What is $E[X | X+Y= 1]$ where $X,Y$ are normally distributed?

Question

I saw this question posted on a non-english website, but the only information provided was $X,Y$ are normal random variables, what is

$E[X | X+Y= 1]$? And some people were saying the answer is 0.5.

You would at least have to know the mean and standard deviation of each $X,Y$ to find a closed form solution for this right? And I think you would need to know whether $X,Y$ are independent?

If we assume that $X,Y \sim N(0,1)$, standard normal, how do we solve for the conditional expectation?

I tried to think about it the following way: $$ E[X | X+Y = 1] = \int_{-\infty}^{\infty} xf_X(x|x+y=1) dx \\ f_X(x|x+y=1) = \frac{f_{X,Y}(x, x+y= 1)}{P(X+Y=1)} $$

But this doesn't really make sense because $X$ and $Y$ are continuous, so $P(X+Y= 1) = 0$. Any suggestions on how to proceed under the assumptions that $X,Y$ are standard normal?

Answer 1

The key is $$E[X+Y \mid X+Y = 1] = E[X \mid X+Y = 1] + E[Y \mid X+Y = 1].$$ This equality requires no distributional assumptions about $X$ and $Y$; they need not be normal, and they can even be dependent.

The left-hand side equals $1$. If you have the assumption that $X$ and $Y$ are exchangeable (i.e. $(X,Y)$ has the same distribution as $(Y,X)$) then the two terms on the right-hand side are equal and must each equal $1/2$.

Answer 2

If $X$ and $Y$ are exchangeable, then the answer is necessarily $1/2$ regardless of the distribution (assuming all the relevant expectations exist). This is because $E(X \mid X + Y = 1) = E(Y \mid X + Y = 1)$ by exchangeability, and you get 1 if you add the left and right side together.

Answer 3

$$E(X+Y|X+Y=1)=1$$ $$E(X|X+Y=1)+E(Y|X+Y=1) = 1$$ By symmetry $$E(X|X+Y=1)=E(Y|X+Y=1) = \frac{1}{2}$$