Imagine I want to compute $E[f(X,Y)]$ where $f$ is a suitable function and $X,Y$ random variables.
If $X$ and $Y$ are correlated, let us say $X=g(Y)$ or even in my case $X_t$ and $Y_t$ are stochastic processes and $Y_t=g(X_{\cdot})$. My question is: am I allowed to do $$E[f(X,Y)]= E[E[f(X,y)|y=Y]]?$$ in my case, now the law of $X$ is known and it is easy to compute $E[f(X,y)|y=Y]$ and then insert $y=Y$ and compute the rest.
The point is, I know the law of $X$ if $Y$ is fixed given, and I fully know the law of $Y$, but I am not 100% sure that this method is correct, I feel I don't use the correlation part. The fact that they are (or not) correlated, does it affect the method of computing it?
Thanks for the support!
Yes, it is correct. It follows from the tower property that
$$\mathbb{E}(f(X,Y))=\mathbb{E}\big( \mathbb{E}(f(X,Y) \mid \mathcal{F}) \big)$$
for any $\sigma$-algebra $\mathcal{F}$. In particular, we can choose $\mathcal{F} := \sigma(Y)$ the $\sigma$-algebra generated by $Y$. Moreover, we know from the factorization lemma that there exists a function $h$ (depending on $f$) such that
$$\mathbb{E}(f(X,Y) \mid Y) := \mathbb{E}(f(X,Y) \mid \sigma(Y)) = h(Y).$$
Usually, this is denoted by
$$h(y) = \mathbb{E}(f(X,Y) \mid Y=y).$$
Combining the equalities yields
$$\mathbb{E}(f(X,Y)) = \mathbb{E}\big( \mathbb{E}(f(X,Y) \mid Y=y) \big|_{y=Y} \big).$$