Isn't $E(Y|X)$ a function of $\omega$ and not of $x$?

Sorry (again) I do not understand you.

actually it is a function of $X(\omega)$.It does not depend on $\omega$ in any other way.

no this time it is an error from me ;)

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 $E(Y)$.

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.

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

Ok, I found it here en.wikipedia.org/wiki/Law_of_total_expectation Did not know this formula before.

So maybe it would be better to define it like $E(Y|X(\omega)):=...$ instead of $E(Y|X)(\omega):=...$?

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.

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

$x$

I get

E(E(Y|X))=\sum_{x}P(X=x)E(Y|X=x)=\sum_{x}\sum_{y}y P(X=x)P(Y=y|X=x)$(

$E(E(Y|X))=\sum_{x}P(X=x)E(Y|X=x)=\sum_{x}\sum_{y}y P(X=x)P(Y=y|X=x)$

And I do not know how to continue now in order to reach $E(Y)$.

If I could write this as

$\sum_y y\sum_x P(X=x)P(Y=y|X=x)$ then this would be $E(Y)$. But I do not have this order of summation - but the other way round.

Okay I can write is this way... makes no difference

this is ok, you just have to apply the Fubuni theorem here.

then switch the order of summation and you are done.

Fubini theorem, which one do you mean? Is there

I only know one for integrals

it is true for any kind of summation

here, it is: if \sum_x \sum_y |f(x,y)| < \infty then \sum_y \sum_x f(x,y) = \sum_x \sum_y f(x,y)

then i have again the problem not to see why $\sum_x \sum_y \lvert y P(X=x)P(Y=y|X=x\rvert <\infty$. :-)

Because I have you here and you can help me so well... I have another question if you agree... i still did not understand why $E(Y|X)$ is a discrete RV. First of all it is $E(Y|X)\colon\Omega\to\mathbb{R}$ or?

yes. and here, Omega is discrete!

And it is measurable because it is a sum of measurable functions?

One last question for now. Why is $E(Y|X)<\infty$? I know you already wrote that it is because $E(|Y|)<\infty$. But can you explain that= Because I did not understand it yet.

Sorry, I mean why $E(Y|X=x)<\infty$.

I only see that $\lvert E(Y|X=x)\rvert\leqslant \sum_y \lvert y\rvert P(Y=y|X=x)$

If you do not mind I write down what I've got now in order to show that $E(Y|X)$ is a discrete RV.

In my opinion (although the exercise does not say it explicitly) it is meant that $X$ and $Y$ are discrete, real-valued RV, i.e. that $X,Y\colon\Omega\to\mathbb{R}$. So if this is meant, then $E(Y|X)\colon\Omega\to\mathbb{R}$.

Now it is to show that $E(Y|X)$ is measurable. I think that's because $E(X|Y)$ is a sum of measurable functions: Each summand is the product of the measurable function $\chi_{\left\{X=x\right\}}$ (which is measurable, because $\left\{X=x\right\}\in\mathcal{A}$ for each $x\in X(\Omega)$) and the constant function which value can be set as being $E(Y|X=x)$.

Moreover $E(Y|X)$ takes $\text{card}(X(\Omega))$ values, because the preimages $X^{-1}(\left\{x\right\})$ are a partition of $\Omega$ and so for each $\omega\in\Omega$ there is exactly one $x\in X(\Omega)$ so that $E(Y|X)(\omega)=E(Y|X=x)$. So $E(Y|X)(\Omega)$ is countable, i.e. $E(Y|X)$ is discrete.

That's my argumentation.

But where do I need here that X and Y are integrable?

Because if $Y$ is real valued then it is automatically $E(Y|X=x)=\sum_y yP(Y=y|X=x)<\infty$, isn't it?