已知概率密度求边缘密度.已知f(x,y)=2/π^2(1+x^2)(1+4y^2)fX(x)=1/π(1+x^2)fY(
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  • fX(x)=∫【-∞,+∞】f(x,y)dy

    =∫【-∞,+∞】2/[π^2(1+x^2)(1+4y^2)]dy

    =1/[π^2(1+x^2)]*∫【-∞,+∞】1/(1+4y^2)d(2y)

    =1/[π^2(1+x^2)]*arctan2y|【-∞,+∞】

    =1/[π^2(1+x^2)]*π

    =1/[π(1+x^2)]

    fY(y)=∫【-∞,+∞】f(x,y)dx

    =∫【-∞,+∞】2/[π^2(1+x^2)(1+4y^2)]dx

    =2/[π^2(1+4y^2)]*∫【-∞,+∞】1/(1+x^2)dx

    =2/[π^2(1+4y^2)]*arctanx|【-∞,+∞】

    =2/[π^2(1+4y^2)]*π

    =2/[π(1+4y^2)]

    解毕

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