4

Assume we have a martingale $((X_n)_{n\ge 0}, (\mathcal{F}_n)_{n\ge 0})$, for which each random variable $X_n$ is bounded (so there is no problem with integrability) and not constant. Is it true that $X_n^2$ cannot be a martingale?

My idea is that by Jensen inequality, if $(X_n)_{n\ge 0}$ is a martingale, then $(X_n^2)_{n\ge 0}$ is a submartingale. If $(X_n^2)_{n\ge 0}$ was a martingale, we would have an equality in Jensen's inequality. However this only occurs if we apply Jensen inequality to affine function (which is not the case here, since we take $f(x) = x^2$) or if the random variable in Jensen's inequality is constant almost surely, which is not a case here due to the assumption.

Is my reasoning correct?

Barabara
  • 690

1 Answers1

2

Your argument, as Kurt pointed out, is correct.

An other way to see this is to define $D_n=X_{n}-X_{n-1}$ and $D_0=0$. Then $$ \mathbb E\left[X_n^2\mid\mathcal F_{n-1}\right]=\mathbb E\left[D_n^2\mid\mathcal F_{n-1}\right]+X_{n-1}^2 $$ hence $X_n^2$ will be a martingale if and only if $\mathbb E\left[D_n^2\mid\mathcal F_{n-1}\right]=0$. Taking $\Omega$ in the definition of conditional expectation gives that $D_n=0$ almost surely hence the only possibility is a constant martingale.

Davide Giraudo
  • 172,925