Alternative Proof of Hoeffding's Lemma / solve the equation $E[1_A | X] = (1+X)/2$ given $X$

Question

In the context of proving the Hoeffding Lemma I came across a slightly weaker statement in the form of an exercise:

"If $X$ is a real valued random variable and $|X| \leq 1$ a.s. then there exists a random variable $Y$ with values in $\{ -1, +1 \}$ such that \begin{equation} E[Y|X] = X \qquad a.s. \end{equation} "

I haven't been able to prove it, and the only ansatz that I have is to try to find $A,B \in \mathcal{F}$ with $Y = 1_{A} - 1_{B} $ such that

\begin{equation} P(A|X) = E[1_A|X] = (1+X)/2 \qquad P(B|X) = E[1_B|X] = (1-X)/2 \end{equation}

However I do not know if I can find such measurable sets $A,B$ ? (Maybe some more stronger assumptions need to be imposed ? )

Answer 1

You totally have the right idea with your ansatz for $Y$ !

Indeed, when looking at the equations you obtained for $P(A\mid X)$ and $P(B\mid X)$, one can see that both expressions look suspiciously similar to the CDF of the uniform distribution. And indeed, if we let $$A :=\{U \leq \eta(X)\},\quad B :=\{U > \eta(X)\}$$

for $U\sim\text{Uniform}([-1,1])$ independent of $X$, such that $$Y = \mathbf{1}[U \leq \eta(X)] - \mathbf{1}[U > \eta(X)] = \begin{cases} 1, & U \leq \eta(X), \\ -1, & U > \eta(X), \end{cases} $$

we indeed find by construction that $$E[Y\mid X] = (1+X)/2 - (1-X)/2 = X \quad \text{a.s.} $$

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Disclaimer : this question is in fact a special case of a question I asked some time ago, and this answer is basically an adaptation of Sangchul Lee's answer to the original one.