Series in Probability Theory

概率论级数公式 | Series in Probability Theory

均匀分布 | Uniform Distribution

概率质量函数 | Probability Mass Function

$$p_X(x)=\frac{1}{n}\text{for}x=1,2,\dots ,n$$

期望值 | Expectation

$$\mathbb{E}(X)=\sum _{x=1}^{n}x{p}_X(x)=\sum _{x=1}^{n}x\cdot \frac{1}{n}=\frac{1}{n}\sum _{x=1}^{n}x$$

使用等差数列求和公式：

$$\sum _{x=1}^{n}x=\frac{n(n+1)}{2}$$

因此，

$$\mathbb{E}(X)=\frac{1}{n}\cdot \frac{n(n+1)}{2}=\frac{n+1}{2}$$

方差 | Variance

$$\mathrm{Var}(X)=\sum _{x=1}^n(x-\mathbb{E}(X){)}^2{p}_X(x)$$

使用离散型均匀分布的方差公式：

$$\mathrm{Var}(X)=\frac{(n^2-1)}{12}$$

伯努利分布 | Bernoulli Distribution

概率质量函数 | Probability Mass Function

$$p_X(x)=\{\begin{array}{ll}p& \text{if}x=1\\ 1-p& \text{if}x=0\end{array}$$

其中，$0\le p\le 1$

where $0\le p\le 1$

期望值 | Expectation

$$\mathbb{E}(X)=\sum _{x\in \{0,1\}}x{p}_X(x)=0\cdot (1-p)+1\cdot p=p$$

方差 | Variance

$$\mathrm{Var}(X)=\sum _{x\in \{0,1\}}(x-\mathbb{E}(X){)}^2{p}_X(x)=(0-p{)}^2(1-p)+(1-p{)}^{2}p=p(1-p)$$

二项分布 | Binomial Distribution

概率质量函数 | Probability Mass Function

$$p_X(k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}\text{for}k=0,1,2,\dots ,n$$

其中，$\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$

where $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$

期望值 | Expectation

$$\mathbb{E}(X)=\sum _{k=0}^{n}k\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}=np$$

方差 | Variance

$$\mathrm{Var}(X)=\sum _{k=0}^n(k-np{)}^2(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}=np(1-p)$$

几何分布 | Geometric Distribution

概率质量函数 | Probability Mass Function

$$p_X(k)=(1-p{)}^{k-1}p\text{for}k=1,2,3,\dots$$

期望值 | Expectation

$$\mathbb{E}(X)=\sum _{k=1}^{\infty }k(1-p{)}^{k-1}p$$

求和过程：

$$\begin{array}{rl}\mathbb{E}(X)& =\sum _{k=1}^{\infty }k(1-p{)}^{k-1}p\\ & =p\sum _{k=1}^{\infty }k(1-p{)}^{k-1}\\ \text{令}q& =1-p\\ \mathbb{E}(X)& =p\sum _{k=1}^{\infty }k{q}^{k-1}\\ \text{使用等比级数求和公式}\sum _{k=1}^{\infty }k{q}^{k-1}& =\frac{1}{(1-q{)}^2}\\ \mathbb{E}(X)& =p\cdot \frac{1}{(1-q{)}^2}=p\cdot \frac{1}{p^2}=\frac{1}{p}\end{array}$$

方差 | Variance

$$\mathrm{Var}(X)=\sum _{k=1}^{\infty }{\left(k-\mathbb{E}(X)\right)}^2(1-p{)}^{k-1}p=\sum _{k=1}^{\infty }{\left(k-\frac{1}{p}\right)}^2(1-p{)}^{k-1}p$$

计算过程 | Calculation Process

1. 展开平方项

$$\begin{array}{rl}\mathrm{Var}(X)& =\sum _{k=1}^{\infty }\left(k^2-2k\cdot \frac{1}{p}+\frac{1}{p^2}\right)(1-p{)}^{k-1}p\\ & =\sum _{k=1}^{\infty }\left(k^2(1-p{)}^{k-1}p-2k\cdot \frac{1}{p}(1-p{)}^{k-1}p+\frac{1}{p^2}(1-p{)}^{k-1}p\right)\\ & =p\left[\sum _{k=1}^{\infty }{k}^2(1-p{)}^{k-1}-2\cdot \frac{1}{p}\sum _{k=1}^{\infty }k(1-p{)}^{k-1}+\frac{1}{p^2}\sum _{k=1}^{\infty }(1-p{)}^{k-1}\right]\end{array}$$

2. 分别计算三个部分

第一部分：

$$\sum _{k=1}^{\infty }{k}^2(1-p{)}^{k-1}$$

利用公式 ${\sum }_{k=1}^{\infty }{k}^2{q}^{k-1}=\frac{1+q}{(1-q{)}^3}$ 其中 $q=1-p$

$$\sum _{k=1}^{\infty }{k}^2(1-p{)}^{k-1}=\frac{1+(1-p)}{(1-(1-p){)}^3}=\frac{2-p}{p^3}$$

第二部分：

$$\sum _{k=1}^{\infty }k(1-p{)}^{k-1}$$

利用公式 ${\sum }_{k=1}^{\infty }k{q}^{k-1}=\frac{1}{(1-q{)}^2}$ 其中 $q=1-p$

$$\sum _{k=1}^{\infty }k(1-p{)}^{k-1}=\frac{1}{p^2}$$

第三部分：

$$\sum _{k=1}^{\infty }(1-p{)}^{k-1}$$

这是一个无穷等比数列，求和结果为：

$$\sum _{k=1}^{\infty }(1-p{)}^{k-1}=\frac{1}{1-(1-p)}=\frac{1}{p}$$

3. 综合计算方差

$$\begin{array}{rl}\mathrm{Var}(X)& =p\left[\frac{2-p}{p^3}-2\cdot \frac{1}{p}\cdot \frac{1}{p^2}+\frac{1}{p^2}\cdot \frac{1}{p}\right]\\ & =p\left[\frac{2-p}{p^3}-\frac{2}{p^3}+\frac{1}{p^3}\right]\\ & =p\left[\frac{2-p-2+1}{p^3}\right]\\ & =p\left[\frac{-p+1}{p^3}\right]\\ & =\frac{1-p}{p^2}\end{array}$$

综上所述，几何分布的方差为：

$$\mathrm{Var}(X)=\frac{1-p}{p^2}$$

泊松分布 | Poisson Distribution

概率质量函数 | Probability Mass Function

$$p_X(k)=\frac{{\lambda }^k{e}^{-\lambda }}{k!}\text{for}k=0,1,2,\dots$$

期望值 | Expectation

$$\mathbb{E}(X)=\sum _{k=0}^{\infty }k\frac{{\lambda }^k{e}^{-\lambda }}{k!}=\lambda$$

求和过程：

$$\begin{array}{rl}\mathbb{E}(X)& =\sum _{k=0}^{\infty }k\frac{{\lambda }^k{e}^{-\lambda }}{k!}\\ \text{令}p(k)& =\frac{{\lambda }^k{e}^{-\lambda }}{k!}\\ \mathbb{E}(X)& =\sum _{k=1}^{\infty }kp(k)\\ & =\sum _{k=1}^{\infty }k\frac{{\lambda }^k{e}^{-\lambda }}{k!}\\ & =\sum _{k=1}^{\infty }\lambda \frac{{\lambda }^{k-1}{e}^{-\lambda }}{(k-1)!}\\ & =\lambda \sum _{k=0}^{\infty }\frac{{\lambda }^k{e}^{-\lambda }}{k!}\\ & =\lambda \cdot 1=\lambda \end{array}$$

方差 | Variance

$$\mathrm{Var}(X)=\sum _{k=0}^{\infty }(k-\lambda {)}^2\frac{{\lambda }^k{e}^{-\lambda }}{k!}=\lambda$$

求和过程：

$$\begin{array}{rl}\mathrm{Var}(X)& =\sum _{k=0}^{\infty }(k-\lambda {)}^2\frac{{\lambda }^k{e}^{-\lambda }}{k!}\\ & =\sum _{k=0}^{\infty }(k^2-2k\lambda +{\lambda }^2)\frac{{\lambda }^k{e}^{-\lambda }}{k!}\\ & =\sum _{k=0}^{\infty }{k}^2\frac{{\lambda }^k{e}^{-\lambda }}{k!}-2\lambda \sum _{k=0}^{\infty }k\frac{{\lambda }^k{e}^{-\lambda }}{k!}+{\lambda }^2\sum _{k=0}^{\infty }\frac{{\lambda }^k{e}^{-\lambda }}{k!}\\ \text{第一项:}& \\ \sum _{k=0}^{\infty }{k}^2\frac{{\lambda }^k{e}^{-\lambda }}{k!}& =\lambda \sum _{k=1}^{\infty }k\frac{{\lambda }^{k-1}{e}^{-\lambda }}{(k-1)!}\\ & =\lambda \sum _{k=0}^{\infty }(k+1)\frac{{\lambda }^k{e}^{-\lambda }}{k!}\\ & =\lambda \left[\sum _{k=0}^{\infty }k\frac{{\lambda }^k{e}^{-\lambda }}{k!}+\sum _{k=0}^{\infty }\frac{{\lambda }^k{e}^{-\lambda }}{k!}\right]\\ & =\lambda (\lambda +1)\\ & ={\lambda }^2+\lambda \\ \text{第二项:}& \\ -2\lambda \sum _{k=0}^{\infty }k\frac{{\lambda }^k{e}^{-\lambda }}{k!}& =-2\lambda \lambda \\ & =-2{\lambda }^2\\ \text{第三项:}& \\ {\lambda }^2\sum _{k=0}^{\infty }\frac{{\lambda }^k{e}^{-\lambda }}{k!}& ={\lambda }^2\cdot 1\\ & ={\lambda }^2\\ \mathrm{Var}(X)& ={\lambda }^2+\lambda -2{\lambda }^2+{\lambda }^2\\ & =\lambda \end{array}$$