Consider two independent random variables $X$ and $Y$, where $Y$ is standard Gaussian noise. I am interested in analyzing $\mathbb{E}[X|X+Y]$. I know that
$\mathbb{E}[X|X+Y] = \frac{\int_{-\infty}^{\infty} x f_{X,Y}(x,y) dx}{\int_{-\infty}^{\infty} f_{X,Y}(x,y) dx}$.
If $X$ is measurable with respect to $\mathcal{G}$, then $\mathbb{E}[X|\mathcal{G}] = X$ almost surely.
Since $X$ and $X+Y$ are not independent, I was wondering if $X$ is measurable w.r.t the sigma algebra generated by $X+Y$? Is $X \in \sigma(X+Y)$?
I am having a hard time navigating conditional expectation. Any help would be deeply appreciated!
I do not have the general answer, but in the case where $X$ is also a standard normal random variable, the answer is no. Indeed, since $(X,X+Y)$ and $(Y,X+Y)$ have the same distribution, it follows that $\mathbb E\left[X\mid X+Y\right]=\mathbb E\left[Y\mid X+Y\right]$ hence $\mathbb E\left[X\mid X+Y\right]=(X+Y)/2$. If $X$ was $\sigma(X+Y)$-measurable, the equality $\mathbb E\left[X\mid X+Y\right]=X$ would hold which is not compatible with the previous one.