Question about conditional expectation and sum of random varibles

Question

Suppose that $X,Y,Z$ are random variables and $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ is measurable.

I think that the following is true

$$\mathbb{E}[\mathbb{E}[f(X+Y,Z)|X,Y]X]=\mathbb{E}[\mathbb{E}[f(X+Y,Z)|X+Y]X].$$

How can I prove it?

Answer 1

This is not true. Let $U,V$ be i.i.d standard normal variables, $X=V-U, Y=U,Z=U-V$ and $f(t,s)=s+t$. Then $E(f(X+Y,Z)|X,Y)=E(Y|X,Y)=Y$. Hence LHS is $E(Y|X)=E(U|V-U)$. Now $E(f(X+Y,Z)|X+Y)=E(U|V)=0$. Hence RHS $=0$. It is easy to see that $E(U|V-U) \neq 0$: if $E(U|V-U) = 0$ then we can conclude that $E(U(V-U))=0$ but $E(U(V-U))=-1$.

The result is true when $Z$ is independent of $(X,Y)$. To prove this use the following steps:

a) It is enough to prove this when $f$ is a simple function

b) It is enough to prove this when $f$ is an indicator function

c) It is enough to prove this when $f=I_{A \times B}$ where $A$ and $B$ are Borel sets in $\mathbb R$.

d) For this last case the result is quite trivial by independence.

Answer 2

As proven above, when $Z$ is independent of $X,Y$ then the statement is false.

Next, I will prove that it is true when $Z$ is independent of $X,Y$.

Actually, I claim that $$ \mathbb{E}[f(X+Y,Z)|X,Y]=\mathbb{E}[f(X+Y,Z)|X+Y]. $$ Thus, the result follows by taking conditional expectations w.r.t. $X$.

The proof is based on the following standard result in probability (see for instance Durrett's book, p. 225). Also, it can be easily proved by means of the monotone class theorem applied on $f$.

Lemma Suppose that $W,Z$ are independent random variables. Let $f$ be a function with $\mathbb{E}[f(W,Z)]<\infty$ and let $h(w):=\mathbb{E}[f(w,Z)]$. Then $\mathbb{E}[f(W,Z)|W]=h(W)$.

Let us proof the claim. Since $Z$ is independent of $X+Y$, by the Lemma above we have $$ \mathbb{E}[f(X+Y,Z)|X+Y]=h(X+Y), $$ where $h(w):=\mathbb{E}[f(w,Z)]$.

Since $Z$ is independent of $(X,Y)$, using the same Lemma, we know that $$ \mathbb{E}[f(X+Y,Z)|X,Y]=g(X,Y), $$ where $g(x,y):=\mathbb{E}[f(x+y,Z)]$.

Since $h(x+y)=g(x,y)$, we conclude that $h(X+Y)=g(X,Y)$ and the assertion follows.