{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-conditional-expectation-01", "mko_id": "mko-conditional-expectation", "title": { "zh-Hant": "有限條件期望：由分割生成的 σ 代數上求 E[X|G]，含零質量區塊", "en": "Finite conditional expectation: computing E[X|G] for a partition-generated σ-algebra, including a zero-mass block" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-conditional-expectation-01", "source": { "mko_id": "mko-conditional-expectation", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "35d7f871dac286e45d4661dcffbafe25a02cca5c55a4c58ef60bd41d2408bd95", "english_sha256": "859840df1ea5def7705f723675b947d841b733e899bda95980bcd58b6f62644c" }, "mode": "worked_example", "audience_prerequisites": [ "prob_axioms", "finite_sum_rules", "set_operations_def", "frac_rules", "mul_inverse_nonzero" ], "goal": { "zh-Hant": "在 Ω={0,1,2,3,4} 上，質量依序為 1/8、3/8、1/4、1/4、0，X 的值依序為 4、0、2、6、10，G 由分割 {0,1}、{2,3}、{4} 生成。由條件期望的兩項定義（G 可測、在每個 G 事件上積分相同）求出 E[X|G] 的全部版本，證明它在兩個正質量區塊上必為 1 與 4，在零質量區塊 {4} 上可取任意值（只幾乎處處決定），並說明逐點條件化、只核總期望與不加權平均為何都錯。", "en": "On Ω={0,1,2,3,4} with masses 1/8, 3/8, 1/4, 1/4, 0 and X taking the values 4, 0, 2, 6, 10 in order, let G be generated by the partition {0,1}, {2,3}, {4}. From the two defining conditions of conditional expectation (G-measurable, same integral on every event of G), find all versions of E[X|G]: prove they must equal 1 and 4 on the two positive-mass blocks and may take any value on the zero-mass block {4} (determined only almost surely), and explain why pointwise conditioning, checking only the total expectation, and the unweighted mean are all wrong." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "Ω={0,1,2,3,4}；P({0})=1/8、P({1})=3/8、P({2})=1/4、P({3})=1/4、P({4})=0（總和 1）；X(0)=4、X(1)=0、X(2)=2、X(3)=6、X(4)=10；全部屬於同一空間（MKO 假設 a1）。", "en": "Ω={0,1,2,3,4}; P({0})=1/8, P({1})=3/8, P({2})=1/4, P({3})=1/4, P({4})=0 (summing to 1); X(0)=4, X(1)=0, X(2)=2, X(3)=6, X(4)=10; all on the same space (MKO assumption a1)." } }, { "id": "G2", "statement": { "zh-Hant": "條件分割 B_1={0,1}、B_2={2,3}、B_3={4}；G=σ(B_1,B_2,B_3)。", "en": "The conditioning partition B_1={0,1}, B_2={2,3}, B_3={4}; G=σ(B_1,B_2,B_3)." } }, { "id": "G3", "statement": { "zh-Hant": "定義：Y 是 E[X|G] 的一個版本，若 (i) Y 為 G 可測，且 (ii) 對每個 A∈G，E[Y𝟏_A]=E[X𝟏_A]，其中 E[Z𝟏_A]=Σ_{ω∈A} Z(ω)P({ω})。", "en": "Definition: Y is a version of E[X|G] if (i) Y is G-measurable and (ii) E[Y𝟏_A]=E[X𝟏_A] for every A∈G, where E[Z𝟏_A]=Σ_{ω∈A} Z(ω)P({ω})." } } ], "steps": [ { "id": "S01", "uses": [ "G2", "sigma_partition" ], "before_tex": "\\mathcal{G}=\\sigma(B_1,B_2,B_3)", "after_tex": "\\mathcal{G}=\\{\\varnothing,B_1,B_2,B_3,B_1\\cup B_2,B_1\\cup B_3,B_2\\cup B_3,\\Omega\\}", "relation": "equivalent", "rule": "sigma_partition", "conditions": [], "explanation": { "zh-Hant": "分割生成的 σ 代數恰為區塊的聯集。三個區塊各選或不選，得 2^3=8 個事件（N9），列出如右。事件 {0} 不在 G 中：它不是區塊的聯集。", "en": "The σ-algebra generated by a partition consists exactly of the unions of blocks. Each of the three blocks is either included or not, giving 2^3=8 events (N9), listed on the right. The event {0} is not in G: it is not a union of blocks." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "sigma_partition", "set_operations_def" ], "explanation": { "zh-Hant": "列出的 8 個集合正是區塊的全部聯集，而 sigma_partition 說 G 恰由它們組成；兩種描述互相決定。", "en": "The 8 listed sets are exactly all unions of blocks, and sigma_partition says G consists of exactly these; the two descriptions determine each other." } }, "leaf_basis": "sigma_partition + set_operations_def + numeric_evaluation" }, { "id": "S02", "uses": [ "S01", "G3" ], "before_tex": "Y\\ \\mathcal{G}\\text{-measurable}", "after_tex": "Y=c_1\\ \\text{on}\\ B_1,\\ Y=c_2\\ \\text{on}\\ B_2,\\ Y=c_3\\ \\text{on}\\ B_3", "relation": "equivalent", "rule": "L_meas_const", "conditions": [], "explanation": { "zh-Hant": "依引理 L_meas_const（在 S02.1、S02.2 證明），條件 (i) 恰好表示 Y 在每個區塊上為常數，記其值為 c_1、c_2、c_3。可測性是集合層面的條件，與質量無關，所以零質量區塊 B_3 上 Y 也必須是常數。", "en": "By Lemma L_meas_const (proved in S02.1 and S02.2), condition (i) says exactly that Y is constant on each block; call the values c_1, c_2, c_3. Measurability is a set-level condition independent of the masses, so Y must be constant on the zero-mass block B_3 as well." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_meas_const" ], "explanation": { "zh-Hant": "引理的兩個方向都在子步證明。", "en": "Both directions of the lemma are proved in the sub-steps." } }, "leaf_basis": "meas_def + sigma_partition + set_operations_def" }, { "id": "S02.1", "uses": [ "meas_def", "sigma_partition" ], "before_tex": "Y=c_j\\ \\text{on}\\ B_j\\ (j=1,2,3)", "after_tex": "Y^{-1}(S)=\\bigcup_{j:\\,c_j\\in S}B_j\\in\\mathcal{G}", "relation": "implies", "rule": "meas_def", "conditions": [], "explanation": { "zh-Hant": "（⇐）若 Y 在 B_j 上恆為 c_j，則對任何實數集合 S，Y^{-1}(S) 恰是那些 c_j∈S 的區塊之聯集，因此屬於 G，Y 為 G 可測。", "en": "(⇐) If Y equals c_j throughout B_j, then for any set S of reals, Y^{-1}(S) is exactly the union of the blocks with c_j∈S, so it lies in G and Y is G-measurable." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "meas_def", "sigma_partition" ], "explanation": { "zh-Hant": "只證一個方向；另一方向在 S02.2。", "en": "Only one direction is proved here; the other is S02.2." } }, "leaf_basis": "image_preimage_def + sigma_partition + set_operations_def" }, { "id": "S02.2", "uses": [ "meas_def", "sigma_partition" ], "before_tex": "Y\\ \\mathcal{G}\\text{-measurable},\\ \\omega\\in B_j,\\ y=Y(\\omega)", "after_tex": "Y=y\\ \\text{on all of}\\ B_j", "relation": "implies", "rule": "meas_def", "conditions": [], "explanation": { "zh-Hant": "（⇒）Y^{-1}({y})∈G，所以它是若干區塊的聯集；它含 ω，而 ω 只屬於區塊 B_j，因此 B_j 是組成這個聯集的區塊之一，B_j⊆Y^{-1}({y})，即 Y 在 B_j 上恆為 y。S02.1 與 S02.2 合起來證得引理 L_meas_const。", "en": "(⇒) Y^{-1}({y})∈G, so it is a union of blocks; it contains ω, and ω belongs only to the block B_j, so B_j is one of the blocks in that union and B_j⊆Y^{-1}({y}), i.e. Y equals y throughout B_j. S02.1 and S02.2 together prove Lemma L_meas_const." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "meas_def", "sigma_partition" ], "explanation": { "zh-Hant": "關鍵是區塊互斥：ω 恰屬於一個區塊，所以包含 ω 的區塊聯集必含整個 B_j。", "en": "The key is that blocks are disjoint: ω lies in exactly one block, so a union of blocks containing ω must contain all of B_j." } }, "leaf_basis": "image_preimage_def + sigma_partition + subset_def" }, { "id": "S03", "uses": [ "S01", "G3" ], "before_tex": "\\forall A\\in\\mathcal{G}:\\ E[Y\\mathbf{1}_A]=E[X\\mathbf{1}_A]", "after_tex": "\\forall j:\\ E[Y\\mathbf{1}_{B_j}]=E[X\\mathbf{1}_{B_j}]", "relation": "equivalent", "rule": "L_block_reduce", "conditions": [], "explanation": { "zh-Hant": "依引理 L_block_reduce（在 S03.1、S03.2 證明），8 個事件上的條件 (ii) 化為 3 個區塊上的條件。注意化約後仍是 3 個條件，不是 1 個：只核 A=Ω（總期望）不夠。", "en": "By Lemma L_block_reduce (proved in S03.1 and S03.2), condition (ii) on the 8 events reduces to the condition on the 3 blocks. Note that 3 conditions remain, not 1: checking only A=Ω (the total expectation) is not enough." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_block_reduce" ], "explanation": { "zh-Hant": "兩個方向都在子步證明。", "en": "Both directions are proved in the sub-steps." } }, "leaf_basis": "finite_sum_rules + sigma_partition + forall_elim" }, { "id": "S03.1", "uses": [ "S01" ], "before_tex": "\\forall A\\in\\mathcal{G}:\\ E[Y\\mathbf{1}_A]=E[X\\mathbf{1}_A]", "after_tex": "\\forall j:\\ E[Y\\mathbf{1}_{B_j}]=E[X\\mathbf{1}_{B_j}]", "relation": "implies", "rule": "forall_elim", "conditions": [], "explanation": { "zh-Hant": "（⇒）每個區塊 B_j 都屬於 G（S01），所以全稱條件的特例給出區塊上的等式。", "en": "(⇒) Each block B_j belongs to G (S01), so instances of the universal condition give the equalities on the blocks." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S01", "forall_elim" ], "explanation": { "zh-Hant": "只需要區塊在 G 中。", "en": "It needs only that the blocks lie in G." } }, "leaf_basis": "forall_elim" }, { "id": "S03.2", "uses": [ "S01", "finite_sum_rules" ], "before_tex": "\\forall j:\\ E[Y\\mathbf{1}_{B_j}]=E[X\\mathbf{1}_{B_j}]", "after_tex": "\\forall A\\in\\mathcal{G}:\\ E[Y\\mathbf{1}_A]=E[X\\mathbf{1}_A]", "relation": "implies", "rule": "finite_sum_rules", "conditions": [], "explanation": { "zh-Hant": "（⇐）任一 A∈G 是互斥區塊的聯集，所以 Σ_{ω∈A} 可按區塊拆開：E[Z𝟏_A]=Σ_{B_j⊆A} E[Z𝟏_{B_j}]，對 Z=X 與 Z=Y 都成立；右邊逐項相等，所以左邊相等。A=∅ 時兩邊都是空和 0。S03.1 與 S03.2 合起來證得引理 L_block_reduce。", "en": "(⇐) Any A∈G is a union of disjoint blocks, so the sum over ω∈A splits by blocks: E[Z𝟏_A]=Σ_{B_j⊆A} E[Z𝟏_{B_j}], for Z=X and for Z=Y; the right sides agree term by term, so the left sides agree. For A=∅ both sides are the empty sum 0. S03.1 and S03.2 together prove Lemma L_block_reduce." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S01", "finite_sum_rules", "set_operations_def" ], "explanation": { "zh-Hant": "需要區塊互斥，才能把和拆開而不重複計算。", "en": "It needs the blocks to be disjoint, so that the sum splits without double counting." } }, "leaf_basis": "finite_sum_rules + set_operations_def + sigma_partition" }, { "id": "S04", "uses": [ "S02", "S03", "G3" ], "before_tex": "E[Y\\mathbf{1}_{B}]=E[X\\mathbf{1}_{B}],\\ Y=c\\ \\text{on}\\ B,\\ P(B)>0", "after_tex": "c=\\frac{E[X\\mathbf{1}_B]}{P(B)}", "relation": "equivalent", "rule": "mul_inverse_nonzero", "conditions": [ "P(B)>0" ], "explanation": { "zh-Hant": "Y 在 B 上恆為 c，所以 E[Y𝟏_B]=Σ_{ω∈B} c·P({ω})=c·P(B)（分配律）。條件變成 c·P(B)=E[X𝟏_B]。若 P(B)>0，兩邊同乘 1/P(B) 得 c=E[X𝟏_B]/P(B)：區塊上的質量加權平均。若 P(B)=0，不能除，見 S07。", "en": "Y equals c on B, so E[Y𝟏_B]=Σ_{ω∈B} c·P({ω})=c·P(B) (distributivity). The condition becomes c·P(B)=E[X𝟏_B]. If P(B)>0, multiplying both sides by 1/P(B) gives c=E[X𝟏_B]/P(B): the mass-weighted mean over the block. If P(B)=0 no division is possible; see S07." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mul_inverse_nonzero", "distributive" ], "explanation": { "zh-Hant": "乘以 1/P(B) 與乘以 P(B) 互為反運算，前提 P(B)≠0；所以在正質量區塊上 c 唯一。", "en": "Multiplying by 1/P(B) and by P(B) undo each other, given P(B)≠0; so on a positive-mass block c is unique." } }, "leaf_basis": "distributive + finite_sum_rules + mul_inverse_nonzero" }, { "id": "S05", "uses": [ "S04", "G1", "G2" ], "before_tex": "B_1=\\{0,1\\}", "after_tex": "P(B_1)=\\tfrac12,\\ E[X\\mathbf{1}_{B_1}]=\\tfrac12,\\ c_1=1", "relation": "implies", "rule": "numeric_evaluation", "conditions": [ "P(B_1)>0" ], "explanation": { "zh-Hant": "P(B_1)=1/8+3/8=1/2（N1）；E[X𝟏_{B_1}]=4·1/8+0·3/8=1/2（N2）；P(B_1)>0，依 S04，c_1=(1/2)/(1/2)=1（N3）。", "en": "P(B_1)=1/8+3/8=1/2 (N1); E[X𝟏_{B_1}]=4·1/8+0·3/8=1/2 (N2); P(B_1)>0, so by S04, c_1=(1/2)/(1/2)=1 (N3)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "G1", "numeric_evaluation" ], "explanation": { "zh-Hant": "可重算；反向核對 1·(1/2)=1/2=E[X𝟏_{B_1}]。", "en": "Recomputable; backward check 1·(1/2)=1/2=E[X𝟏_{B_1}]." } }, "leaf_basis": "numeric_evaluation + frac_rules + prob_axioms" }, { "id": "S06", "uses": [ "S04", "G1", "G2" ], "before_tex": "B_2=\\{2,3\\}", "after_tex": "P(B_2)=\\tfrac12,\\ E[X\\mathbf{1}_{B_2}]=2,\\ c_2=4", "relation": "implies", "rule": "numeric_evaluation", "conditions": [ "P(B_2)>0" ], "explanation": { "zh-Hant": "P(B_2)=1/4+1/4=1/2；E[X𝟏_{B_2}]=2·1/4+6·1/4=2（N4）；依 S04，c_2=2/(1/2)=4（N5）。", "en": "P(B_2)=1/4+1/4=1/2; E[X𝟏_{B_2}]=2·1/4+6·1/4=2 (N4); by S04, c_2=2/(1/2)=4 (N5)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "G1", "numeric_evaluation" ], "explanation": { "zh-Hant": "反向核對 4·(1/2)=2=E[X𝟏_{B_2}]。", "en": "Backward check 4·(1/2)=2=E[X𝟏_{B_2}]." } }, "leaf_basis": "numeric_evaluation + frac_rules + prob_axioms" }, { "id": "S07", "uses": [ "S02", "S03", "G1", "G2" ], "before_tex": "B_3=\\{4\\},\\ P(B_3)=0", "after_tex": "c_3\\cdot 0=10\\cdot 0\\ \\text{for every}\\ c_3\\in\\mathbb{R}", "relation": "implies", "rule": "mul_zero", "conditions": [ "P(B_3)=0" ], "explanation": { "zh-Hant": "在零質量區塊上，E[Y𝟏_{B_3}]=c_3·0=0，E[X𝟏_{B_3}]=10·0=0，所以條件 (ii) 對每個實數 c_3 都成立。這裡不做除法：E[X𝟏_{B_3}]/P(B_3)=0/0 沒有定義。c_3 只需讓 Y 在 B_3 上為常數（S02），其值不受限制。", "en": "On the zero-mass block, E[Y𝟏_{B_3}]=c_3·0=0 and E[X𝟏_{B_3}]=10·0=0, so condition (ii) holds for every real c_3. No division is performed: E[X𝟏_{B_3}]/P(B_3)=0/0 is undefined. c_3 only has to make Y constant on B_3 (S02); its value is unrestricted." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "mul_zero", "mul_inverse_nonzero" ], "explanation": { "zh-Hant": "S04 的等價需要 P(B)≠0，這裡不成立，所以不能沿用 S04 的公式；只用到 u·0=0。", "en": "The equivalence of S04 needs P(B)≠0, which fails here, so S04's formula cannot be reused; only u·0=0 is used." } }, "leaf_basis": "mul_zero + finite_sum_rules" }, { "id": "S08", "uses": [ "S02", "S03", "S04", "S05", "S06", "S07" ], "before_tex": "c_1=1,\\ c_2=4,\\ c_3\\ \\text{arbitrary}", "after_tex": "E[X|\\mathcal{G}]=Y_c,\\ Y_c=\\begin{cases}1&\\text{on }B_1\\\\4&\\text{on }B_2\\\\c&\\text{on }B_3\\end{cases},\\ c\\in\\mathbb{R}", "relation": "equivalent", "rule": "mko_condexp_def", "conditions": [], "explanation": { "zh-Hant": "由 S02 與 S03，Y 是一個版本當且僅當它在區塊上為常數且三個區塊條件都成立；由 S05、S06（以 S04 的等價），c_1=1、c_2=4 是唯一可能；由 S07，c_3 任意。所以版本恰為 Y_c（c∈ℝ）。任兩個版本只在 B_3 上可能不同，而 P(B_3)=0，所以它們幾乎處處相等：E[X|G] 在正質量區塊上精確決定，在零質量區塊上只幾乎處處決定。", "en": "By S02 and S03, Y is a version exactly when it is constant on the blocks and all three block conditions hold; by S05 and S06 (through the equivalence of S04), c_1=1 and c_2=4 are the only possibilities; by S07, c_3 is arbitrary. So the versions are exactly the Y_c with c∈ℝ. Any two versions can differ only on B_3, and P(B_3)=0, so they agree almost surely: E[X|G] is determined exactly on the positive-mass blocks and only almost surely on the zero-mass block." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "S02", "S03", "S04", "S07" ], "explanation": { "zh-Hant": "⇒：任何版本都必須是某個 Y_c（S02、S04）。⇐：每個 Y_c 都滿足 (i)、(ii)（S02、S03、S05 到 S07）。兩個方向都有理由。", "en": "⇒: every version must be some Y_c (S02, S04). ⇐: every Y_c satisfies (i) and (ii) (S02, S03, S05 to S07). Both directions have reasons." } }, "leaf_basis": "and_intro_elim + iff_intro + def_unfold" }, { "id": "S09", "uses": [ "S08", "S01", "G1" ], "before_tex": "A=B_1\\cup B_2,\\ A=\\Omega", "after_tex": "E[Y_c\\mathbf{1}_A]=E[X\\mathbf{1}_A]=\\tfrac52", "relation": "implies", "rule": "finite_sum_rules", "conditions": [], "explanation": { "zh-Hant": "反向核對兩個非區塊事件：A=B_1∪B_2 時 E[Y_c𝟏_A]=1·1/2+4·1/2=5/2，E[X𝟏_A]=4·1/8+0·3/8+2·1/4+6·1/4=5/2；A=Ω 時再加上 B_3 的 c·0=0 與 10·0=0，兩邊仍為 5/2（N6）。這與 S03 的化約一致，但只是兩個事件的核對，證明由 S03 給出。", "en": "Backward check on two non-block events: for A=B_1∪B_2, E[Y_c𝟏_A]=1·1/2+4·1/2=5/2 and E[X𝟏_A]=4·1/8+0·3/8+2·1/4+6·1/4=5/2; for A=Ω, the contributions c·0=0 and 10·0=0 from B_3 are added and both sides stay 5/2 (N6). This agrees with the reduction of S03, but it is a check on two events only; the proof is S03." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S08", "S03", "numeric_evaluation" ], "explanation": { "zh-Hant": "數值可重算；這一步不宣稱取代 S03 對全部 8 個事件的論證。", "en": "The numbers can be recomputed; this step does not claim to replace the argument of S03 for all 8 events." } }, "leaf_basis": "finite_sum_rules + numeric_evaluation" }, { "id": "S10", "uses": [ "S02", "G1" ], "before_tex": "Z=X", "after_tex": "Z\\ \\text{not}\\ \\mathcal{G}\\text{-measurable}", "relation": "implies", "rule": "L_meas_const", "conditions": [], "explanation": { "zh-Hant": "對照（逐點條件化）：把 X 本身當作答案。X(0)=4≠0=X(1)，而 0、1 在同一區塊 B_1，所以 X 在 B_1 上不是常數；依 L_meas_const，X 不是 G 可測，(i) 失敗。G 只知道「落在哪個區塊」，不知道區塊內的確切結果。", "en": "Contrast (pointwise conditioning): take X itself as the answer. X(0)=4≠0=X(1), while 0 and 1 lie in the same block B_1, so X is not constant on B_1; by L_meas_const, X is not G-measurable and (i) fails. G knows only which block occurred, not the exact outcome inside it." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "L_meas_const" ], "explanation": { "zh-Hant": "一個區塊內的兩個不同值就足以否定可測性。", "en": "Two different values inside one block suffice to refute measurability." } }, "leaf_basis": "meas_def + sigma_partition" }, { "id": "S11", "uses": [ "S03", "S05", "G1" ], "before_tex": "W\\equiv\\tfrac52", "after_tex": "E[W]=E[X]\\ \\text{but}\\ E[W\\mathbf{1}_{B_1}]=\\tfrac54\\neq\\tfrac12", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "對照（只核總期望）：常數 W≡5/2 是 G 可測，且 E[W]=5/2=E[X]，所以 A=Ω 的條件成立。但 E[W𝟏_{B_1}]=(5/2)(1/2)=5/4，而 E[X𝟏_{B_1}]=1/2（N7），所以 (ii) 在 B_1 上失敗，W 不是 E[X|G]。", "en": "Contrast (checking only the total expectation): the constant W≡5/2 is G-measurable and E[W]=5/2=E[X], so the condition for A=Ω holds. But E[W𝟏_{B_1}]=(5/2)(1/2)=5/4 while E[X𝟏_{B_1}]=1/2 (N7), so (ii) fails on B_1 and W is not E[X|G]." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S05", "numeric_evaluation" ], "explanation": { "zh-Hant": "一個失敗的區塊就足以否定；這說明 S03 化約後仍需每個區塊。", "en": "One failing block suffices to refute; this shows that every block is still needed after the reduction of S03." } }, "leaf_basis": "numeric_evaluation + forall_elim" }, { "id": "S12", "uses": [ "S04", "S05", "G1" ], "before_tex": "c_1'=\\tfrac{4+0}{2}=2", "after_tex": "E[c_1'\\mathbf{1}_{B_1}]=1\\neq\\tfrac12", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "對照（不加權平均）：在 B_1 上取 X 值的普通平均 (4+0)/2=2。則 2·P(B_1)=1≠1/2=E[X𝟏_{B_1}]（N8），條件失敗。因為 1 的質量 3/8 大於 0 的質量 1/8，正確的加權平均 1 比 2 更靠近 X(1)=0。", "en": "Contrast (unweighted mean): on B_1 take the plain average of the X values, (4+0)/2=2. Then 2·P(B_1)=1≠1/2=E[X𝟏_{B_1}] (N8), so the condition fails. Because outcome 1 has mass 3/8, larger than the mass 1/8 of outcome 0, the correct weighted mean 1 lies closer to X(1)=0 than 2 does." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "S05", "numeric_evaluation" ], "explanation": { "zh-Hant": "對正質量區塊 B，『對 B 上所有實值 X，加權平均都等於普通平均』當且僅當條件點權重 P({ω})/P(B) 對每個 ω∈B 均為 1/|B|：必要性取各單點指示函數，充分性直接代入。權重不等時，特定 X 仍可能使兩平均相等；本題 B₁ 的普通平均為 2，加權平均為 1。", "en": "On a block B of positive probability, the weighted and plain means agree for every real-valued X on B if and only if each conditional point weight P({ω})/P(B) equals 1/|B|. Necessity follows by taking the point indicators, and sufficiency by substitution. Unequal weights may still give agreement for particular X; on B₁ here the plain mean is 2 and the weighted mean is 1." } }, "leaf_basis": "numeric_evaluation + frac_rules" }, { "id": "S13", "uses": [ "S08", "S10", "S11", "S12" ], "before_tex": "\\text{S08, S10 to S12}", "after_tex": "E[X|\\mathcal{G}]=1\\ \\text{on}\\ B_1,\\ 4\\ \\text{on}\\ B_2,\\ \\text{arbitrary on the null block}\\ B_3", "relation": "implies", "rule": "mko_condexp_def", "conditions": [], "explanation": { "zh-Hant": "E[X|G] 的版本恰為 Y_c：在 {0,1} 上為 1、在 {2,3} 上為 4、在零質量的 {4} 上為任意常數 c；所有版本幾乎處處相等。逐點的 X、常數總平均與不加權平均都不是條件期望。", "en": "The versions of E[X|G] are exactly the Y_c: 1 on {0,1}, 4 on {2,3} and an arbitrary constant c on the zero-mass {4}; all versions agree almost surely. The pointwise X, the constant overall mean and the unweighted mean are not conditional expectations." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S08", "S10", "S11", "S12" ], "explanation": { "zh-Hant": "目標 G：求出全部版本並否定三種錯誤做法。足夠條件 H：S08 與 S10 到 S12；H⇒G 由合取。H 由 G1 到 G3 正向證成，沒有把 G 當作前提，也沒有在零質量區塊做除法。", "en": "Goal G: find all versions and refute the three wrong methods. Sufficient condition H: S08 together with S10 to S12; H⇒G by conjunction. H is proved forwards from G1 to G3; G is never used as a premise, and no division is done on the zero-mass block." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "E[X|G] 的版本恰為 Y_c：在 B_1={0,1} 上為質量加權平均 (4·1/8+0·3/8)/(1/2)=1，在 B_2={2,3} 上為 (2·1/4+6·1/4)/(1/2)=4，在零質量區塊 B_3={4} 上可取任意常數 c，不做 0/0 除法；所有版本幾乎處處相等，總期望 5/2 保持不變。逐點條件化的候選 Y=X 在 G 下不可測，但其積分等式本身成立，並非在 B_1 上積分失敗。另兩種候選（只核總期望的 W≡5/2、區塊不加權平均）才在 B_1 上違反積分條件。", "en": "The versions of E[X|G] are exactly the Y_c: on B_1={0,1} the mass-weighted mean (4·1/8+0·3/8)/(1/2)=1, on B_2={2,3} the value (2·1/4+6·1/4)/(1/2)=4, and on the zero-mass block B_3={4} any constant c, with no 0/0 division; all versions agree almost surely, and the total expectation 5/2 is preserved. The pointwise candidate Y=X is not G-measurable, although its integral equality itself holds. The other two candidates, W≡5/2 based only on the total expectation and the unweighted block mean, violate the integral condition on B_1." }, "negative_controls": [ { "claim": { "zh-Hant": "X 的每個值都已知，所以 E[X|G]=X", "en": "Every value of X is known, so E[X|G]=X" }, "expected": "reject", "reason": { "zh-Hant": "X 在區塊 B_1 上不是常數，不是 G 可測（S10）。", "en": "X is not constant on the block B_1, so it is not G-measurable (S10)." } }, { "claim": { "zh-Hant": "在 B_3 上 E[X|G]=E[X𝟏_{B_3}]/P(B_3)=0/0=0", "en": "On B_3, E[X|G]=E[X𝟏_{B_3}]/P(B_3)=0/0=0" }, "expected": "reject", "reason": { "zh-Hant": "0/0 沒有定義；任何常數都滿足條件，值只幾乎處處決定（S07）。", "en": "0/0 is undefined; every constant satisfies the condition, and the value is determined only almost surely (S07)." } }, { "claim": { "zh-Hant": "W≡5/2 滿足 E[W]=E[X]，所以 W=E[X|G]", "en": "W≡5/2 satisfies E[W]=E[X], so W=E[X|G]" }, "expected": "reject", "reason": { "zh-Hant": "E[W𝟏_{B_1}]=5/4≠1/2（S11）。", "en": "E[W𝟏_{B_1}]=5/4≠1/2 (S11)." } }, { "claim": { "zh-Hant": "在 B_1 上 E[X|G]=(4+0)/2=2", "en": "On B_1, E[X|G]=(4+0)/2=2" }, "expected": "reject", "reason": { "zh-Hant": "要用質量加權：2·(1/2)=1≠1/2；正確值是 1（S05、S12）。", "en": "Masses must be used: 2·(1/2)=1≠1/2; the correct value is 1 (S05, S12)." } }, { "claim": { "zh-Hant": "條件期望逐點唯一，所以在結果 4 的值必為 X(4)=10", "en": "Conditional expectation is unique pointwise, so its value at outcome 4 must be X(4)=10" }, "expected": "reject", "reason": { "zh-Hant": "Y_0（在 B_3 上取 0）也是版本；版本只幾乎處處唯一（S08）。", "en": "Y_0 (taking 0 on B_3) is also a version; versions are unique only almost surely (S08)." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "prob_axioms", "kind": "prerequisite", "statement": { "zh-Hant": "機率公理：P(A)≥0；P(Ω)=1；兩兩互斥事件的（可數）聯集機率等於各機率之和。由此 P(∅)=0、P(Aᶜ)=1−P(A)。", "en": "Probability axioms: P(A)≥0; P(Ω)=1; the probability of a (countable) union of pairwise disjoint events is the sum of their probabilities. Hence P(∅)=0 and P(Aᶜ)=1−P(A)." } }, { "id": "finite_sum_rules", "kind": "prerequisite", "statement": { "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。", "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i." } }, { "id": "set_operations_def", "kind": "prerequisite", "statement": { "zh-Hant": "集合運算：x∈A∪B ⇔ x∈A 或 x∈B；x∈A∩B ⇔ x∈A 且 x∈B；x∈A∖B ⇔ x∈A 且 x∉B；在全集 U 中 Aᶜ=U∖A。", "en": "Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A." } }, { "id": "frac_rules", "kind": "prerequisite", "statement": { "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。", "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0." } }, { "id": "mul_inverse_nonzero", "kind": "prerequisite", "statement": { "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。", "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined." } }, { "id": "numeric_evaluation", "kind": "prerequisite", "statement": { "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。", "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable." } }, { "id": "image_preimage_def", "kind": "prerequisite", "statement": { "zh-Hant": "像與原像：f(S)={f(a) ∣ a∈S}；f⁻¹(T)={a∈A ∣ f(a)∈T}。原像記號不預設 f 有反函數。", "en": "Image and preimage: f(S)={f(a) ∣ a∈S}; f⁻¹(T)={a∈A ∣ f(a)∈T}. The preimage notation does not presuppose that f has an inverse." } }, { "id": "subset_def", "kind": "prerequisite", "statement": { "zh-Hant": "子集：A⊆B 表示 A 的每個元素都是 B 的元素，即 ∀x (x∈A⇒x∈B)。", "en": "Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B)." } }, { "id": "forall_elim", "kind": "prerequisite", "statement": { "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。", "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds." } }, { "id": "distributive", "kind": "prerequisite", "statement": { "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。", "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu." } }, { "id": "mul_zero", "kind": "prerequisite", "statement": { "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。", "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0." } }, { "id": "and_intro_elim", "kind": "prerequisite", "statement": { "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。", "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q." } }, { "id": "iff_intro", "kind": "prerequisite", "statement": { "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。", "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold." } }, { "id": "def_unfold", "kind": "prerequisite", "statement": { "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。", "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true." } }, { "id": "mko_condexp_def", "kind": "cited", "source": { "mko_id": "mko-conditional-expectation", "sha256": "35d7f871dac286e45d4661dcffbafe25a02cca5c55a4c58ef60bd41d2408bd95", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "在同一有限機率空間上，條件期望 E[X|G] 須 G 可測，且在每個 G 可測事件上與 X 積分相同。對分割生成的 G，它在每個正質量區塊為加權平均；零質量區塊只幾乎處處決定，不除以零。", "en": "On the same finite probability space, E[X|G] must be G-measurable and have the same integral as X on every G-measurable event. When G is generated by a partition, its value is the weighted mean on each positive-mass block. On zero-mass blocks it is determined only almost surely; division by zero is not performed." } }, { "id": "sigma_partition", "kind": "prerequisite", "statement": { "zh-Hant": "有限集合 Ω 的分割 𝒫 生成的 σ 代數 σ(𝒫) 恰由區塊的聯集組成（含空聯集 ∅）；共有 2^{|𝒫|} 個事件。", "en": "The σ-algebra σ(𝒫) generated by a partition 𝒫 of a finite set Ω consists exactly of the unions of blocks (including the empty union ∅); it has 2^{|𝒫|} events." } }, { "id": "meas_def", "kind": "prerequisite", "statement": { "zh-Hant": "在有限 Ω 上，函數 Y 為 G 可測，當且僅當對每個實數集合 S，Y^{-1}(S)∈G。", "en": "On a finite Ω, a function Y is G-measurable exactly when Y^{-1}(S)∈G for every set S of reals." } }, { "id": "L_meas_const", "kind": "lemma", "proved_in": [ "S02.1", "S02.2" ], "statement": { "zh-Hant": "Y 為 σ(𝒫) 可測，當且僅當 Y 在 𝒫 的每個區塊上為常數。", "en": "Y is σ(𝒫)-measurable exactly when Y is constant on each block of 𝒫." } }, { "id": "L_block_reduce", "kind": "lemma", "proved_in": [ "S03.1", "S03.2" ], "statement": { "zh-Hant": "對 σ(𝒫)，條件 (ii) 對所有 A∈σ(𝒫) 成立，當且僅當它對每個區塊成立。", "en": "For σ(𝒫), condition (ii) holds for every A∈σ(𝒫) exactly when it holds for every block." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S05", "lhs": "1/8+3/8", "rhs": "1/2" }, { "id": "N2", "step": "S05", "lhs": "4*(1/8)+0*(3/8)", "rhs": "1/2" }, { "id": "N3", "step": "S05", "lhs": "(1/2)/(1/2)", "rhs": "1" }, { "id": "N4", "step": "S06", "lhs": "2*(1/4)+6*(1/4)", "rhs": "2" }, { "id": "N5", "step": "S06", "lhs": "2/(1/2)", "rhs": "4" }, { "id": "N6", "step": "S09", "lhs": "1*(1/2)+4*(1/2)", "rhs": "4*(1/8)+0*(3/8)+2*(1/4)+6*(1/4)+10*0" }, { "id": "N7", "step": "S11", "lhs": "(5/2)*(1/2)", "relation": "!=", "rhs": "1/2" }, { "id": "N8", "step": "S12", "lhs": "((4+0)/2)*(1/2)", "relation": "!=", "rhs": "1/2" }, { "id": "N9", "step": "S01", "lhs": "2**3", "rhs": "8" }, { "id": "N10", "lhs": "1/8+3/8+1/4+1/4+0", "rhs": "1" } ] }, "presentation": { "md_scope": { "zh-Hant": "只在 MKO 範圍內工作：完整有限樣本、精確有理質量、X 值與條件分割屬於同一機率空間（假設 a1）；主要計算用正質量區塊，所有版本幾乎處處唯一，而零質量區塊保留逐點非唯一的自由，不強迫點值（假設 a2）。不處理一般（非離散）條件期望的存在性、filtration 或 martingale；不改動 canonical MKO 或英文 edition。", "en": "Works inside the MKO's scope: the complete finite outcomes, exact rational masses, X values and conditioning partition all belong to one probability space (assumption a1); the main computation uses positive-mass blocks, and versions are unique almost surely, while the zero-mass block retains pointwise freedom without a forced value (assumption a2). It does not treat existence of conditional expectation in general (non-discrete) spaces, filtrations or martingales, and does not change the canonical MKO or its English edition." }, "md_symbols": [ [ "\\Omega,\\ P", "樣本空間與機率", "the sample space and the probability" ], [ "B_1,B_2,B_3", "條件分割的三個區塊 {0,1}、{2,3}、{4}", "the three blocks {0,1}, {2,3}, {4} of the conditioning partition" ], [ "\\mathcal{G}=\\sigma(B_1,B_2,B_3)", "由分割生成的 σ 代數", "the σ-algebra generated by the partition" ], [ "E[Z\\mathbf{1}_A]", "Z 在事件 A 上的積分 Σ_{ω∈A} Z(ω)P({ω})", "the integral of Z over A, Σ_{ω∈A} Z(ω)P({ω})" ], [ "Y_c", "E[X|G] 的一個版本，在 B_3 上取值 c", "a version of E[X|G] taking the value c on B_3" ] ], "md_error_controls": [ { "title": "逐點條件化與零質量除法 / Pointwise conditioning and division by zero mass", "zh": "MKO 的誤解之一是逐點條件化：答案 X 本身在 B_1 上取 4 與 0，不是 G 可測（S10）。另一個是在零質量區塊做除法：E[X𝟏_{B_3}]/P(B_3)=0/0 沒有定義；正確結論是任何常數都可以，版本只幾乎處處唯一（S07、S08）。", "en": "One MKO misconception is pointwise conditioning: the answer X itself takes the values 4 and 0 on B_1 and is not G-measurable (S10). Another is dividing on a zero-mass block: E[X𝟏_{B_3}]/P(B_3)=0/0 is undefined; the correct conclusion is that any constant works and versions are unique only almost surely (S07, S08)." }, { "title": "只核總期望不夠 / Checking the total expectation is not enough", "zh": "MKO 的另一個誤解：只核 E[Y]=E[X] 就宣稱完整條件期望。常數 5/2 滿足總期望，卻在 B_1 上積分 5/4≠1/2（S11）。條件 (ii) 要對每個 G 事件成立，化約後仍是每個區塊（S03）。", "en": "Another MKO misconception: declaring a full conditional expectation after checking only E[Y]=E[X]. The constant 5/2 matches the total expectation but integrates to 5/4≠1/2 over B_1 (S11). Condition (ii) must hold on every event of G, which after reduction still means every block (S03)." }, { "title": "要用質量加權 / Weight by mass", "zh": "正質量區塊上的值是 E[X𝟏_B]/P(B)，一般不能直接用 X 值的普通平均取代。對該區塊上所有實值 X 都保證兩平均一致，當且僅當各條件點權重都為 1/|B|；特定 X 則可能在不等權時偶然一致。本題 B₁ 上普通平均是 2，正確值是 1（S12）。零質量區塊仍不做除法，其版本如 S07、S08 所述。", "en": "On a block of positive probability, the value is E[X𝟏_B]/P(B); it cannot in general be replaced by the plain average. Agreement for every real-valued X on that block holds if and only if all conditional point weights equal 1/|B|. For a particular X the means may coincide even with unequal weights. On B₁ here the plain mean is 2 and the correct value is 1 (S12). No division is made on a zero-probability block, whose versions remain as in S07 and S08." } ], "md_open_items": [ { "zh-Hant": "本稿只處理分割生成的 σ 代數；一般機率空間上條件期望的存在性（例如經由 Radon–Nikodym）與 filtration、martingale（MKO 的下一步）不在範圍內。", "en": "This guide treats only partition-generated σ-algebras; existence of conditional expectation on general probability spaces (for example via Radon–Nikodym) and filtrations and martingales (the MKO's next step) are outside its scope." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "fb6d81a6503a7c7ed987151c144cb532f48bda0ba9ff318c35b0da5747915922", "original_source_sha256": "f1f1276883bf28f45ac381b84049dc540bebe522aa5588bd14a827b966580991", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [ "BR-M093", "BR-M094" ], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }