# Find version of conditional expectation

I'm struggling with the concept of conditional expectation. We didn't cover it in my probability theory class, yet it's required for my statistics course. I'm basically having no idea how to solve any of our homework problems where conditional expectation is involved which really bothers me - the other problems are usually not that hard, but I'm currently losing it as soon as conditional expectation is involved...

I'm currently working this problem, quite unsuccessfully though:

Let $X$ be an integrable, real valued random variable and $f$ it's density with respect to the Lebesgue measure. Let $Y=g(X)$ where $g: \mathbb{R} \rightarrow \mathbb{R}$ has a positive derivative on $(0,\infty)$ and is symmetrical around $0$, i.e. $g(-x)=g(x)$. Explicitly give a function $\psi$ (depending on $f$ and $g$) such that $\psi(X)$ is a version of the conditional expectation $\mathbb{E}[X|Y]$.

I have the definition of the conditional expectation given like here:

http://en.wikipedia.org/wiki/Conditional_expectation#Formal_definition

This means, I have to choose $\psi$ such that it satisfies these two conditions:

a) $\psi(X)$ is $\sigma(Y)$-measurable.

b) $\mathbb{E}[X1_B]=\mathbb{E}[\psi(X)1_B]$, i.e. $\int_{B} X d\mathbb{P}=\int_{B} \psi(X) d\mathbb{P}$ for each $B \in \mathfrak{B}$.

The density property of $f$ means $\mathbb{P}(X\in A)=\int_{A}f d\lambda=\int_{X^{-1}(A)}1 d\mathbb{P}$ for each $A\in \mathfrak{B}$. But how does it relate to these conditions a) and b)? I can't figure it out, not even to speak of making use of the special form of $g$...

Can anyone help me to solve this problem?

I would also be thankful for links to similar problems and examples, as I know that conditional expectations will be crucial in statistics. So I would like to achieve a thorough understanding of the concept.

Thanks in advance for any help whatsoever!

-
This is a wonderful exercise (and I know at least one source for it). My advice is to try very seriously to solve the case $g(x)=|x|$ (the general case being an easy consequence). –  Did Apr 25 '13 at 20:31

As explained in the comments, we solve the question when $g(x)=|x|$. Then $E[X\mid Y]=h(Y)$ for some function $h$ to be found, such that, for every bounded function $u$, $E[Xu(Y)]=E[h(Y)u(Y)]$, that is, $E[Xu(|X|)]=E[h(|X|)u(|X|)]$.
The distribution of $|X|$ has density $\bar f:y\mapsto f(y)+f(-y)$ on $y\geqslant0$ hence one asks that $$\int_\mathbb R xu(|x|)f(x)\mathrm dx=\int_0^\infty h(y)u(y)\bar f(y)\mathrm dy.$$ Note that the LHS is also $$\int_0^\infty y(f(y)-f(-y))u(y)\mathrm dy.$$ By identification, this proves that, for every $y\geqslant0$, $$h(y)=y\frac{f(y)-f(-y)}{f(y)+f(-y)},$$ hence $$E[X\mid |X|]=|X|\frac{f(|X|)-f(-|X|)}{f(|X|)+f(-|X|)}.$$ Edit: For a thorough understanding of the concept of conditional expectation, I recommend the small and excellent book Probability with martingales by David Williams.
I didn't know that criteria: $E[Xu(Y)]=E[h(Y)u(Y)]$ for every bounded function $h$. Is it equivalent to the one I used above? It's kind of what I was looking for because it allows to state condition b) in the form $E[v(X)]=E[w(X)]$ where $v$ and $w$ are functions of $X$ - unlike the definition using the indicator functions that map from $\Omega$ to $\mathbb{R}$. The one point that puzzles me is >>The distribution of $|X|$ has density $\bar f:y\mapsto f(y)+f(-y)$ on $y\geqslant0$<<. I'm still trying to see why that holds. But thanks, that answer is almost too much, but I'll study it first. –  Amarus Apr 25 '13 at 23:15
I guess the density of $|X|$ is derived like this? Didn't know that either... I wonder if I'll run into problems in applying it to the general case though. –  Amarus Apr 25 '13 at 23:42