Prove $X^2$ is a submartingale given X is a martingale

Question

As stated in the question, I have an exam tomorrow and am looking to prove this result.

I know:

$$E(|X_n^2|) = E(X_n^2)$$

I don't know how to advance showing that this is finite. If anyone could help that would be great. Thanks.

Edit: Is $$E(X^2)\geq(E(X))^2$$ always true?

Answer 1

You need to know a priori that $X_n \in L^2$. Then, for any convex function $f$ such that $E |f(X_n)| < \infty$ and $X_n$ is a martingale wrt $\mathcal{F}_n$ , then $f(X_n)$ is a submartingale wrt $\mathcal{F}_n$. This is just a consequence of Jensen's inequality: $E[f(X_{n+1})|\mathcal{F}_n] \geq f(E[X_{n+1}|\mathcal{F}_n]) = f(X_n)$. Take $f(x) = x^2$ in this case.

Regarding the edit, you should know for $X \in L^2$, $var (X) \geq 0$, and using the fact that $E[X^2] - (E[X])^2 = var (X)$, you should know $E[X^2] \geq (E[X])^2$. (Alternatively, Jensen's inequality with the squaring function).