[概率论] - 常见统计分布，指数

0-1分布

• 符号：$\color{red}{X \sim{} 0 - 1(p)}$
• 分布律

$P(X=k) = p^k(1-p)^{1-k} \quad k = 0, 1$

• 期望值

$\sum^1_{k=0} k \cdot p^k(1-p)^{1-k} = 0(1-p) + p = p$

• 方差

$\begin{aligned} D(X) &= \sum^1_{i=0} \frac{(x_i - \mu)^2}{1} = p(1-p) \\ &\text{或者} \\ D(X) &= E(X^2) - E(X) \\ &= 0^2 \cdot (1-p) + 1^2 \cdot p - (0 \cdot (1-p) + 1 \cdot p)^2 \\ &= p(1-p) \end{aligned}$

二项分布

• 符号：$\color{red}{X \sim{} b(n, p)}$
• 分布律

$\begin{aligned} P(X=k) = &\binom{n}{k}p^k(1-p)^{n-k} \quad k = 1, 2, \cdots, n \\ \text{其中}&\binom{n}{k} = C^k_n = \frac{n!}{k!(n-k)!} \end{aligned}$

• 期望值

$\begin{aligned} E(X) &= \sum^n_{k=0} k \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k} \\ &= \sum^n_{k=1} k \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k} \\ &= \sum^n_{k=1} k \frac{n(n-1)!}{k(k-1)!(n-k)!}p^{k}(1-p)^{n-k} \\ &= np\color{red}{\sum^n_{k=1} \frac{(n-1)!}{(k-1)!(n-k)!}p^{k-1}(1-p)^{n-k}} \\ &= np\color{red}{\sum^n_{k=1} \frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}p^{k-1}(1-p)^{(n-1)-(k-1)}} \\ &= np\color{red}{\sum^{n-1}_{(k-1)=0}\binom{n-1}{k-1}p^{k-1}(1-p)^{(n-1)-(k-1)}} \\ &= np\color{red}{\sum^b_{a=0}\binom{b}{a}p^a(1-p)^{b-a} \quad (a = k -1, b = n - 1)} \\ &= np \end{aligned}$

• 方差

$\begin{aligned} E(X^2) &= E(X(X - 1) + X) \\ &= E(X(X - 1)) + E(X) \\ &= \sum^n_{k=0} k(k-1)\binom{k}{n}p^k(1-p)^{n-k} + np \\ &= \sum^n_{k=1} k(k-1) \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k} + np \\ &= \sum^n_{k=2} k(k-1) \frac{n(n-1)(n-2)!}{k(k-1)(k-2)!(n-k)!}p^k(1-p)^{n-k} + np \\ &= n(n-1)\color{red}\sum^n_{k=2} \frac{(n-2)!}{(k-2)!(n-k)!}p^k(1-p)^{n-k}\color{black} + np \\ &= n(n-1)np + np \\ &= (n^2 - n)p^2 + np \\ \\ D(X) &= E(X^2) - E(X)^2 \\ &= (n^2 - n)p^2 + np - (np)^2 \\ &= np(1 - p) \end{aligned}$

几何分布

• 符号：$\color{red}{X \sim{} G(p)}$
• 分布律

$P(X=k) = (1-p)^{k-1}p \quad k=1,2,\cdots$

• 期望值

$\begin{aligned} E(X) &= \sum^n_{k=0} k (1-p)^{k-1}p \\ &= p\sum^n_{k=0} k (1-p)^{k-1} \\ &= p(1 + 2q +3q^2 + \cdots + nq^{n+1}), \quad q=(1-p) \\ \\ S_k &= 1 + 2q +3q^2 + \cdots + nq^{n+1} \\ &\ \downarrow \\ (1-q)S_k &= 1 + q +q^2 + \cdots + q^{n+1} \\ &= \frac{1-q^n}{1-q} - nq^n \\ &\ \downarrow \\ S_k &= \frac{1-q^n}{(1-q)^2} - \frac{nq^n}{1-q} \\ &\ \downarrow \\ \lim_{n\rightarrow+\infty}{q^n} = 0 &\Rightarrow \lim_{n\rightarrow+\infty}{S_k} = \frac{1}{(1-q)^2} = 0 \\ &\ \downarrow \\ E(X) &= \frac{p}{(1-q)^2} = \frac{1}{p} \end{aligned}$

• 方差

$\begin{aligned} E(X^2) &= \sum^n_{k=0} k^2 (1-p)^{k-1}p \\ &= p\sum^n_{k=0} k^2 (1-p)^{k-1} \\ &= p(1 + 2^2q +3^2q^2 + \cdots + n^2q^{n-1}) \\ &= p(q + 2q^2 +3q^3 + \cdots + nq^{n})' \\ &= p[\frac{q}{(1-q)^2}]' \\ &= p\frac{1+q}{(1-q)^3} \\ &= \frac{1+q}{(1-q)^2} \\ \\ D(X) &= E(X^2) - E(X)^2 \\ &= \frac{1+q}{(1-q)^2} - \frac{1}{p} \\ &= \frac{1-p}{p^2} \end{aligned}$

超几何分布

• 符号：$\color{red}{X \sim{} H(n, K, N)}$
• 分布律

$\begin{aligned} P(X=k, n, K, N) &= \frac{C^k_MC^{n-k}_{N-M}}{C^n_N} \\ \frac{\binom{M}{k}\binom{N-M}{n-k}}{\frac{N}{k}} \end{aligned}$

• 期望值

$\begin{aligned} E(X) &= \sum^l_{k=0} k \frac{\binom{M}{k}\binom{N-M}{n-k}}{\frac{N}{k}} \\ &= \frac{M}{\binom{N}{n}} \sum^l_{k=0} \binom{M - 1}{k - 1}\binom{N-M}{n-k} \\ &= \frac{M}{\binom{N}{n}} \binom{N - 1}{n - 1} \\ &= \frac{nM}{N} \end{aligned}$

• 方差

$\begin{aligned} E(X^2) &= E(X(X - 1) + X) \\ &= E(X(X - 1)) + E(X) \\ &= \sum^l_{k=0} k^2 \frac{\binom{M}{k}\binom{N-M}{n-k}}{\frac{N}{k}} - (\frac{nM}{N})^2 \\ &= \frac{M}{\binom{N}{n}} \sum^l_{k=0} k \binom{M - 1}{k - 1}\binom{N-M}{n-k} - (\frac{nM}{N})^2 \\ &= \frac{M}{\binom{N}{n}} \sum^l_{k=1} (i - 1)\binom{M - 1}{k - 1}\binom{N-M}{n-k} + \frac{M}{\binom{N}{n}}\sum^l_{k=1}\binom{M - 1}{k - 1}\binom{N-M}{n-k} - (\frac{nM}{N})^2 \\ &= \frac{M}{\binom{N}{n}} (M-1) \sum^l_{k=2} \binom{M - 2}{k - 2}\binom{N - M}{n - i} + \frac{M}{\binom{N}{n}}\sum^l_{k=1} \binom{M - 1}{k - 2}\binom{N - M}{n - i} - (\frac{nM}{N})^2 \\ &= \frac{M}{\binom{N}{n}} (M - 1)\binom{N - 2}{n - 2} + \frac{M}{\binom{N}{n}} \binom{N - 1}{n - 1} - (\frac{nM}{N})^2 \\ &= n\frac{M}{N}(1 - \frac{M}{N})(1 - \frac{n-1}{N-1}) \end{aligned}$

泊松分布

• 符号：$\color{red}{X \sim{} \pi(\lambda)}$
• 分布律

$P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} \quad (k = 1, 2, \cdots, n)$

• 期望值

$\begin{aligned} E(X) &= \sum^\infty_{k=0} k\frac{\lambda^k e^{-\lambda}}{k!} \\ &= \lambda{}e^{-\lambda}\sum^\infty_{k=1}\frac{\lambda^{k-1}}{(k-1)!} \\ &= \lambda{}e^{-\lambda} \cdot e^\lambda \\ &= \lambda \end{aligned}$

• 方差

$\begin{aligned} E(X^2) &= E(X(X - 1) + X) \\ &= E(X(X - 1)) + E(X) \\ &= \sum^\infty_{k=0} k(k-1) \frac{\lambda^ke^{-\lambda}}{k!} + \lambda \\ &= \lambda^2e^{-\lambda}\sum^\infty_{k=2}\frac{\lambda^{k-2}}{(k-2)!}+\lambda \\ &=\lambda^2e^{-\lambda}e^\lambda + \lambda \\ &= \lambda^2 + \lambda \\ \\ D(X) &= E(X^2) - E(X)^2 \\ &= \lambda^2 + \lambda - \lambda^2 \\ &= \lambda \end{aligned}$

泊松定理

设$\lambda > 0$是一个常数，$n$是任意正整数，设$np_n = \lambda$，则对于任一固定的非负整数$k$，有

$\lim_{n \to \infty}\binom{n}{k}p^k_n(1-p_n)^{n-k} = \frac{\lambda^k e^{-\lambda}}{k!}$

定理的条件$np_n = \lambda$（常数）意味着当$n$很大时$p_n$必定很小，因此，上述定理表明当$n$很大，$p$很小时（$np=\lambda$）有以下近似式

$\binom{n}{k}p^k(1-p)^{n-k} \approx \frac{\lambda^k e^{-\lambda}}{k!}$

均匀分布

• 符号：$\color{red}{X \sim{} U(a, b)}$
• 概率密度函数

$f(x) = \begin{cases} \frac{1}{b-a} \quad & a < x < b \\ 0 \quad & \text{其他} \end{cases}$

• 期望值

$\begin{aligned} E(X) &= \int^b_axf(x)dx \\ &= \int^b_a\frac{x}{b-a}dx \\ &= [\frac{x^2}{2(b-a)}]^b_a \\ &=\frac{b^2}{2(b-a)} - \frac{a^2}{2(b-a)} \\ &=\frac{a+b}{2} \end{aligned}$

• 方差

$\begin{aligned} E(X^2) &= \int^b_ax^2\frac{1}{b-a}dx \\ &= [\frac{x^3}{3(b-a)}]^b_a \\ &= \frac{b^3 - a^3}{3(b-a)} \\ &= \frac{b^2 + ab + a^2}{3} \\ \\ D(X) &= E(X^2) - E(X)^2 \\ &= \frac{b^2 + ab + a^2}{3} - \frac{a^2 + 2ab + b^2}{4} \\ &= \frac{(b-a)^2}{12} \end{aligned}$

指数分布

• 符号：$\color{red}{X \sim{} Exp(\lambda)}$
• 概率密度函数

$f(x) = \begin{cases} \lambda{}e^{-\lambda{}x} \quad & x \geq 0 \\ 0 \quad & x < 0 \end{cases}$

• 分布函数

$F(x) = \begin{cases} 1 - e^{-\lambda{}x} \quad & x \geq 0 \\ 0 \quad & x < 0 \end{cases}$

• 期望值

$\begin{aligned} E(X) &= \int^\infty_{-\infty}xf(x)dx \\ &= \int^\infty_0 x\lambda{}e^{-\lambda{}x}dx \\ &= [-xe^{-\lambda x}]^\infty_0 + \int^\infty_0 e^{-\lambda{}x}dx \\ &= \frac{1}{\lambda} \end{aligned}$

• 方差

$\begin{aligned} E(X^2) &= \int^\infty_{-\infty}x^2f(x)dx \\ &= \int^\infty_0 x^2\lambda{}e^{-\lambda{}x}dx \\ &= [-x^2e^{-\lambda x}]^\infty_0 + \int^\infty_0 2xe^{-\lambda{}x}dx \\ &=\frac{2}{\lambda^2} \\ \\ D(X) &= E(X^2) - E(X)^2 \\ &= \frac{2}{\lambda^2} - \frac{1}{\lambda^2} \\ &= \frac{1}{\lambda^2} \end{aligned}$

正态分布

• 符号：$\color{red}{X \sim{} N(\mu, \sigma^2)}$
• 概率密度函数

$f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-u)^2}{2\sigma^2}} \quad -\infty < x < \infty$

• 期望值

取$Z = \dfrac{X - \mu}{\lambda}$

$\begin{aligned} E(Z) &= \frac{1}{\sqrt{2\pi}}\int^\infty_\infty te^{-t^2/2}dt \\ &= \frac{-1}{\sqrt{2\pi}}e^{-t^2/2}|^\infty_\infty \\ &= 0 \\ \\ E(X) &= E(\mu + \lambda Z) = \mu \end{aligned}$

• 方差

$\begin{aligned} D(Z) &= E(Z^2) - E(Z)^2 \\ &= E(Z^2) \\ &= \frac{1}{\sqrt{2\pi}}\int^\infty_{-\infty} t^2e^{-t^2/2}dt \\ &= \frac{-1}{\sqrt{2\pi}}te^{-t^2/2}|^\infty_{-\infty} + \frac{1}{\sqrt{2\pi}}\int^\infty_{-\infty} e^{-t^2/2} \\ &= 1 \\ \\ D(X) &= D(\mu + \lambda Z) = D(\lambda Z) = \lambda^2D(Z) = \lambda^2 \end{aligned}$

常见指数

• 拉氏价格指数

$P_L = \frac{\sum^n_{i=1}\text{比较年价格} \times \text{\textbf{基准}年数量}}{\sum^n_{i=1}\text{\textbf{基准}年价格} \times \text{\textbf{基准}年数量}} \times 100$

• 帕氏指数

$P_P = \frac{\sum^n_{i=1}\text{\textbf{比较}年价格} \times \text{\textbf{比较}年数量}}{\sum^n_{i=1}\text{基准年价格} \times \text{\textbf{比较}年数量}} \times 100$

• 费舍理想指数

$F_P = \sqrt{P_L \times P_P}$

总结

分布	参数	符号	分布律/概率密度函数	期望	方差
$0-1$分布	$0<p<1$	$\color{red}{X \sim{} 0 - 1(p)}$	$P(X=k) = p^k(1-p)^{1-k}$	$p$	$p(1-p)$
二项分布	$\begin{aligned}n \geq 1\quad\\0<p<1\end{aligned}$	$\color{red}{X \sim{} b(n, p)}$	$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$	$np$	$np(1 - p)$
几何分布	$0<p<1$	$\color{red}{X \sim{} G(p)}$	$P(X=k) = (1-p)^{k-1}p \quad k=1,2,\cdots$	$\frac{1}{p}$	$\frac{1-p}{p^2}$
超几何分布	$\begin{aligned}N,M,n\\(M \leq N)\\(n \leq N)\end{aligned}$	$\color{red}{X \sim{} H(n, K, N)}$	$P(X=k, n, K, N) = \frac{\binom{M}{k}\binom{N-M}{n-k}}{\frac{N}{k}}$	$\frac{nM}{N}$	$n\frac{M}{N}(1 - \frac{M}{N})(1 - \frac{n-1}{N-1})$
泊松分布	$\lambda > 0$	$\color{red}{X \sim{} \pi(\lambda)}$	$P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$	$\lambda$	$\lambda$
均匀分布	$a<b$	$\color{red}{X \sim{} U(a, b)}$	$f(x) =\begin{cases}\frac{1}{b-a} \quad & a < x < b\\0 \quad & \text{其他}\end{cases}$	$\frac{a+b}{2}$	$\frac{(b-a)^2}{12}$
指数分布	$0<\lambda<1$	$\color{red}{X \sim{} Exp(\lambda)}$	$f(x) =\begin{cases}\lambda{}e^{-\lambda{}x} \quad & x \geq 0\\0 \quad & x < 0\end{cases}$	$\frac{1}{\lambda}$	$\frac{1}{\lambda^2}$
正态分布	$u.\sigma>0$	$\color{red}{X \sim{} N(\mu, \sigma^2)}$	$f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-u)^2}{2\sigma^2}} \quad -\infty < x < \infty$	$\mu$	$\lambda$