# Expectation of independent normal distribution

I have a question attached below, and I really do't know how to solve it. Is there anyone can help me with this? $x,y\sim N(0,1)$, independent, what is $E(x\mid x+y=1)$, what about variance? What I am thinking is to calculate $P(x\mid x+y=1)$ first, $P(x\mid x+y=1)=1=P(x+y=1\mid x)\cdot P(x)/P(x+y=1)$, but I don't know how to express it...

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I do not understand (nor agree) with the equation in the last line. Calculating $E[X| X+Y=1] = \int xP(x|x+y=1)\, dx$ from the definition of expected value will be sufficient (if you can carry out the integration). – Calvin Lin Jan 20 '13 at 16:25
I think $P(X+Y=1) = 0$ since $X$, $Y$ are continuous variables. – Patrick Li Jan 20 '13 at 17:07
What do you mean by what about variance? Which variance? – Did Jan 20 '13 at 19:48
Hello Calvin, but how to get P(x|x+y=1) if x,y∼N(0,1). This is what I am confused with. – Hello_World Jan 21 '13 at 21:18
It should be Var(x|x+y=1) – Hello_World Jan 21 '13 at 21:19

As is the case as soon as $x$ and $y$ are i.i.d., the symmetry $(x,y)\to(y,x)$ shows that $\mathbb E(x\mid x+y)=\mathbb E(y\mid x+y)=\frac12\mathbb E(x+y\mid x+y)=\frac12(x+y)$. In particular, $\mathbb E(x\mid x+y=1)=\frac12$.