Question

设X的分布律为P(X=k)=1/2^k,k=1,2,3,……,求Y=sin(πX/2)的分布律?

Answer 1

Y=sin(πX/2)
Y=1,πX/2=π/2+2mπ=(4m+1)*π/2,m=0,1,2,3,……
k=4m+1,1/2^k=1/2^(4m+1)=(1/2)*(1/16^m)
Σ(m:0→∞)(1/2)*(1/16^m)=lim(m→∞)(1/2)*[1-1/16^(m+1)]/(1-1/16)=(1/2)/(1-1/16)=8/15
Y=0,πX/2=π+mπ=(m+1)*π=(2m+2)*π/2,m=0,1,2,3,……
k=2m+2,1/2^k=1/2^(2m+2)=(1/4)*(1/4^m)
Σ(m:0→∞)(1/4)*(1/4^m)=lim(m→∞)(1/4)*[1-1/4^(m+1)]/(1-1/4)=(1/4)/(1-1/4)=1/3=5/15
Y=-1,πX/2=3π/2+2mπ=(4m+3)*π/2,m=0,1,2,3,……
k=4m+3,1/2^k=1/2^(4m+3)=(1/8)*(1/16^m)
Σ(m:0→∞)(1/8)*(1/16^m)=lim(m→∞)(1/8)*[1-1/16^(m+1)]/(1-1/16)=(1/8)/(1-1/16)=2/15
所以Y的分布律：
P(Y=-1)=2/15
P(Y=0)=5/15
P(Y=1)=8/15