Let XXX and YYY have joint PDF f(x,y)=e−(x+y)f(x, y) = e^{-(x+y)}f(x,y)=e−(x+y) for x,y≥0x, y \ge 0x,y≥0. Are XXX and YYY independent?
Yes, because f(x,y)=fX(x)fY(y)f(x, y) = f_X(x)f_Y(y)f(x,y)=fX(x)fY(y)
No, because f(x,y)≠fX(x)fY(y)f(x, y) \neq f_X(x)f_Y(y)f(x,y)=fX(x)fY(y)
Yes, because E[XY]=E[X]E[Y]E[XY] = E[X]E[Y]E[XY]=E[X]E[Y]
No, because Cov(X,Y)≠0Cov(X, Y) \neq 0Cov(X,Y)=0