Conditional expectation with more than one condition

Question

We know that $E(Y), E(X), E(YZ)$ exist and $E(Z|X,Y) = E(Z|X)$. We have prove that $E(YZ|X) = E(Y|Z) E(Z|X)$.

So the solution goes: \begin{align*} E(YZ|X) & = E[E(YZ|X,Y)|X] \\ & = E[Y E(Z|X,Y) | X]\\ & = E[Y E(Z|X) | X]\\ & = E(Y|Z) E(Z|X) \end{align*}

My question is: why does the first equality $E(YZ|X) = E[E(YZ|X,Y)|X]$ (and than the second) hold?

Thank you.

Answer 1

Proof: Given (|,)=(|)

this means that P(Z|X,Y)=P(Z|X).

Now,

P(Z|X,Y)=$\frac{P(Z,Y|X)}{P(Y|X)}$

for this to be equal to P(Z|X),

P(Z,Y|X) should be equal to P(Z|X)*P(Y|X),

this implies that Y and Z are independent.

If two rvs A,B are independent then E(A,B)=E(A)*E(B)

Therefore E(Z,Y|X)=E(Z|X)*E(Y|X).

In your case, the first equality holds because of Law of total expectation. According to LOTUS (Law of total expectation),

E(Y) = $E_y(E(Y|X))$

Let YZ|X = A E(YZ|X) = E(A) = $E_y(E(A|Y)) = E_y(E(YZ|XY)) = E_y((Y|X) * E(Z|XY)) = E_y((Y|X) * E(Z|X)) = E(Y|X) * E(Z|X)$

Answer 2

The first equality is the tower rule. Assume $\mathcal{H}\subseteq\mathcal{G}\subseteq\mathcal{F}$ are sigma-algebras and $X$ is a random variable with respect to $\mathcal{F}$. Then $E[X|\mathcal{H}]=E[E[X|\mathcal{G}]|\mathcal{H}]$. The proof is simple. Let $Y=E[X|\mathcal{G}]$ and $Z=E[X|\mathcal{H}]$. By definition it means that:

$E[Y\mathbb{1}_G]=E[X\mathbb{1}_G]$ for all $G\in\mathcal{G}$.

$E[Z\mathbb{1}_H]=E[X\mathbb{1}_H]$ for all $H\in\mathcal{H}$.

Since $\mathcal{H}\subseteq\mathcal{G}$ we conclude that $E[Y\mathbb{1}_H]=E[Z\mathbb{1}_H]$ for all $H\in\mathcal{H}$. Hence $Z=E[Y|\mathcal{H}]$.

As for the second equality: since $Y$ is measurable with respect to the sigma algebra $\sigma(X,Y)$ you can take it out of the conditional expectation.