因而E(x)=(xb=[xb=b2-a2 b 2(b-a)2 Example3.9设随机变量X服从柯西分布,其密度函数为 f(x)= (-∞<x<+∞),由于积分 发散,因而E(X)不存在 三、随机变量的函数的数学期望( Mathematical expectation of random variable function) Theorem3.1设Y为随机变量X的函数:Y=g(X)(g是连续函数,(1)X是离散型 随机变量,分布律为P=P(X=x)k=1,2,…;若级数∑g(xk)Pk绝对收敛,则有 E()=Eg(=∑g(x)P(2)X是连续型随机变量,它的分布密度为f(x),若积 分g(x)f(x绝对收敛,则有E(Y)=Eg(X=g(x)f(x)t.( ( Suppose Y is a function of random variable X, r=g(r(g is a continuous function),(I)X is a discrete random variable, distribution law is P=P(X=xx), k=1, 2,., if series, >g(xk)Pk,is absolutely convergent, then E(Y)=E[g(X)]=>8(xx)Pr.(2)X is a continuous random variable,its probability distribution density function is f(x), if integral g(x)f(xddx is absolutely convergent, then E(Y)=Elg(x)]=28(xk )Pk) (证明略) 定理31告诉我们:求E(Y)时,不必知道Y的分布,而只需知道X的分布就可以了 Theorem3.2设Z是随机变量(x,Y)的连续函数Z=g(X,Y),(1)(X,F)是二维离 散型随机变量,联合分布律为P=P(X=x,Y=y)1,j=12,…:则有 E(Z)=E(X,=∑∑g(x,y)P4·(设该级数绝对收敛)(2)(X,)是二维连续型 随机变量 联合分布密度为f(x,y) 则有 E(Z)=Eg(X,=g(xy)(xytd.(设该积分绝对收敛)( Suppose z is a continuous function of random vector(x,y,z=g(x,,(1)(x,r)are discrete random vector of two dimensions, its joint distribution law is Pui=P(x=x,r=y),i,j=1, 2 then E(Z)=E[g(X,Y)]=2>g(x, y, )Pu (Suppose this series is absolutely convergent) (2)(X, r)are continuous random vector of two dimensions, its joint distribution density function is f(x, y), then E(Z)=Eg(x, y]= g(x, y)f(x, ykdxdy (Suppose this integral is absolutely convergent) (证明略) Example310设随机变量X服从正态分布N(A,a2),求(1)E(X2):(2)E(e2)39 因而 2( ) 2 ( ) ( ) 2 2 a b b a b a dx b a x E X xf x dx b a b a + = − − = − = =   ． Example 3.9 设随机变量 X 服 从 柯 西 分 布 ， 其 密 度 函 数 为 (1 ) 1 ( ) 2 x f x + =  (−  x  +) ，由于积分  + − (1+ ) 2 x x dx  发散，因而 E(X ) 不存在． 三、 随机变量的函数的数学期望(Mathematical expectation of random variable function) Theorem 3.1 设 Y 为随机变量 X 的函数： Y = g(X ) (g 是连续函数)，（1） X 是离散型 随机变量，分布律为 pk = P(X = xk ), k =1,2,  ；若级数   =1 ( ) k k pk g x 绝对收敛，则有 E(Y) = E[g(X )] =   =1 ( ) k k pk g x ．（2） X 是连续型随机变量，它的分布密度为 f (x) ，若积 分 g x f x dx  + − ( ) ( ) 绝对收敛，则有 E(Y) = E[g(X )] = g x f x dx  + − ( ) ( ) ．(Suppose Y is a function of random variable X ,Y = g(X ) (g is a continuous function), (1) X is a discrete random variable, distribution law is pk = P(X = xk ), k =1,2,  ; if series,   =1 ( ) k k pk g x , is absolutely convergent, then E(Y) = E[g(X )] =   =1 ( ) k k pk g x . (2) X is a continuous random variable, its probability distribution density function is f (x) , if integral g x f x dx  + − ( ) ( ) is absolutely convergent, then E(Y) = E[g(X )] =   =1 ( ) k k pk g x .) （证明略） 定理 3.1 告诉我们：求 E(Y) 时，不必知道 Y 的分布，而只需知道 X 的分布就可以了． Theorem 3.2 设 Z 是随机变量 (X,Y) 的连续函数 Z = g(X ,Y) ，（1） (X,Y) 是二维离 散型随机变量，联合分布律为 pij = P(X = xi ,Y = y j ),i, j = 1,2,  ； 则 有 E(Z) = E[g(X,Y)] =   =  1 =1 ( , ) i i j ij j g x y p ．（设该级数绝对收敛）（2） (X,Y) 是二维连续型 随 机 变 量 ， 联 合 分 布 密 度 为 f (x, y) ，则有 E(Z) = E[g(X,Y)] = g x y f x y dxdy   + − + − ( , ) ( , ) ．（设该积分绝对收敛）(Suppose Z is a continuous function of random vector (X,Y) , Z = g(X ,Y) , (1) (X,Y) are discrete random vector of two dimensions, its joint distribution law is pij = P(X = xi ,Y = y j ),i, j = 1,2,  ; then E(Z) = E[g(X,Y)] =   =  1 =1 ( , ) i i j ij j g x y p . (Suppose this series is absolutely convergent) (2) (X,Y) are continuous random vector of two dimensions, its joint distribution density function is f (x, y) , then E(Z) = E[g(X,Y)] = g x y f x y dxdy   + − + − ( , ) ( , ) . (Suppose this integral is absolutely convergent) （证明略） Example 3.10 设随机变量 X 服从正态分布 ( , ) 2 N   ，求 (1) ( ) 2 E X ；(2) ( ) X E e ．