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
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
数学人气:520 ℃时间:2019-10-23 07:41:15
优质解答
F(x,y)=A(B+arctanx/2)(C+arctany/3)F(-∞,-∞)=A(B-π/2)(C-π/2)=0F(-∞,+∞)=A(B-π/2)(C+π/2)=0F(+∞,-∞)=A(B+π/2)(C-π/2)=0F(+∞,+∞)=A(B+π/2)(C+π/2)=1解得:A=1/π^2,B=π/2,C=π/2F(+∞,y)=1/2+1/π*...
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