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I have problems in understanding a small part in a proof, which is, however, a really important part.

Given:

  • $X,Y,Z$ are random variables such that $\sigma(X,Y)$ and $\sigma(Z)$ are independent
  • $h: \mathbb{R} \rightarrow \mathbb{R}^+$ is a Borel function which we assume to be bounded
  • (*) By the uniqueness theorem of probability measures, we know that for all $A \in \mathcal{B}(\mathbb{R})$, $\mathbb{E}[h(X)\mathbb{I}_{\{(Y,Z)\in A\}}] = \mathbb{E}[\mathbb{E}[h(X)\mid Y]\mathbb{I}_{\{(Y,Z)\in A\}}] $ holds

What the proof states (not relevant for the question, only if someone is interested :-) ):

$\mathbb{P}$ a. s. $\mathbb{E}[h(X)\mid\sigma(Y,Z)] = \mathbb{E}[h(X)\mid Y]$

My question:

After (*), it is stated that by the uniqueness of the conditional expectation$^1$, $$\mathbb{E}[h(X)\mid\sigma(Y,Z)] = \mathbb{E}[h(X)\mid Y] \hspace{2cm} \tag 1$$ is implied $\mathbb{P}$ a.s.. I don't understand where the equation (1) comes from...

My attempt of explanation:

If I look at (*) and consider the uniqueness theorem of conditional expectation, I would say that $$\mathbb{E}[h(X)\mid Y] \mbox{ is a version of the conditional expectation of } h(X) \hspace{2cm} \tag 2$$ $\mathbb{P}$ a.s.. Now, I try to come from $(2)$ to $(1).$ But that does not really work. So, I think I miss something?

A further question (not so important)

Does someone may have an intuitive explanation about why $h$ maps to $\mathbb{R}^+$ and not $\mathbb{R}$? I don't see why this general case should not hold. (But this is not my main worry :-) )

Thanks a million in advance for your help!

$^1$The uniqueness theorem of conditional expectations sates that if (*) holds for two random variables $X_0$ and $\tilde X_0$, then $X_0$ = $\tilde X_0$ $\mathbb{P}$ a.s.

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  • $\begingroup$ $h(X)$ is not the same as its conditional expectation given $Y$. So your equation 2 is incorrect. $\endgroup$
    – Michael
    Commented Jul 6, 2020 at 21:23
  • $\begingroup$ I thought that this is implied by the uniqueness theorem? If (*) holds for two random variables (which I considered as h(X) and the conditional expectation), then those are the same? $\endgroup$
    – user787885
    Commented Jul 6, 2020 at 21:26
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    $\begingroup$ Let $h(X)=X$ and let $Y=Z=0$. Then you are claiming $X$ is almost surely equal to its expectation, which cannot be true in general. $\endgroup$
    – Michael
    Commented Jul 6, 2020 at 21:29
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    $\begingroup$ You can use the required properties of a conditional expectation $E[h(X)|Y,Z]$, show that $E[h(X)|Y]$ satisfies it. By the way , not that it seems to matter here, but what precisely do you mean by “the uniqueness theorem”? $\endgroup$
    – Michael
    Commented Jul 6, 2020 at 21:39
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    $\begingroup$ (There isn’t much to show as equation * is essentially it already) $\endgroup$
    – Michael
    Commented Jul 6, 2020 at 21:48

2 Answers 2

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Definitions

  • A version of $E[h(X)|Y]$ is a random variable of the form $g(Y)$ (for some function $g$) that satisfies $$ E[h(X)1_{\{Y \in A\}}] = E[g(Y)1_{\{Y \in A\}}] \quad \forall A \in B(\mathbb{R})$$

  • A version of $E[h(X)|Y, Z]$ is a random variable of the form $r(Y,Z)$ (for some function $r$) that satisfies $$ E[h(X)1_{\{(Y,Z) \in B\}}] = E[r(Y,Z)1_{\{(Y,Z) \in B\}}] \quad \forall B \in B(\mathbb{R}^2)$$

The given problem

Now I think your problem is this: You are told $E[h(X)|Y]$ is a particular version of the conditional expectation of $h(X)$ given $Y$ (so you can call it $g(Y)$ for some function $g$ if you want, that is, $g(Y)=E[h(X)|Y]$). You are also told $E[h(X)|Y]$ satisfies an additional property called Property *:

\begin{align} &\mbox{Property *}: \\ &E[h(X)1_{\{(Y,Z) \in B\}}] = E\left[\underbrace{E[h(X)|Y]}_{g(Y)}1_{\{(Y,Z) \in B\}}\right] \quad \forall B \in B(\mathbb{R}^2) \end{align}

Finally, from this property, you are being asked to verify that your random variable $g(Y)$ is also a version of $E[h(X)|Y,Z]$.

Solution

What do we need to show to conclude that $g(Y)$ is a version of $E[h(X)|Y,Z]$? Well, we would need to show $g(Y)$ is a pure function of $(Y,Z)$ (which it is, since it is in fact a pure function of $Y$ alone) and $$ E[h(X)1_{\{(Y,Z) \in B\}}] = E[g(Y)1_{\{(Y,Z)\in B\}}] \quad \forall B \in B(\mathbb{R}^2)$$ But the given Property * itself is the same as this equation that we need to show. So this problem really just asks to verify that property * immediately implies that $g(Y)$ is a version of $E[h(X)|Y,Z]$.

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  • $\begingroup$ Thank you very much! I was totally confused although it was "only" verifying the definition closely. Now, it is clear to me :-) $\endgroup$
    – user787885
    Commented Jul 7, 2020 at 14:43
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    $\begingroup$ You are welcome. Note that I did not say "by the uniqueness theorem" anywhere because we never needed it. We were only being asked to verify that a given function satisfied certain requirements. $\endgroup$
    – Michael
    Commented Jul 7, 2020 at 14:45
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I think I have figured it out. Thanks for the thought provoking impulses @Michael.

So, as mentioned in the question, we know that $\mathbb{E}[h(X)\mid Y]$ is a version of the conditional expectation of $h(X) $.

However, we also know that for all

$$A \in \mathcal{B}(\mathbb{R}), \quad\mathbb{E}[h(X)\mathbb{I}_{\{(Y,Z)\in A\}}] = \mathbb{E}[\mathbb{E}[h(X)\mid \sigma(Y,Z)]\mathbb{I}_{\{(Y,Z)\in A\}}] $$

simply by definition of the conditional expectation.

Now, the uniqueness theorem for conditional expectation yields that

$\mathbb{P}$ a. s. $\mathbb{E}[h(X)\mid\sigma(Y,Z)] = \mathbb{E}[h(X)\mid Y]$

since the conditional expectation of $h(X)$ is unique.

Would be great if one could leave a short comment if I did not miss anything in the answer :-)

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  • $\begingroup$ I do not follow. You start with $E[h(X)|Y]$, then you state a fact for $E[h(X)|Y, Z]$. At no point do you mention the equation (*) from your question, which I thought you were supposed to use. I do not know where the "Now the uniqueness theorem" comes in to your argument. I think you are saying "by the uniqueness theorem" too much. $\endgroup$
    – Michael
    Commented Jul 7, 2020 at 14:05

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