There are two random variables $X$ and $Y$, both of them distributed standard normal. Based on $X$ and $Y$, define $Z = X + a Y$ for $a$ a known parameter.
I'm interested in $E(X | Z)$.
It seems to me that a starting point could be the approach taken here. So I start with:
$ Z = E( X | Z) + a E(Y | Z) $
and then I write out the expectation terms explicitly, hoping to find an easy substitution:
$ E(X | Z) = \frac{\int_{-\infty}^{\infty}x f(x)f(\frac{Z-x}{a})dx}{\int_{-\infty}^{\infty} f(x)f(\frac{Z-x}{a})dx} $
and
$ E(Y | Z) = \frac{\int_{-\infty}^{\infty}y f(y)f(Z-ay)dy}{\int_{-\infty}^{\infty} f(y)f(Z-ay)dy} $
But here I'm stuck, I can't find an substition that would allow me to express $E(Y|Z)$ in terms of $E(X|Z)$, in order to plug in above. So how to continue from here?
Edit:
For $a =1$, one can simply use $E(X|Z) = E(Y | Z)$, plug in above, and obtain $E(X|Z) = Z/2$.
I conjecture that in general, it holds that $$E(X|Z) = \frac{Z}{1 + a^2}.$$ This is supported by several numerical simulations I made.
This would require showing that $E(Y|Z) = a E(X | Z)$. In case substituting in the integral does not help, how else could I show that?
We first need to find the joint distribution of $X$ and $Z$. This can easily be found since the $cov(X,Z)=1$ and hence
$$ \left(\begin{array}{cc} X \\ Z \end{array} \right) \sim \mathcal{N}\left(\left(\begin{array}{cc} 0 \\ 0 \end{array} \right) , \left(\begin{array}{cc} 1 & 1 \\ 1 & 1+a^2 \end{array} \right) \right)$$
Now we can simply apply the formula for the conditional expectation from a bivariate normal distribution and obtain
$$E(X | Z) = \frac{Z}{1+a^2}.$$
Some background notes:
Lecture notes on multivariate normal distribution
Marginal and conditional distributions of multivariate normal distribution
and Chapter 3 (Theorem 3.3.4) in Tong, Y.L., 2012. The multivariate normal distribution. Springer Science & Business Media.