摘要訊息 : 一些初等機率論中的經典模型和分佈.
0. 前言
在《【機率論】初等機率論——有限種結局試驗的機率模型》 中, 我們對有限種結局試驗建立了機率論 體系. 在本節中, 我們將要討論一些經典的機率模型和分佈.
更新紀錄 :
2022 年 6 月 6 日進行第一次更新和修正.
1. 二項分佈
假設將一枚硬幣接連擲 n n n 次, 觀測結果用有序陣列 ( a 1 , a 2 , . . . , a n ) (a_{1}, a_{2}, ..., a_{n}) ( a 1 , a 2 , . . . , a n ) 表示. 其中, 當第 i i i 次擲出現正面時, a i = 1 a_{i} = 1 a i = 1 ; 當第 i i i 次擲出現反面時, a i = 0 a_{i} = 0 a i = 0 , i = 1 , 2 , . . . , n i = 1, 2, ..., n i = 1 , 2 , . . . , n . 那麼基本事件空間具有如下形式 : Ω = { ω : ω = ( a 1 , a 2 , . . . , a n ) , a i = 0 或者 a i = 1 ( i = 1 , 2 , . . . , n ) } . \displaystyle {\Omega = \{ \omega : \omega = (a_{1}, a_{2}, ..., a_{n}), a_{i} = 0 \text { 或者 } a_{i} = 1\ (i = 1, 2, ..., n) \}}. Ω = { ω : ω = ( a 1 , a 2 , . . . , a n ) , a i = 0 或者 a i = 1 ( i = 1 , 2 , . . . , n ) } . 賦予每一個基本事件 ω = ( a 1 , a 2 , . . . , a n ) \omega = (a_{1}, a_{2}, ..., a_{n}) ω = ( a 1 , a 2 , . . . , a n ) 以機率 p ( ω ) = p ∑ i : { a i = 1 } a i ⋅ q n − ∑ i : { a i = 1 } a i , \displaystyle {p(\omega) = p^{\sum \limits_{i : \left \{ a_{i} = 1 \right \}}a_{i}} \cdot q^{n - \sum \limits_{i : \left \{ a_{i} = 1 \right \}}a_{i}}}, p ( ω ) = p i : { a i = 1 } ∑ a i ⋅ q n − i : { a i = 1 } ∑ a i , 其中 p p p 與 q q q 為非負實數且 p + q = 1 p + q = 1 p + q = 1 . 首先, 我們需要說明這樣去定義機率 p ( ω ) p(\omega) p ( ω ) 的合理性. 為此, 我們只需要驗證 ∑ ω ∈ Ω p ( ω ) = 1 . \displaystyle {\sum \limits_{\omega \in \Omega}p(\omega) = 1}. ω ∈ Ω ∑ p ( ω ) = 1 .
考慮所有滿足 ∑ i : { a i = 1 } a i = k \displaystyle {\sum \limits_{i\ :\ \{ a_{i} = 1 \}}a_{i} = k} i : { a i = 1 } ∑ a i = k 的基本事件 ω = ( a 1 , a 2 , . . . , a n ) \omega = (a_{1}, a_{2}, ..., a_{n}) ω = ( a 1 , a 2 , . . . , a n ) . 其中, k = 0 , 1 , 2 , . . . , n k = 0, 1, 2, ..., n k = 0 , 1 , 2 , . . . , n . 這等同於將 k k k 個 1 1 1 分配至 n n n 個位置上, 屬於不放回無序抽樣. 因此, 對於任意的正整數 k ( 0 ≤ k ≤ n ) k\ (0 \leq k \leq n) k ( 0 ≤ k ≤ n ) , 這樣的基本事件個數共有 ( k n ) \binom {k}{n} ( n k ) . 於是, ∑ ω ∈ Ω p ( ω ) = ∑ k = 0 n ( k n ) p k q n − k = ( 0 n ) p 0 q n + ( 1 n ) p 1 q n − 1 + . . . + ( n − 1 n ) p n − 1 q 1 + ( n n ) p n q 0 = q n + n p q n − 1 + . . . + n p n − 1 q + p n = ( p + q ) n = 1 n = 1. \displaystyle {\begin {aligned} \sum \limits_{\omega \in \Omega}p(\omega) &= \sum \limits_{k = 0}^{n} \binom {k}{n}p^{k}q^{n - k} \\ &= \binom {0}{n}p^{0}q^{n} + \binom {1}{n}p^{1}q^{n - 1} + ... + \binom {n - 1}{n}p^{n - 1}q^{1} + \binom {n}{n}p^{n}q^{0} \\ &= q^{n} + npq^{n - 1} + ... + np^{n - 1}q + p^{n} \\ &= (p + q)^{n} \\ &= 1^{n} = 1. \end {aligned}} ω ∈ Ω ∑ p ( ω ) = k = 0 ∑ n ( n k ) p k q n − k = ( n 0 ) p 0 q n + ( n 1 ) p 1 q n − 1 + . . . + ( n n − 1 ) p n − 1 q 1 + ( n n ) p n q 0 = q n + n p q n − 1 + . . . + n p n − 1 q + p n = ( p + q ) n = 1 n = 1 . 設 A \mathscr {A} A 為空間 Ω \Omega Ω 的一切子集的代數, 在 A \mathscr {A} A 上定義了機率 P ( A ) = ∑ ω ∈ Ω p ( ω ) , A ∈ A . \displaystyle {\mathop {\mathbf {P}}(A) = \sum \limits_{\omega \in \Omega}p(\omega), A \in \mathscr {A}}. P ( A ) = ω ∈ Ω ∑ p ( ω ) , A ∈ A . 其中, p ( { ω } ) = p ( ω ) , ω ∈ Ω p(\{ \omega \}) = p(\omega), \omega \in \Omega p ( { ω } ) = p ( ω ) , ω ∈ Ω . 這樣, 我們就定義了描繪 n n n 次擲硬幣的機率模型.
事件 A k = { ω : ω = ( a 1 , a 2 , . . . , a n ) , a 1 + a 2 + . . . + a n = k } \displaystyle {A_{k} = \{ \omega : \omega = (a_{1}, a_{2}, ..., a_{n}), a_{1} + a_{2} + ... + a_{n} = k \}} A k = { ω : ω = ( a 1 , a 2 , . . . , a n ) , a 1 + a 2 + . . . + a n = k } 表示恰好有 k k k 次 "成功". 其中, k = 1 , 2 , . . . , n k = 1, 2, ..., n k = 1 , 2 , . . . , n . 由之前的討論, 我們知道 P ( A k ) = ( k n ) p k q n − k \mathop {\mathbf {P}}(A_{k}) = \binom {k}{n}p^{k}q^{n - k} P ( A k ) = ( n k ) p k q n − k 且 ∑ k = 0 n P ( A k ) = 1 \sum \limits_{k = 0}^{n}\mathop {\mathbf {P}}(A_{k}) = 1 k = 0 ∑ n P ( A k ) = 1 . 機率組 ( P ( A 0 ) , P ( A 1 ) , P ( A 2 ) , . . . , P ( A n ) ) (\mathop {\mathbf {P}}(A_{0}), \mathop {\mathbf {P}}(A_{1}), \mathop {\mathbf {P}}(A_{2}), ..., \mathop {\mathbf {P}}(A_{n})) ( P ( A 0 ) , P ( A 1 ) , P ( A 2 ) , . . . , P ( A n ) ) 稱作二項分佈 (binomial distribution), 也稱在容量為 n n n 的樣本中, "成功" 的次數服從二項分佈.
例題 1. 對於投擲對稱硬幣的情形, 我們一般假定 p = 1 2 p = \frac {1}{2} p = 2 1 . 若我們投擲 n n n 次硬幣, 記 A k A_{k} A k 為 k k k 次正面朝上的情形 (k = 0 , 1 , 2 , . . . , n k = 0, 1, 2, ..., n k = 0 , 1 , 2 , . . . , n ), 則 n n n 次投擲中共有 k k k 次正面朝上的機率即為 P n ( k ) = P ( A k ) P_{n}(k) = \mathop {\mathbf {P}}(A_{k}) P n ( k ) = P ( A k ) .
解 解 解 :當 n = 5 n = 5 n = 5 時, 有 { P 5 ( 0 ) = ( 0 5 ) ⋅ 1 2 0 ⋅ 1 2 5 = 1 32 , P 5 ( 1 ) = ( 1 5 ) ⋅ 1 2 1 ⋅ 1 2 4 = 5 32 , P 5 ( 2 ) = ( 2 5 ) ⋅ 1 2 2 ⋅ 1 2 3 = 10 32 , P 5 ( 3 ) = ( 3 5 ) ⋅ 1 2 3 ⋅ 1 2 2 = 10 32 , P 5 ( 4 ) = ( 4 5 ) ⋅ 1 2 4 ⋅ 1 2 1 = 5 32 , P 5 ( 5 ) = ( 5 5 ) ⋅ 1 2 5 ⋅ 1 2 0 = 1 32 . \displaystyle {\begin {cases} P_{5}(0) = \binom {0}{5} \cdot \frac {1}{2^{0}} \cdot \frac {1}{2^{5}} = \frac {1}{32}, \\ P_{5}(1) = \binom {1}{5} \cdot \frac {1}{2^{1}} \cdot \frac {1}{2^{4}} = \frac {5}{32}, \\ P_{5}(2) = \binom {2}{5} \cdot \frac {1}{2^{2}} \cdot \frac {1}{2^{3}} = \frac {10}{32}, \\ P_{5}(3) = \binom {3}{5} \cdot \frac {1}{2^{3}} \cdot \frac {1}{2^{2}} = \frac {10}{32}, \\ P_{5}(4) = \binom {4}{5} \cdot \frac {1}{2^{4}} \cdot \frac {1}{2^{1}} = \frac {5}{32}, \\ P_{5}(5) = \binom {5}{5} \cdot \frac {1}{2^{5}} \cdot \frac {1}{2^{0}} = \frac {1}{32}. \end {cases}} ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ P 5 ( 0 ) = ( 5 0 ) ⋅ 2 0 1 ⋅ 2 5 1 = 3 2 1 , P 5 ( 1 ) = ( 5 1 ) ⋅ 2 1 1 ⋅ 2 4 1 = 3 2 5 , P 5 ( 2 ) = ( 5 2 ) ⋅ 2 2 1 ⋅ 2 3 1 = 3 2 1 0 , P 5 ( 3 ) = ( 5 3 ) ⋅ 2 3 1 ⋅ 2 2 1 = 3 2 1 0 , P 5 ( 4 ) = ( 5 4 ) ⋅ 2 4 1 ⋅ 2 1 1 = 3 2 5 , P 5 ( 5 ) = ( 5 5 ) ⋅ 2 5 1 ⋅ 2 0 1 = 3 2 1 .
當 n = 10 n = 10 n = 1 0 時, 有 { P 10 ( 0 ) = ( 0 10 ) ⋅ 1 2 0 ⋅ 1 2 10 = 1 1024 , P 10 ( 1 ) = ( 1 10 ) ⋅ 1 2 1 ⋅ 1 2 9 = 10 1024 , P 10 ( 2 ) = ( 2 10 ) ⋅ 1 2 2 ⋅ 1 2 8 = 45 1024 , P 10 ( 3 ) = ( 3 10 ) ⋅ 1 2 3 ⋅ 1 2 7 = 120 1024 , P 10 ( 4 ) = ( 4 10 ) ⋅ 1 2 4 ⋅ 1 2 6 = 210 1024 , P 10 ( 5 ) = ( 5 10 ) ⋅ 1 2 5 ⋅ 1 2 5 = 252 1024 , P 10 ( 6 ) = ( 6 10 ) ⋅ 1 2 6 ⋅ 1 2 4 = 210 1024 , P 10 ( 7 ) = ( 7 10 ) ⋅ 1 2 7 ⋅ 1 2 3 = 120 1024 , P 10 ( 8 ) = ( 8 10 ) ⋅ 1 2 8 ⋅ 1 2 2 = 45 1024 , P 10 ( 9 ) = ( 9 10 ) ⋅ 1 2 9 ⋅ 1 2 1 = 10 1024 , P 10 ( 10 ) = ( 10 10 ) ⋅ 1 2 10 ⋅ 1 2 0 = 1 1024 . \displaystyle {\begin {cases} P_{10}(0) = \binom {0}{10} \cdot \frac {1}{2^{0}} \cdot \frac {1}{2^{10}} = \frac {1}{1024}, \\ P_{10}(1) = \binom {1}{10} \cdot \frac {1}{2^{1}} \cdot \frac {1}{2^{9}} = \frac {10}{1024}, \\ P_{10}(2) = \binom {2}{10} \cdot \frac {1}{2^{2}} \cdot \frac {1}{2^{8}} = \frac {45}{1024}, \\ P_{10}(3) = \binom {3}{10} \cdot \frac {1}{2^{3}} \cdot \frac {1}{2^{7}} = \frac {120}{1024}, \\ P_{10}(4) = \binom {4}{10} \cdot \frac {1}{2^{4}} \cdot \frac {1}{2^{6}} = \frac {210}{1024}, \\ P_{10}(5) = \binom {5}{10} \cdot \frac {1}{2^{5}} \cdot \frac {1}{2^{5}} = \frac {252}{1024}, \\ P_{10}(6) = \binom {6}{10} \cdot \frac {1}{2^{6}} \cdot \frac {1}{2^{4}} = \frac {210}{1024}, \\ P_{10}(7) = \binom {7}{10} \cdot \frac {1}{2^{7}} \cdot \frac {1}{2^{3}} = \frac {120}{1024}, \\ P_{10}(8) = \binom {8}{10} \cdot \frac {1}{2^{8}} \cdot \frac {1}{2^{2}} = \frac {45}{1024}, \\ P_{10}(9) = \binom {9}{10} \cdot \frac {1}{2^{9}} \cdot \frac {1}{2^{1}} = \frac {10}{1024}, \\ P_{10}(10) = \binom {10}{10} \cdot \frac {1}{2^{10}} \cdot \frac {1}{2^{0}} = \frac {1}{1024}. \end {cases}} ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ P 1 0 ( 0 ) = ( 1 0 0 ) ⋅ 2 0 1 ⋅ 2 1 0 1 = 1 0 2 4 1 , P 1 0 ( 1 ) = ( 1 0 1 ) ⋅ 2 1 1 ⋅ 2 9 1 = 1 0 2 4 1 0 , P 1 0 ( 2 ) = ( 1 0 2 ) ⋅ 2 2 1 ⋅ 2 8 1 = 1 0 2 4 4 5 , P 1 0 ( 3 ) = ( 1 0 3 ) ⋅ 2 3 1 ⋅ 2 7 1 = 1 0 2 4 1 2 0 , P 1 0 ( 4 ) = ( 1 0 4 ) ⋅ 2 4 1 ⋅ 2 6 1 = 1 0 2 4 2 1 0 , P 1 0 ( 5 ) = ( 1 0 5 ) ⋅ 2 5 1 ⋅ 2 5 1 = 1 0 2 4 2 5 2 , P 1 0 ( 6 ) = ( 1 0 6 ) ⋅ 2 6 1 ⋅ 2 4 1 = 1 0 2 4 2 1 0 , P 1 0 ( 7 ) = ( 1 0 7 ) ⋅ 2 7 1 ⋅ 2 3 1 = 1 0 2 4 1 2 0 , P 1 0 ( 8 ) = ( 1 0 8 ) ⋅ 2 8 1 ⋅ 2 2 1 = 1 0 2 4 4 5 , P 1 0 ( 9 ) = ( 1 0 9 ) ⋅ 2 9 1 ⋅ 2 1 1 = 1 0 2 4 1 0 , P 1 0 ( 1 0 ) = ( 1 0 1 0 ) ⋅ 2 1 0 1 ⋅ 2 0 1 = 1 0 2 4 1 .
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我們發現, 不論 n n n 的值如何, P n ( k ) P_{n}(k) P n ( k ) 的值會在 k k k 接近 1 2 \frac {1}{2} 2 1 時達到頂峰. 下面是二項分佈下, 幾個幾個不同的 n n n 最終在圖像上的表現 :
Figure 1-1. n = 5 n = 5 n = 5
Figure 1-2. n = 10 n = 10 n = 1 0
■ \blacksquare ■
例題 2. 設有一質點在直角座標系統的原點出發, 每經過單位時間就向上或者向下移動一步 :
Figure 2. 軌道 ω \omega ω
於是, 經過 n n n 個單位時間後, 質點移動 n n n 步. 顯然, 質點運動的每一條軌跡 ω \omega ω 可以用序列 ( s 1 , s 2 , . . . , s n ) (s_{1}, s_{2}, ..., s_{n}) ( s 1 , s 2 , . . . , s n ) 來表示. 對於第 i i i 步, 若質點向上移動, 則 s i = 1 s_{i} = 1 s i = 1 ; 若質點向下移動, 則 s i = − 1 s_{i} = -1 s i = − 1 . 其中, i = 1 , 2 , . . . , n i = 1, 2, ..., n i = 1 , 2 , . . . , n . 現賦予每一條軌跡 ω \omega ω 以機率 p ( ω ) = p υ ( ω ) q n − υ ( ω ) , \displaystyle {p(\omega) = p^{\upsilon(\omega)}q^{n - \upsilon(\omega)}}, p ( ω ) = p υ ( ω ) q n − υ ( ω ) , 其中 υ ( ω ) \upsilon(\omega) υ ( ω ) 表示序列 ω \omega ω 中 s i = 1 s_{i} = 1 s i = 1 的個數, 即 υ ( ω ) = ∑ i = 1 n s i + n 2 \upsilon(\omega) = \frac {\sum \limits_{i = 1}^{n}s_{i} + n}{2} υ ( ω ) = 2 i = 1 ∑ n s i + n . 除此之外, p p p 和 q q q 都為非負實數且 p + q = 1 p + q = 1 p + q = 1 . 由於 ∑ ω ∈ Ω p ( ω ) = 1 \sum \limits_{\omega \in \Omega}p(\omega) = 1 ω ∈ Ω ∑ p ( ω ) = 1 , 可見機率組 { p ( ω ) } \left \{ p(\omega) \right \} { p ( ω ) } 連同軌跡 ω = ( s 1 , s 2 , . . . , s n ) \omega = (s_{1}, s_{2}, ..., s_{n}) ω = ( s 1 , s 2 , . . . , s n ) 的空間 Ω \Omega Ω 及其子集確定了質點運動的機率模型. 那麼事件 A k = { 質點經過 n 步到達縱座標為 k 的點 } A_{k} = \left \{ \text {質點經過 } n \text { 步到達縱座標為 } k \text { 的點} \right \} A k = { 質點經過 n 步到達縱座標為 k 的點 } 的機率如何?
解 解 解 :一切滿足 v ( ω ) − ( n − v ( ω ) ) = k v(\omega) - (n - v(\omega)) = k v ( ω ) − ( n − v ( ω ) ) = k , 即 v ( ω ) = n + k 2 v(\omega) = \frac {n + k}{2} v ( ω ) = 2 n + k 的軌跡都符合問題描述. 那麼上述問題可以化約為將 n + k 2 \frac {n + k}{2} 2 n + k 個 "1 1 1 " 放置到 n n n 個位置上, 這樣, 符合描述的軌跡共有 ( n + k 2 2 ) \begin {pmatrix} \frac {n + k}{2} \\ 2 \end {pmatrix} ( 2 n + k 2 ) 條, 故有 P ( A k ) = ( n + k 2 n ) p n + k 2 q n − n + k 2 = ( n + k 2 n ) p n + k 2 q n − k 2 . \displaystyle {\mathop {\mathbf {P}}(A_{k}) = \binom {\frac {n + k}{2}}{n}p^{\frac {n + k}{2}}q^{n - \frac {n + k}{2}} = \binom {\frac {n + k}{2}}{n}p^{\frac {n + k}{2}}q^{\frac {n - k}{2}}}. P ( A k ) = ( n 2 n + k ) p 2 n + k q n − 2 n + k = ( n 2 n + k ) p 2 n + k q 2 n − k .
■ \blacksquare ■
在例題 2 中, 我們可以認為二項分佈 { P ( A − n ) , P ( A − ( n − 1 ) ) , . . . , P ( A − 1 ) , P ( A 0 ) , P ( A 1 ) , P ( A 2 ) , . . . , P ( A n ) } \displaystyle {\left \{ \mathop {\mathbf {P}}(A_{-n}), \mathop {\mathbf {P}}(A_{-(n - 1)}), ..., \mathop {\mathbf {P}}(A_{-1}), \mathop {\mathbf {P}}(A_{0}), \mathop {\mathbf {P}}(A_{1}), \mathop {\mathbf {P}}(A_{2}), ..., \mathop {\mathbf {P}}(A_{n}) \right \}} { P ( A − n ) , P ( A − ( n − 1 ) ) , . . . , P ( A − 1 ) , P ( A 0 ) , P ( A 1 ) , P ( A 2 ) , . . . , P ( A n ) } 描繪了質點移動 n n n 步後位置的機率分佈. 特別地, 對於 p = 1 2 p = \frac {1}{2} p = 2 1 的情形, 每條軌跡的機率為 1 2 n \frac {1}{2^{n}} 2 n 1 , 於是有 P ( A k ) = ( n + k 2 n ) 2 n . \displaystyle {\mathop {\mathbf {P}}(A_{k}) = \frac {\binom {\frac {n + k}{2}}{n}}{2^{n}}}. P ( A k ) = 2 n ( n 2 n + k ) . 其中, − n ≤ k ≤ n -n \leq k \leq n − n ≤ k ≤ n . 接著, 我們來討論當 n → ∞ n \to \infty n → ∞ 時, 這些機率的漸進性質.
若令移動步伐為 2 n 2n 2 n , 由二項式係數的性質可見, 在機率 P ( A k ) ( ∣ k ∣ ≤ 2 n ) \mathop {\mathbf {P}}(A_{k})\ (|k| \leq 2n) P ( A k ) ( ∣ k ∣ ≤ 2 n ) 中的最大機率為 P ( A 0 ) = ( n 2 n ) 1 2 2 n = ( 2 n ) ! ( n ! ) 2 ⋅ 1 2 2 n . \displaystyle {\mathop {\mathbf {P}}(A_{0}) = \binom {n}{2n}\frac {1}{2^{2n}} = \frac {(2n)!}{(n!)^{2}} \cdot \frac {1}{2^{2n}}}. P ( A 0 ) = ( 2 n n ) 2 2 n 1 = ( n ! ) 2 ( 2 n ) ! ⋅ 2 2 n 1 . 根據 Stirling 公式, 我們知道 n ! ≐ 2 π n ( n e ) n n! \doteq \sqrt {2\pi n}\left ( \frac {n}{e} \right )^{n} n ! ≐ 2 π n ( e n ) n . 因此, 對於充分大的 n n n , 有 P ( A 0 ) ≐ 1 π n . \displaystyle {\mathop {\mathbf {P}}(A_{0}) \doteq \frac {1}{\sqrt {\pi n}}}. P ( A 0 ) ≐ π n 1 .
2. 多項分佈
現在, 我們推廣二項 分佈 的機率模型. 假設基本事件空間有如下構造 : Ω = { ω : ω = ( a 1 , a 2 , . . . , a n ) , a i = b 1 , b 2 , . . . , b r ( i = 1 , 2 , . . . , n ) } . \displaystyle {\Omega = \left \{ \omega : \omega = (a_{1}, a_{2}, ..., a_{n}), a_{i} = b_{1}, b_{2}, ..., b_{r}\ (i = 1, 2, ..., n) \right \}}. Ω = { ω : ω = ( a 1 , a 2 , . . . , a n ) , a i = b 1 , b 2 , . . . , b r ( i = 1 , 2 , . . . , n ) } . 其中, b 1 , b 2 , . . . , b r b_{1}, b_{2}, ..., b_{r} b 1 , b 2 , . . . , b r 都是給定的數. 設 υ i ( ω ) \upsilon_{i}(\omega) υ i ( ω ) 是序列 ω = ( a 1 , a 2 , . . . , a n ) \omega = (a_{1}, a_{2}, ..., a_{n}) ω = ( a 1 , a 2 , . . . , a n ) 中等於 b i ( i = 1 , 2 , . . . , r ) b_{i}\ (i = 1, 2, ..., r) b i ( i = 1 , 2 , . . . , r ) 元素的數量, 而基本事件 ω \omega ω 有機率 p ( ω ) = p 1 υ 1 ( ω ) p 2 v 2 ( ω ) . . . p r υ r ( ω ) \displaystyle {p(\omega) = p_{1}^{\upsilon_{1}(\omega)}p_{2}^{v_{2}(\omega)}...p_{r}^{\upsilon_{r}(\omega)}} p ( ω ) = p 1 υ 1 ( ω ) p 2 v 2 ( ω ) . . . p r υ r ( ω ) 其中, p i ≥ 0 ( i = 1 , 2 , . . . , r ) , p 1 + p 2 + . . . + p r = 1 p_{i} \geq 0\ (i = 1, 2, ..., r), p_{1} + p_{2} + ... + p_{r} = 1 p i ≥ 0 ( i = 1 , 2 , . . . , r ) , p 1 + p 2 + . . . + p r = 1 . 為了說明推廣後的情形仍然合理, 我們首先注意到 ∑ ω ∈ Ω p ( ω ) = ∑ { n 1 ≥ 0 , n 2 ≥ 0 , . . . , n r ≥ 0 n 1 + n 2 + . . . + n r = n } ( n 1 , n 2 , . . . , n r n ) p 1 n 1 p 2 n 2 . . . p r n r . \displaystyle {\sum \limits_{\omega \in \Omega}p(\omega) = \sum \limits_{ \left \{ \begin {gathered} \scriptsize {n_{1} \geq 0, n_{2} \geq 0, ..., n_{r} \geq 0} \\ \scriptsize {n_{1} + n _{2} + ... + n_{r} = n} \end {gathered} \right \}} \binom {n_{1}, n_{2}, ..., n_{r}}{n}p_{1}^{n_{1}}p_{2}^{n_{2}}...p_{r}^{n_{r}}}. ω ∈ Ω ∑ p ( ω ) = { n 1 ≥ 0 , n 2 ≥ 0 , . . . , n r ≥ 0 n 1 + n 2 + . . . + n r = n } ∑ ( n n 1 , n 2 , . . . , n r ) p 1 n 1 p 2 n 2 . . . p r n r . 上式中, ( n 1 , n 2 , . . . , n r n ) \binom {n_{1}, n_{2}, ..., n_{r}}{n} ( n n 1 , n 2 , . . . , n r ) 表示有序序列 ( a 1 , a 2 , . . . , a n ) (a_{1}, a_{2}, ..., a_{n}) ( a 1 , a 2 , . . . , a n ) 的數量. 其中, 元素 b 1 b_{1} b 1 重複 n 1 n_{1} n 1 次, 元素 b 2 b_{2} b 2 重複 n 2 n_{2} n 2 次, ..., 元素 b r b_{r} b r 重複 n r n_{r} n r 次. 由於共有 ( n 1 n ) \binom {n_{1}}{n} ( n n 1 ) 種方法將元素 b 1 b_{1} b 1 放置在 n n n 個位置上, 共有 ( n 2 n ) \binom {n_{2}}{n} ( n n 2 ) 種方法將元素 b 2 b_{2} b 2 放置在 n n n 個位置上, ..., 共有 ( n r n ) \binom {n_{r}}{n} ( n n r ) 種方法將元素 b r b_{r} b r 放置在 n n n 個位置上, 所以 ( n 1 , n 2 , . . . , n r n ) = ( n 1 n ) ( n 2 n − n 1 ) . . . ( n r n − ( n 1 + n 2 + . . . n r − 1 ) ) = n ! n 1 ! ( n − n 1 ) ! × n ! n 2 ! ( n − n 1 − n 2 ) ! × . . . × 1 = n ! n 1 ! n 2 ! . . . n r ! . \displaystyle {\begin {aligned} \binom {n_{1}, n_{2}, ..., n_{r}}{n} &= \binom {n_{1}}{n}\binom {n_{2}}{n - n_{1}}...\binom {n_{r}}{n - (n_{1} + n_{2} + ... n_{r - 1})} \\ &= \frac {n!}{n_{1}!(n - n_{1})!} \times \frac {n!}{n_{2}!(n - n_{1} - n_{2})!} \times ... \times 1 \\ &= \frac {n!}{n_{1}!n_{2}!...n_{r}!}. \end {aligned}} ( n n 1 , n 2 , . . . , n r ) = ( n n 1 ) ( n − n 1 n 2 ) . . . ( n − ( n 1 + n 2 + . . . n r − 1 ) n r ) = n 1 ! ( n − n 1 ) ! n ! × n 2 ! ( n − n 1 − n 2 ) ! n ! × . . . × 1 = n 1 ! n 2 ! . . . n r ! n ! . 故 ∑ ω ∈ Ω p ( ω ) = ∑ { n 1 ≥ 0 , n 2 ≥ 0 , . . . , n r ≥ 0 n 1 + n 2 + . . . + n r = n } ( n 1 , n 2 , . . . , n r n ) p 1 n 1 p 2 n 2 . . . p r n r = ( p 1 + p 2 + . . . + p r ) n = 1. \displaystyle {\begin {aligned} \sum \limits_{\omega \in \Omega}p(\omega) &= \sum \limits_{ \left \{ \begin {gathered} \scriptsize {n_{1} \geq 0, n_{2} \geq 0, ..., n_{r} \geq 0} \\ \scriptsize {n_{1} + n _{2} + ... + n_{r} = n} \end {gathered} \right \}} \binom {n_{1}, n_{2}, ..., n_{r}}{n}p_{1}^{n_{1}}p_{2}^{n_{2}}...p_{r}^{n_{r}} \\ &= (p_{1} + p_{2} + ... + p_{r})^{n} = 1. \end {aligned}} ω ∈ Ω ∑ p ( ω ) = { n 1 ≥ 0 , n 2 ≥ 0 , . . . , n r ≥ 0 n 1 + n 2 + . . . + n r = n } ∑ ( n n 1 , n 2 , . . . , n r ) p 1 n 1 p 2 n 2 . . . p r n r = ( p 1 + p 2 + . . . + p r ) n = 1 . 因此, 針對二項 分佈推廣後的 分佈仍然是機率 分佈.
設 A n 1 , n 2 , . . . , n r = { ω : υ 1 ( ω ) = n 1 , υ 2 ( ω ) = n 2 , . . . , υ r ( ω ) = n r } , \displaystyle {A_{n_{1}, n_{2}, ..., n_{r}} = \left \{ \omega : \upsilon_{1}(\omega) = n_{1}, \upsilon_{2}(\omega) = n_{2}, ..., \upsilon_{r}(\omega) = n_{r} \right \}}, A n 1 , n 2 , . . . , n r = { ω : υ 1 ( ω ) = n 1 , υ 2 ( ω ) = n 2 , . . . , υ r ( ω ) = n r } , 則 P ( A n 1 , n 2 , . . . , n r ) = ( n 1 , n 2 , . . . , n r n ) p 1 n 1 p 2 n 2 . . . p r n r . \displaystyle {\mathop {\mathbf {P}}(A_{n_{1}, n_{2}, ..., n_{r}}) = \binom {n_{1}, n_{2}, ..., n_{r}}{n}p_{1}^{n_{1}}p_{2}^{n_{2}}...p_{r}^{n_{r}}}. P ( A n 1 , n 2 , . . . , n r ) = ( n n 1 , n 2 , . . . , n r ) p 1 n 1 p 2 n 2 . . . p r n r . 機率組 { P ( A n 1 , n 2 , . . . , n r ) } \{ \mathop {\mathbf {P}}(A_{n_{1}, n_{2}, ..., n_{r}}) \} { P ( A n 1 , n 2 , . . . , n r ) } 稱為多項分佈 (multinomial distribution). 而二項分佈是 r = 2 r = 2 r = 2 時多項分佈的特例.
3. 多元超幾何分佈
例題 3. 假設一箱子中有編號為 1 , 2 , . . . , n 1, 2, ..., n 1 , 2 , . . . , n 的 n n n 個不同的球. 其中, n 1 n_{1} n 1 個球具有顏色 c 1 c_{1} c 1 , n 2 n_{2} n 2 個球具有顏色 c 2 c_{2} c 2 , ..., n r n_{r} n r 個球具有顏色 c r c_{r} c r , n 1 + n 2 + . . . + n r = n n_{1} + n_{2} + ... + n_{r} = n n 1 + n 2 + . . . + n r = n . 現在從箱中進行 m m m 次不放回抽樣 (m < n m < n m < n ). 基本事件空間為 Ω = { ω : ω = ( a 1 , a 2 , . . . , a m ) , a k ≠ a l , k , l = 1 , 2 , . . . , m , k ≠ l , a i = 1 , 2 , . . . , n ( i = 1 , 2 , . . . , m ) } . \displaystyle {\begin {aligned} \Omega = \{ &\omega : \omega = (a_{1}, a_{2}, ..., a_{m}), a_{k} \neq a_{l}, k, l = 1, 2, ..., m, k \neq l, \\ &\ \ \ \ a_{i} = 1, 2, ..., n \ (i = 1, 2, ..., m) \}. \end {aligned}} Ω = { ω : ω = ( a 1 , a 2 , . . . , a m ) , a k = a l , k , l = 1 , 2 , . . . , m , k = l , a i = 1 , 2 , . . . , n ( i = 1 , 2 , . . . , m ) } . 顯然, c a r d Ω = ( n ) m \mathop {\mathrm {card}}{\Omega} = (n)_{m} c a r d Ω = ( n ) m . 假設基本事件都是等可能的, 而事件 B m 1 , m 2 , . . . , m r = { m 1 個球具有顏色 c 1 , m 2 個球具有顏色 c 2 , . . . , m r 個球具有顏色 c r , m 1 + m 2 + . . . + m r = m } B_{m_{1}, m_{2}, ..., m_{r}} = \{ m_{1} \text { 個球具有顏色 } c_{1}, m_{2} \text { 個球具有顏色 } c_{2}, ..., m_{r} \text { 個球具有顏色 } c_{r}, m_{1} + m_{2} + ... + m_{r} = m \} B m 1 , m 2 , . . . , m r = { m 1 個球具有顏色 c 1 , m 2 個球具有顏色 c 2 , . . . , m r 個球具有顏色 c r , m 1 + m 2 + . . . + m r = m } 的機率如何?
解 解 解 :從 n i n_{i} n i 個具有顏色 c i c_{i} c i 的球中抽出 m i m_{i} m i 個, 共有 ( m i n i ) \binom {m_{i}}{n_{i}} ( n i m i ) 種可能的結局. 其中, i = 1 , 2 , . . . , r i = 1, 2, ..., r i = 1 , 2 , . . . , r . 故有 c a r d { B m 1 , m 2 , . . . , m r } = ( m 1 n 1 ) ( m 2 n 2 ) . . . ( m r n r ) . \displaystyle {\mathop {\mathrm {card}}{\left \{ B_{m_{1}, m_{2}, ..., m_{r}} \right \}} = \binom {m_{1}}{n_{1}}\binom {m_{2}}{n_{2}}...\binom {m_{r}}{n_{r}} }. c a r d { B m 1 , m 2 , . . . , m r } = ( n 1 m 1 ) ( n 2 m 2 ) . . . ( n r m r ) . 因此, P ( B m 1 , m 2 , . . . , m r ) = c a r d { B m 1 , m 2 , . . . , m r } c a r d Ω = ( m 1 n 1 ) ( m 2 n 2 ) . . . ( m r n r ) ( m n ) \displaystyle {\mathop {\mathbf {P}}(B_{m_{1}, m_{2}, ..., m_{r}}) = \frac {\mathop {\mathrm {card}}{\left \{ B_{m_{1}, m_{2}, ..., m_{r}} \right \}}}{\mathop {\mathrm {card}}{\Omega}} = \frac {\binom {m_{1}}{n_{1}}\binom {m_{2}}{n_{2}}...\binom {m_{r}}{n_{r}}}{\binom {m}{n}}} P ( B m 1 , m 2 , . . . , m r ) = c a r d Ω c a r d { B m 1 , m 2 , . . . , m r } = ( n m ) ( n 1 m 1 ) ( n 2 m 2 ) . . . ( n r m r )
■ \blacksquare ■
機率組 { B m 1 , m 2 , . . . , m r } \{ B_{m_{1}, m_{2}, ..., m_{r}} \} { B m 1 , m 2 , . . . , m r } 稱為多元超幾何分布 (multivariate hypergeometric distribution). 當 r = 2 r = 2 r = 2 時, 多元超幾何分布退化為超幾何分布 (hypergeometric distribution), 其母函數為超幾何函數.
多元超幾何分布的構造相當複雜, 當 r = 2 r = 2 r = 2 時, P ( B m 1 , B m 2 ) = ( m 1 n 1 ) ( m 2 n 2 ) ( m 1 + m 2 n 1 + n 2 ) , n 1 + n 2 = n , m 1 + m 2 = m . \displaystyle {\mathop {\mathbf {P}}(B_{m_{1}}, B_{m_{2}}) = \frac {\binom {m_{1}}{n_{1}}\binom {m_{2}}{n_{2}}}{\binom {m_{1} + m _{2}}{n_{1} + n_{2}}}, n_{1} + n_{2} = n, m_{1} + m_{2} = m}. P ( B m 1 , B m 2 ) = ( n 1 + n 2 m 1 + m 2 ) ( n 1 m 1 ) ( n 2 m 2 ) , n 1 + n 2 = n , m 1 + m 2 = m . 其中包含了九階乘數. 當 n → ∞ n \to \infty n → ∞ , n 1 → ∞ n_{1} \to \infty n 1 → ∞ 且 n 1 n → p \frac {n_{1}}{n} \to p n n 1 → p 時, 有 n 2 n → 1 − p \frac {n_{2}}{n} \to 1 - p n n 2 → 1 − p . 根據斯特靈公式, 可以得到 P ( B m 1 , B m 2 ) ≐ ( m 2 m 1 + m 2 ) p m 1 ( 1 − p ) m 2 , \displaystyle {\mathop {\mathbf {P}}(B_{m_{1}}, B_{m_{2}}) \doteq \binom {m_{2}}{m_{1} + m_{2}} p^{m_{1}}(1 - p)^{m_{2}}}, P ( B m 1 , B m 2 ) ≐ ( m 1 + m 2 m 2 ) p m 1 ( 1 − p ) m 2 , 即當 n → ∞ , n 1 → ∞ n \to \infty, n_{1} \to \infty n → ∞ , n 1 → ∞ 且 n 1 n → p \frac {n_{1}}{n} \to p n n 1 → p 時, 超幾何分布逼近二項分布. 從直觀上來說, 這也是明顯的. 因為當 n n n 和 n 1 n_{1} n 1 充分大但是有限時, 不放回抽樣得到的結果幾乎和放回抽樣時一樣的.
4. 練習題
自主習題 1. 證明 : 對於多項分布的機率, 若且唯若點 ( k 1 , k 2 , . . . , k r ) (k_{1}, k_{2}, ..., k_{r}) ( k 1 , k 2 , . . . , k r ) 在滿足 n p i − 1 < k i ≤ ( n + r − 1 ) p i \displaystyle {np_{i} - 1 <k_{i} \leq (n + r - 1)p_{i}} n p i − 1 < k i ≤ ( n + r − 1 ) p i 時達到最大值. 其中, i = 1 , 2 , . . . , r i = 1, 2, ..., r i = 1 , 2 , . . . , r .
自主習題 2. 假設 N N N 是某個總體的容量, 要求在對總體的全部元素沒有簡單重複計數的情況下, 以最少的成本去估計 N N N 的大小. 例如, 在估計某個地區或者國家的人口等類似問題. Pierre-Simon Laplace 在 1786 年法國人口為 N N N 時, 提出過以下方法 : 從總體中選擇 m m m 個元素, 並且做上標記. 然後將這 m m m 個元素放回原總體, 並且與無標記的元素均衡混合. 然後從混合後的總體中再抽取 n n n 個元素, 其中有 x x x 個元素帶有標記.
證明 : 由超幾何分布的公式 P ( B m 1 , B m 2 ) = ( m 1 n 1 ) ( m 2 n 2 ) ( m 1 + m 2 n 1 + n 2 ) \mathop {\mathbf {P}}(B_{m_{1}}, B_{m_{2}}) = \frac {\binom {m_{1}}{n_{1}}\binom {m_{2}}{n_{2}}}{\binom {m_{1} + m _{2}}{n_{1} + n_{2}}} P ( B m 1 , B m 2 ) = ( n 1 + n 2 m 1 + m 2 ) ( n 1 m 1 ) ( n 2 m 2 ) , n 1 + n 2 = n , m 1 + m 2 = m n_{1} + n_{2} = n, m_{1} + m_{2} = m n 1 + n 2 = n , m 1 + m 2 = m , 相應的機率 P N , m , n { X = m } \mathop {\mathbf {P}}_{N, m, n} \{ X = m \} P N , m , n { X = m } 可以表示為 P N , m , n { X = M } = ( n m ) ( n − M N − M ) ( n N ) . \displaystyle {\mathop {\mathbf {P}}_{N, m, n} \left \{ X = M \right \} = \frac {\binom {n}{m}\binom {n - M}{N - M}}{\binom {n}{N}}}. P N , m , n { X = M } = ( N n ) ( m n ) ( N − M n − M ) . 假設 m , n m, n m , n 和 m m m 固定, 對 N N N 求上面機率的最大值, 即求總體的極大概似 (maximum likelihood) 容量 N N N , 使得對於給定的 m m m 和 n n n , 有標記的元素個數 X = M X = M X = M . 證明 : 對總體容量的極大概似估計值, 不妨記為 N ^ \hat {N} N ^ , 有 N ^ = ⌊ n m M ⌋ . \displaystyle {\hat {N} = \left \lfloor \frac {nm}{M} \right \rfloor}. N ^ = ⌊ M n m ⌋ .
這樣得到的估計量 N ^ \hat {N} N ^ 稱作 N N N 的極大概似估計 (maximum likelihood estimation).