If $|X| \le 1,$ there is a binary random variable $Y \in \{-1,+1\}$ such that $E[Y|X] = X$

Question

I am stuck with Exercise 9.2.4 (i) in Achim Klenke's Probability theory:

Show the following: 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 $E[Y|X] = X.$

My thoughts so far:

• Doesn't this depend on the underlying $\sigma$-algebra? Clearly there is no $\sigma(X)$-measurable $Y$ that satisfies the equation.
• If I define $A = \{Y = +1\},$ the condition reads $X = E[1_A - 1_{A^C}|X] = 2 P(A|X) - 1,$ or equivalently, $(X+1)/2 = P(A|X)$.

Answer 1

This is possible since $[-1,1]$ is the convex hull of $\{-1,1 \}$, i.e. for $x\in[-1,1]$ there is $p_x\in[0,1]$ such that $ x = -1\cdot (1-p_x)+ 1\cdot p_x $.

So, define $Y$ to be $1$ with probability $p_X$ and $-1$ with probability $1-p_X$. Then, $E(Y|X=x)= -1\cdot (1-p_x)+ 1\cdot p_x =x$. Therefore, $E(Y|X)=X$.