设二维连续型随机变量(X,Y)的联合分布函数为F(x,y)=A(B+arctanx/2)(C+arctany/3),判断
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设二维连续型随机变量(X,Y)的联合分布函数为F(x,y)=A(B+arctanx/2)(C+arctany/3),判断X和Y的独立性
其中A=1/π^2,B=π/2,C=π/2
其中A=1/π^2,B=π/2,C=π/2
F(x,y)=A(B+arctanx/2)(C+arctany/3)
F(-∞,-∞)=A(B-π/2)(C-π/2)=0
F(-∞,+∞)=A(B-π/2)(C+π/2)=0
F(+∞,-∞)=A(B+π/2)(C-π/2)=0
F(+∞,+∞)=A(B+π/2)(C+π/2)=1
解得:A=1/π^2,B=π/2,C=π/2
F(+∞,y)=1/2+1/π*arctan(y/3)
F(x,+∞)=1/2+1/π*arctan(x/2)
F(x,y)=F(+∞,y)×F(x,+∞)
X和Y相互独立.
F(-∞,-∞)=A(B-π/2)(C-π/2)=0
F(-∞,+∞)=A(B-π/2)(C+π/2)=0
F(+∞,-∞)=A(B+π/2)(C-π/2)=0
F(+∞,+∞)=A(B+π/2)(C+π/2)=1
解得:A=1/π^2,B=π/2,C=π/2
F(+∞,y)=1/2+1/π*arctan(y/3)
F(x,+∞)=1/2+1/π*arctan(x/2)
F(x,y)=F(+∞,y)×F(x,+∞)
X和Y相互独立.
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