{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-variance-01", "mko_id": "mko-variance", "title": { "zh-Hant": "有限機率模型的變異數：平移、縮放與零變異數的刻畫", "en": "Variance in finite probability models: translation, scaling and the zero-variance characterization" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-variance-01", "source": { "mko_id": "mko-variance", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "282681c1ed2d55f1c853c9759022470c991dd46c2b150626fa4bf654642032bc", "english_sha256": "a07a03bffcd0bef45c5514c2b4ba2db48202bf94f336f0d8589082871649306d" }, "mode": "theorem_proof", "audience_prerequisites": [ "finite_sum_rules", "square_nonneg", "order_basics", "distributive", "power_laws", "numeric_evaluation" ], "goal": { "zh-Hant": "在有限合法機率模型中證明 Var(X)≥0、Var(X+c)=Var(X)、Var(kX)=k²Var(X)，以及 Var(X)=0 ⇔ X 在所有正權重結果上取同一個值。再以 S={1,2,3,4}、w₁=1/2、w₂=1/4、w₃=1/4、w₄=0、X=(1,3,−1,10) 計算 μ=1、Var(X)=2、Var(X+5)=2、Var(−3X)=18，並以 Z=(2,2,2,7) 說明零權重結果上的不同值不影響零變異數。", "en": "In a valid finite probability model, prove Var(X)≥0, Var(X+c)=Var(X), Var(kX)=k²Var(X), and Var(X)=0 ⇔ X takes one common value at every positive-weight outcome. Then with S={1,2,3,4}, w₁=1/2, w₂=1/4, w₃=1/4, w₄=0 and X=(1,3,−1,10), compute μ=1, Var(X)=2, Var(X+5)=2 and Var(−3X)=18, and use Z=(2,2,2,7) to show that a different value at a zero-weight outcome does not prevent zero variance." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "一般模型：S 有限非空，對所有 s 有 w_s≥0 且 Σ_{s∈S} w_s=1；X:S→ℝ 為總函數；μ=E[X]=Σ w_s X(s)；c、k 為實數。", "en": "General model: S is finite and nonempty, w_s≥0 for all s with Σ_{s∈S} w_s=1; X:S→ℝ is total; μ=E[X]=Σ w_s X(s); c and k are reals." } }, { "id": "G2", "statement": { "zh-Hant": "實例：S={1,2,3,4}，w₁=1/2、w₂=1/4、w₃=1/4、w₄=0。", "en": "Instance: S={1,2,3,4}, w₁=1/2, w₂=1/4, w₃=1/4, w₄=0." } }, { "id": "G3", "statement": { "zh-Hant": "X(1)=1、X(2)=3、X(3)=−1、X(4)=10；Z(1)=Z(2)=Z(3)=2、Z(4)=7。", "en": "X(1)=1, X(2)=3, X(3)=−1, X(4)=10; Z(1)=Z(2)=Z(3)=2, Z(4)=7." } } ], "steps": [ { "id": "S01", "uses": [ "G2", "support_def" ], "before_tex": "w=(\\tfrac12,\\tfrac14,\\tfrac14,0)", "after_tex": "w_s\\ge0,\\ \\sum_s w_s=1,\\ \\mathrm{supp}=\\{1,2,3\\}", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "先驗模型：四個權重都非負，總和 1/2+1/4+1/4+0=1，所以合法。正權重支持是 {1,2,3}；結果 4 屬於樣本空間 S，但權重為 0，不在支持中。MKO 假設 a2 要求把這兩者分開。", "en": "Validate the model first: the four weights are nonnegative and 1/2+1/4+1/4+0=1, so they are valid. The positive-weight support is {1,2,3}; outcome 4 belongs to the sample space S but has weight 0 and is not in the support. MKO assumption a2 asks for these two sets to be kept apart." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G2", "support_def", "numeric_evaluation" ], "explanation": { "zh-Hant": "精確計算與定義判讀，可獨立重算。", "en": "Exact arithmetic and reading a definition; independently recomputable." } }, "leaf_basis": "numeric_evaluation + support_def" }, { "id": "S02", "uses": [ "S01", "G3", "mko_expectation_lin" ], "before_tex": "\\mu=E[X]", "after_tex": "\\tfrac12\\cdot1+\\tfrac14\\cdot3+\\tfrac14\\cdot(-1)+0\\cdot10=1", "relation": "equivalent", "rule": "mko_expectation_lin", "conditions": [], "explanation": { "zh-Hant": "依期望值定義：1/2+3/4−1/4+0=1。結果 4 的值 10 乘上權重 0，貢獻為 0。", "en": "By the definition of expectation: 1/2+3/4−1/4+0=1. The value 10 at outcome 4 is multiplied by weight 0 and contributes 0." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mko_expectation_lin", "numeric_evaluation" ], "explanation": { "zh-Hant": "定義展開加精確計算，兩邊同值。", "en": "Unfolding the definition plus exact arithmetic; both sides have the same value." } }, "leaf_basis": "mko_expectation_lin + numeric_evaluation" }, { "id": "S03", "uses": [ "S02", "G3" ], "before_tex": "X(s)-\\mu\\ \\ (s=1,2,3,4)", "after_tex": "0,\\ 2,\\ -2,\\ 9", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "逐點計算偏差：1−1=0、3−1=2、−1−1=−2、10−1=9。偏差對每個結果都有定義，包括零權重的結果 4。", "en": "Compute the deviations pointwise: 1−1=0, 3−1=2, −1−1=−2, 10−1=9. The deviation is defined at every outcome, including the zero-weight outcome 4." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "numeric_evaluation" ], "explanation": { "zh-Hant": "整數減法，可逆向核對（偏差加 μ 回到原值）。", "en": "Integer subtraction, checkable backwards (deviation plus μ returns the value)." } }, "leaf_basis": "numeric_evaluation" }, { "id": "S04", "uses": [ "S03", "S01", "mko_variance_def" ], "before_tex": "\\mathrm{Var}(X)=\\sum_s w_s(X(s)-\\mu)^2", "after_tex": "\\tfrac12\\cdot0^2+\\tfrac14\\cdot2^2+\\tfrac14\\cdot(-2)^2+0\\cdot9^2=2", "relation": "equivalent", "rule": "mko_variance_def", "conditions": [], "explanation": { "zh-Hant": "代入定義：0+1+1+0=2。零權重結果 4 的偏差 9 很大，但乘上權重 0 後貢獻為 0。", "en": "Substitute into the definition: 0+1+1+0=2. The zero-weight outcome 4 has a large deviation 9, but multiplied by weight 0 it contributes 0." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mko_variance_def", "numeric_evaluation" ], "explanation": { "zh-Hant": "定義展開與精確計算，兩邊同值。", "en": "Unfolding the definition and exact arithmetic; both sides have the same value." } }, "leaf_basis": "mko_variance_def + power_laws + numeric_evaluation" }, { "id": "S05", "uses": [ "G1", "mko_variance_def" ], "before_tex": "w_s\\ge0,\\ (X(s)-\\mu)^2\\ge0", "after_tex": "\\mathrm{Var}(X)\\ge0", "relation": "implies", "rule": "square_nonneg", "conditions": [], "explanation": { "zh-Hant": "一般性質。每一項 t_s=w_s(X(s)−μ)²：平方非負；若 w_s=0，t_s=0；若 w_s>0，把 0≤(X(s)−μ)² 兩邊乘以正數 w_s 保持順序，得 t_s≥0。非負數的有限和非負，所以 Var(X)≥0。", "en": "General property. Each term t_s=w_s(X(s)−μ)²: the square is nonnegative; if w_s=0 then t_s=0; if w_s>0, multiplying 0≤(X(s)−μ)² by the positive w_s keeps the order, so t_s≥0. A finite sum of nonnegative numbers is nonnegative, so Var(X)≥0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "square_nonneg", "order_basics", "finite_sum_rules" ], "explanation": { "zh-Hant": "需要權重非負；若允許負權重，結論可能失敗。這是單向推論，不宣稱 Var(X)>0。", "en": "It needs nonnegative weights; with a negative weight the conclusion could fail. This is one-way and does not claim Var(X)>0." } }, "leaf_basis": "square_nonneg + order_basics + mul_zero + finite_sum_rules" }, { "id": "S06", "uses": [ "G1", "mko_variance_def" ], "before_tex": "\\mathrm{Var}(X+c)", "after_tex": "\\mathrm{Var}(X)", "relation": "equivalent", "rule": "L_shift", "conditions": [], "explanation": { "zh-Hant": "依引理 L_shift：平移不改變變異數。證明在 S06.1、S06.2：平均也平移 c，所以偏差不變。", "en": "By Lemma L_shift, translation does not change variance. The proof is in S06.1 and S06.2: the mean also shifts by c, so the deviations are unchanged." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_shift" ], "explanation": { "zh-Hant": "兩邊是同一個數（引理），可雙向替換。", "en": "Both sides are the same number (by the lemma) and can replace each other." } }, "leaf_basis": "mko_expectation_lin + eq_replace" }, { "id": "S06.1", "uses": [ "G1", "mko_expectation_lin" ], "before_tex": "E[X+c]", "after_tex": "\\mu+c", "relation": "equivalent", "rule": "mko_expectation_lin", "conditions": [], "explanation": { "zh-Hant": "依期望值線性性（a=1、b=c）：E[X+c]=E[X]+c=μ+c。這需要權重總和為 1。", "en": "By linearity of expectation (a=1, b=c): E[X+c]=E[X]+c=μ+c. This needs the weights to sum to 1." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mko_expectation_lin" ], "explanation": { "zh-Hant": "線性性是等式，可雙向使用。", "en": "Linearity is an equation and can be used either way." } }, "leaf_basis": "mko_expectation_lin" }, { "id": "S06.2", "uses": [ "S06.1", "eq_replace" ], "before_tex": "\\sum_s w_s\\bigl((X(s)+c)-(\\mu+c)\\bigr)^2", "after_tex": "\\sum_s w_s(X(s)-\\mu)^2", "relation": "equivalent", "rule": "eq_replace", "conditions": [], "explanation": { "zh-Hant": "對每個 s：(X(s)+c)−(μ+c)=X(s)−μ（加法結合、交換與反元素）。每一項都換成相等的項，和不變。S06.1 與 S06.2 合起來證得引理 L_shift。", "en": "For each s: (X(s)+c)−(μ+c)=X(s)−μ (associativity, commutativity and inverses for addition). Each term is replaced by an equal term, so the sum is unchanged. S06.1 and S06.2 together prove Lemma L_shift." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "eq_replace", "add_inverse", "add_comm" ], "explanation": { "zh-Hant": "逐項等值替換，雙向成立。", "en": "Term-by-term replacement of equals; it holds both ways." } }, "leaf_basis": "eq_replace + add_assoc + add_comm + add_inverse + add_zero" }, { "id": "S07", "uses": [ "G1", "mko_variance_def" ], "before_tex": "\\mathrm{Var}(kX)", "after_tex": "k^2\\,\\mathrm{Var}(X)", "relation": "equivalent", "rule": "L_scale", "conditions": [], "explanation": { "zh-Hant": "依引理 L_scale：縮放 k 倍使變異數乘以 k²。證明在 S07.1 至 S07.3；對所有實數 k 成立，包括 k=0 與負 k。", "en": "By Lemma L_scale, scaling by k multiplies the variance by k². The proof is in S07.1 to S07.3 and holds for every real k, including k=0 and negative k." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_scale" ], "explanation": { "zh-Hant": "兩邊是同一個數（引理），可雙向替換。", "en": "Both sides are the same number (by the lemma) and can replace each other." } }, "leaf_basis": "mko_expectation_lin + power_laws + finite_sum_rules" }, { "id": "S07.1", "uses": [ "G1", "mko_expectation_lin" ], "before_tex": "E[kX]", "after_tex": "k\\mu", "relation": "equivalent", "rule": "mko_expectation_lin", "conditions": [], "explanation": { "zh-Hant": "依期望值線性性（a=k、b=0）：E[kX]=kμ。", "en": "By linearity of expectation (a=k, b=0): E[kX]=kμ." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mko_expectation_lin" ], "explanation": { "zh-Hant": "線性性是等式，可雙向使用。", "en": "Linearity is an equation and can be used either way." } }, "leaf_basis": "mko_expectation_lin" }, { "id": "S07.2", "uses": [ "S07.1", "distributive", "power_laws" ], "before_tex": "(kX(s)-k\\mu)^2", "after_tex": "k^2(X(s)-\\mu)^2", "relation": "equivalent", "rule": "power_laws", "conditions": [], "explanation": { "zh-Hant": "先用分配律提出 k：kX(s)−kμ=k(X(s)−μ)；再用 (ab)²=a²b²。偏差乘 k，平方後帶出 k²，負號在平方中消失。", "en": "First factor out k by distributivity: kX(s)−kμ=k(X(s)−μ); then use (ab)²=a²b². The deviation is multiplied by k, and squaring brings out k², so a negative sign disappears." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "power_laws" ], "explanation": { "zh-Hant": "兩條都是對所有實數成立的恆等式，雙向成立。", "en": "Both are identities valid for all reals, so the step holds both ways." } }, "leaf_basis": "distributive + power_laws + mul_comm" }, { "id": "S07.3", "uses": [ "S07.2", "finite_sum_rules" ], "before_tex": "\\sum_s w_s\\,k^2(X(s)-\\mu)^2", "after_tex": "k^2\\sum_s w_s(X(s)-\\mu)^2=k^2\\,\\mathrm{Var}(X)", "relation": "equivalent", "rule": "finite_sum_rules", "conditions": [], "explanation": { "zh-Hant": "常數 k² 可從有限和中提出，剩下的和依定義是 Var(X)。S07.1 至 S07.3 合起來證得引理 L_scale。k=0 時得 Var(0)=0，與常數變數的變異數一致。", "en": "The constant k² factors out of the finite sum, and what remains is Var(X) by definition. S07.1 to S07.3 together prove Lemma L_scale. For k=0 this gives Var(0)=0, consistent with the variance of a constant variable." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "finite_sum_rules", "def_unfold" ], "explanation": { "zh-Hant": "提出常數與折回定義都是等式，雙向成立。", "en": "Factoring out a constant and folding back the definition are equations and hold both ways." } }, "leaf_basis": "finite_sum_rules + def_unfold" }, { "id": "S08", "uses": [ "S06", "S04", "G3" ], "before_tex": "\\mathrm{Var}(X+5),\\ X+5=(6,8,4,15)", "after_tex": "\\mathrm{Var}(X+5)=\\mathrm{Var}(X)=2", "relation": "implies", "rule": "L_shift", "conditions": [], "explanation": { "zh-Hant": "依 S06（c=5），Var(X+5)=2。直接核對：X+5 的值為 6、8、4、15，平均 1/2·6+1/4·8+1/4·4+0·15=6=μ+5，偏差 0、2、−2、9 與原來相同，變異數仍為 2。", "en": "By S06 (c=5), Var(X+5)=2. Direct check: X+5 has values 6, 8, 4, 15, mean 1/2·6+1/4·8+1/4·4+0·15=6=μ+5, deviations 0, 2, −2, 9 as before, so the variance is still 2." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06", "S04" ], "explanation": { "zh-Hant": "結論由引理得出；直接計算只是交叉核對。", "en": "The conclusion comes from the lemma; the direct computation is only a cross-check." } }, "leaf_basis": "mko_variance_def + numeric_evaluation" }, { "id": "S09", "uses": [ "S07", "S04", "G3" ], "before_tex": "\\mathrm{Var}(-3X),\\ -3X=(-3,-9,3,-30)", "after_tex": "\\mathrm{Var}(-3X)=(-3)^2\\cdot2=18", "relation": "implies", "rule": "L_scale", "conditions": [], "explanation": { "zh-Hant": "依 S07（k=−3），Var(−3X)=9·2=18。直接核對：−3X 的值為 −3、−9、3、−30，平均 −3=−3μ，偏差 0、−6、6、−27，變異數 1/4·36+1/4·36+0·729=18。負係數不使變異數變負。", "en": "By S07 (k=−3), Var(−3X)=9·2=18. Direct check: −3X has values −3, −9, 3, −30, mean −3=−3μ, deviations 0, −6, 6, −27, and variance 1/4·36+1/4·36+0·729=18. A negative factor does not make the variance negative." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S07", "S04" ], "explanation": { "zh-Hant": "結論由引理得出；直接計算只是交叉核對。", "en": "The conclusion comes from the lemma; the direct computation is only a cross-check." } }, "leaf_basis": "mko_variance_def + numeric_evaluation" }, { "id": "S10", "uses": [ "finite_sum_rules", "order_basics" ], "before_tex": "t_s\\ge0,\\ \\sum_s t_s=0", "after_tex": "t_s=0\\ \\ \\forall s", "relation": "implies", "rule": "L_nonneg_sum_zero", "conditions": [], "explanation": { "zh-Hant": "準備零變異數刻畫所需的引理 L_nonneg_sum_zero：非負數的有限和為零，則每一項都是零。證明在 S10.1、S10.2。", "en": "Prepare Lemma L_nonneg_sum_zero for the zero-variance characterization: if a finite sum of nonnegative numbers is zero, every term is zero. The proof is in S10.1 and S10.2." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_nonneg_sum_zero" ], "explanation": { "zh-Hant": "需要每項非負；若允許負項，例如 1+(−1)=0，結論不成立。", "en": "It needs every term nonnegative; with negative terms allowed, for example 1+(−1)=0, the conclusion fails." } }, "leaf_basis": "order_basics + finite_sum_rules + contradiction" }, { "id": "S10.1", "uses": [ "finite_sum_rules", "order_basics" ], "before_tex": "t_r>0\\ \\text{for some}\\ r", "after_tex": "\\sum_s t_s=t_r+\\sum_{s\\neq r}t_s\\ge t_r>0", "relation": "implies", "rule": "order_basics", "conditions": [ "t_s\\ge0" ], "explanation": { "zh-Hant": "暫時假設某一項 t_r>0。把和拆成 t_r 加上其餘各項；其餘各項非負，其和也非負，所以總和 ≥t_r>0。", "en": "Assume for the moment that some term t_r>0. Split the sum into t_r plus the remaining terms; the remaining terms are nonnegative, so is their sum, and the total is ≥t_r>0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "finite_sum_rules", "order_basics" ], "explanation": { "zh-Hant": "只用到其餘項非負與加法保序；這是暫時假設下的推論，在 S10.2 釋放。", "en": "Only the nonnegativity of the other terms and order-preservation under addition are used; this is reasoning under a temporary assumption, discharged in S10.2." } }, "leaf_basis": "finite_sum_rules + order_basics" }, { "id": "S10.2", "uses": [ "S10.1", "contradiction" ], "before_tex": "\\sum_s t_s>0\\ \\text{and}\\ \\sum_s t_s=0", "after_tex": "\\neg\\exists r:\\ t_r>0,\\ \\text{so}\\ t_s=0", "relation": "implies", "rule": "contradiction", "conditions": [], "explanation": { "zh-Hant": "S10.1 的結論與假設 Σ t_s=0 矛盾，所以沒有任何 t_r>0。每個 t_s 既 ≥0 又不 >0，依三分律 t_s=0。S10.1 與 S10.2 合起來證得引理 L_nonneg_sum_zero。", "en": "The conclusion of S10.1 contradicts the hypothesis Σ t_s=0, so no t_r>0. Each t_s is ≥0 and not >0, so by trichotomy t_s=0. S10.1 and S10.2 together prove Lemma L_nonneg_sum_zero." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "contradiction", "order_basics" ], "explanation": { "zh-Hant": "矛盾法釋放 S10.1 的暫時假設；三分律把「≥0 且不 >0」化為「=0」。", "en": "Proof by contradiction discharges the temporary assumption of S10.1; trichotomy turns '≥0 and not >0' into '=0'." } }, "leaf_basis": "contradiction + order_basics" }, { "id": "S11", "uses": [ "G1", "mko_variance_def", "support_def" ], "before_tex": "\\mathrm{Var}(X)=0", "after_tex": "X(s)=\\mu\\ \\ \\forall s\\in\\mathrm{supp}", "relation": "equivalent", "rule": "L_var_zero", "conditions": [], "explanation": { "zh-Hant": "依引理 L_var_zero：變異數為零恰好表示 X 在每個正權重結果上等於平均。條件只涉及支持，零權重結果的值不受限制。證明在 S11.1（⇒）與 S11.2（⇐）。", "en": "By Lemma L_var_zero, zero variance means exactly that X equals the mean at every positive-weight outcome. The condition involves only the support; values at zero-weight outcomes are unrestricted. The proof is in S11.1 (⇒) and S11.2 (⇐)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_var_zero" ], "explanation": { "zh-Hant": "兩個方向分別在子步中證明，所以是等價。", "en": "The two directions are proved in the sub-steps, so this is an equivalence." } }, "leaf_basis": "L_nonneg_sum_zero + zero_product + square_nonneg + case_analysis" }, { "id": "S11.1", "uses": [ "S05", "S10", "zero_product", "square_nonneg" ], "before_tex": "\\mathrm{Var}(X)=0,\\ s\\in\\mathrm{supp}", "after_tex": "X(s)=\\mu", "relation": "implies", "rule": "L_nonneg_sum_zero", "conditions": [ "w_s>0" ], "explanation": { "zh-Hant": "（⇒）S05 中已知每項 t_s=w_s(X(s)−μ)²≥0，而它們的和 Var(X)=0，依 L_nonneg_sum_zero 每項 t_s=0。對 s∈supp，w_s>0 故 w_s≠0，由零積性質得 (X(s)−μ)²=0，再由平方為零 ⇔ 底數為零得 X(s)=μ。", "en": "(⇒) By S05 each term t_s=w_s(X(s)−μ)²≥0, and their sum Var(X)=0, so by L_nonneg_sum_zero every t_s=0. For s∈supp, w_s>0 so w_s≠0, and the zero-product property gives (X(s)−μ)²=0; since a square is zero exactly when its base is zero, X(s)=μ." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S05", "L_nonneg_sum_zero", "zero_product", "square_nonneg" ], "explanation": { "zh-Hant": "對 w_s=0 的結果，t_s=0 不提供任何關於 X(s) 的資訊，所以結論只對支持成立。", "en": "For outcomes with w_s=0, t_s=0 gives no information about X(s), so the conclusion holds only on the support." } }, "leaf_basis": "L_nonneg_sum_zero + zero_product + square_nonneg" }, { "id": "S11.2", "uses": [ "G1", "support_def", "case_analysis" ], "before_tex": "X(s)=\\mu\\ \\ \\forall s\\in\\mathrm{supp}", "after_tex": "\\mathrm{Var}(X)=\\sum_s 0=0", "relation": "implies", "rule": "case_analysis", "conditions": [], "explanation": { "zh-Hant": "（⇐）每個 s 不是在支持內就是在支持外。在支持內：X(s)−μ=0，t_s=w_s·0=0。在支持外：w_s=0，t_s=0·(X(s)−μ)²=0。所有項都是 0，和為 0。S11.1 與 S11.2 合起來證得引理 L_var_zero。", "en": "(⇐) Each s is either in the support or not. In the support: X(s)−μ=0 and t_s=w_s·0=0. Outside: w_s=0 and t_s=0·(X(s)−μ)²=0. Every term is 0, so the sum is 0. S11.1 and S11.2 together prove Lemma L_var_zero." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "case_analysis", "support_def", "mul_zero" ], "explanation": { "zh-Hant": "兩個情形窮盡 S（support_def 保證支持外即 w_s=0）。", "en": "The two cases exhaust S (support_def guarantees w_s=0 outside the support)." } }, "leaf_basis": "case_analysis + support_def + mul_zero + finite_sum_rules" }, { "id": "S12", "uses": [ "S11" ], "before_tex": "\\mathrm{Var}(X)=0", "after_tex": "\\exists c\\ \\forall s\\in\\mathrm{supp}:\\ X(s)=c", "relation": "equivalent", "rule": "L_const_support", "conditions": [], "explanation": { "zh-Hant": "把 S11 改寫成 MKO 的說法：變異數為零恰好表示 X 在所有正權重結果上取同一個值。依引理 L_const_support，證明在 S12.1、S12.2。", "en": "Rewrite S11 in the MKO's words: zero variance means exactly that X takes one common value at all positive-weight outcomes. This is Lemma L_const_support, proved in S12.1 and S12.2." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_const_support" ], "explanation": { "zh-Hant": "兩個方向都在子步中證明。", "en": "Both directions are proved in the sub-steps." } }, "leaf_basis": "L_var_zero + exists_intro + finite_sum_rules" }, { "id": "S12.1", "uses": [ "S11", "exists_intro" ], "before_tex": "\\mathrm{Var}(X)=0", "after_tex": "X(s)=\\mu\\ \\ \\forall s\\in\\mathrm{supp}\\ \\Rightarrow\\ c=\\mu", "relation": "implies", "rule": "L_var_zero", "conditions": [], "explanation": { "zh-Hant": "（⇒）由 L_var_zero 的 ⇒ 方向，X 在支持上處處等於 μ，取 c=μ 即得共同值。", "en": "(⇒) By the ⇒ direction of L_var_zero, X equals μ throughout the support; taking c=μ gives the common value." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_var_zero", "exists_intro" ], "explanation": { "zh-Hant": "存在引入只需見證 c=μ。", "en": "Existential introduction needs only the witness c=μ." } }, "leaf_basis": "L_var_zero + exists_intro" }, { "id": "S12.2", "uses": [ "S11", "G1", "support_def" ], "before_tex": "X(s)=c\\ \\ \\forall s\\in\\mathrm{supp}", "after_tex": "\\mu=\\sum_{s\\in\\mathrm{supp}}w_s c+\\sum_{s\\notin\\mathrm{supp}}0\\cdot X(s)=c,\\ \\text{so}\\ \\mathrm{Var}(X)=0", "relation": "implies", "rule": "L_var_zero", "conditions": [], "explanation": { "zh-Hant": "（⇐）把 μ 的和式分成支持內外兩部分：支持外的權重都是 0，貢獻 0，也使支持內權重之和為 1；所以 μ=c·1=c。於是 X(s)=c=μ 對每個 s∈supp 成立，由 L_var_zero 的 ⇐ 方向得 Var(X)=0。S12.1 與 S12.2 合起來證得引理 L_const_support。", "en": "(⇐) Split the sum for μ into the parts inside and outside the support: outside, every weight is 0 and contributes 0, which also makes the weights inside sum to 1; so μ=c·1=c. Then X(s)=c=μ for every s∈supp, and the ⇐ direction of L_var_zero gives Var(X)=0. S12.1 and S12.2 together prove Lemma L_const_support." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_var_zero", "finite_sum_rules", "support_def", "mul_zero" ], "explanation": { "zh-Hant": "關鍵是 μ 只由支持上的值決定；支持外的值（例如 Z(4)=7）乘以 0 而消失。", "en": "The key point is that μ is determined by the values on the support alone; values outside (such as Z(4)=7) are multiplied by 0 and vanish." } }, "leaf_basis": "L_var_zero + finite_sum_rules + support_def + mul_zero" }, { "id": "S13", "uses": [ "S12", "S01", "G3" ], "before_tex": "Z=2\\ \\text{on}\\ \\mathrm{supp}=\\{1,2,3\\},\\ Z(4)=7", "after_tex": "\\mathrm{Var}(Z)=0", "relation": "implies", "rule": "L_const_support", "conditions": [], "explanation": { "zh-Hant": "Z 在支持 {1,2,3} 上都等於 2，依 S12 得 Var(Z)=0，雖然 Z 在整個 S 上不是常數（Z(4)=7）。直接核對：μ_Z=2·(1/2+1/4+1/4)+7·0=2，Var(Z)=0+0+0+0·25=0。", "en": "Z equals 2 throughout the support {1,2,3}, so by S12 Var(Z)=0, even though Z is not constant on all of S (Z(4)=7). Direct check: μ_Z=2·(1/2+1/4+1/4)+7·0=2 and Var(Z)=0+0+0+0·25=0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S12", "S01" ], "explanation": { "zh-Hant": "只用到 S12 的 ⇐ 方向與支持的判定；直接計算是交叉核對。", "en": "Only the ⇐ direction of S12 and the identification of the support are used; the direct computation is a cross-check." } }, "leaf_basis": "L_const_support + numeric_evaluation" }, { "id": "S14", "uses": [ "S12", "S01", "G3", "S04" ], "before_tex": "X(1)=1\\neq3=X(2),\\ 1,2\\in\\mathrm{supp}", "after_tex": "\\mathrm{Var}(X)\\neq0", "relation": "implies", "rule": "contrapositive", "conditions": [], "explanation": { "zh-Hant": "X 在支持中的兩個結果取不同值 1 與 3，所以沒有共同值；依 S12 的 ⇒ 方向取逆否，Var(X)≠0。這與 S04 的 Var(X)=2>0 一致。", "en": "X takes the different values 1 and 3 at two outcomes of the support, so it has no common value there; by the contrapositive of the ⇒ direction of S12, Var(X)≠0. This agrees with Var(X)=2>0 from S04." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S12", "contrapositive" ], "explanation": { "zh-Hant": "兩個支持內的不同值就足以否定共同值；結論只說不為零，數值由 S04 給出。", "en": "Two different values inside the support suffice to rule out a common value; the conclusion only says nonzero, and the value comes from S04." } }, "leaf_basis": "contrapositive + L_const_support" }, { "id": "S15", "uses": [ "S04", "S05", "S06", "S07", "S08", "S09", "S12", "S13" ], "before_tex": "\\text{S04 to S14}", "after_tex": "\\mathrm{Var}\\ge0,\\ \\mathrm{Var}(X+c)=\\mathrm{Var}(X),\\ \\mathrm{Var}(kX)=k^2\\mathrm{Var}(X),\\ \\mathrm{Var}=0\\iff\\text{const on supp}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "整理：四個一般性質都已證明；實例中 μ=1、Var(X)=2、Var(X+5)=2、Var(−3X)=18，Var(Z)=0 而 Z 只在零權重結果上不同。", "en": "Summary: the four general properties are proved; in the instance μ=1, Var(X)=2, Var(X+5)=2, Var(−3X)=18, and Var(Z)=0 although Z differs at a zero-weight outcome." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S04", "S05", "S06", "S07", "S08", "S09", "S12", "S13", "and_intro_elim" ], "explanation": { "zh-Hant": "目標 G 為 goal 的證明與計算；足夠條件 H 為各步結論的合取，皆已正向證成，H⇒G 只用合取引入。", "en": "Goal G is the proofs and computations in the goal; the sufficient condition H is the conjunction of the step results, all proved forwards, and H⇒G uses only conjunction introduction." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "在任何有限合法模型中 Var(X)≥0、Var(X+c)=Var(X)、Var(kX)=k²Var(X)，且 Var(X)=0 恰在 X 於正權重支持上取同一值時成立。實例：μ=1、Var(X)=2、Var(X+5)=2、Var(−3X)=18；Z=(2,2,2,7) 在支持 {1,2,3} 上恆為 2，所以 Var(Z)=0，Z(4)=7 不影響結果。", "en": "In every valid finite model Var(X)≥0, Var(X+c)=Var(X) and Var(kX)=k²Var(X), and Var(X)=0 exactly when X takes one common value throughout the positive-weight support. Instance: μ=1, Var(X)=2, Var(X+5)=2, Var(−3X)=18; Z=(2,2,2,7) is constantly 2 on the support {1,2,3}, so Var(Z)=0 and Z(4)=7 does not matter." }, "negative_controls": [ { "claim": { "zh-Hant": "Z(4)=7≠2，所以 Var(Z)>0", "en": "Z(4)=7≠2, so Var(Z)>0" }, "expected": "reject", "reason": { "zh-Hant": "w₄=0，結果 4 的項為 0·25=0；Z 在支持上恆為 2，Var(Z)=0。", "en": "w₄=0, so the term of outcome 4 is 0·25=0; Z is constantly 2 on the support and Var(Z)=0." } }, { "claim": { "zh-Hant": "Var(−3X)=−3Var(X)=−6", "en": "Var(−3X)=−3Var(X)=−6" }, "expected": "reject", "reason": { "zh-Hant": "變異數非負；正確為 (−3)²·2=18。", "en": "Variance is nonnegative; the correct value is (−3)²·2=18." } }, { "claim": { "zh-Hant": "Var(X+5)=Var(X)+5=7", "en": "Var(X+5)=Var(X)+5=7" }, "expected": "reject", "reason": { "zh-Hant": "平移使平均也加 5，偏差不變，Var(X+5)=2。", "en": "The shift also adds 5 to the mean, the deviations are unchanged, and Var(X+5)=2." } }, { "claim": { "zh-Hant": "X 的變異數是 1、3、−1、10 的樣本變異數 275/12", "en": "The variance of X is the sample variance 275/12 of 1, 3, −1, 10" }, "expected": "reject", "reason": { "zh-Hant": "本模型的變異數以權重 (1/2,1/4,1/4,0) 加權，等於 2；樣本估計是另一回事。", "en": "This model's variance is weighted by (1/2,1/4,1/4,0) and equals 2; a sample estimator is a different thing." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "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": "square_nonneg", "kind": "prerequisite", "statement": { "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。", "en": "Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=0." } }, { "id": "order_basics", "kind": "prerequisite", "statement": { "zh-Hant": "實數的序：任兩數恰有 uv 之一；u0 推出 uwvw。", "en": "Order of the reals: exactly one of uv holds; u0 give uwvw." } }, { "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": "power_laws", "kind": "prerequisite", "statement": { "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。", "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1." } }, { "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": "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": "eq_replace", "kind": "prerequisite", "statement": { "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。", "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true." } }, { "id": "add_assoc", "kind": "prerequisite", "statement": { "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。", "en": "Associativity of addition: (u+v)+w=u+(v+w)." } }, { "id": "add_comm", "kind": "prerequisite", "statement": { "zh-Hant": "加法交換律：u+v=v+u。", "en": "Commutativity of addition: u+v=v+u." } }, { "id": "add_inverse", "kind": "prerequisite", "statement": { "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。", "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)." } }, { "id": "add_zero", "kind": "prerequisite", "statement": { "zh-Hant": "加法單位元：u+0=u。", "en": "Additive identity: u+0=u." } }, { "id": "mul_comm", "kind": "prerequisite", "statement": { "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。", "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)." } }, { "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": "contradiction", "kind": "prerequisite", "statement": { "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。", "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds." } }, { "id": "zero_product", "kind": "prerequisite", "statement": { "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。", "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0." } }, { "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": "exists_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。", "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, 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": "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_variance_def", "kind": "cited", "source": { "mko_id": "mko-variance", "sha256": "282681c1ed2d55f1c853c9759022470c991dd46c2b150626fa4bf654642032bc", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "令 μ=E[X]，對有限合法機率模型定義 V(X)=Var(X)=Σ_s w_s(X(s)−μ)²≥0。平移不改變變異數，Var(kX)=k²Var(X)；Var(X)=0 當且僅當 X 在所有正權重的基本結果上取同一個值。", "en": "Put μ=E[X]. In a valid finite probability model, define V(X)=Var(X)=Σ_s w_s(X(s)−μ)²≥0. Translation leaves variance unchanged and Var(kX)=k²Var(X). Variance is zero exactly when X has one common value at every positive-weight outcome." } }, { "id": "mko_expectation_lin", "kind": "cited", "source": { "mko_id": "mko-expectation", "sha256": "71bc68faf69ca0df9cdd11d093055262aa277d07abd77982d0aa1afd849daf3c", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "對非空有限 S、歸一化非負權重 w_s 與實值隨機變數 X，定義 E[X]=Σ_(s∈S)w_s X(s)。在同一機率模型上 E[aX+b]=aE[X]+b、E[X+Y]=E[X]+E[Y]，不需假設 X、Y 獨立；常數 c 的期望為 c。", "en": "For nonempty finite S, normalized nonnegative weights w_s, and a real random variable X, define E[X]=Σ_(s∈S)w_s X(s). On the same probability model, E[aX+b]=aE[X]+b and E[X+Y]=E[X]+E[Y] without assuming independence of X and Y; a constant c has expectation c." } }, { "id": "support_def", "kind": "prerequisite", "statement": { "zh-Hant": "正權重支持：supp={s∈S ∣ w_s>0}。因 w_s≥0，s∉supp 恰表示 w_s=0。", "en": "Positive-weight support: supp={s∈S ∣ w_s>0}. Since w_s≥0, s∉supp means exactly w_s=0." } }, { "id": "L_shift", "kind": "lemma", "proved_in": [ "S06.1", "S06.2" ], "statement": { "zh-Hant": "在 G1 的模型中，對任意實數 c：Var(X+c)=Var(X)。", "en": "In the model of G1, for every real c: Var(X+c)=Var(X)." } }, { "id": "L_scale", "kind": "lemma", "proved_in": [ "S07.1", "S07.2", "S07.3" ], "statement": { "zh-Hant": "在 G1 的模型中，對任意實數 k：Var(kX)=k²Var(X)。", "en": "In the model of G1, for every real k: Var(kX)=k²Var(X)." } }, { "id": "L_nonneg_sum_zero", "kind": "lemma", "proved_in": [ "S10.1", "S10.2" ], "statement": { "zh-Hant": "若有限多個實數 t_s 都 ≥0 且 Σ_s t_s=0，則每個 t_s=0。", "en": "If finitely many reals t_s are all ≥0 and Σ_s t_s=0, then every t_s=0." } }, { "id": "L_var_zero", "kind": "lemma", "proved_in": [ "S11.1", "S11.2" ], "statement": { "zh-Hant": "在 G1 的模型中：Var(X)=0 ⇔ 對每個 s∈supp 有 X(s)=μ。", "en": "In the model of G1: Var(X)=0 ⇔ X(s)=μ for every s∈supp." } }, { "id": "L_const_support", "kind": "lemma", "proved_in": [ "S12.1", "S12.2" ], "statement": { "zh-Hant": "在 G1 的模型中：Var(X)=0 ⇔ 存在實數 c 使得對每個 s∈supp 有 X(s)=c。", "en": "In the model of G1: Var(X)=0 ⇔ there is a real c with X(s)=c for every s∈supp." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S01", "lhs": "1/2+1/4+1/4+0", "rhs": "1" }, { "id": "N2", "step": "S02", "lhs": "1/2*1+1/4*3+1/4*(-1)+0*10", "rhs": "1" }, { "id": "N3", "step": "S04", "lhs": "1/2*(1-1)**2+1/4*(3-1)**2+1/4*(-1-1)**2+0*(10-1)**2", "rhs": "2" }, { "id": "N4", "step": "S08", "lhs": "1/2*6+1/4*8+1/4*4+0*15", "rhs": "6" }, { "id": "N5", "step": "S08", "lhs": "1/2*(6-6)**2+1/4*(8-6)**2+1/4*(4-6)**2+0*(15-6)**2", "rhs": "2" }, { "id": "N6", "step": "S09", "lhs": "1/2*(-3)+1/4*(-9)+1/4*3+0*(-30)", "rhs": "-3" }, { "id": "N7", "step": "S09", "lhs": "1/2*(-3+3)**2+1/4*(-9+3)**2+1/4*(3+3)**2+0*(-30+3)**2", "rhs": "18" }, { "id": "N8", "step": "S09", "lhs": "(-3)**2*2", "rhs": "18" }, { "id": "N9", "step": "S07.2", "vars": [ "k", "x", "m" ], "degree": 2, "lhs": "(k*x-k*m)**2", "rhs": "k**2*(x-m)**2" }, { "id": "N10", "step": "S06.2", "vars": [ "x", "m", "c" ], "degree": 2, "lhs": "((x+c)-(m+c))**2", "rhs": "(x-m)**2" }, { "id": "N11", "step": "S13", "lhs": "1/2*2+1/4*2+1/4*2+0*7", "rhs": "2" }, { "id": "N12", "step": "S13", "lhs": "1/2*(2-2)**2+1/4*(2-2)**2+1/4*(2-2)**2+0*(7-2)**2", "rhs": "0" }, { "id": "N13", "step": "S04", "lhs": "((1-13/4)**2+(3-13/4)**2+(-1-13/4)**2+(10-13/4)**2)/3", "rhs": "275/12" }, { "id": "N14", "step": "S04", "lhs": "275/12", "relation": "!=", "rhs": "2" }, { "id": "N15", "step": "S14", "lhs": "1", "relation": "!=", "rhs": "3" } ] }, "presentation": { "md_scope": { "zh-Hant": "限定 MKO：有限非空樣本空間、非負歸一化權重、實值總函數，使用機率分布變異數而非樣本估計，並區分正權重支持與全樣本空間（a1、a2）。全部以精確有理數計算。期望值的線性性引自 mko-expectation。不處理 Var(X+Y)、協方差、標準差或 E[X²]−μ² 公式；不改動 canonical MKO 或英文 edition。", "en": "Stays inside the MKO: finite nonempty sample space, nonnegative normalized weights, a total real-valued function, the probability-distribution variance rather than a sample estimator, and a clear distinction between the positive-weight support and the whole sample space (a1, a2). All arithmetic is exact and rational. Linearity of expectation is cited from mko-expectation. Var(X+Y), covariance, standard deviation and the formula E[X²]−μ² are not treated; the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "\\mu=E[X]", "X 的期望值", "the expectation of X" ], [ "\\mathrm{Var}(X)=\\sum_s w_s(X(s)-\\mu)^2", "X 的變異數", "the variance of X" ], [ "\\mathrm{supp}=\\{s\\in S\\mid w_s>0\\}", "正權重支持", "the positive-weight support" ], [ "c,k", "平移量與縮放係數（實數）", "translation amount and scaling factor (reals)" ], [ "t_s", "變異數和式中的第 s 項", "the s-th term of the variance sum" ] ], "md_error_controls": [ { "title": "零變異數只要求正權重結果同值 / Zero variance constrains only positive-weight outcomes", "zh": "MKO 的誤解之一是要求零變異數的變數在零機率結果上也同值。Z 在 S 上不是常數（Z(4)=7），但 w₄=0，Var(Z)=0（S13）。零變異數的正確刻畫是在支持上常值（S12）。", "en": "One misconception identified in the MKO is requiring a zero-variance variable to agree at zero-probability outcomes as well. Z is not constant on S (Z(4)=7), but w₄=0 and Var(Z)=0 (S13). The correct characterization of zero variance is constancy on the support (S12)." }, { "title": "機率分布變異數不是樣本變異數 / Distribution variance is not the sample variance", "zh": "另一個誤解是混用樣本變異數。若把 1、3、−1、10 當成四筆資料、用 n−1=3 為分母，得 275/12，而不是本模型的 Var(X)=2。本稿的變異數以權重加權，不用 n−1 分母；標準差 √2 是另一個量，本稿未處理。", "en": "The other misconception is to mix in the sample variance. Treating 1, 3, −1, 10 as four data points with denominator n−1=3 gives 275/12, not this model's Var(X)=2. Variance here is weighted by the probabilities and uses no n−1 denominator; the standard deviation √2 is a different quantity not treated here." }, { "title": "縮放因子要平方，平移不加到變異數上 / The factor is squared, and a shift is not added to the variance", "zh": "Var(−3X)=(−3)²·2=18，不是 −3·2=−6，也不是 3·2=6；Var(X+5)=2，不是 2+5=7。原因分別是 S07.2 的平方與 S06.2 的偏差不變。", "en": "Var(−3X)=(−3)²·2=18, not −3·2=−6 nor 3·2=6; Var(X+5)=2, not 2+5=7. The reasons are the square in S07.2 and the unchanged deviations in S06.2." } ], "md_open_items": [ { "zh-Hant": "Var(X+Y)、協方差與標準差不在本稿範圍；本稿也不使用 E[X²]−μ² 公式。", "en": "Var(X+Y), covariance and standard deviation are outside this guide; the formula E[X²]−μ² is not used either." }, { "zh-Hant": "期望值線性性引自 mko-expectation，其證明見該 MKO 的教學稿，本稿不重證。", "en": "Linearity of expectation is cited from mko-expectation; its proof belongs to that MKO's guide and is not repeated here." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "ce0f84e07a301a2959c247b101c66b419e062d8b1325ebf04f27a9c4914677a1", "original_source_sha256": "ce0f84e07a301a2959c247b101c66b419e062d8b1325ebf04f27a9c4914677a1", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }