题目
31.设随机变量(X,Y)具有概率密度 f(x,y)= ) 1,|y|lt x,0lt xlt 1 0, .-|||-,-|||-求E(X),E(Y),Cov(X,Y).

题目解答
答案

解析
步骤 1:计算E(X)
E(X) = ∫∫ x f(x,y) dxdy
根据题目中的概率密度函数,我们有:
E(X) = ∫_{0}^{1} ∫_{-x}^{x} x dxdy
= ∫_{0}^{1} x [y]_{-x}^{x} dx
= ∫_{0}^{1} x (x - (-x)) dx
= ∫_{0}^{1} 2x^2 dx
= [2/3 x^3]_{0}^{1}
= 2/3
步骤 2:计算E(Y)
E(Y) = ∫∫ y f(x,y) dxdy
根据题目中的概率密度函数,我们有:
E(Y) = ∫_{0}^{1} ∫_{-x}^{x} y dxdy
= ∫_{0}^{1} [y^2/2]_{-x}^{x} dx
= ∫_{0}^{1} (x^2/2 - (-x)^2/2) dx
= ∫_{0}^{1} 0 dx
= 0
步骤 3:计算E(XY)
E(XY) = ∫∫ xy f(x,y) dxdy
根据题目中的概率密度函数,我们有:
E(XY) = ∫_{0}^{1} ∫_{-x}^{x} xy dxdy
= ∫_{0}^{1} x [y^2/2]_{-x}^{x} dx
= ∫_{0}^{1} x (x^2/2 - (-x)^2/2) dx
= ∫_{0}^{1} 0 dx
= 0
步骤 4:计算Cov(X,Y)
Cov(X,Y) = E(XY) - E(X)E(Y)
根据步骤 1 至 3 的计算结果,我们有:
Cov(X,Y) = 0 - (2/3) * 0
= 0
E(X) = ∫∫ x f(x,y) dxdy
根据题目中的概率密度函数,我们有:
E(X) = ∫_{0}^{1} ∫_{-x}^{x} x dxdy
= ∫_{0}^{1} x [y]_{-x}^{x} dx
= ∫_{0}^{1} x (x - (-x)) dx
= ∫_{0}^{1} 2x^2 dx
= [2/3 x^3]_{0}^{1}
= 2/3
步骤 2:计算E(Y)
E(Y) = ∫∫ y f(x,y) dxdy
根据题目中的概率密度函数,我们有:
E(Y) = ∫_{0}^{1} ∫_{-x}^{x} y dxdy
= ∫_{0}^{1} [y^2/2]_{-x}^{x} dx
= ∫_{0}^{1} (x^2/2 - (-x)^2/2) dx
= ∫_{0}^{1} 0 dx
= 0
步骤 3:计算E(XY)
E(XY) = ∫∫ xy f(x,y) dxdy
根据题目中的概率密度函数,我们有:
E(XY) = ∫_{0}^{1} ∫_{-x}^{x} xy dxdy
= ∫_{0}^{1} x [y^2/2]_{-x}^{x} dx
= ∫_{0}^{1} x (x^2/2 - (-x)^2/2) dx
= ∫_{0}^{1} 0 dx
= 0
步骤 4:计算Cov(X,Y)
Cov(X,Y) = E(XY) - E(X)E(Y)
根据步骤 1 至 3 的计算结果,我们有:
Cov(X,Y) = 0 - (2/3) * 0
= 0