Conditional expectation of $E[X|Y-X,Z+X]$ given three independent Gaussian RVs $(X,Y,Z)$?

Question

Let $X \sim (0,\sigma_x^2), Y \sim (0,\sigma_y^2), Z \sim (0,\sigma_z^2)$ be independent Gaussian RVs. I'm trying to compute $E[X|Y-X,Z+X]$. Here is what I've tried.

Set $A=Y-X, B=Z+X$, then, $X = \frac{B-A-Z+Y}{2}$,

\begin{align} E[X|Y-X,Z+X]&=E[\frac{B-A-Z+Y}{2}|A,B]\\ &=\frac{1}{2}[E[B|A,B]-E[A|A,B]-E[Z|A,B]+E[Y|A,B]]\\ &=\frac{1}{2}[B-A-E[Z|Z+X]+E[Y|Y-X]]\\ &=\frac{1}{2}[Z+X-Y+X-\frac{\sigma_z^2}{\sigma_z^2+\sigma_x^2}(Z+X)+\frac{\sigma_y^2}{\sigma_y^2+\sigma_x^2}(Y-X)]\\ &=\frac{1}{2}[\frac{\sigma_x^2}{\sigma_z^2+\sigma_x^2}(Z+X) - \frac{\sigma_x^2}{\sigma_y^2+\sigma_x^2}(Y-X)]. \end{align} Is this proof correct?

Answer 1

As pointed out by William M, the problem in your attempt is that $E(Z \mid A, B) = E(Z \mid B)$ is not true in general.

Here is a way to find $\mathbb E[X|Y-X,Z+X]$: for each $a,b\in\mathbb R$, $$ \mathbb E[X|Y-X,Z+X]=\mathbb E[X-a(Y-X)-b(Z+X)|Y-X,Z+X]+a(Y-X)+b(Z+X). $$ If we choose $a_0$ and $b_0$ such that $X-a(Y-X)-b(Z+X)$ is independent of the vector $(Y-X,Z+X)$, then we will get $$ \mathbb E[X|Y-X,Z+X]=a_0(Y-X)+b_0(Z+X). $$ In order to find $(a_0,b_0)$, note that since the vector $(X,Y,Z)$ is Gaussian, it suffices to solve in $(a,b)$ $$ \operatorname{Cov}\left(X-a(Y-X)-b(Z+X),Y-X\right)=\operatorname{Cov}\left(X-a(Y-X)-b(Z+X),Z+X\right)=0, $$ which will give a linear system in $(a,b)$.

Answer 2

Hint: find the joint distribution of $(X, Y - X, Z + X)$ using the affine transformation property of multivariate distribution, then apply the conditional distribution theorem for multivariate normal random vectors.

Answer 3

This is mechanized in Mathematica 13.2:

Expectation[x \[Conditioned] y - x == t && z + x == s, {x \[Distributed] 
NormalDistribution[0, \[Sigma][x]],  z \[Distributed] NormalDistribution[0, \[Sigma][z]], 
y \[Distributed] NormalDistribution[0, \[Sigma][y]]}]

$$\frac{t \sigma (x)^2 \sigma (y)^2-s \sigma (x)^2 \sigma (z)^2}{\sigma (x)^2 \sigma (y)^2+\sigma (x)^2 \sigma (z)^2+\sigma (y)^2 \sigma (z)^2} $$