泊松分布的期望和方差推导

泊松分布是一个离散型随机变量分布，其分布律是：

P (X = k) = λ k e - λ k !

$P(X=k)=\frac{\lambda ^{k}e^{-\lambda }}{k!}$
根据离散型随机变量分布的期望定义，泊松分布的期望：

E (X) = \sum k = 0 \infty k \cdot λ k e - λ k !

$E(X)=\sum_{k=0}^{\infty }k\cdot \frac{\lambda ^{k}e^{-\lambda }}{k!}$
因为k=0时：

k \cdot λ k e - λ k ! = 0

$k\cdot \frac{\lambda ^{k}e^{-\lambda }}{k!}=0$
所以：

E (X) = \sum k = 1 \infty k \cdot λ k e - λ k !

$E(X)=\sum_{k=1}^{\infty }k\cdot \frac{\lambda ^{k}e^{-\lambda }}{k!}$
做一下变换：

E (X) = \sum k = 1 \infty k \cdot λ k e - λ k ! = \sum k = 1 \infty λ k e - λ ( k - 1 ) ! = \sum k = 1 \infty λ k - 1 λ e - λ ( k - 1 ) ! = λ e - λ \sum k = 1 \infty λ k - 1 ( k - 1 ) !

$E(X)=\sum_{k=1}^{\infty }k\cdot \frac{\lambda ^{k}e^{-\lambda }}{k!}=\sum_{k=1}^{\infty } \frac{\lambda ^{k}e^{-\lambda }}{(k-1)!}=\sum_{k=1}^{\infty } \frac{\lambda ^{k-1}\lambda e^{-\lambda }}{(k-1)!}=\lambda e^{-\lambda }\sum_{k=1}^{\infty } \frac{\lambda ^{k-1}}{(k-1)!}$
这里需要用到泰勒展开式，我们知道常用的泰勒展开式中：

e x = 1 + x + x 2 2 ! + x 3 3 ! + . . . + x n n ! + . . . = \sum k = 1 \infty x k - 1 ( k - 1 ) !

$e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...+\frac{x^{n}}{n!}+...=\sum_{k=1}^{\infty } \frac{x ^{k-1}}{(k-1)!}$
因此，泊松分布的期望为:

E (X) = λ e - λ \sum k = 1 \infty λ k - 1 ( k - 1 ) ! = λ e - λ e λ = λ

$E(X)=\lambda e^{-\lambda }\sum_{k=1}^{\infty } \frac{\lambda ^{k-1}}{(k-1)!}=\lambda e^{-\lambda }e^{\lambda }=\lambda$
对于方差

D(X) $D(X)$ ，先求出

E(X2) $E(X^2)$ :

E (X 2) = \sum k = 0 \infty k 2 \cdot λ k e - λ k ! = λ e - λ \sum k = 1 \infty k λ k - 1 ( k - 1 ) ! = λ e - λ \sum k = 1 \infty ( k - 1 + 1 ) λ k - 1 ( k - 1 ) !

$E(X^2)=\sum_{k=0}^{\infty }k^2 \cdot \frac{\lambda ^{k}e^{-\lambda }}{k!}=\lambda e^{-\lambda} \sum_{k=1}^{\infty } \frac{k \lambda ^{k-1}}{(k-1)!}=\lambda e^{-\lambda} \sum_{k=1}^{\infty } \frac{(k-1+1) \lambda ^{k-1}}{(k-1)!}$

= λ e - λ (\sum m = 0 \infty m \cdot λ m m ! + \sum m = 0 \infty λ m m !) (m = k - 1)

$=\lambda e^{-\lambda} (\sum_{m=0}^{\infty } \frac{m \cdot \lambda ^{m}}{m!}+\sum_{m=0}^{\infty } \frac{ \lambda ^{m}}{m!}) (m=k-1)$

= λ e - λ (λ \cdot \sum m = 1 \infty λ m - 1 ( m - 1 ) ! + \sum m = 0 \infty λ m m !)

$=\lambda e^{-\lambda} ( \lambda \cdot \sum_{m=1}^{\infty } \frac{\lambda ^{m-1}}{(m-1)!}+\sum_{m=0}^{\infty } \frac{ \lambda ^{m}}{m!})$

= λ e - λ (λ e λ + e λ) = λ (λ + 1)

$=\lambda e^{-\lambda}(\lambda e^{\lambda}+e^\lambda)=\lambda(\lambda+1)$
所以：

D (X) = E (X 2) - (E (X)) 2 = λ (λ + 1) - λ 2 = λ

$D(X)=E(X^2)-(E(X))^2=\lambda(\lambda+1)-\lambda^2=\lambda$
因此，泊松分布的期望和方差为：

E (X) = λ

$E(X)=\lambda$

D (X) = λ

$D(X)=\lambda$

泊松分布的期望和方差推导

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