# Conditional Expectation | Independent sigma-fields and uniqueness theorem

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 :-) )

$$^1$$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.

• $h(X)$ is not the same as its conditional expectation given $Y$. So your equation 2 is incorrect. Jul 6, 2020 at 21:23
• 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?
– user787885
Jul 6, 2020 at 21:26
• 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. Jul 6, 2020 at 21:29
• 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”? Jul 6, 2020 at 21:39
• (There isn’t much to show as equation * is essentially it already) Jul 6, 2020 at 21:48

### 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 $$E[h(X)|Y,Z]$$.

• Thank you very much! I was totally confused although it was "only" verifying the definition closely. Now, it is clear to me :-)
– user787885
Jul 7, 2020 at 14:43
• 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. Jul 7, 2020 at 14:45

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 :-)

• 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. Jul 7, 2020 at 14:05