离散随机变量X的分布列 P = ( X = x i ) = p ( x i ) i = 1 , 2 , . . . , n P = (X = x_i) = p(x_i) \qquad i = 1,2,...,n P=(X=xi​)=p(xi​)i=1,2,...,n 则X的数学期望为： ∑ i = 1 n x i p ( x i ) \sum_{i=1}^{n} x_i p(x_i) ∑i=1n​xi​p(xi​)，记为; E ( X ) = ∑ i = 1 n x i p ( x i ) E(X) = \sum_{i=1}^{n} x_i p(x_i) E(X)=i=1∑n​xi​p(xi​) 若X的取值可列，且无穷级数 ∑ i = 1 ∞ x i p ( x i ) \sum_{i=1}^{\infty} x_i p(x_i) ∑i=1∞​xi​p(xi​)收敛，则 E ( X ) = ∑ i = 1 ∞ x i p ( x i ) E(X) = \sum_{i=1}^{\infty} x_i p(x_i) E(X)=∑i=1∞​xi​p(xi​).

p ( x ) = P ( X = x ) = p ( 1 − p ) x − 1 , x = 1 , 2 , . . . . p(x) = P(X = x) = p(1 - p)^{x - 1},x = 1,2,.... p(x)=P(X=x)=p(1−p)x−1,x=1,2,.....

E ( X ) = 1 p E(X) = \frac{1}{p} E(X)=p1​.

令 q = 1 − p q = 1 -p q=1−p，则有; E ( X ) = ∑ x = 1 ∞ x p q x − 1 = p ∑ x = 1 ∞ x q x − 1 = p ∑ x = 1 ∞ d ( q x ) d q = p d d q ∑ x = 0 ∞ q x = p d d q ( 1 1 − q ) = 1 p E(X) = \sum_{x=1}^{\infty} x p q^{x -1} \\ = p \sum_{x=1}^{\infty} x q^{x -1} \\ = p \sum_{x=1}^{\infty} \frac{d(q^x)}{dq} \\ = p \frac{d}{dq} \sum_{x=0}^{\infty} q^x \\ = p \frac{d}{dq}(\frac{1}{1-q}) = \frac{1}{p} E(X)=x=1∑∞​xpqx−1=px=1∑∞​xqx−1=px=1∑∞​dqd(qx)​=pdqd​x=0∑∞​qx=pdqd​(1−q1​)=p1​

V a r ( X ) = 1 − p p 2 Var(X) = \frac{1 - p}{p^2} Var(X)=p21−p​

E ( X 2 ) = ∑ x = 1 ∞ x 2 p q x − 1 = p ∑ x = 1 ∞ x 2 q x − 1 = p ∑ x = 1 ∞ x d ( q x ) d q = p d d q ∑ x = 0 ∞ x q x E(X^2) =\sum_{x=1}^{\infty} x^2 p q^{x -1} \\ = p \sum_{x=1}^{\infty} x^2 q^{x -1} \\ = p \sum_{x=1}^{\infty} x \frac{d(q^x)}{dq} \\ = p \frac{d}{dq} \sum_{x=0}^{\infty} x q^x E(X2)=x=1∑∞​x2pqx−1=px=1∑∞​x2qx−1=px=1∑∞​xdqd(qx)​=pdqd​x=0∑∞​xqx

由无穷级数的理论可知 ∑ x = 0 ∞ x q x = q ( 1 − q ) 2 \sum_{x=0}^{\infty} x q^x = \frac{q}{(1-q)^2} ∑x=0∞​xqx=(1−q)2q​

从而：

接上式: E ( X 2 ) = p d d q ( q ( 1 − q ) 2 ) = 2 p − p 2 p 3 = 2 − p p 2 E(X^2) = p \frac{d}{dq} (\frac{q}{(1-q)^2)} = \frac{2p-p^2}{p^3} = \frac{2-p}{p^2} E(X2)=pdqd​((1−q)2)q​=p32p−p2​=p22−p​.从而得到方差。

p ( x ) = P ( X = x ) = ( n x ) p x ( 1 − p ) n − x x = 0 , 1 , . . . . , n p(x) = P(X = x) = \begin{pmatrix} n \\ x \end{pmatrix} p^x (1 - p)^{n -x} \qquad x = 0,1,....,n p(x)=P(X=x)=(nx​)px(1−p)n−xx=0,1,....,n

n=1,时的二项分布b(1,p),又称为两点分布或0-1分布。

E ( X ) = n p E(X) = np E(X)=np

E ( X ) = ∑ x = 0 n x ( n x ) p x ( 1 − p ) n − x = n p ∑ x = 0 n ( n − 1 x − 1 ) p x − 1 ( 1 − p ) n − x = n p ∑ x = 0 n ( n − 1 x ) p x ( 1 − p ) n − 1 − x = n p [ p + ( 1 − p ) ] n − 1 = n p E(X) = \sum_{x = 0}^{n} x \begin{pmatrix} n \\ x \end{pmatrix} p^x (1 - p)^{n -x} \\ =np \sum_{x = 0}^{n} \begin{pmatrix} n - 1 \\ x - 1 \end{pmatrix} p^{x -1} (1 - p)^{n -x} \\ = np \sum_{x = 0}^{n} \begin{pmatrix} n - 1 \\ x \end{pmatrix} p^{x} (1 - p)^{n - 1 - x} \\ =np[p + (1 - p)]^{n -1} = np E(X)=x=0∑n​x(nx​)px(1−p)n−x=npx=0∑n​(n−1x−1​)px−1(1−p)n−x=npx=0∑n​(n−1x​)px(1−p)n−1−x=np[p+(1−p)]n−1=np

V a r ( X ) = n p ( 1 − p ) Var(X) = np(1-p) Var(X)=np(1−p)

E ( X 2 ) = ∑ x = 0 n x 2 ( n x ) p x ( 1 − p ) n − x = ∑ x = 2 n x ( x − 1 ) ( n x ) p x ( 1 − p ) n − x + ∑ x = 1 n x ( n x ) p x ( 1 − p ) n − x = n ( n − 1 ) p 2 ∑ x = 2 n ( n − 2 x − 2 ) p x − 2 ( 1 − p ) n − x + n p = n ( n − 1 ) p 2 + n p = n 2 p 2 + n p ( 1 − p ) E(X^2) = \sum_{x = 0} ^ n x ^2 \begin{pmatrix} n \\ x \end{pmatrix} p ^ x (1-p)^{n-x} \\ = \sum_{x = 2} ^ n x (x-1) \begin{pmatrix} n \\ x \end{pmatrix} p ^ x (1-p)^{n-x} + \sum_{x = 1} ^ n x \begin{pmatrix} n \\ x \end{pmatrix} p ^ x (1-p)^{n-x} \\ = n (n-1) p^2 \sum_{x = 2} ^ n \begin{pmatrix} n-2 \\ x-2 \end{pmatrix} p^{x-2} (1-p)^{n-x} + np \\ = n(n-1) p^2 + np = n^2p^2 + np(1-p) E(X2)=x=0∑n​x2(nx​)px(1−p)n−x=x=2∑n​x(x−1)(nx​)px(1−p)n−x+x=1∑n​x(nx​)px(1−p)n−x=n(n−1)p2x=2∑n​(n−2x−2​