因而E(X)=∑k=∑ kCn p'q k-1k np k=0 Example3.4设随机变量X服从参数为λ的泊松分布,求它的数学期望 Solution由于phe 因而E(X)=∑k=∑ke=∑ Example35已知离散型随机变量的概率分布为P(X=1)=0.2,P(x=2)=0.3, P(X=3)=0.5,求E(X) Solution E(X)=1×0.2+2×0.3+3×0.5=23 二、连续型随机变量的数学期望( Mathematical expectation of a continual random variable) Defi2设连续型随机变量x的分布密度函数为f(x),若积分x(x)绝对 收敛,则称其为X的数学期望或均值.记为E(X),E(X)=[xf(xhx.( Suppose X is a continuous random variable, which its probability density function is f(x). if integral f(xx, is absolutely convergent, then it is called mathematical expectation(or random variable X, which is written E(X), and E(X)=xf(xxx) Example3.6设随机变量X服从正态分布N(,a2),求E(x) Solution于正态分布N(A,a2)的密度函数为f(x)=- 因而 E(X)= xf(x)dx r 今 t,则E(X)= dt+ dt Example3.7设随机变量X服从参数为λ的指数分布,求E(X) x≥0 Solution由于指数分布的密度函数为f(x)= 因而E(X)=xf(x)tx=axe-ax= =--e Example3.8设随机变量X服从[a,b上的均匀分布,求E(X) a≤X Solution由于均匀分布的密度函数为f(x)={b-a 0 其他38 因而  = − = = = n k k k n k n n k E X k pk kC p q 0 0 ( ) np C p q np p q np n n k k k n k = n = + = − = − − − − −  − 1 0 1 1 ( 1) ( 1) 1 ( ) Example 3.4 设随机变量 X 服从参数为  的泊松分布，求它的数学期望． Solution 由于  − = e k p k k ! 因而            = = − = − = = = −  =  = − − −  = −  =     e e k e e k e k E X k p k k k k k k k k k 1 1 1 1 1 ! ( 1)! ( 1)! ( ) Example 3.5 已知离散型随机变量的概率分布为 P(X = 1) = 0.2,P(x = 2) = 0.3, P(X = 3) = 0.5 ，求 E(X ) ． Solution E(X ) = 1 0.2 + 2 0.3 + 3 0.5 = 2.3 二、 连续型随机变量的数学期望(Mathematical expectation of a continual random variable) Definition 3.2 设连续型随机变量 X 的分布密度函数为 f (x) ，若积分 xf x dx  + − ( ) 绝对 收敛，则称其为 X 的数学期望或均值．记为 E(X ) ，E X xf x dx  + − ( ) = ( ) ．(Suppose X is a continuous random variable, which its probability density function is f (x) . if integral, xf x dx  + − ( ) , is absolutely convergent, then it is called mathematical expectation (or average) of random variable X , which is written E(X ) , and E X xf x dx  + − ( ) = ( ) .) Example 3.6 设随机变量 X 服从正态分布 ( , ) 2 N   ，求 E(X ) ． Solution 由于正态分布 ( , ) 2 N   的密度函数为 2 2 2 ( ) 2 1 ( )    − − = x f x e 因而   + − − + − − = = e dx x E X xf x dx x 2 2 2 ( ) 2 ( ) ( )    令 t x = −   ，则   + − + − − − = + =      e dt t E X e dt t t 2 2 2 2 2 2 1 ( ) ． Example 3.7 设随机变量 X 服从参数为  的指数分布，求 E(X ) ． Solution 由于指数分布的密度函数为 , 0 ( ) 0, 0 x e x f x x   −   =    因而     + − + − + − + − + = = = − = − + 0 0 0 0 0 E(X) x f(x)dx x e dx xde xe e dx x x x x    1  1 0 = − = + − x e ． Example 3.8 设随机变量 X 服从 [a,b] 上的均匀分布，求 E(X ) ． Solution 由于均匀分布的密度函数为 1 , ( ) 0, a x b f x b a     =  −   其他