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

???

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- Feb 23rd 2010, 09:55 PMsezminComputing conditional expectation...
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

??? - Feb 24th 2010, 09:02 AMsezmin
Also,

I think I heard somewhere that if you have X and Y are two random variables, both of which have the standard normal distribution N(0,1), then X and Y are independent...

This sounds strange to me, is it true? - Feb 24th 2010, 10:49 AMmatheagle
no and no

you need the covariance to be zero in the second question. - Feb 24th 2010, 02:34 PMsezmin
i see, how

*do*i find the first one then?

i see, for the second one, thanks! - Feb 24th 2010, 03:03 PMmatheagle
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.