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I was recently asked a question about Exercise 9.2.4(i) in Klenke's Probability theory. First I state the exercise itself.

Let $X$ be a real-valued random variable (on $(\Omega,\mathcal F,\mathbb P)$) with $|X|\leq 1$ almost surely. Then there exists a random variable $Y:\Omega\rightarrow \{-1,1\}$ such that $E[Y|X]=X$.

Now the question is regarding the following counter-example. Consider two coin tosses, which leads to the event space $$\Omega=\{HH,HT,TH,TT\},$$ and the random variable $X:\Omega\rightarrow\mathbb R$ with $$X(HH)=-0.5, \, X(HT)=X(TH)=0, \, X(TT)=0.5.$$ Using the definition of conditional expectation $E[Y|X]$ is a random variable defined w.r.t. $\sigma(X)$ and satisfies
$$ E[Y|X](TT)\, \mathbb P(\{TT\}) = \int_{\{TT\}} E[Y|X](\omega)d\mathbb P(\omega) = \int_{\{TT\}} Y (\omega) d\mathbb P(\omega) = Y(TT)\mathbb P(\{TT\}), $$ and therefore $E[Y|X]=X$ would mean that $$0.5=X(TT)= E[Y|X](TT)=Y(TT),$$ which means that $Y$ has to have values other than $\{-1,1\}$. But this contradicts the exercise which states that $Y$ should only have values in $\{-1,1\}$.

Question: What is the issue with the argument in this counterexample.

Its possible that I missing something simple with conditional expectations. Also any hints on the original question in Klenke's book.

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  • $\begingroup$ I checked all 3 editions of Klenke's "Probability Theory: A Comprehensive Course" and Exercise 9.2.4 is something different. I also could not find this in Chapter 8 (Conditional Expectations). Where did you find this exercise? $\endgroup$
    – angryavian
    Jun 3, 2021 at 16:41
  • $\begingroup$ Its the first question in the exercise titled 'Azuma's inequality' at the end of Section 9.2 called Martingales. Its possible the pdf I have is messed up. $\endgroup$
    – UPS
    Jun 3, 2021 at 18:19
  • $\begingroup$ Thanks, I've updated my answer. You modified the statement of the exercise in a subtle way that makes the claim false. The original exercise is true. $\endgroup$
    – angryavian
    Jun 3, 2021 at 18:28

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I'm not sure where the original exercise comes from, but it seems to be stated incorrectly for a minor technical reason.

I think the intended idea is to define a random variable $Y$ such that $P(Y=1 \mid X=x) = \frac{1+x}{2}$ and $P(Y=-1 \mid X=x) = \frac{1-x}{2}$ so that $E[Y \mid X=x] = \frac{1+x}{2} \cdot 1 + \frac{1-x}{2} \cdot (-1) = x$.

But if $\sigma(X)=\mathcal{F}$ like in your counterexample, there may not be enough "room" in the probability space $(\Omega, \mathcal{F})$ to define this new random variable $Y$ on the same probability space.

Edit: The original problem statement did not explicitly specify probability spaces $(\Omega, \mathcal{F})$ so the above solution will work.

Here is the actual statement of the exercise:

If $X$ is a random variable with $|X| \le 1$ a.s., then there is a random variable $Y$ with values in $\{-1, 1\}$ and with $E[Y \mid X] = X$.


The error in your counterexample is $E[Y \mid X](TT) = Y(TT)$, which I think comes from your imposition that $Y$ must be measurable with respect to $\sigma(X)$. This is not required by the original problem statement. You should not have modified the statement of the exercise with explicit mentions of the $(\Omega, \mathcal{F})$; as your attempted counterexample shows, $Y$ generally should not be $\sigma(X)$-measurable for the exercise's claim to hold. In your counterexample, you would need to "augment" the probability space $\Omega$ in order to specify $Y$ (e.g., $\Omega' = \Omega \times \{-1, 1\}$).

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    $\begingroup$ Thanks for your answer. Don't we have 𝐸[π‘Œβˆ£π‘‹](𝑇𝑇)=π‘Œ(𝑇𝑇) by the definition of conditional expectation where TT belongs to the sigma algebra generated by X? The equality ∫𝐸[π‘Œ|𝑋](πœ”)𝑑ℙ(πœ”)=βˆ«π‘Œ(πœ”)𝑑ℙ(πœ”) holds for any set in sigma algebra of X and we make the specific choice with the set {TT}. Sorry if this is a naive question. $\endgroup$
    – UPS
    Jun 3, 2021 at 18:59
  • $\begingroup$ @UPS This is true if you force $Y$ to be measurable on $(\Omega, \mathcal{F})$. In essence, you are forcing $Y$ to be $\sigma(X)$-measurable, which leads to the well-known property $E[Y \mid \mathcal{A}]=Y$ when $Y$ is $\mathcal{A}$-measurable. So your counterexample correctly shows that proving the exercise's claim is impossible if $Y$ is forced to be $\sigma(X)$-measurable. But the exercise does not require this. $\endgroup$
    – angryavian
    Jun 3, 2021 at 20:08
  • $\begingroup$ @UPS In your particular example, you can redefine the same random variable $X$ on a slightly larger probability space $\Omega \times \{-1, 1\}$ (e.g., with $X((HH, -1)) = -0.5$) and still have the same distribution. You can then define another random variable $Y$ with $Y((t, s)) = s$. In this case, the integral you were considering $\int Y(\omega) dP(\omega)$ will be over two atoms $(TT, -1)$ and $(TT, 1)$. $\endgroup$
    – angryavian
    Jun 3, 2021 at 20:08
  • $\begingroup$ Regarding your second comment above, if you extend the state space as you suggest how exactly do you define X? Following your suggestion do we define it as X((HH,-1))=-0.5=X((HH,-1)) (and similarly for other events)? I mean we need to prescribe X for all values in the even space. Also why do you say that the distribution will be the same? Could you please elaborate on this comment in the solution? I feel its very insightful, although I don't understand it yet. $\endgroup$
    – UPS
    Jun 4, 2021 at 9:07
  • $\begingroup$ @UPS Yes, $X(((HH, -1))=X((HH, 1))=-0.5$ and so on. You then need to assign probabilities for each outcome in the event space. You can do so such that (a) the distribution of $X$ is the same (specifically, $P(X=-0.5)=1/4$, $P(X=0.5)=1/4$, and $P(X=0)=1/2$), and (b) such that $P(Y=1 \mid X=x)=\frac{1+x}{2}$. $\endgroup$
    – angryavian
    Jun 4, 2021 at 15:20

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