if I have X1,X2 iid normal with N(0,1)
and I want to find E(X1*X2 | X1 + X2 = x)
Can I simply say: X1 = x - X2 and thus
E(X1*X2 | X1 + X2 = x) =
E[ (x - X2)*X2) = E[ (x * X2) - ((X2)^2) ] <=>
x*E[X] - E[X2^2] =
0 - 1 =
-1
???
if I have X1,X2 iid normal with N(0,1)
and I want to find E(X1*X2 | X1 + X2 = x)
Can I simply say: X1 = x - X2 and thus
E(X1*X2 | X1 + X2 = x) =
E[ (x - X2)*X2) = E[ (x * X2) - ((X2)^2) ] <=>
x*E[X] - E[X2^2] =
0 - 1 =
-1
???
I'm bet there is a clever way to do the first, but the straight fowrard way will work.
Obtain the conditional density.
let W=XY and Z=X+Y.
From the joint density of X and Y you can obtain the joint density of W and Z.
Then $\displaystyle f(W|Z)={f(W,Z)\over f(Z)}$
So $\displaystyle E(W|Z)=\int wf(w|Z)dw$
The marginal density of Z is obvious.