Let $X,Y \sim N(0, 1)$. Find $E[X\mid X+Y=1]$. Assume the joint distribution for X, Y is also normal. I have no clue how to approach this problem. 
Follow up question: Without knowing the joint distribution of X, Y can you still calculate it? 
 A: $\begin{align}\mathsf E(X\mid X+Y=1) & = \dfrac{\int_\Bbb R x f_{X,Y}(x,1-x)\operatorname d x}{\int_\Bbb R f_{X,Y}(x,1-x)\operatorname d x} \\[1ex] & = \dfrac{\int_\Bbb R xe^{-x^2/2}e^{-(1-x)^2/2}\operatorname d x }{\int_\Bbb R e^{-x^2/2}e^{-(1-x)^2/2}\operatorname d x} \\[1ex] & =\dfrac{\Big[-\tfrac 1 2 e^{-(x^2+(1-x)^2)/2} -\sqrt\pi\operatorname {erf}(\tfrac 1 2-x)\big/2\sqrt[4]e\Big]_{x=-\infty}^{x=+\infty}}{\Big[-\sqrt\pi\operatorname {erf}(\tfrac 1 2-x)\big/2\sqrt[4]e\Big]_{x=-\infty}^{x=+\infty}} \\[1ex] & = \dfrac{1}{2}\end{align}$
Which can more easily be obtained by symmetry.
A: If we assume that


*

*$X,Y$ are independent; or that

*$X,Y$ are jointly normal and $(X,Y)$ has the same distribution as $(Y,X)$ (a weaker assumption consistent with nonzero correlation); or that

*$(X,Y)$ has the same distribution as $(Y,X)$ (which can be true with each of them normally distributed, even if they're not jointly normal);


then
$$
\operatorname{E}(X\mid X+Y=1) = \operatorname{E}(Y\mid X+Y=1)
$$
and
$$
\operatorname{E}(X\mid X+Y=1) + \operatorname{E}(Y\mid X+Y=1) = \operatorname{E}(X+Y\mid X+Y=1) = 1,
$$
so
$$
\operatorname{E}(X\mid X+Y=1) = \frac 1 2.
$$
