Prove that $\mathbb{E}(Y|X)=\mathbb{E}[\mathbb{E}(Y|W)|X]$ and $\mathbb{E}(Y|X)=\mathbb{E}[\mathbb{E}(Y|X)|W]$

Question

Suppose that $W$ and $Y$ are random variables.
Let $X$ be a random variable that is some function of $W$, say $X=f(W)$.
How can one show that
$$\mathbb{E}(Y|X)=\mathbb{E}[\mathbb{E}(Y|W)|X]\text{ and }\mathbb{E}(Y|X)=\mathbb{E}[\mathbb{E}(Y|X)|W]?$$

Answer 1

There is no non measure theoratic way to "prove" this. However one can obviously get an intuitive idea by considering some easy cases and examples.

Here's the formal proof anyway

First notice that $\sigma(X)\subseteq \sigma(W)$ for all $f$ measurable(you can think this to be nice enough functions which are not very very very difficult to construct)

First let $A\in\sigma(X)\subseteq\sigma(W)$

Then $\int_A E(E(Y|W)|X)dP=\int_A E(Y|W) dP=\int_A Y dP=\int_A E(Y|X) dP$ This holds for all such $A$

Hence $E(E(Y|W)|X)=E(Y|X)$

Also for $E(E(Y|X)|W))=E(Y|X)$ as $E(Y|X)$ is measurable with respect to $X$ and hence with respect to $W$ .

Another interpretation is that conditional expectation , in case of square integrable random variables is nothing but projection to smaller subspaces. So think of X as something smaller(i.e. information contained in X is already contained in W) . Then projecting to this smaller subspace is equivalent to projecting to a larger subspace containing it and then again projecting to the smaller space.

That is if you project a point to a plane and then to a line in a plane , it is same as projecting directly to the line.

This is however only true in the Hilbert Space $L^2$ but it serves as THE motivation for defining conditional expectation the way its defined.

So one intuitive notion to keep in mind that the expected value of the expected value of the random variable $Y$ given certain information , given even lesser information is the same as the expected value of $Y$ given the lesser information.

Answer 2

The most intuitive way is to see that: $$P_{X/W}(A)=P[X\in A\mid W]=I_A(X)\quad and$$ $$P_{W/X}(f^{-1}(A))=P[W\in f^{-1}(A)\mid X]=I_A(X)$$ Then: $$E\big[E[Y\mid X]\mid W\big]=\int E[Y\mid X]\,\,dP_{X/W}=E[Y\mid X]$$ And: $$E\big[E[Y\mid W]\mid X\big]=\int E[Y\mid W]\,\,dP_{W/X}=E[Y\mid X]$$