分析

简答题:已知离散型随机变量X的概率分布为
 
且E(X)=0.
(1)求a,b;
(2)求E[X(X+1)].

正确答案
(1)由概率的性质可知a+0.5+b=1,
又E(X)=0,得-1×a+0×0.5+2×b=0,
故有a=,b=
(2)E[X(X+1)]=E(X2+X)=E(X2)+E(X),
而E(X2)=D(X)+[E(X)]2
=·(-1-0)2+·(0-0)2+·(2-0)2
=1。
因此E[X(X+1)]=1+0=1.
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