Conditional expectation given sum of weighted average Suppose X, Y are i.i.d standard normal (mean 0, standard deviation 1) random variables, a, b, c are constant scalars.
$$Z = a X + b Y$$
How to express $E[X|Z=c]$ using $a,b,c$?
 A: Still searching for a quick solution, but anyway I solved the problem in brute force.

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*Prove that $f_{X,Z}(x_0,c) = \frac{1}{b} f_X(x_0) f_Y(\frac{c - ax_0}{b})$, where $f_{X,Z}$ is PDF of joint distribution of $X$ and $Z$
(hint: use Leibniz integral rule and Fundamental theorem of calculus)


*Prove $E[X|Z=c]=\frac{ac}{a^2+b^2}$ (hint: use definition of conditional expectation and PDF of standard normal random variable)
A: You could express the relationship as $X = Z/a - bY/a$, which gives you that $X\mid Z=c$ is distributed as $c/a - bY/a$ which means it is $\mathcal{N}(c/a, b^2/a^2)$ therefore you get
$$\mathrm{E}[X\mid Z=c] = c/a$$
The absence of $b$ here comes from the fact $Y\sim\mathcal{N}(0,1)$.
A: a)$E(aX+bY|aX+bY)=aX+bY$,
so
$$aE(X|aX+bY)+bE(Y|aX+bY)=aX+bY \tag{1}$$
b)Find $d$ such that $X-dY$ be independent(Multivariate_normal_distribution#Correlations_and_independence) of $aX+bY$ so
$E(X-dY|aX+bY)=E(X-dY)=0$ and hence
$$E(X|aX+bY)=dE(Y|aX+bY) \tag{2}$$
$d$ can be computed as follows:
$$0=\text{cou}(X-dY, aX+bY)=a\text{Var}(X)-db \text{Var}(Y)=a-db$$
so $d=\frac{a}{b}$
Now use (1) and (2)
$$aE(X|aX+bY)+b(\frac{b}{a}E(X|aX+bY))=aX+bY$$ so
$$E(X|aX+bY)=(\frac{a}{a^2+b^2})(aX+bY)$$
