Suppose $X, Y, Z$ are random variables. How to write $E(X)$ as a function of $E(X|Y, Z)$?

Question

Suppose $X, Y, Z$ are random variables. Suppose I'm interested in the quantity $E(X)$. How can I obtain $E(X)$ from $E(X|Y, Z)$ with law of iterated expectations?

I know that $E(X) = E(E(X|Y))$ is true. Moreover, $E(X|Y) = E(E(X|Y, Z)|Y)$ is true. Therefore, is it true that

$$E(X) = E(E(E(X|Y, Z)|Y))?$$

The most outer expectation on the RHS is wrt $Y$, the second expectation is wrt $Y, Z$? And the most inner expectation is wrt to X|Y, Z? That is, we have:

$$E(X) = E_Y(E_{Y,Z}(E_{X|Y,Z}(X|Y, Z)|Y))?$$

Answer 1

I know that $E(X) = E(E(X|Y))$ is true. Moreover, $E(X|Y) = E(E(X|Y, Z)|Y)$ is true. Therefore, is it true that

$$E(X) = E(E(E(X|Y, Z)|Y))?$$

Yes. The tower property sais that: $$\begin{align}\mathsf E(X)&=\mathsf E(\mathsf E(X\mid Y, Z))\\&=\mathsf E(\mathsf E(\mathsf E(X\mid Y,Z)\mid Y)) \end{align}$$

The most outer expectation on the RHS is wrt $Y$, the second expectation is wrt $Y, Z$? And the most inner expectation is wrt to X|Y, Z? That is, we have:

$$E(X) = E_Y(E_{Y,Z}(E_{X|Y,Z}(X|Y, Z)|Y))?$$

That is a depreciated notation. If $X,Y,Z$ were continuous random variables with appropriate conditional probability density functions we would have

$$\begin{align}\iiint_{\Bbb R^3} x\, f_{X,Y,Z}(x,y,z)\,\mathrm d\langle y,z,x\rangle &=\iint_{\Bbb R^2} f_{Y,Z}(y,z)\int_\Bbb R x~ f_{X\mid Y,Z}(x\mid y,z)\,\mathrm d x\,\mathrm d\langle y,z\rangle\\&=\int_\Bbb R f_Y(y)\,\int_\Bbb R f_{Z\mid Y}(z\mid y)\,\int_\Bbb R x~f_{X\mid Y,Z}(x\mid y,z)\,\mathrm d x\,\mathrm d z\,\mathrm d y\end{align}$$

So you would present this as: $$\mathsf E_X(X)=\mathsf E_Y(\mathsf E_{Z\mid Y}(\mathsf E_{X\mid Y,Z}(X\mid Y,Z)\mid Y))$$

Answer 2

"How can I obtain $E(X)$ from $E(X|Y, Z)$ with law of iterated expectations?"

By definition $E(X|Y, Z)$ is $E(X|\mathcal{G})$ where the sigma-algebra is generated by $(Y,Z)$. It's well known that $EX = E E(X|\mathcal{G})$.

Hence the answer is $EX = E ( E(X|Y, Z) )$.

R.h.s. of $E(X) = E_Y(E_{Y,Z}(E_{X|Y,Z}(X|Y, Z)|Y))$ doesn't have sence. Because $E(\xi|\eta)$ has sense, and $E_{\eta}(\xi|\eta)$ doesn't: why should we write $\eta$ twice? If we get rid of this problem, we will have $$E(X) = E(E(E(X|Y, Z)|Y)).$$ It's true, but it's a bit strange, because $E ( E(X|Y, Z) ) = EX$ so we don't need to take expectation one more time. Of course, we may write $$E(X) = E\Bigg( E(E(E \Big( E( E(X|Y,Z) |Y) |X \Big)|Z)|Y) \Bigg)$$ as it's true, but it's not needed.