Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $E[X|Y=y]=y$, does $E[X]=E[Y]$? Similarly, if $E[X|Y=y]=y^2$, does $E[X]=E[Y^2]$?

I'm having some trouble with this conditional expectation concept, although it seems intuitively true

share|cite|improve this question
The notation $E[X\mid Y=y]=y$ means that the conditional expectation of $X$ with respect to $Y$ is $Y$. Since $\Omega$ in $\sigma(Y)$, we have $E[X]=\int_{\Omega}E[X\mid Y]dP=\int_{\Omega}YdP=E[Y]$, if these two random variables are integrable. – Davide Giraudo Dec 18 '11 at 17:47
Thank you Davide! Does that also hold for the Y^2 case? – DumbQuestion Dec 18 '11 at 17:58
$E[X]=\int_{\Omega}E[X\mid Y]dP=\int_{\Omega}Y^2dP=E[Y^2]$, so these results are true if $X$ is integrable and $Y$ square integrable. – Davide Giraudo Dec 18 '11 at 18:02

We assume that $X$ is integrable and $Y$ is square integrable. The notation $E[X\mid Y=y]=g(y)$ means that $E[X\mid Y]=g(Y)$ (Doob's theorem ensures us it's possible, since $E[X\mid Y]$ is $\sigma(Y)$-measurable.

If $E[X\mid Y=y]=g(y)$, since $\Omega\in\sigma(Y)$, we have $$E[X]=\int_{\Omega}E[X\mid Y]\mathrm{d}\mathbb P=\int_{\Omega}g(Y)\mathrm{d}\mathbb P=E[g(Y)],$$ and applying it to $g(x)=x$ and $g(x)=x^2$, we get your results.

share|cite|improve this answer

Yes, you’re right. Davide gave you a precise and quick answer, I’ll try to elaborate a little bit (edit: when I started, Davide just wrote a comment, his answer was written a moment later).

First, to better understand the conditional expectation concept, you may think to this situation as follows : we first draw the value y of Y, following the law of Y, and then the value x of X, using a law for X which depends on which y was drawn.

I will use discrete notations but for the general case you can just replace everything by fancy integrals. The law of X, given Y = y, the law used to draw a value of X knowing that Y = y, is denoted $P(X = x\ |\ Y = y)$. The conditional expectation $E(X \ |\ Y=y)$ is of course $\sum_x P(X = x\ |\ Y = y)$.

Now turn to the law of X: $P(X = x) = \sum_y P(X=x\ |\ Y=y) P(Y=y)$ (this is simply the formula of total probability). The expectation of X is then $$E(X) = \sum_x x P(X = x) = \sum_x x \left(\sum_y P(X = x\ |\ Y=y) P(Y=y)\right)$$ $${} = \sum_y \left( \sum_x x P(X=x\ |\ Y= y) \right) P(Y=y) = \sum_y E(X\ |\ Y =y) P(Y=y).$$ I hope that this makes the sense of this formula clear.

Know denote $E(X | Y=y) = f(y)$. Then $E(X) =\sum_y f(y) P(Y = y) = E\bigl( f(Y) \bigr)$, which answers your question.


share|cite|improve this answer
Typo? You wrote The conditional expectation $E(X | Y=y)$ is of course $\sum_x P(X=x | Y=y)$. You must have meant $\sum_x xP(X=x | Y=y)$. – Michael Hardy Dec 18 '11 at 19:10
But then you repeated it: $E(X)=\sum_xP(X=x)$. That's wrong. It should say $E(X)=\sum_x xP(X=x)$. – Michael Hardy Dec 18 '11 at 19:11
I've downvoted for now; I'll rescind that if this gets fixed. – Michael Hardy Dec 18 '11 at 19:12
Oups, of course it’s a typo – I wrote too fast. It’s fixed. – Elvis Dec 18 '11 at 20:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.