$X_i,i=1,2,3$ are i.i.d standard normal random variables. Then $\mathbb{E}(2X_1+3X_2\mid X_1+3X_2-X_3=4)$=?
(Source: UOH PhD Entrance 2017)
I am not able to proceed too much with the problem except thinking of lengthy calculation of finding the joint density of $2X_1+3X_2$ and $X_1+3X_2-X_3$. I am getting difficulty in finding the appropriate transformation. This being an multiple choice question, carrying only $1$ mark, I think I am not expected to use this method. Even if there is some nice idea to proceed with the problem, I suspect that I am missing that due to lack of adequate background. Hope I would be provided with hint.
Let $Z:=X_1+3X_2-X_3$ and $Y:= 2X_1+3X_2$. Then $Z\sim \mathcal N(0,11)$ and $Y=Z+X_1+X_3$. Therefore $$\mathbb P\{Y\leq y\mid Z=z\}=\mathbb P\{X_1+X_3\leq y-z\mid Z=z\}$$ Now, by independence of $Z$ and $X_1+X_3$ (hint: correlation...), $$\mathbb P\{Y\leq y\mid Z=z\}=\mathbb P\{X_1+X_3\leq y-z\}$$ Since $X_1+X_3\sim \mathcal N(0,2)$, this proves that, for every $z$, $Y\mid Z=z\sim \mathcal N(z,2)$.
It follows that the desired conditional PDF is $$f_{Y\mid Z=z}(y)=\frac{1}{2\sqrt{\pi}}e^{-\frac{(y-z)^2}{4}}$$ and that the desired conditional expectation is $$\mathbb E[Y\mid Z=z]=z$$