概率统计的一道题,设二维随机变量(X,Y)在x轴,y轴及直线x+y+1=0所围成的区域D上服从均匀分布,求相关系数.

f(x,y)=2
E(X)=∫[-1,0]dx∫[-1-x,0]2xdy
=∫[-1,0]2x(1+x)dx=(x^2+2/3*x^3)|[-1,0]=-1/3
同理：E(Y)=-1/3
E(XY)=∫[-1,0]dx∫[-1-x,0]2xydy
=∫[-1,0]xy^2|[-1-x,0]dx=-∫[-1,0]x(1+x)^2dx
=-(1/4*x^4+2/3*x^3+1/2*x^2)|[-1,0]=1/12
COV(X,Y)=E(XY)-EX*EY=-1/36
E(X^2)=∫[-1,0]dx∫[-1-x,0]2x^2dy
=∫[-1,0]2x^2(1+x)dx=(2/3*x^3+1/2*x^4)|[-1,0]=1/6
D(X)=E(X^2)-(EX)^2=1/18
同理：D(Y)=1/18

x=56
y=60

sorry!
f(x,y)=2
fx(x)=2(x+1),fy(y)=2(y+1)
EX=∫xfx(x)dx=-1/3=EY
E(XY)=∫∫2xydxdy=1/12
cov(x,y)=E[(x-Ex)*E(y-Ey)]=-1/36
E(x^2)==∫x^2fx(x)dx=1/6
DX=E(x^2)-(Ex)^2=1/6-1/9=1/18=Dy
相关系数r=cov(x,y)/(DxDy)^(1/2)=(-1/36)/(1/18)=-1/2