第一題
解法一：分布函數(shù)法
F(y)=P(Y<=y)=P(arctanX<=y)
當(dāng)y<-π/2時,F(y)=0
當(dāng)-π/2<=y<π/2時,F(y)=P(X<=tany)=∫f(x)dx=∫1/[π（1+x^2）]dx=y/π+1/2
當(dāng)y>=π/2時,F(y)=1
所以f(y)=1/π,-π/2<=y<π/2；f(y)=0,y取其它值
解法二：公式法
x=tany,在-π/2<=y<π/2,單調(diào)增
f(y)=1/{π[1+(tany)^2}*x'=1/π*1/(secy)^2*(secy)^2=1/π,
所以
f(y)=1/π,-π/2<=y<π/2；f(y)=0,y取其它值
第二小題
已知A與B相互獨(dú)立,概率密度函數(shù)都是f(x)=e^-x(X>=0),f(x)=0(x<0),求P=min(A,B)的概率密度函數(shù)
F(p)=P(P<=p)=P(min(A,B)<=p)
當(dāng)p<0時,F(p)=0
當(dāng)p>=0時,F(p)=P(min(A,B)<=p)=1-P(min(A,B)>p)=1-P(A>p)*P(B>p)=1-e^(-2p)
所以f(p)=2e^(-2p),p>=0;f(p)=0,p<0
解畢