Martingale whose distribution is independent of n

Question

Claim: A one-point stationary sequence $(X_n)_{n\in\mathbb{N}}$ of square integrable r.v's is a martingale w.r.t the natural filtration iff it is surely constant in time.

I am convinced that this is a false statement, and therefore I'd like to provide a counterexample. As a simple counterexample I wanted to take a sequence of $i.i.d. ~ Bernoulli(p)$.
Then $E[X_{n+1} | \mathcal{F}_n$] = $E[X_{n+1}] = E[X_n] = p$, independent of $n$, and therefore one-point stationary. On the other hand, $P(X_{n+1} = X_{n}) \neq 1$.

As I'm not very familiar yet with martingales, I am wondering if the sequence of Bernoulli i.i.d. r.v's is indeed a martingale, and if it suffices as a counterexample to disprove the claim.

EDIT: $E[X_{n+1} | \mathcal{F}_n] = E[X_n] = p \neq X_n$. Therefore, this sequence is not a martingale. What is an appropriate counterexample?

Answer 1

$E(X_{n+1}|\mathcal F_n) =X_n$. So $X_n^{2}=(E(X_{n+1}|\mathcal F_n))^{2} \leq E(X^{2}_{n+1}|\mathcal F_n)$. Since both sides have the same expecatation (by stationarity) we get $X_n^{2}= E(X^{2}_{n+1}|\mathcal F_n)$. Now $E((X_{n+1}-X_n)^{2} |\mathcal F_n)=E(X^{2}_{n+1} |\mathcal F_n)-2X_nE(X_{n+1} |\mathcal F_n)+X_n^{2}=0$ and taking expectation we get $X_{n+1}=X_n$ a.s.

Answer 2

If they are one-point stationary, there are a lot of things we can do with them to show there invariance.
In this case, you have further that they're square-integrable, which makes thing even easier.
Just observe that:
$E[X_{n+1}^2]= E[X_n^2]+E[ (X_n-X_{n+1})^2]$
Done

Answer 3

You have $$ \eqalign{ E[(X_{n+1}-X_n)^2] &= E[X_{n+1}^2]+E[X_n^2]-2E[X_{n+1}X_n]\cr &=2E[X_n^2]-2E[X_{n+1}X_n]\cr &=2E[X_n^2]-2E[X_nX_n]=0,\cr } $$ the second equality because of 1-point stationarity and the third because $X_n$ is a martingale.