随机信号处理题目,正态分布.y(t)=1+x(t),x(t)是服从正态分布的随机过程,正太分布平均值为0,令y(t)值小

此题等同于,求y ~ N(1,0.1),N(1,0.2), 和 N（1,1）随机出10000个值.其中E(y|y>0) 等于多少?
y(t)=0分别对应上述三个分布的几率是多少.P1 = Pr(y(t)=0|N(1,0.1)) = 0P2 = Pr(y(t)=0|N(1,0.5)) = 0.0228P3 = Pr(y(t)=0|N(1,1)) = 0.1587
然后expected value of N 就是 10000*（1-P)N1 = 10000*(1-P1) = 10000N2 = 10000*(1-P2) = 9772
N3 = 10000*(1-P3) = 8413MATLAB CODE 如下：sum((randn(10000,1)*0.5+1)>0) 这个是检验N2的程序.N1和N3楼主就照着改就行了.