I am trying to solve the following exercise on martingale: Let $X_n$ be the recursive process defined as follows: $X_0 \sim \mathcal{U}([-1,+1]), X_n = \sim \mathcal{U}([-|X_{n-1}|,+|X_{n-1}|])$
is the process a martingale? Honestly I think it is a martingale, but how do I compute $E(X_{n+1}|\mathcal{F}_n) = E[\mathcal{U}([-|X_{n}|,+|X_{n}|])|\mathcal{F}_n]$?
$X_{n+1}$ should take any value in the interval $[-|X_n|,+|X_n|]$ with probability $\sim \frac{1}{2|X_n|}$ , but I wouldn't Know how to compute the conditional expectation in this case, what is the proper way to formalize this?
$\newcommand{\E}{\mathbb E}\newcommand{\F}{\mathcal F}$It is not a martingale. We prove it by contradiction, asssume it is a martingale. I assume $\F_n = \sigma(X_1,...,X_n)$. Then one has: \begin{align*}\tag{1} \E[X_{n+1} \mid \F_n ] =X_n \end{align*} Notice that $\sigma(X_n)\subset \F_n$, so: \begin{align} \E[\E[X_{n+1}\mid \F_n ] \mid X_n]= \E[X_{n+1}\mid X_n]=0 \end{align} Since the "smaller $\sigma$-algebra wins". Moreover: $$\E[X_n\mid X_n ] =X_n$$ Now condition on $\sigma(X_n)$ on both sides of $(1)$ to get: \begin{align} 0 = X_n \ \ \text{ a.s.} \end{align} This is clearly wrong, hence a contradiction.