
Before
p(1,1)+p(2,1)
After
\tfrac{1}{12}+\tfrac{2}{12}=\tfrac14Conditions: none
Rule: frac_rules · Uses: S10.1, G1
Leaf basis: frac_rules + numeric_evaluation
1/12+1/6=1/12+2/12=3/12=1/4.
S13.1 · implies
Before
Q(x,y)=p_X(x)\,p_Y(y),\ (x,y)\in S_X\times S_Y
After
\text{distinct pairs},\ Q(x,y)\ge 0\ \text{exact rational}Conditions: p_X(x)\ge 0; p_Y(y)\ge 0
Rule: cartesian_product_def · Uses: cartesian_product_def, order_basics
Leaf basis: cartesian_product_def + order_basics + frac_rules
The elements of S_X×S_Y are listed by the choice of first and second coordinate, and each choice gives a different ordered pair, so the pairs are distinct. The product of two nonnegative exact rationals is again a nonnegative exact rational. This gives requirements (i) and (ii) of a joint law.
S13.2 · equivalent
Before
\sum_{x\in S_X}\sum_{y\in S_Y}p_X(x)\,p_Y(y)After
\sum_{x\in S_X}p_X(x)\cdot\sum_{y\in S_Y}p_Y(y)=1\cdot 1=1Conditions: \sum_x p_X(x)=1; \sum_y p_Y(y)=1
Rule: distributive · Uses: S13.1, finite_sum_rules, distributive, finite_sum_regroup
Leaf basis: distributive + finite_sum_rules + finite_sum_regroup + mul_one
Regroup the sum over S_X×S_Y by first coordinate into a double sum; in the inner sum Σ_y p_X(x)p_Y(y) the factor p_X(x) is constant and comes out, giving p_X(x)Σ_y p_Y(y)=p_X(x)·1; the outer sum then gives Σ_x p_X(x)=1. Requirement (iii) holds.
S13.3 · equivalent
Before
\sum_{y\in S_Y}Q(x,y),\ \sum_{x\in S_X}Q(x,y)After
p_X(x)\sum_{y}p_Y(y)=p_X(x),\ p_Y(y)\sum_{x}p_X(x)=p_Y(y)Conditions: \sum_x p_X(x)=1; \sum_y p_Y(y)=1
Rule: finite_sum_rules · Uses: S13.1, finite_sum_rules, marginal_def
Leaf basis: finite_sum_rules + mul_one + marginal_def
The X-marginal of Q fixes x and sums over y: the constant p_X(x) comes out and the remaining Σ_y p_Y(y) is 1, so it equals p_X(x); the Y-marginal equals p_Y(y) in the same way. S13.1 to S13.3 together prove Lemma L_product_law.