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$X,Y$ discrete, integrable Random variables on $(\Omega,A,P)$.

Show that $E(E(Y|X))=E(Y)$.

First of all in another task it was to show that $E(Y|X)$ is a discrete Random variable. So it is $$ E(E(Y|X))=\sum_{z}zP(E(Y|X)=z), $$ Moreover it is $$ E(Y|X)(\omega):=\sum_x E(Y|X=x)1_{X=x}(\omega) $$ and $$ E(Y|X=x):=\sum_y y P(Y=y|X=x). $$

So putting this all together I get $$ E(E(Y|X))=\sum_z zP\left(\sum_x\sum_y yP(Y=y|X=x)1_{X=x}=z\right) $$

Oh my god. What can I do now to get the result?

mathfemi
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1 Answers1

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A general thing to remember is not to introduce $\omega$ unless you absolutely have to.

Just stay with $$ E(Y|X=x)=\sum_y y P(Y=y|X=x). $$ Use $$ P(Y=y|X=x)P(X=x) = {P(Y=y,X=x)}$$ Now remember that $E(Y|X)$ is a function of $X(\omega)$, so $$ E[E(Y|X)] = \sum P(X=x) E(Y|X=x) $$ and you will be fine.

mookid
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  • Isn't $E(Y|X)$ a function of $\omega$ and not of $x$? – mathfemi Oct 18 '14 at 12:52
  • Sorry (again) I do not understand you. – mathfemi Oct 18 '14 at 12:57
  • actually it is a function of $X(\omega)$.It does not depend on $\omega$ in any other way. – mookid Oct 18 '14 at 12:58
  • no this time it is an error from me ;) – mookid Oct 18 '14 at 12:58
  • I do understand the first line and the second line. But not why $E(Y|X)$ is a function of $X(\omega)$ because in the task it is defined $E(Y|X)(\omega):=...$ so it is a function of $\omega$. And even if it is indeed a function of $X(\omega)$ then I cannot see how you get the last formula and why this is $E(Y)$. – mathfemi Oct 18 '14 at 13:00
  • this is a function of $\omega$, of course: this is a random variable. But it does depend on $\omega$ only through $X(\omega)$: this is a deterministic function of $X(\omega)$. That is why the 3rd line. – mookid Oct 18 '14 at 13:30
  • I see now, thank you. Noe the remaining question is why the 3rd line is equal to E(Y). I thought og total probability but cannot finish it – mathfemi Oct 18 '14 at 13:31
  • Ok, I found it here http://en.wikipedia.org/wiki/Law_of_total_expectation Did not know this formula before. – mathfemi Oct 18 '14 at 13:33
  • So maybe it would be better to define it like $E(Y|X(\omega)):=...$ instead of $E(Y|X)(\omega):=...$? – mathfemi Oct 18 '14 at 13:34
  • No, because it would mean that $E[Y|.]$ is a function, which does not make sense. Sometimes, this function is noted $E[Y|X = X(\omega)]$. But I don't like this one. – mookid Oct 18 '14 at 13:38
  • Hm do not see clearly yet... but ok. And is the sum in the third line over x or y? I think x. But then its different than in the lnk – mathfemi Oct 18 '14 at 13:49
  • Let us continue this discussion in chat. – mathfemi Oct 18 '14 at 13:52