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n重伯努利试验与二项分布

2022-01-26 12:45 作者:匆匆-cc  | 我要投稿

        伯努利试验(Bernoulli experiment):在同样的条件下重复地、相互独立地进行的一种随机试验。该随机试验只有两种可能结果:发生或者不发生。

        比如说,掷一枚硬币,其结果必然是正面朝上,或者反面朝上(即不是正面朝上),即为一种伯努利试验

        我们假设该项试验独立重复地进行了n次,那么就称这一系列独立重复的随机试验为n重伯努利试验。

        n重伯努利试验具有以下基本特征:

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

        该公式含义为:

        从n次试验中选出k次为发生该事件,由于事件间相互独立,因此乘上每一个事件发生的概率。

        显然,由二项式定理,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

        这也符合分布列的基本要求:

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

        即:所有基本事件发生的可能性之和为1。

        下面推导n重伯努利试验的期望(也叫做均值)。

        根据期望的定义,

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

        为对此项求和,引入公式:

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

        代数证明如下:

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

        组合意义证明如下:

            考虑一个n个人的队伍,需要从中选出k个人,并且其中1个为队长。

            解法一:先选出k个人,再从这k个人中选出1个队长,第一步有C_%7Bn%7D%5Ek%20种,第二步有k种,由乘法原理,共有kC%5Ek_n种。

            解法二:先选出1个队长,再从剩下(n-1)个人中选出(k-1)个队员,第一步有n种,第二步有C%5E%7Bk-1%7D_%7Bn-1%7D种,由乘法原理,共有nC_%7Bn-1%7D%5E%7Bk-1%7D种。

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

        接着证明。

%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,根据方差公式

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

## 简单推导如下:

%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

        我们有

%5Cbegin%7Balign%7D%0AD_%7B(%5Cxi)%7D%26%3D%5Csum_%7Bk%3D%5Ccolor%7Bred%7D%7B0%7D%7D%5En%20k%5E2%20C_%7Bn%7D%5Ek%20p%5Ek%20(1-p)%5E%7Bn-k%7D-E%5E2_%7B(%5Cxi)%7D%0A%5C%5C%26%3D%5Csum_%7Bk%3D%5Ccolor%7Bred%7D%7B1%7D%7D%5En%20%5Ccolor%7Bblue%7D%7Bk%5E2%20C_%7Bn%7D%5Ek%7D%20p%5Ek%20(1-p)%5E%7Bn-k%7D-E%5E2_%7B(%5Cxi)%7D%0A%5C%5C%26%3D%5Csum_%7Bk%3D1%7D%5En%20%5Ccolor%7Bblue%7D%7Bn%20C_%7Bn-1%7D%5E%7Bk-1%7D%20k%7D%20p%5Ek%20(1-p)%5E%7Bn-k%7D-E%5E2_%7B(%5Cxi)%7D%0A%5C%5C%26%3Dn%5Csum_%7Bk%3D1%7D%5En%20kC_%7Bn-1%7D%5E%7Bk-1%7D%20p%5Ek%20(1-p)%5E%7Bn-k%7D-E%5E2_%7B(%5Cxi)%7D%0A%5C%5C%26%3Dn%5Csum_%7Bk%3D%5Ccolor%7Bred%7D%7B1%7D%7D%5En%20(k-1)C_%7Bn-1%7D%5E%7Bk-1%7D%20p%5Ek%20(1-p)%5E%7Bn-k%7D%2Bn%5Csum_%7Bk%3D1%7D%5En%20C_%7Bn-1%7D%5E%7Bk-1%7D%20p%5Ek%20(1-p)%5E%7Bn-k%7D-E%5E2_%7B(%5Cxi)%7D%0A%5C%5C%26%3Dn%5Csum_%7Bk%3D%5Ccolor%7Bred%7D%7B2%7D%7D%5En%20(k-1)C_%7Bn-1%7D%5E%7Bk-1%7D%20p%5Ek%20(1-p)%5E%7Bn-k%7D%2Bnp%5Csum_%7Bk%3D1%7D%5En%20C_%7Bn-1%7D%5E%7Bk-1%7D%20p%5E%7Bk-1%7D%20(1-p)%5E%7B(n-1)-(k-1)%7D-E%5E2_%7B(%5Cxi)%7D%0A%5C%5C%26%3Dn%5Csum_%7Bk%3D2%7D%5En%20(n-1)C_%7Bn-2%7D%5E%7Bk-2%7D%20p%5Ek%20(1-p)%5E%7Bn-k%7D%2Bnp%5Bp%2B(1-p)%5D%5E%7Bn-1%7D-E%5E2_%7B(%5Cxi)%7D%0A%5C%5C%26%3Dn(n-1)%5Csum_%7Bk%3D2%7D%5En%20C_%7Bn-2%7D%5E%7Bk-2%7D%20p%5Ek%20(1-p)%5E%7Bn-k%7D%2Bnp-E%5E2_%7B(%5Cxi)%7D%0A%5C%5C%26%3Dn(n-1)p%5E2%5Csum_%7Bk%3D2%7D%5En%20C_%7Bn-2%7D%5E%7Bk-2%7D%20p%5E%7Bk-2%7D%20(1-p)%5E%7B(n-2)-(k-2)%7D%2Bnp-E%5E2_%7B(%5Cxi)%7D%0A%5C%5C%26%3Dn(n-1)p%5E2%5Bp%2B(1-p)%5D%5E%7Bn-2%7D%2Bnp-E%5E2_%7B(%5Cxi)%7D%0A%5C%5C%26%3Dn(n-1)p%5E2%2Bnp-n%5E2p%5E2%0A%5C%5C%26%3D%5Ccancel%7Bn%5E2p%5E2%7D-np%5E2%2Bnp-%5Ccancel%7Bn%5E2p%5E2%7D%0A%5C%5C%26%3Dnp(1-p)%0A%5Cend%7Balign%7D

        思路很简单,围绕一个组合恒等式进行证明。

        看起来很长,仔细琢磨就是不断从未知套入已知。

        特别的,当n%3D1时,二项分布退化为两点分布

        相关期望与方差只要在二项分布中取n%3D1即可。

E_%7B(%5Cxi%20)%7D%3Dp

D_%7B(%5Cxi)%7D%3Dp(1-p)

        两点分布典型的例子有:

            抛掷硬币的正反面,明天是否下雨,etc.

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