Showing $E(X_T ^2)\le E([X]_T)$ for bounded stopping times $T$

Question

$X=(X_t)$ is a continuous local martingale (that is also adapted with continuous sample paths) and $X_0=0$. Let $[X]_t$ denote the quadratic variation of $X$ over $[0,t]$. How is $E(X_T^2)\le E([X]_T)$ for every bounded stopping time $T$?

Given $T$, I can prove that $X_T^2-[X]_T$ is a continuous local martingale, and that it is a uniformly integrable if $X$ is unformly bounded in $L^2$. If I was given that $X_t$ was in $L^2$ or bounded by an $L^2$ random variable, I could apply Jensen's inequality to say $E(X_T ^2)\le E(X_0 ^2)=0\le E([X]_T)$, but how can I show this without using this additional assumption?

Answer 1

Hints: Let $T$ be a bounded stopping time.

1. Fix a localizing sequence $(\sigma_n)_{n \in \mathbb{N}}$ of stopping times. Sinc $(X_t)_{t \geq 0}$ has continuous sample paths, we can choose the sequence in such a way that $|X_{t \wedge \sigma_n}| \leq n$ for all $t \geq 0$. Show (or recall) that $$M_t := X_{t \wedge \sigma_n \wedge T}^2 - [X]_{t \wedge \sigma_n \wedge T}$$ is a martingale. In particular, $$\mathbb{E}(X_{t \wedge \sigma_n \wedge T}^2) = \mathbb{E}([X]_{t \wedge \sigma_n \wedge T}).$$
2. Choose $t>0$ sufficiently large such that $T(\omega) \leq t$ for all $\omega$ and conclude $$\mathbb{E}(X_{\sigma_n \wedge T}^2) = \mathbb{E}([X]_{\sigma_n \wedge T}).$$
3. Use Fatou's lemma, the monotone convergence theorem and the fact that $t \mapsto [X]_t$ is non-decreasing to show that $$\mathbb{E}(X_T^2) \leq \mathbb{E}([X]_T).$$