Finite second moment implies convergence

Question

Let $X$ be a martingale such that $\sup _{n>0} \mathbb{E} [X^2] < \infty$. Show that $X_n$ converges to a random variable in probability and also in $L^2$.

I've played around with trying to get convergence in probability but Im not really sure why it would converge or what to. Any help would be much appreciated.

Answer 1

Hints:

1. $L^2$-convergence implies convergence in probability, and therefore it enough to show that $X_n$ converges in $L^2$.
2. $L^2$ is a complete space and therefore it suffices to prove that $(X_n)_{n \in \mathbb{N}}$ is an $L^2$-Cauchy sequence, i.e. $$\|X_n-X_m\|_{L^2}^2 = \mathbb{E}((X_n-X_m)^2) \to 0 \quad \text{as $n,m \to \infty$}.\tag{1}$$
3. Check (or recall) that $X^2$ is a submartingale, and conclude that $$M := \sup_{n \geq 1} \|X_n\|_{L^2}^2 = \lim_{n \to \infty} \mathbb{E}(X_n^2).$$ In particular, $(\mathbb{E}(X_n^2))_{n \geq 1}$ is a Cauchy sequence.
4. Use the martingale property to show that $$\mathbb{E}((X_n-X_m)^2) = \mathbb{E}(X_n^2)-\mathbb{E}(X_m^2) \quad \text{for all $m \leq n$}.$$ Hint: Write $$\mathbb{E}((X_n-X_m)^2) = \mathbb{E} \bigg[ \mathbb{E}((X_n-X_m)^2 \mid \mathcal{F}_m) \bigg]$$ and expand the square.
5. Conclude from step 3 and 4 that $(X_n)_{n \in \mathbb{N}}$ is an $L^2$-Cauchy sequence.