Is $E[X\vert Y,Z]$ necessarily a function of $Y,Z$ only? $X,Y,Z$ are random variables.

Question

I am wondering whether $E[X\vert Y,Z]$ is a function of $Y,Z$ only? (That is, given that we don't know anything about $X$ besides that it is a R.V).

I am thinking yes, and my logic is that if $X$ is some function (a random variable), then it is a function of $Y,Z$ and possible other stuff. Once I condition on $Y,Z$ and take the expectation, though, anything that is not $Y,Z$ (the "possible other stuff") does not matter anymore, because we take the expected value (which is a constant).

That is, we integrate over all possible values of whatever is not $Y,Z$, and therefore all terms besides $Y,Z$ go away when we take $E[X\vert Y,Z]$

My motivation for the question is that if, for example $X= 2Y + Z +W^2$ where $W$ is also an R.V, then $$E[X\vert Y,Z] =E[2Y+Z+W^2\vert Y,Z] = 2Y + Z +E[W^2\vert Y,Z]$$ and I think $E[W^2\vert Y,Z]$ is just a constant after we compute the expectation (sum or integrate).

What makes me second guess that the answer is "yes" is that I am unsure whether the bounds of the integral $$E[W^2\vert Y,Z]=\int_a^b w^2 f_{W\vert \{Y,Z\}} dw$$ can maybe depend on some variable besides $Y,Z$ values.

Thanks.

Answer 1

The answer is yes, of course. Recall the definition of conditional expectation:

Definition: Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space, $X:\Omega \to \mathbb R$ a random variable and $\mathcal A\subset \mathcal F$ a $\sigma$-algebra. We define the conditional expectation $Y=\mathbb E[X \mid \mathcal A]$ as the only random variable satisfying

1. $Y$ is $\mathcal A$-measurable.

2. For every $A\in \mathcal A$, $$\mathbb E[X1_A ]=\mathbb E[Y1_A]$$

Existence and uniqueness in the $\mathcal L^2_+$ case are a consequence of the existence of orthogonal projections, while the general case is obtained by approximation.

The other crucial part to show what you are saying is the following theorem:

Theorem: Let $(E, \mathcal E)$ be a measurable space, $X: \Omega \to E$ a function. Endow $\Omega$ with the $\sigma$-algebra $\sigma(X)=X^\leftarrow(\mathcal E)$. Let $Y:(\Omega, \sigma(X)) \to (\mathbb R, \mathcal B(\mathbb R))$ be a measurable function. Then there exists a measurable function $f:(E, \mathcal E)\to(\mathbb R, \mathcal B(\mathbb R))$ such that $$Y=f \circ X$$

To prove your claim for one random variable, just observe that if you set $\mathcal A=\sigma (Y)$, then $\mathbb E[X\mid \mathcal A]$ is $\sigma(Y)$-measurable by definition. As a consequence of the theorem there exists a function $f$ such that $$\mathbb E[X\mid \mathcal A]=\mathbb E[X\mid Y]=f \circ Y$$ so the conditional expectation depends solely on $Y$.