Conditional expectation of $\sin S_n$ given $S_n^2$

Question

Consider the following exercise from Lawler's Stochastic Calculus:

Suppose $X_1 , X_2 , \dots$ are independent random variables with $$ P\{X_j = 1\} = P\{X_j = −1\} = \frac12. $$ Let $S_n = X_1 + \dots + X_n$. Find $$ E(\sin S_n | S_n^2). $$

Let's start from a few simple cases:

1. $n=1$. Then certainly $S_n^2=1$, and $$ E(\sin S_1|S_1^2) = E(\sin S_1) = \frac12 \sin(1) + \frac12 \sin(-1) = \frac12\left(\sin(1) - \sin(1)\right)=0. $$

2. $n=2$. We may have $S_2^2=0$ or $S_2^2=4$. Both cases have probability 1/2. Then, $$ E(\sin S_2|S_2^2=0) = 0 $$ because in this case $S_2=1 + (-1)=0$, and $$ \begin{align} E(\sin S_2|S_2^2=4) &= \sin(2)\cdot P\{X_1=X_2=1|S_2^2=4\} + \sin(-2)\cdot P\{X_1=X_2=-1|S_2^2=4\} \\ &=\sin(2)\left(\frac12 - \frac12\right) =0. \end{align} $$

Thus I am tempted to say that $$ E(\sin S_n|S_n^2)=0, $$ but I do not know how to prove this rigorously.

Answer 1

Hint: By flipping the sign of $X_j$, all $j=1,2,\dots,n$, we have $\mathbb{P}(S_n=s\mid S_n^2=s^2)=\mathbb{P}(S_n=-s\mid S_n=s^2)$ for all $s$, so ...