Question

概率论大题，急求，谢谢

Answer 1

2(1)kS(1,+oo)e^(-2x)dxS(2,+oo)e^(-y)dx=1/2ke^(-2)e^(-2)=1/2ke^(-4) =1
k=2e^4
代入原式f(x,y)=2e^(-2x-y+4)
(2)f(x)=S(2,+oo)2e^(-2x-y+4)dy=2e^(-2x+2)
f(y)=S(1,+oo)2e^(-2x-y+4)dx=e^(-y+2)
3(1)X -1,1/2; 0,1/2.
Y -1,1/4;0,1/4;1,1/2;
(2)EX=-1/2 ,EX^2=1/2 ,DX=1/4
EY=1/4,EY^2=3/4,DY=11/16
EXY=-1/4
COVXY= -1/8,相关系数=-1/8/根号下(11/16 *1/4)=-1/根号下11
4EX=S(0,o)x/2o dx+S(o,1)x/2(1-o) dx=o/4+1/4(1-o)-o^2/4(1-o)=[o(1-o)+1-o^2]/[4(1-o)]=[o+1-2o^2]/[4(1-o)]=X~
解方程即得矩估计值