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Let $X$ be in integrable, with density $f$ with respect to the Lebesgue measure. Compute the conditional expectation : $ \operatorname{E} \left[ X\, \Big|\, |X| \,\right] $

My ansatz was : $ \operatorname{E} \left[ X\, \Big|\, |X| \,\right] = \operatorname{E} \left[ \operatorname{sgn}(X)\cdot |X| \, \Big|\, |X| \,\right] = |X| \operatorname{E} \left[ \operatorname{sgn}(X) \, \Big|\, |X| \,\right] $

At this point I am stuck since one cannot assume in general that $\operatorname{sgn}(X)$ is independent of $|X|$ nor measurable.

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    The set $\{0\}$ has Lebesgue measure $0$, so you treat this as if $X:\Omega\to\mathbb R \setminus \{0\},$ and then $x\mapsto\operatorname{sgn}(x)$ is an everywhere continuous function, so for $\omega\in\Omega,$ the mapping $\omega \mapsto X(\omega) \mapsto \operatorname{sgn}(X(\omega))$ is measurable. $\qquad$ – 2017-01-21
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    Thanks for your answer Michael, but i do not fully understand how do you get from the continouity of $x \rightarrow sgn(x)$ to the measurability w.r.t $\mathcal{\sigma}(|X|)$ – 2017-01-21
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    So what you meant was you don't know if $\operatorname{sgn}(X)$ is measurable with respect to $\sigma(|X|)\text{?}\qquad$ – 2017-01-21
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    If $\operatorname{sgn}(X)$ were measurable with respect to $\sigma(|X|)$ then the conditional expected value of $\operatorname{sgn}(X)$ given $|X|$ would just be $\operatorname{sgn}(X).$ But maybe I haven't yet understood your question. $\qquad$ – 2017-01-21
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    Dear Michael, I thought you claim that $sgn(X)$ is measurable w.r.t $\sigma(|X|)$ and i didn't understand why. My original question is to compute the conditional expectation: $\mathbb{E}[X | |X|]$ – 2017-01-21
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    For example if X is uniform on [-1,1] then I calculated this cond. expectation as follows : $\mathbb{E}[X| |X|]=|X|\cdot\mathbb{E}[sgn(X)| |X| ]=|X|\cdot\mathbb{E}[sgn(X)|]=|X|\cdot \frac{1}{2}$. Where I used that sgn(X) is indipendent of |X| because of the symmetry of the distribution around 0. Now I wanted to know if this cond. expectation can be also simplified in the case of a general $X\sim f$. In this case I can't use the independence of sgn(X) and |X|. However my ansatz might be wrong. Hopefully you understand now my question. – 2017-01-21

1 Answers 1

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Disclaimer: This answer is currently downvoted. Needless to say, it is perfectly correct, and it answers the question, apparently to the OP's satisfaction. The downvote might be due to extra-mathematical reasons not worthy of being further commented upon. Happy reading!

Let $Y=|X|$. When $X$ has a PDF $f$, you might convince yourself that $$E(X\mid Y)=g(Y)$$ where $g(y)$ is defined on $y\geqslant0$ by

$$g(y)=y\cdot\frac{f(y)-f(-y)}{f(y)+f(-y)}$$

In the general case, some more measure theoretical machinery is needed. Consider the measures defined by $$\mu_+(B)=P(X\in B)\quad \mu_-(B)=P(-X\in B)\quad \mu(B)=P(Y\in B)$$ for every Borel set $B\subset\mathbb R_+$. Then $\mu_\pm\leqslant\mu$ hence $\mu_\pm\ll\mu$ hence there exists some measurable functions $f_\pm$ such that $$\mu_\pm(B)=\int_Bf_\pm d\mu$$ for every Borel set $B$. Then $E(X\mid Y)=g(Y)$ where $g(y)$ is defined on $y\geqslant0$ by

$$g(y)=y\cdot\frac{f_+(y)-f_-(-y)}{f_+(y)+f_-(-y)}$$

Edit 1: This conditional expectation reflects the fact that, for every $y>0$, the conditional distribution of $X$ conditionally on $Y=y$ is given by $$P(X=y\mid Y=y)=\frac{f_+(y)}{f_+(y)+f_-(-y)}$$ and $$P(X=-y\mid Y=y)=\frac{f_-(y)}{f_+(y)+f_-(-y)}$$ which can be summarized as $$P(X=Y\mid Y)=\frac{f_+(Y)}{f_+(Y)+f_-(-Y)}=1-P(X=-Y\mid Y)$$ Edit 2: As often on this site, one meets a specific problem when answering questions involving conditional expectations, which is that every nontrivial one requires a solid definition of the concept. To understand why the OPs interested in such questions regularly lack any such definition would require a separate analysis, hence, for the time being, we will concentrate on adding some sketchy explanations on this definition, referring the interested reader to some accessible and rigorous source (such as the small blue textbook Probability with martingales, by David Williams).

Thus, in full generality, one considers random variables $X$ and $Y$ defined on the same probability space with $X$ integrable, then $E(X\mid Y)$ is defined as the (class of) random variable(s) $u(Y)$, for some measurable function $u$ such that $u(Y)$ is integrable and, for every bounded measurable function $w$, $$E(Xw(Y))=E(u(Y)w(Y))$$ Equivalently, one can ask that $u(Y)$ is integrable and that, for every event $A$ in $\sigma(Y)$, $$E(X\mathbf 1_A)=E(u(Y)\mathbf 1_A) $$ In the present case, to prove the formulas proposed in this post when $X$ has PDF $f$, one should find $g$ such that, for every bounded measurable function $w$, $$E(Xw(Y))=E(g(Y)w(Y))$$ that is, since $Y=|X|$ and $X$ has PDF $f$, $$\int xw(|x|)f(x)dx=\int g(|x|)w(|x|)f(x)dx$$ which is equivalent to $$\int_{y>0}yw(y)(f(y)-f(-y))dy=\int_{y>0}g(y)w(y)(f(y)+f(-y))dy$$ This identity holds for every function $w$ if and only if $$y\cdot(f(y)-f(-y))=g(y)\cdot(f(y)+f(-y))$$ almost everywhere for $y>0$, and $g$ follows.

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    First of all thank you Did for your detailed explanation. I have just one more question. For the part $E(X|Y)=g(Y)$: I need the conditional distribution $ X | Y=y = \begin{cases} y, & \text{with probability p}\\ -y ,& \text{with probability 1-p} \end{cases}$. As I can conclude after seeing your result you claim that $p=\frac{f(y)}{f(y)+f(-y)}$. How did you got that ? – 2017-01-22
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    I know that the density of Y is $f_{y}(y)=f(y)+f(-y) \quad ,y>0$. I tried $\mathbb{P}[X=y | Y=y ]=\frac{\mathbb{P}[X=y,Y=y]}{P[Y=y]}=\frac{\mathbb{P}[X=y]}{P[Y=y]}=\frac{f(y)}{f(y)+f(-y)}$. But I do not fully understand why the last equality should hold? For example if X has a cont. distribution. Thank you in advance for your support. – 2017-01-22
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    Of course, when $X$ has a PDF, $P(Y=y)=0$ for every $y$ hence none of the formulas in your comment makes any sense. I added a brief reminder of the definition of conditional expectations in the general case. – 2017-01-23
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    Thank you for the solution Did ! – 2017-01-23
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    @Did Let $X$ be standard normal, then $f_x(y)-f_x(-y)=0$ and hence $E(X | |X|)=0$, this looks like an interesting result and a bit funny ... can it actually be? – 2017-01-23
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    @Math-fun Yes, it is quite natural if you think about it... and it holds for every symmetric random variable. – 2017-01-23
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    @Did I think your result is correct for more general case. Let $\mu_X,\mu_{-X}$ be the distributions of $X, -X$ on $\mathbb{R}$, and $f_+=\frac{d\mu_X}{d(\mu_X+\mu_{-X})}$ be the Radon-Nikodym derivative, then $E(X| |X|)=|X|(2f_+(|X|)-1)$. – 2017-01-24
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    @JGWang The "more general case" you describe is exactly the case solved in my answer, starting at "In the general case, some more measure theoretical machinery is needed". Not sure I am following your point... – 2017-01-24
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    @Did：“more general case": The random variable X is discrete distributed, the distribution function of X is absolutely continuous and other cases. – 2017-01-24
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    @JGWang All dealt with in my answer (did you read it?). – 2017-01-24
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    @Did: In your answer, I read the following sentence: "to prove the formulas proposed in this post when $X$ has PDF $f$''. Now, if the $X$ doesn't have a PDF, is the conclusion still effective. In my comment, I just say that, "$E(X| |X|)=|X|(2f_+(|X|)-1)$, $f_+=\frac{d\mu_X}{d(\mu_X+\mu_{-X})}$. It is effective in any case, even $X$ doesn't have a PDF. – 2017-01-25
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    @JGWang Yeah, my answer explicitely states that the conclusion holds in the general case, in the general formulation given using $f_\pm$. But perhaps what you mean is that the **proof** (not only the formula) in this general case is similar to the proof I explain in the restricted case where $X$ has a PDF $f$? Then you are right. – 2017-01-25
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    @Did Thanks for your answer. – 2017-01-26