Given two unknown functions $f,g : \mathbb{R} \to \mathbb{R}$, how would I go about to compute the expression for
$h(X) = \mathbb{E}[g(f(X)+V)|X]$ where $V \sim \mathcal{N}(0,\sigma^2)$ and $X$ and $V$ are independent.
From all the reading I now know how to compute $\mathbb{E}[f(X)|f(X)+V]$. That is, if $Y = f(X)+V$
$\mathbb{E}[f(X)|Y] = \frac{\int_{\mathbb{R}}f(x) f_V(z-f(x))f_X(x) dx}{\int_{\mathbb{R}} f_V(y-f(x))f_X(x) dx}$
But I am unable to figure out $\mathbb{E}[g(Y)|X]$.
I tried using the following identity:
$\mathbb{E}[f(X)+V | X] = f(X) + \mathbb{E}[V|X] = f(X) + \mathbb{E}[V] = f(X)$.
I tried computing the required expectation in the following way:
$\mathbb{E}[g(Y)|X] = \frac{\int_{\mathbb{R}}g(y)f_{x,y}(x,y) dy}{\int_{\mathbb{R}}f_{x,y}(x,y) dy}$ where $f_{x,y}(x,y)$ is the joint density of $X$ and $Y$. To compute the joint density, I started with the joint distribution:
$\mathbb{P}(X\leq x, Y \leq y) = \int_{\mathbb{R}} \mathbb{P}(X\leq x, Y \leq y | Y=y) f_Y(y) dy \\ = \int_{\mathbb{R}} \mathbb{P}(X\leq x) f_Y(y) dy \\ = \int_{\mathbb{R}} F_X(x) f_Y(y) dy\\$
where $F_X$ is the distribution function of X and $f_Y$ is the density of $Y$.
Since the distribution of $V$ is known, it would be helpful to get the expression in terms of the density of $V,$ i.e., $f_V(v)$. But I am not able to figure it out. Any help or suggestion is deeply appreciated!