n重伯努利試驗(yàn)與二項(xiàng)分布

2022-01-26 12:45 作者:匆匆-cc 0人讀過 | 我要投稿

? ? ? ? 伯努利試驗(yàn)（Bernoulli experiment）：在同樣的條件下重復(fù)地、相互獨(dú)立地進(jìn)行的一種隨機(jī)試驗(yàn)。該隨機(jī)試驗(yàn)只有兩種可能結(jié)果：發(fā)生或者不發(fā)生。

? ? ? ? 比如說，擲一枚硬幣，其結(jié)果必然是正面朝上，或者反面朝上（即不是正面朝上），即為一種伯努利試驗(yàn)。

? ? ? ??我們假設(shè)該項(xiàng)試驗(yàn)獨(dú)立重復(fù)地進(jìn)行了 $n$ 次，那么就稱這一系列獨(dú)立重復(fù)的隨機(jī)試驗(yàn)為n重伯努利試驗(yàn)。

? ? ? ? n重伯努利試驗(yàn)具有以下基本特征：

$P_%7B(%5Cxi%20%3Dk)%7D%3DC_%7Bn%7D%5Ek%20p%5Ek%20(1-p)%5E%7Bn-k%7D$

? ? ? ? 該公式含義為：

? ? ? ? 從 $n$ 次試驗(yàn)中選出 $k$ 次為發(fā)生該事件，由于事件間相互獨(dú)立，因此乘上每一個(gè)事件發(fā)生的概率。

? ? ? ? 顯然，由二項(xiàng)式定理，有：

$%5Csum_%7Bk%3D0%7D%5En%20P_%7B(%5Cxi%20%3Dk)%7D%20%3D%20%5Csum_%7Bk%3D0%7D%5En%20C_%7Bn%7D%5Ek%20p%5Ek%20(1-p)%5E%7Bn-k%7D%20%3D%20%20%5Csum_%7Bk%3D0%7D%5En%20C_%7Bn%7D%5E%7Bn-k%7D%20p%5Ek%20(1-p)%5E%7Bn-k%7D%20%3D%20%5Bp%20%2B%20(1-p)%5D%5En%3D1$

? ? ? ? 這也符合分布列的基本要求：

$%5Csum_%7Bi%3D1%7D%5En%20Pi%20%3D%201%20$

? ? ? ? 即：所有基本事件發(fā)生的可能性之和為1。

? ? ? ? 下面推導(dǎo)n重伯努利試驗(yàn)的期望（也叫做均值）。

? ? ? ? 根據(jù)期望的定義，

$E_%7B(%5Cxi)%7D%3D%5Csum_%7Bk%3D0%7D%5En%20C_%7Bn%7D%5Ek%20p%5Ek%20(1-p)%5E%7Bn-k%7D%20%5Ccdot%20k%20%3D%20%5Csum_%7Bk%3D0%7D%5En%20k%20C_%7Bn%7D%5Ek%20p%5Ek%20(1-p)%5E%7Bn-k%7D$

? ? ? ? 為對(duì)此項(xiàng)求和，引入公式：

$kC_%7Bn%7D%5Ek%20%3D%20nC_%7Bn-1%7D%5E%7Bk-1%7D$

? ? ? ? 代數(shù)證明如下：

$kC_%7Bn%7D%5Ek%20%3D%20k%5Ccdot%20%5Cfrac%7Bn!%7D%7Bk!(n-k)!%7D%3Dn%5Ccdot%20%5Cfrac%7B(n-1)!%7D%7B(k-1)!(n-k)!%7D%3DnC_%7Bn-1%7D%5E%7Bk-1%7D$

? ? ? ? 組合意義證明如下：

? ? ? ? ? ? 考慮一個(gè) $n$ 個(gè)人的隊(duì)伍，需要從中選出 $k$ 個(gè)人，并且其中 $1$ 個(gè)為隊(duì)長(zhǎng)。

? ? ? ? ? ? 解法一：先選出 $k$ 個(gè)人，再?gòu)倪@ $k$ 個(gè)人中選出 $1$ 個(gè)隊(duì)長(zhǎng)，第一步有 $C_%7Bn%7D%5Ek%20$ 種，第二步有 $k$ 種，由乘法原理，共有 $kC%5Ek_n$ 種。

????????????解法二：先選出 $1$ 個(gè)隊(duì)長(zhǎng)，再?gòu)氖Ｏ拢?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=n-1" alt="n-1">）個(gè)人中選出（ $k-1$ ）個(gè)隊(duì)員，第一步有 $n$ 種，第二步有 $C%5E%7Bk-1%7D_%7Bn-1%7D$ 種，由乘法原理，共有 $nC_%7Bn-1%7D%5E%7Bk-1%7D$ 種。

????????????故 $kC_%7Bn%7D%5Ek%20%3D%20nC_%7Bn-1%7D%5E%7Bk-1%7D$ 。

????????接著證明。

$%5Cbegin%7Balign%7D%0AE_%7B(%5Cxi)%7D%26%3D%5Csum_%7Bk%3D%5Ccolor%7Bred%7D%7B0%7D%7D%5En%20k%20C_%7Bn%7D%5Ek%20p%5Ek%20(1-p)%5E%7Bn-k%7D%0A%5C%5C%26%3D%5Csum_%7Bk%3D%5Ccolor%7Bred%7D%7B1%7D%7D%5En%20%5Ccolor%7Bblue%7D%7Bk%20C_%7Bn%7D%5Ek%7D%20p%5Ek%20(1-p)%5E%7Bn-k%7D%0A%5C%5C%26%3D%5Csum_%7Bk%3D1%7D%5En%20%5Ccolor%7Bblue%7D%7Bn%20C_%7Bn-1%7D%5E%7Bk-1%7D%7D%20p%5Ek%20(1-p)%5E%7Bn-k%7D%0A%5C%5C%26%3Dn%5Csum_%7Bk%3D1%7D%5En%20C_%7Bn-1%7D%5E%7Bk-1%7D%20p%5Ek%20(1-p)%5E%7Bn-k%7D%0A%5C%5C%26%3Dnp%5Csum_%7Bk%3D1%7D%5En%20C_%7Bn-1%7D%5E%7Bk-1%7D%20p%5Ek%20(1-p)%5E%7B(n-1)-(k-1)%7D%0A%5C%5C%26%3Dnp%5Bp%2B(1-p)%5D%5E%7Bn-1%7D%0A%5C%5C%26%3Dnp%0A%5Cend%7Balign%7D$

????????同理，根據(jù)方差公式

$D_%7B(%5Cxi%20)%7D%3DE_%7B(%5Cxi%5E2)%7D-E%5E2_%7B(%5Cxi)%7D$

## 簡(jiǎn)單推導(dǎo)如下：

$%5Cbegin%7Balign%7D%0A%5Ccolor%7BGray%7D%7BD_%7B(%5Cxi)%7D%7D%26%5Ccolor%7BGray%7D%7B%3D%7D%5Ccolor%7BGray%7D%7B%5Csum_%7Bi%3D1%7D%5En(%5Cxi_i-E_%7B(%5Cxi)%7D)%5E2p_i%7D%0A%5C%5C%26%5Ccolor%7BGray%7D%7B%3D%7D%5Ccolor%7BGray%7D%7B%5Csum_%7Bi%3D1%7D%5En%5Cxi_i%5E2p_i-2E_%7B(%5Cxi)%7D%5Csum_%7Bi%3D1%7D%5En%5Cxi_ip_i%2BE%5E2_%7B(%5Cxi)%7D%7D%0A%5C%5C%26%5Ccolor%7BGray%7D%7B%3D%7D%5Ccolor%7BGray%7D%7B%5Csum_%7Bi%3D1%7D%5En%5Cxi_i%5E2p_i-E%5E2_%7B(%5Cxi)%7D%7D%0A%5C%5C%26%5Ccolor%7BGray%7D%7B%3D%7D%5Ccolor%7BGray%7D%7BE_%7B(%5Cxi%5E2)%7D-E%5E2_%7B(%5Cxi)%7D%7D%0A%5Cend%7Balign%7D$

????????我們有

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