{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-discrete-stochastic-process-01", "mko_id": "mko-discrete-stochastic-process", "title": { "zh-Hant": "離散隨機過程：由完整路徑分布導出邊際與兩時刻聯合分布，並檢驗 Markov 性與平穩性", "en": "Discrete-time stochastic process: marginals and two-time joint laws from a complete path distribution, with Markov and stationarity checks" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-discrete-stochastic-process-01", "source": { "mko_id": "mko-discrete-stochastic-process", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "1b73d458f056e14bbaaf318ebfecf7e93214b9895dae967bfcffc48910854832", "english_sha256": "e70a7222e382b4caea1df2ed7b1f73c9fda919ac5d2f73ad35983043f2a80bd7" }, "mode": "worked_example", "audience_prerequisites": [ "prob_axioms", "finite_sum_rules", "image_preimage_def", "cartesian_product_def", "function_def" ], "goal": { "zh-Hant": "時域 {0,1,2}、狀態 {0,1}。完整路徑支撐與精確 PMF 為 (0,0,0):1/4、(0,1,1):1/4、(1,0,1):1/8、(1,1,0):1/8、(1,1,1):1/4。(1) 核對支撐與 PMF，建立機率空間並說明 X_0、X_1、X_2 是同一空間上的隨機變數；(2) 求三個時刻的邊際分布與 (X_1,X_2) 的聯合分布；(3) 證明這個過程不平穩、不具 Markov 性；(4) 構造一個邊際相同但聯合分布不同的路徑律，說明邊際表不等於完整過程。", "en": "Horizon {0,1,2}, states {0,1}. The complete path support with exact PMF is (0,0,0):1/4, (0,1,1):1/4, (1,0,1):1/8, (1,1,0):1/8, (1,1,1):1/4. (1) Check the support and the PMF, build the probability space and explain why X_0, X_1, X_2 are random variables on the same space; (2) find the marginal law at each time and the joint law of (X_1,X_2); (3) prove that this process is neither stationary nor Markov; (4) construct a path law with the same marginals but a different joint law, showing that the marginal tables are not the whole process." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "時域 {0,1,2}，狀態 {0,1}。支撐列出 5 條長度 3 的路徑：ω_1=(0,0,0)、ω_2=(0,1,1)、ω_3=(1,0,1)、ω_4=(1,1,0)、ω_5=(1,1,1)（MKO 假設 a1：互異等長的完整有限路徑支撐）。", "en": "Horizon {0,1,2}, states {0,1}. The support lists 5 paths of length 3: ω_1=(0,0,0), ω_2=(0,1,1), ω_3=(1,0,1), ω_4=(1,1,0), ω_5=(1,1,1) (MKO assumption a1: complete support of distinct equal-length finite paths)." } }, { "id": "G2", "statement": { "zh-Hant": "精確 PMF：p(ω_1)=1/4、p(ω_2)=1/4、p(ω_3)=1/8、p(ω_4)=1/8、p(ω_5)=1/4；支撐外的路徑機率為 0。", "en": "Exact PMF: p(ω_1)=1/4, p(ω_2)=1/4, p(ω_3)=1/8, p(ω_4)=1/8, p(ω_5)=1/4; paths outside the support have probability 0." } }, { "id": "G3", "statement": { "zh-Hant": "比較律 Q：在 {0,1}^3 上定義 Q(x_0,x_1,x_2)=p_0(x_0)p_1(x_1)p_2(x_2)，其中 p_n 是 G2 之律下 X_n 的邊際（在 S07 到 S09 求得）。", "en": "Comparison law Q: on {0,1}^3 define Q(x_0,x_1,x_2)=p_0(x_0)p_1(x_1)p_2(x_2), where p_n is the marginal of X_n under the law of G2 (found in S07 to S09)." } } ], "steps": [ { "id": "S01", "uses": [ "mko_process_def" ], "before_tex": "(X_n)_{n\\in\\{0,1,2\\}}\\ \\text{process}", "after_tex": "X_0,X_1,X_2:\\Omega\\to\\{0,1\\}\\ \\text{on one}\\ (\\Omega,\\mathcal{F},P);\\ \\text{law}=\\text{path PMF}", "relation": "definition", "rule": "mko_process_def", "conditions": [], "explanation": { "zh-Hant": "依 MKO 展開：離散時間隨機過程是同一機率空間上的隨機變數族。在有限時域上，完整路徑分布（每條路徑的機率）就是過程的聯合律；邊際與兩時刻聯合都由它導出，但 Markov 性與平穩性不會自動成立。", "en": "Unfold by the MKO: a discrete-time stochastic process is a family of random variables on one probability space. On a finite horizon, the complete path distribution (the probability of each path) is the joint law of the process; marginals and two-time joints are derived from it, but the Markov property and stationarity do not follow automatically." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_process_def" ], "explanation": { "zh-Hant": "只展開定義；「同一個機率空間」是關鍵：各 X_n 若各自定義在不同空間上，就沒有時間間的依賴可談。", "en": "Only unfolds the definition; 'one probability space' is the key point: if each X_n lived on a different space, there would be no dependence across time to speak of." } }, "leaf_basis": "mko_process_def + def_unfold" }, { "id": "S02", "uses": [ "G1", "cartesian_product_def" ], "before_tex": "\\omega_1,\\omega_2,\\omega_3,\\omega_4,\\omega_5\\in\\{0,1\\}^3", "after_tex": "\\omega_k\\ \\text{pairwise distinct},\\ \\text{each of length}\\ 3", "relation": "implies", "rule": "cartesian_product_def", "conditions": [], "explanation": { "zh-Hant": "每條路徑都是 {0,1} 的三元組（時刻 0、1、2）。有序組相等需每個坐標相等，所以只要找到一個不同坐標就相異。系統地分組：時刻 0 把 {ω_1,ω_2}（值 0）與 {ω_3,ω_4,ω_5}（值 1）分開；ω_1、ω_2 在時刻 1 不同；ω_3（時刻 1 為 0）與 ω_4、ω_5（時刻 1 為 1）分開；ω_4、ω_5 在時刻 2 不同。所以 10 對都相異。", "en": "Each path is a triple over {0,1} (times 0, 1, 2). Ordered triples are equal only if every coordinate agrees, so one differing coordinate makes them distinct. Group systematically: time 0 separates {ω_1,ω_2} (value 0) from {ω_3,ω_4,ω_5} (value 1); ω_1 and ω_2 differ at time 1; ω_3 (0 at time 1) is separated from ω_4, ω_5 (1 at time 1); ω_4 and ω_5 differ at time 2. So all 10 pairs are distinct." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "cartesian_product_def" ], "explanation": { "zh-Hant": "分組論證涵蓋全部 10 對：跨組的對由較早的坐標分開，組內的對由後面的坐標分開。互異性保證每條路徑只被計一次質量。", "en": "The grouping covers all 10 pairs: pairs in different groups are separated by an earlier coordinate and pairs in the same group by a later one. Distinctness ensures each path's mass is counted once." } }, "leaf_basis": "cartesian_product_def + case_analysis" }, { "id": "S03", "uses": [ "G2" ], "before_tex": "p=(\\tfrac14,\\tfrac14,\\tfrac18,\\tfrac18,\\tfrac14)", "after_tex": "p(\\omega_k)\\ge0,\\ \\ \\sum_{k=1}^{5}p(\\omega_k)=1", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "五個值都是正有理數；總和 1/4+1/4+1/8+1/8+1/4=1（N1），是精確等式。所以 p 是 Ω 上合法的 PMF。", "en": "All five values are positive rationals; the total 1/4+1/4+1/8+1/8+1/4=1 (N1) is an exact equality. So p is a valid PMF on Ω." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G2", "numeric_evaluation" ], "explanation": { "zh-Hant": "非負與總和為 1 兩個條件都要核；總和只近似為 1 不算合法 PMF。", "en": "Both nonnegativity and total 1 must be checked; a total only approximately 1 would not be a valid PMF." } }, "leaf_basis": "numeric_evaluation + finite_sum_rules" }, { "id": "S04", "uses": [ "S02", "S03", "prob_axioms" ], "before_tex": "\\Omega,\\ p", "after_tex": "(\\Omega,\\mathcal{P}(\\Omega),P),\\ \\ P(A)=\\sum_{\\omega\\in A}p(\\omega)\\ \\text{satisfies the axioms}", "relation": "implies", "rule": "prob_axioms", "conditions": [], "explanation": { "zh-Hant": "取事件族為 Ω 的冪集，P(A) 為 A 中路徑質量之和。非負項的和非負；P(Ω)=1（S03）；不交事件的可加性在 S04.1 證明。Ω 有限，所以可數可加化為有限可加。", "en": "Take the event family to be the power set of Ω and P(A) the sum of path masses in A. A sum of nonnegative terms is nonnegative; P(Ω)=1 (S03); additivity for disjoint events is proved in S04.1. Ω is finite, so countable additivity reduces to finite additivity." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "prob_axioms", "finite_sum_rules" ], "explanation": { "zh-Hant": "需要 S02 的互異性，否則同一路徑會被重複計算；有限性讓可數可加化為有限可加（至多 5 個非空互斥事件）。", "en": "Needs the distinctness of S02, otherwise one path could be counted twice; finiteness reduces countable additivity to finite additivity (at most 5 nonempty disjoint events)." } }, "leaf_basis": "prob_axioms + finite_sum_rules" }, { "id": "S04.1", "uses": [ "finite_sum_rules" ], "before_tex": "A\\cap B=\\emptyset", "after_tex": "P(A\\cup B)=\\sum_{\\omega\\in A\\cup B}p(\\omega)=\\sum_{\\omega\\in A}p(\\omega)+\\sum_{\\omega\\in B}p(\\omega)", "relation": "implies", "rule": "finite_sum_rules", "conditions": [ "A\\cap B=\\emptyset" ], "explanation": { "zh-Hant": "A、B 不交時，A∪B 中每條路徑恰在 A 或 B 之一，所以有限和可拆成兩部分。這就是 P 的有限可加性。", "en": "When A and B are disjoint, each path of A∪B lies in exactly one of A, B, so the finite sum splits into two parts. This is finite additivity of P." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "finite_sum_rules", "set_operations_def" ], "explanation": { "zh-Hant": "不交是必要條件；若有重疊，重疊路徑的質量會被算兩次。", "en": "Disjointness is necessary; with overlap the mass of shared paths would be counted twice." } }, "leaf_basis": "finite_sum_rules + set_operations_def" }, { "id": "S05", "uses": [ "S04", "function_def", "image_preimage_def" ], "before_tex": "X_n(\\omega)=\\omega_n", "after_tex": "X_0,X_1,X_2\\ \\text{random variables on the same}\\ (\\Omega,\\mathcal{P}(\\Omega),P)", "relation": "implies", "rule": "function_def", "conditions": [], "explanation": { "zh-Hant": "X_n 把每條路徑送到它在時刻 n 的坐標，對每個 ω 給出恰好一個值，是 Ω→{0,1} 的總函數。事件族是冪集，所以每個原像 {X_n=s} 都是事件，X_n 是隨機變數。三個 X_n 定義在同一個 Ω 上，這正是 S01 要求的「同一機率空間上的隨機變數族」。", "en": "X_n sends each path to its coordinate at time n, giving exactly one value for each ω, so it is a total function Ω→{0,1}. The event family is the power set, so every preimage {X_n=s} is an event and X_n is a random variable. All three X_n live on the same Ω, which is exactly the 'family of random variables on one probability space' required by S01." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "function_def", "image_preimage_def" ], "explanation": { "zh-Hant": "冪集事件族讓可測性自動成立；在一般事件族下要另外核原像。", "en": "The power-set event family makes measurability automatic; with a general event family the preimages would need to be checked separately." } }, "leaf_basis": "function_def + image_preimage_def" }, { "id": "S06", "uses": [ "S04", "S05", "L_marginal" ], "before_tex": "P(X_n=s)", "after_tex": "\\sum_{\\omega\\in\\Omega,\\ \\omega_n=s}p(\\omega)", "relation": "equivalent", "rule": "L_marginal", "conditions": [], "explanation": { "zh-Hant": "引理 L_marginal（S06.1 證明）：時刻 n 的邊際機率，是在時刻 n 取值 s 的所有路徑質量之和（MKO 段落 1：固定時刻聚合路徑質量）。兩時刻聯合同理，只是同時固定兩個坐標。", "en": "Lemma L_marginal (proved in S06.1): the marginal probability at time n is the sum of the masses of all paths taking the value s at time n (MKO paragraph 1: aggregate path mass at a fixed time). Two-time joints are the same with two coordinates fixed." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_marginal", "image_preimage_def" ], "explanation": { "zh-Hant": "兩邊是同一個數的兩種寫法：左邊是事件 {X_n=s} 的機率，右邊是依 P 的定義展開的和。", "en": "Both sides are two expressions of one number: the left is the probability of the event {X_n=s}, the right is the sum obtained by unfolding the definition of P." } }, "leaf_basis": "L_marginal + finite_sum_rules" }, { "id": "S06.1", "uses": [ "S04", "image_preimage_def" ], "before_tex": "\\{X_n=s\\}=X_n^{-1}(\\{s\\})", "after_tex": "\\{\\omega\\in\\Omega\\mid\\omega_n=s\\},\\ \\ P(X_n=s)=\\sum_{\\omega_n=s}p(\\omega)", "relation": "equivalent", "rule": "image_preimage_def", "conditions": [], "explanation": { "zh-Hant": "依原像定義，{X_n=s} 是使 X_n(ω)=s 的路徑集合，也就是 ω_n=s 的路徑。把 P 的定義（S04）用在這個事件上，就是這些路徑質量之和。對兩時刻，{X_m=s, X_n=t} 是兩個原像之交，同樣展開。這證得引理 L_marginal。", "en": "By the definition of preimage, {X_n=s} is the set of paths with X_n(ω)=s, that is the paths with ω_n=s. Applying the definition of P (S04) to this event gives the sum of their masses. For two times, {X_m=s, X_n=t} is the intersection of two preimages and unfolds the same way. This proves Lemma L_marginal." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "image_preimage_def", "set_operations_def" ], "explanation": { "zh-Hant": "原像與集合建構式雙向對應；不需要 X_n 可逆。", "en": "The preimage and the set-builder description correspond both ways; X_n need not be invertible." } }, "leaf_basis": "image_preimage_def + set_builder + def_unfold" }, { "id": "S07", "uses": [ "S06", "G1", "G2" ], "before_tex": "P(X_0=0),\\ P(X_0=1)", "after_tex": "p_0=(\\tfrac12,\\tfrac12)", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "時刻 0 為 0 的路徑是 ω_1、ω_2：1/4+1/4=1/2（N2）；為 1 的是 ω_3、ω_4、ω_5：1/8+1/8+1/4=1/2（N3）。兩者和為 1，與 S03 一致。", "en": "The paths with value 0 at time 0 are ω_1, ω_2: 1/4+1/4=1/2 (N2); those with value 1 are ω_3, ω_4, ω_5: 1/8+1/8+1/4=1/2 (N3). The two add to 1, consistent with S03." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "S06", "numeric_evaluation" ], "explanation": { "zh-Hant": "每個數都可由 N2、N3 重算；兩組路徑恰好分割 Ω，所以和為 1 是一個獨立核對。", "en": "Each number is recomputed by N2, N3; the two groups of paths partition Ω exactly, so their total of 1 is an independent check." } }, "leaf_basis": "numeric_evaluation + finite_sum_rules" }, { "id": "S08", "uses": [ "S06", "G1", "G2" ], "before_tex": "P(X_1=0),\\ P(X_1=1)", "after_tex": "p_1=(\\tfrac38,\\tfrac58)", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "時刻 1 為 0 的路徑是 ω_1、ω_3：1/4+1/8=3/8（N4）；為 1 的是 ω_2、ω_4、ω_5：1/4+1/8+1/4=5/8（N5）。", "en": "The paths with value 0 at time 1 are ω_1, ω_3: 1/4+1/8=3/8 (N4); those with value 1 are ω_2, ω_4, ω_5: 1/4+1/8+1/4=5/8 (N5)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "S06", "numeric_evaluation" ], "explanation": { "zh-Hant": "3/8+5/8=1，兩組路徑分割 Ω。", "en": "3/8+5/8=1, and the two groups of paths partition Ω." } }, "leaf_basis": "numeric_evaluation + finite_sum_rules" }, { "id": "S09", "uses": [ "S06", "G1", "G2" ], "before_tex": "P(X_2=0),\\ P(X_2=1)", "after_tex": "p_2=(\\tfrac38,\\tfrac58)", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "時刻 2 為 0 的路徑是 ω_1、ω_4：1/4+1/8=3/8（N6）；為 1 的是 ω_2、ω_3、ω_5：1/4+1/8+1/4=5/8（N7）。注意 X_1 與 X_2 的邊際相同，但由不同的路徑組成。", "en": "The paths with value 0 at time 2 are ω_1, ω_4: 1/4+1/8=3/8 (N6); those with value 1 are ω_2, ω_3, ω_5: 1/4+1/8+1/4=5/8 (N7). Note that X_1 and X_2 have the same marginal, made up of different paths." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "S06", "numeric_evaluation" ], "explanation": { "zh-Hant": "與 S08 相同的計算方式；邊際相同不表示 X_1=X_2（見錯誤控制 3）。", "en": "Computed as in S08; equal marginals do not mean X_1=X_2 (see error control 3)." } }, "leaf_basis": "numeric_evaluation + finite_sum_rules" }, { "id": "S10", "uses": [ "S06", "G1", "G2" ], "before_tex": "P(X_1=s,\\ X_2=t)", "after_tex": "(0,0):\\tfrac14,\\ (0,1):\\tfrac18,\\ (1,0):\\tfrac18,\\ (1,1):\\tfrac12", "relation": "equivalent", "rule": "L_marginal", "conditions": [], "explanation": { "zh-Hant": "依 L_marginal 的兩時刻形式，固定時刻 1 與 2 的值：(0,0) 只有 ω_1，得 1/4；(0,1) 只有 ω_3，得 1/8；(1,0) 只有 ω_4，得 1/8；(1,1) 是 ω_2、ω_5，得 1/4+1/4=1/2（N9）。四格和為 1（N8）；行（row）和 1/4+1/8=3/8 回到 S08，列（column）和 1/8+1/2=5/8 回到 S09。", "en": "By the two-time form of L_marginal, fix the values at times 1 and 2: (0,0) is only ω_1, giving 1/4; (0,1) only ω_3, giving 1/8; (1,0) only ω_4, giving 1/8; (1,1) is ω_2, ω_5, giving 1/4+1/4=1/2 (N9). The four cells add to 1 (N8); the row sum 1/4+1/8=3/8 recovers S08 and the column sum 1/8+1/2=5/8 recovers S09." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_marginal", "numeric_evaluation" ], "explanation": { "zh-Hant": "聯合表的行（row）和與列（column）和回到邊際，是一致性核對；反過來，邊際無法唯一決定聯合表（S15）。", "en": "Row and column sums of the joint table return the marginals, a consistency check; conversely the marginals do not determine the joint table uniquely (S15)." } }, "leaf_basis": "L_marginal + finite_sum_rules + numeric_evaluation" }, { "id": "S11", "uses": [ "S07", "S08", "p_stationary_nec" ], "before_tex": "p_0=(\\tfrac12,\\tfrac12),\\ p_1=(\\tfrac38,\\tfrac58)", "after_tex": "p_0\\neq p_1\\ \\Rightarrow\\ \\text{not stationary}", "relation": "implies", "rule": "contrapositive", "conditions": [], "explanation": { "zh-Hant": "平穩的必要條件是所有時刻的邊際相同（p_stationary_nec）。這裡 P(X_0=0)=1/2 而 P(X_1=0)=3/8（N10），條件失敗；由逆否命題，這個過程不平穩。MKO 強調平穩性不會由路徑律自動得到。", "en": "A necessary condition for stationarity is that all one-time marginals agree (p_stationary_nec). Here P(X_0=0)=1/2 while P(X_1=0)=3/8 (N10), so the condition fails; by contraposition the process is not stationary. The MKO stresses that stationarity does not follow automatically from a path law." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "p_stationary_nec", "contrapositive" ], "explanation": { "zh-Hant": "只用到必要條件的逆否；反向（邊際相同推出平穩）不宣稱，也一般不成立。", "en": "Uses only the contrapositive of a necessary condition; the converse (equal marginals imply stationarity) is not claimed and fails in general." } }, "leaf_basis": "contrapositive + numeric_evaluation" }, { "id": "S12", "uses": [ "S06", "p_cond_prob", "G2" ], "before_tex": "P(X_2=1\\mid X_1=1,\\ X_0=0)", "after_tex": "\\frac{P(\\{\\omega_2\\})}{P(\\{\\omega_2\\})}=\\frac{1/4}{1/4}=1", "relation": "equivalent", "rule": "p_cond_prob", "conditions": [ "P(X_0=0,X_1=1)=\\tfrac14>0" ], "explanation": { "zh-Hant": "條件事件 {X_0=0, X_1=1} 只含 ω_2（S12.1），機率 1/4>0，所以條件機率有定義。其中 X_2=1 的路徑也只有 ω_2，所以條件機率是 (1/4)/(1/4)=1（N11）。", "en": "The conditioning event {X_0=0, X_1=1} contains only ω_2 (S12.1), with probability 1/4>0, so the conditional probability is defined. The paths in it with X_2=1 are again only ω_2, so the conditional probability is (1/4)/(1/4)=1 (N11)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "p_cond_prob", "numeric_evaluation" ], "explanation": { "zh-Hant": "條件機率的定義在分母為正時是一個等式，兩邊同值；分母為 0 時這一步不能做。", "en": "With a positive denominator the definition of conditional probability is an equality, both sides having the same value; with denominator 0 the step could not be taken." } }, "leaf_basis": "p_cond_prob + numeric_evaluation" }, { "id": "S12.1", "uses": [ "G1", "set_operations_def" ], "before_tex": "\\{X_0=0\\}\\cap\\{X_1=1\\}", "after_tex": "\\{\\omega_2\\};\\quad \\{X_0=1\\}\\cap\\{X_1=1\\}=\\{\\omega_4,\\omega_5\\};\\quad \\{X_1=1\\}=\\{\\omega_2,\\omega_4,\\omega_5\\}", "relation": "equivalent", "rule": "set_operations_def", "conditions": [], "explanation": { "zh-Hant": "逐條檢查五條路徑的前兩個坐標：(0,1) 只出現在 ω_2；(1,1) 出現在 ω_4、ω_5；時刻 1 為 1 的是 ω_2、ω_4、ω_5。這些集合是 S12 到 S14 的分母與分子所需的事件。", "en": "Check the first two coordinates of the five paths one by one: (0,1) occurs only in ω_2; (1,1) occurs in ω_4, ω_5; value 1 at time 1 occurs in ω_2, ω_4, ω_5. These sets are the events needed for the numerators and denominators of S12 to S14." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "set_operations_def", "G1" ], "explanation": { "zh-Hant": "完整列舉 5 條路徑，交集定義雙向成立。", "en": "All 5 paths are enumerated, and the definition of intersection holds both ways." } }, "leaf_basis": "set_operations_def + set_builder" }, { "id": "S13", "uses": [ "S12.1", "p_cond_prob", "G2" ], "before_tex": "P(X_2=1\\mid X_1=1,\\ X_0=1)", "after_tex": "\\frac{P(\\{\\omega_5\\})}{P(\\{\\omega_4,\\omega_5\\})}=\\frac{1/4}{3/8}=\\frac23", "relation": "equivalent", "rule": "p_cond_prob", "conditions": [ "P(X_0=1,X_1=1)=\\tfrac38>0" ], "explanation": { "zh-Hant": "條件事件是 {ω_4,ω_5}，機率 1/8+1/4=3/8>0。其中 X_2=1 的只有 ω_5（ω_4 的時刻 2 為 0），所以條件機率是 (1/4)/(3/8)=2/3（N12）。", "en": "The conditioning event is {ω_4,ω_5}, with probability 1/8+1/4=3/8>0. Within it only ω_5 has X_2=1 (ω_4 has 0 at time 2), so the conditional probability is (1/4)/(3/8)=2/3 (N12)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "p_cond_prob", "numeric_evaluation" ], "explanation": { "zh-Hant": "分母為正，定義是等式；數值可由 N12 重算。", "en": "The denominator is positive, so the definition is an equality; the value is recomputed by N12." } }, "leaf_basis": "p_cond_prob + frac_rules + numeric_evaluation" }, { "id": "S14", "uses": [ "S10", "S12", "S13", "p_markov_def" ], "before_tex": "P(X_2=1\\mid X_1=1)=\\frac{1/2}{5/8}=\\frac45", "after_tex": "1\\neq\\tfrac23:\\ \\text{the Markov condition fails at time 1}", "relation": "implies", "rule": "p_markov_def", "conditions": [], "explanation": { "zh-Hant": "P(X_2=1∣X_1=1)=P(X_1=1,X_2=1)/P(X_1=1)=(1/2)/(5/8)=4/5（N13，用 S10 與 S08）。Markov 性要求 S12 與 S13 的兩個值都等於 4/5，但它們是 1 與 2/3，而且彼此不等。所以在 X_1=1 的條件下，X_2 仍依賴 X_0，過程不具 Markov 性。", "en": "P(X_2=1∣X_1=1)=P(X_1=1,X_2=1)/P(X_1=1)=(1/2)/(5/8)=4/5 (N13, using S10 and S08). The Markov property requires both values of S12 and S13 to equal 4/5, but they are 1 and 2/3, and they even differ from each other. So given X_1=1, X_2 still depends on X_0, and the process is not Markov." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "p_markov_def", "S12", "S13" ], "explanation": { "zh-Hant": "否定 Markov 性只需一組 (h,i,j) 失敗；兩個條件事件的機率都為正（1/4 與 3/8），所以比較是合法的。", "en": "One failing triple (h,i,j) refutes the Markov property; both conditioning events have positive probability (1/4 and 3/8), so the comparison is legitimate." } }, "leaf_basis": "p_markov_def + p_cond_prob + quantifier_negation" }, { "id": "S15", "uses": [ "G3", "S07", "S08", "S09", "S10", "L_product_marg" ], "before_tex": "Q(x_0,x_1,x_2)=p_0(x_0)p_1(x_1)p_2(x_2)", "after_tex": "Q\\ \\text{has marginals}\\ p_0,p_1,p_2,\\ \\text{but}\\ Q(X_1=1,X_2=1)=\\tfrac{25}{64}\\neq\\tfrac12", "relation": "implies", "rule": "L_product_marg", "conditions": [], "explanation": { "zh-Hant": "依引理 L_product_marg（S15.1），Q 的三個邊際正是 p_0、p_1、p_2，與原過程完全相同。但 Q 下 (X_1,X_2)=(1,1) 的機率是 Σ_{x_0} p_0(x_0)·(5/8)·(5/8)=25/64（S15.2），而原律下是 1/2（S10）；25/64≠1/2（N15）。所以兩個不同的聯合律有相同的邊際：邊際表不決定過程。", "en": "By Lemma L_product_marg (S15.1), the three marginals of Q are exactly p_0, p_1, p_2, the same as the original process. But under Q the probability of (X_1,X_2)=(1,1) is Σ_{x_0} p_0(x_0)·(5/8)·(5/8)=25/64 (S15.2), whereas under the original law it is 1/2 (S10); 25/64≠1/2 (N15). So two different joint laws share the same marginals: the marginal tables do not determine the process." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_product_marg", "S10" ], "explanation": { "zh-Hant": "只需一個邊際相同而聯合不同的例子，就否定「邊際決定過程」。Q 本身的 Markov 性或其他性質在這裡不需要。", "en": "One example with equal marginals and different joints refutes 'marginals determine the process'. Whether Q is Markov, or any other property of Q, is not needed here." } }, "leaf_basis": "L_product_marg + numeric_evaluation" }, { "id": "S15.1", "uses": [ "distributive", "finite_sum_rules" ], "before_tex": "\\sum_{x_1,x_2}q_0(x_0)q_1(x_1)q_2(x_2)", "after_tex": "q_0(x_0)\\Big(\\sum_{x_1}q_1(x_1)\\Big)\\Big(\\sum_{x_2}q_2(x_2)\\Big)=q_0(x_0)\\cdot1\\cdot1", "relation": "equivalent", "rule": "distributive", "conditions": [ "\\sum q_1=\\sum q_2=1" ], "explanation": { "zh-Hant": "固定 x_0，對 x_1、x_2 求和；分配律把 q_0(x_0) 提出，四項和因式分解成兩個和的乘積（N14 驗證這個多項式恆等式）。每個 q_n 總和為 1，所以 Q 的時刻 0 邊際是 q_0(x_0)。時刻 1、2 同理。Q 的每個值非負，總和為 1·1·1=1，所以 Q 是合法 PMF。這證得引理 L_product_marg。", "en": "Fix x_0 and sum over x_1, x_2; the distributive law factors out q_0(x_0), and the four-term sum factors as a product of two sums (N14 checks this polynomial identity). Each q_n sums to 1, so the time-0 marginal of Q is q_0(x_0). Times 1 and 2 are the same. Every value of Q is nonnegative and the total is 1·1·1=1, so Q is a valid PMF. This proves Lemma L_product_marg." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "finite_sum_rules" ], "explanation": { "zh-Hant": "展開與因式分解互逆，雙向成立；最後一步需要各 q_n 的總和為 1。", "en": "Expanding and factoring undo each other, so both directions hold; the final step needs each q_n to sum to 1." } }, "leaf_basis": "distributive + finite_sum_rules + mul_one" }, { "id": "S15.2", "uses": [ "S15.1", "S07", "S08", "S09" ], "before_tex": "Q(X_1=1,X_2=1)=\\sum_{x_0}p_0(x_0)p_1(1)p_2(1)", "after_tex": "1\\cdot\\tfrac58\\cdot\\tfrac58=\\tfrac{25}{64}", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "同樣提出 p_1(1)p_2(1)=(5/8)(5/8)，剩下 Σ_{x_0}p_0(x_0)=1，得 25/64（N16）。原律的值 1/2=32/64，所以兩者不同。", "en": "Again factor out p_1(1)p_2(1)=(5/8)(5/8); what remains is Σ_{x_0}p_0(x_0)=1, giving 25/64 (N16). The original value is 1/2=32/64, so the two differ." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "numeric_evaluation" ], "explanation": { "zh-Hant": "精確有理數計算，可由 N15、N16 重算。", "en": "An exact rational computation, recomputed by N15 and N16." } }, "leaf_basis": "distributive + numeric_evaluation" }, { "id": "S16", "uses": [ "S05", "S07", "S08", "S09", "S10", "S11", "S14", "S15" ], "before_tex": "\\text{S05 to S15}", "after_tex": "p_0,p_1,p_2,\\ \\text{joint}(X_1,X_2);\\ \\text{not stationary};\\ \\text{not Markov};\\ \\text{marginals do not determine the law}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "總結：完整路徑律給出同一空間上的過程（S05）、三個邊際與兩時刻聯合（S07 到 S10）；它既不平穩（S11）也不具 Markov 性（S14）；而且邊際相同的乘積律 Q 有不同的聯合分布（S15）。", "en": "Summary: the complete path law gives a process on one space (S05), three marginals and a two-time joint (S07 to S10); it is neither stationary (S11) nor Markov (S14); and the product law Q with the same marginals has a different joint law (S15)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S05", "S10", "S11", "S14", "S15" ], "explanation": { "zh-Hant": "只是合取已證結論；所有結論都只針對 G1、G2 這個有限時域實例。", "en": "Only conjoins conclusions already proved; every conclusion concerns only the finite-horizon instance of G1 and G2." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "邊際 p_0=(1/2,1/2)、p_1=(3/8,5/8)、p_2=(3/8,5/8)；(X_1,X_2) 的聯合分布為 (0,0):1/4、(0,1):1/8、(1,0):1/8、(1,1):1/2。過程不平穩（p_0≠p_1），也不具 Markov 性（P(X_2=1∣X_1=1,X_0=0)=1、P(X_2=1∣X_1=1,X_0=1)=2/3、P(X_2=1∣X_1=1)=4/5）。邊際相同的乘積律給 P(X_1=1,X_2=1)=25/64≠1/2，所以邊際表不是完整的過程。", "en": "Marginals p_0=(1/2,1/2), p_1=(3/8,5/8), p_2=(3/8,5/8); the joint law of (X_1,X_2) is (0,0):1/4, (0,1):1/8, (1,0):1/8, (1,1):1/2. The process is not stationary (p_0≠p_1) and not Markov (P(X_2=1∣X_1=1,X_0=0)=1, P(X_2=1∣X_1=1,X_0=1)=2/3, P(X_2=1∣X_1=1)=4/5). The product law with the same marginals gives P(X_1=1,X_2=1)=25/64≠1/2, so the marginal tables are not the whole process." }, "negative_controls": [ { "claim": { "zh-Hant": "知道 X_0、X_1、X_2 的三個邊際表就知道整個過程", "en": "Knowing the three marginal tables of X_0, X_1, X_2 determines the whole process" }, "expected": "reject", "reason": { "zh-Hant": "乘積律 Q 有相同邊際，但 Q(X_1=1,X_2=1)=25/64≠1/2（S15）。", "en": "The product law Q has the same marginals but Q(X_1=1,X_2=1)=25/64≠1/2 (S15)." } }, { "claim": { "zh-Hant": "任何有限時域的離散過程都具 Markov 性", "en": "Every finite-horizon discrete process is Markov" }, "expected": "reject", "reason": { "zh-Hant": "本例 P(X_2=1∣X_1=1,X_0=0)=1≠2/3=P(X_2=1∣X_1=1,X_0=1)（S12 到 S14）。", "en": "Here P(X_2=1∣X_1=1,X_0=0)=1≠2/3=P(X_2=1∣X_1=1,X_0=1) (S12 to S14)." } }, { "claim": { "zh-Hant": "X_1 與 X_2 同分布，所以 X_1=X_2 幾乎必然成立", "en": "X_1 and X_2 have the same distribution, so X_1=X_2 almost surely" }, "expected": "reject", "reason": { "zh-Hant": "P(X_1≠X_2)=P({ω_3,ω_4})=1/4>0。", "en": "P(X_1≠X_2)=P({ω_3,ω_4})=1/4>0." } }, { "claim": { "zh-Hant": "由 {0,1,2} 上的路徑律可以讀出一個對所有時刻都適用的轉移律", "en": "From the path law on {0,1,2} one can read off a transition law valid for all times" }, "expected": "reject", "reason": { "zh-Hant": "MKO 假設 a2 明示有限時域聯合律不外推到無限時間或轉移律；而且本例連時刻 1 的轉移都依賴 X_0。", "en": "MKO assumption a2 states that a finite-horizon joint law does not extend to infinite time or a transition law; moreover here even the transition at time 1 depends on X_0." } } ], "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": "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": "cartesian_product_def", "kind": "prerequisite", "statement": { "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。", "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d." } }, { "id": "function_def", "kind": "prerequisite", "statement": { "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。", "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a." } }, { "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": "case_analysis", "kind": "prerequisite", "statement": { "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。", "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility." } }, { "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": "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": "set_builder", "kind": "prerequisite", "statement": { "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。", "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)." } }, { "id": "contrapositive", "kind": "prerequisite", "statement": { "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。", "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q." } }, { "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": "quantifier_negation", "kind": "prerequisite", "statement": { "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。", "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)." } }, { "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_one", "kind": "prerequisite", "statement": { "zh-Hant": "乘法單位元：1·u=u。", "en": "Multiplicative identity: 1·u=u." } }, { "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": "mko_process_def", "kind": "cited", "source": { "mko_id": "mko-discrete-stochastic-process", "sha256": "1b73d458f056e14bbaaf318ebfecf7e93214b9895dae967bfcffc48910854832", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "離散時間隨機過程 (X_n) 是同一機率空間上的隨機變數族。以有限時域完整路徑分布指定 joint law，從而導出各時刻 marginal 與兩時刻 joint；這不自動給 Markov 性或 stationarity。", "en": "A discrete-time stochastic process (X_n) is a family of random variables on the same probability space. A complete path distribution over a finite horizon specifies the joint law, from which time marginals and two-time joint distributions are derived. Neither the Markov property nor stationarity follows automatically." } }, { "id": "p_cond_prob", "kind": "prerequisite", "statement": { "zh-Hant": "條件機率：P(B)>0 時 P(A∣B)=P(A∩B)/P(B)；P(B)=0 時不定義。", "en": "Conditional probability: for P(B)>0, P(A∣B)=P(A∩B)/P(B); it is undefined when P(B)=0." } }, { "id": "p_markov_def", "kind": "prerequisite", "statement": { "zh-Hant": "Markov 性（在時刻 1）：對所有 h、i、j 且 P(X_0=h, X_1=i)>0，P(X_2=j ∣ X_1=i, X_0=h)=P(X_2=j ∣ X_1=i)。一個過程具 Markov 性，必須在每個時刻都滿足這樣的條件。", "en": "Markov property (at time 1): for all h, i, j with P(X_0=h, X_1=i)>0, P(X_2=j ∣ X_1=i, X_0=h)=P(X_2=j ∣ X_1=i). A Markov process must satisfy such a condition at every time." } }, { "id": "p_stationary_nec", "kind": "prerequisite", "statement": { "zh-Hant": "（嚴格）平穩：時間平移不改變任何聯合分布；其必要條件之一是所有單一時刻的邊際分布相同。", "en": "(Strict) stationarity: a time shift changes no joint law; one necessary condition is that all one-time marginal laws coincide." } }, { "id": "L_marginal", "kind": "lemma", "proved_in": [ "S06.1" ], "statement": { "zh-Hant": "對 n∈{0,1,2} 與 s∈{0,1}：P(X_n=s)=Σ_{ω∈Ω, ω_n=s} p(ω)；兩時刻聯合 P(X_m=s, X_n=t) 同理為 Σ_{ω_m=s, ω_n=t} p(ω)。", "en": "For n∈{0,1,2} and s∈{0,1}: P(X_n=s)=Σ_{ω∈Ω, ω_n=s} p(ω); likewise the two-time joint P(X_m=s, X_n=t)=Σ_{ω_m=s, ω_n=t} p(ω)." } }, { "id": "L_product_marg", "kind": "lemma", "proved_in": [ "S15.1" ], "statement": { "zh-Hant": "若 q_0、q_1、q_2 是 {0,1} 上的機率向量，Q(x_0,x_1,x_2)=q_0(x_0)q_1(x_1)q_2(x_2)，則 Q 的時刻 n 邊際是 q_n。", "en": "If q_0, q_1, q_2 are probability vectors on {0,1} and Q(x_0,x_1,x_2)=q_0(x_0)q_1(x_1)q_2(x_2), then the time-n marginal of Q is q_n." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S03", "lhs": "1/4+1/4+1/8+1/8+1/4", "rhs": "1" }, { "id": "N2", "step": "S07", "lhs": "1/4+1/4", "rhs": "1/2" }, { "id": "N3", "step": "S07", "lhs": "1/8+1/8+1/4", "rhs": "1/2" }, { "id": "N4", "step": "S08", "lhs": "1/4+1/8", "rhs": "3/8" }, { "id": "N5", "step": "S08", "lhs": "1/4+1/8+1/4", "rhs": "5/8" }, { "id": "N6", "step": "S09", "lhs": "1/4+1/8", "rhs": "3/8" }, { "id": "N7", "step": "S09", "lhs": "1/4+1/8+1/4", "rhs": "5/8" }, { "id": "N8", "step": "S10", "lhs": "1/4+1/8+1/8+1/2", "rhs": "1" }, { "id": "N9", "step": "S10", "lhs": "1/4+1/4", "rhs": "1/2" }, { "id": "N10", "step": "S11", "lhs": "1/2", "relation": "!=", "rhs": "3/8" }, { "id": "N11", "step": "S12", "lhs": "(1/4)/(1/4)", "rhs": "1" }, { "id": "N12", "step": "S13", "lhs": "(1/4)/(1/8+1/4)", "rhs": "2/3" }, { "id": "N13", "step": "S14", "lhs": "(1/2)/(5/8)", "rhs": "4/5" }, { "id": "N14", "step": "S15.1", "vars": [ "a", "b0", "b1", "c0", "c1" ], "degree": 1, "lhs": "a*b0*c0+a*b0*c1+a*b1*c0+a*b1*c1", "rhs": "a*(b0+b1)*(c0+c1)" }, { "id": "N15", "step": "S15", "lhs": "25/64", "relation": "!=", "rhs": "1/2" }, { "id": "N16", "step": "S15.2", "lhs": "(1/2+1/2)*(5/8)*(5/8)", "rhs": "25/64" }, { "id": "N17", "lhs": "1/8+1/8", "rhs": "1/4" } ] }, "presentation": { "md_scope": { "zh-Hant": "只處理 G1、G2 指定的有限時域路徑律（MKO 假設 a1、a2）。不外推到無限時間、轉移律、平穩性或遍歷性；Markov 性只在時刻 1 以定義檢驗。比較用的乘積律 Q 只用來顯示邊際不決定聯合分布。不改動 canonical MKO 或其英文 edition。", "en": "Treats only the finite-horizon path law specified by G1 and G2 (MKO assumptions a1, a2). Nothing is extended to infinite time, a transition law, stationarity or ergodicity; the Markov property is tested by its definition at time 1 only. The comparison product law Q serves only to show that marginals do not determine the joint law. The canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "\\Omega=\\{\\omega_1,\\omega_2,\\omega_3,\\omega_4,\\omega_5\\}", "完整路徑支撐", "the complete path support" ], [ "p(\\omega)", "精確路徑 PMF", "the exact path PMF" ], [ "X_n(\\omega)=\\omega_n", "時刻 n 的座標（n=0,1,2）", "the coordinate at time n (n=0,1,2)" ], [ "P", "由 p 導出的機率律 P(A)=Σ_{ω∈A}p(ω)", "the probability law induced by p, P(A)=Σ_{ω∈A}p(ω)" ], [ "p_n(s)=P(X_n=s)", "時刻 n 的邊際", "the marginal at time n" ], [ "Q", "邊際的乘積律（比較用）", "the product law of the marginals (for comparison)" ] ], "md_error_controls": [ { "title": "邊際表不是完整過程 / Marginal tables are not the whole process", "zh": "MKO 指出的誤解是把各時刻的邊際分布集合當成完整隨機過程。S15 的乘積律 Q 與原律有完全相同的三個邊際，卻給 P(X_1=1,X_2=1)=25/64 而非 1/2。時間之間的依賴只記錄在完整路徑律中（MKO 段落 0），邊際把它丟掉了。", "en": "The misconception named by the MKO is to take the collection of time marginals as the complete stochastic process. The product law Q of S15 has exactly the same three marginals as the original law, yet gives P(X_1=1,X_2=1)=25/64 instead of 1/2. Dependence across time is recorded only in the complete path law (MKO paragraph 0), and the marginals discard it." }, { "title": "不要自動宣稱 Markov / Do not assert Markov automatically", "zh": "Markov 性是對條件分布的責任，必須用定義檢驗（MKO 段落 2）。這裡 P(X_2=1∣X_1=1,X_0=0)=1 而 P(X_2=1∣X_1=1,X_0=1)=2/3，所以已知現在仍需過去。另外條件事件必須有正機率才可比較；本稿用到的兩個條件事件機率為 1/4 與 3/8。", "en": "The Markov property is an obligation about conditional distributions and must be checked from the definition (MKO paragraph 2). Here P(X_2=1∣X_1=1,X_0=0)=1 but P(X_2=1∣X_1=1,X_0=1)=2/3, so the past still matters given the present. Also the conditioning events must have positive probability to be compared; the two used here have probabilities 1/4 and 3/8." }, { "title": "邊際相同不表示變數相同 / Equal marginals do not mean equal variables", "zh": "X_1 與 X_2 有相同的邊際 (3/8,5/8)，但它們不是同一個隨機變數：在 ω_3 上 X_1=0、X_2=1，在 ω_4 上 X_1=1、X_2=0，所以 P(X_1≠X_2)=1/8+1/8=1/4（N17）。分布相同只是說兩個律相同，不是說逐點相等。", "en": "X_1 and X_2 have the same marginal (3/8,5/8), but they are not the same random variable: on ω_3, X_1=0 and X_2=1, and on ω_4, X_1=1 and X_2=0, so P(X_1≠X_2)=1/8+1/8=1/4 (N17). Equal distributions mean equal laws, not pointwise equality." } ], "md_open_items": [ { "zh-Hant": "Markov 性只在時刻 1 檢驗；本稿不討論三點時域 {0,1,2} 以外的情形。時刻 0 在此時域沒有更早的觀察過去；這不是另作無限時域 Markov 性的宣稱。", "en": "The Markov property is tested at time 1 only; nothing beyond the three-point horizon {0,1,2} is discussed. Time 0 has no earlier observed past within this horizon; no infinite-horizon Markov claim is made." }, { "zh-Hant": "乘積律 Q 是否具 Markov 性與本稿結論無關，未加證明。", "en": "Whether the product law Q is Markov is irrelevant to the conclusions and is not proved." }, { "zh-Hant": "有限 Markov 鏈、轉移矩陣與路徑機率（MKO 的 formal next）另見後續條目。", "en": "Finite Markov chains, transition matrices and path probabilities (the MKO's formal next) belong to a later entry." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "494d4109b4e53c43ae761b91728e2a7845b2d0d29b33ffa8731e9bd3c6b60e80", "original_source_sha256": "0c576baf78d0d575087097afb891ccdf82353c877594afdc55475e74c390656e", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [ "BR-M059" ], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }