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设(X,Y)服从在D上的均匀分布,其中D由x轴、y轴及x+y=1所围成,求D(X)答案:E(X)=1/3E(XX)=1/6D(X)=E(XX)-E(X)E(X)=1/18求解法
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设(X,Y)服从在D上的均匀分布,其中D由x轴、y轴及x+y=1所围成,求D(X)
答案:E(X)=1/3
E(XX)=1/6
D(X)=E(XX)-E(X)E(X)=1/18
求解法
答案:E(X)=1/3
E(XX)=1/6
D(X)=E(XX)-E(X)E(X)=1/18
求解法
▼优质解答
答案和解析
D的面积S=1/2,密度函数f(x,y)=2,(x∈D).
E(X)=∫∫[D]xf(x,y)dxdy
=∫[0,1]2xdx∫[0,1-x]dy
=∫[0,1]2x(1-x)dx
=(x^2-2x^3/3)|[0,1]
=1/3.
E(X^2)=∫∫[D]x^2f(x,y)dxdy
=∫[0,1]2x^2dx∫[0,1-x]dy
=∫[0,1]2x^2(1-x)dx
=(2x^3/3-2x^4/4)|[0,1]
=1/6.
D(X)=E(X^2)-[E(X)]^2=1/6-(1/3)^2=1/18.
E(X)=∫∫[D]xf(x,y)dxdy
=∫[0,1]2xdx∫[0,1-x]dy
=∫[0,1]2x(1-x)dx
=(x^2-2x^3/3)|[0,1]
=1/3.
E(X^2)=∫∫[D]x^2f(x,y)dxdy
=∫[0,1]2x^2dx∫[0,1-x]dy
=∫[0,1]2x^2(1-x)dx
=(2x^3/3-2x^4/4)|[0,1]
=1/6.
D(X)=E(X^2)-[E(X)]^2=1/6-(1/3)^2=1/18.
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