概率论与数理统计问题:已知X~N(μ,1),即EX=μ,求E(e^X),
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由定义
f(x)=1/根号(2pi)exp(-(x-mu)^2/竖核1^2)
E(e^X)=积分(-无穷,无穷)e^x *f(x) dx
=积分(-无穷,无穷碰岩)e^x *1/根号(2pi)exp(-(x-mu)^2/1^2) dx
=1/根号(2pi)积分(-无穷,无穷)exp(x)*exp(-(x-mu)^2) dx
=1/根号(2pi)积分(-无穷,无穷)exp(x-(x-mu)^2) dx
配方x-(x-mu)^2=-x^2+(2mu+1)x-mu^2
=-(x-mu-1/2)^2+1/2*(2mu+1/2)
=1/根号(2pi)积笑纤御分(-无穷,无穷)exp(-(x-mu-1/2)^2) exp(1/2*(2mu+1/2))dx
换元y=x-1/2
=exp(1/2*(2mu+1/2))*1/根号(2pi)积分(-无穷,无穷)exp(-(y-mu)^2) dy
=exp(1/2*(2mu+1/2))*1
=exp(mu+1/4)
f(x)=1/根号(2pi)exp(-(x-mu)^2/竖核1^2)
E(e^X)=积分(-无穷,无穷)e^x *f(x) dx
=积分(-无穷,无穷碰岩)e^x *1/根号(2pi)exp(-(x-mu)^2/1^2) dx
=1/根号(2pi)积分(-无穷,无穷)exp(x)*exp(-(x-mu)^2) dx
=1/根号(2pi)积分(-无穷,无穷)exp(x-(x-mu)^2) dx
配方x-(x-mu)^2=-x^2+(2mu+1)x-mu^2
=-(x-mu-1/2)^2+1/2*(2mu+1/2)
=1/根号(2pi)积笑纤御分(-无穷,无穷)exp(-(x-mu-1/2)^2) exp(1/2*(2mu+1/2))dx
换元y=x-1/2
=exp(1/2*(2mu+1/2))*1/根号(2pi)积分(-无穷,无穷)exp(-(y-mu)^2) dy
=exp(1/2*(2mu+1/2))*1
=exp(mu+1/4)
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