0

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

FD_bfa
  • 3,989
zyy
  • 977
  • It depends on whether you know the rigorous definition(measure theoratic) of conditional expectation. If yes then it is just step and is known as the tower property of conditional expectation. – Mr.Gandalf Sauron Apr 28 '23 at 14:46
  • Firstly, I guess not all functions $f(x)$ are valid. Secondly, you need to verify that $\int_{A} Y d \mathbb{P} = \int_{A} \mathbb{E}(Y|W) d \mathbb{P}$ for $A \in \sigma(X^{-1})$. But notice that $\int_{A} \mathbb{E}(Y|W) d \mathbb{P} = \int_{A} Y d \mathbb{P}$ for $A \in \sigma(W^{-1})$. Can you deduce from here? – openspace Apr 28 '23 at 14:47
  • @Mr.GandalfSauron I know little about the rigorous foundation of probability theory. Can it be shown by elementary methods? – zyy Apr 28 '23 at 14:52
  • @openspace I know little about the rigorous foundation of probability theory. Can it be shown by elementary methods? – zyy Apr 28 '23 at 14:53
  • @zyy I guess this can be usefull – openspace Apr 28 '23 at 14:55
  • @openspace I think I know how to prove $E(Y) = E[E(Y|X)]$, but it seems the same method does not apply to my problem. Also, by saying "rigorous foundation of probability", I mean "measure theoretic foundation of probability". – zyy Apr 28 '23 at 15:12
  • There is an issue with $E(Y\mid X)=?E[E(Y\mid W) \mid X]$ if $f$ is not a 1-1 (injective) function. – Henry Apr 28 '23 at 15:12

2 Answers2

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.

Mr.Gandalf Sauron
  • 13,906
1

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: $$E\big[E[Y\mid W]\mid X\big]=\int E[Y\mid W]\,\,dP_{W/X}=E[Y\mid X]$$

Speltzu
  • 543