Show that $M-M_0$ is an $L^2$ bounded martingale if $E([M]_\infty)<\infty$.

Question

Let $M$ be a continuous local martingale, and let $[M]_\infty=\lim\limits_{t\rightarrow \infty}[M]_t,$ and if $E([M]_\infty)<\infty$. Show that $M-M_0$ is an $L^2$ bounded martingale.

Since $M$ is a local martingale, there is a localizing sequence $\{T_n\}_{n\in\mathbb{N}}$ such that $T_n$ increases to $\infty$ and $M_{t\wedge T_n}-M_0$ is a martingale for all $n$. Given $t\in\mathbb{R}^+,$ we have $$ E(M_t-M_0)^2=E(\lim\limits_{n\rightarrow \infty}M_{t\wedge T_n}-M_0)\leq \liminf\limits_{n\rightarrow \infty} E(M_{t\wedge T_n}-M_0)^2. $$ Since we also know that $$ E(M_{t\wedge T_n}-M_0)^2=E([M]_{t\wedge T_n})+2E \left[\int_{0}^{t\wedge T_n}(M-M_0)d(M-M_0) \right]. \tag{$\star$} $$ My idea to bound the first term of right hand side by the following: $$ E([M]_{t\wedge T_n})\leq E([M]_{t})\leq E([M]_{\infty})<\infty. $$ However, I am not sure whether we have the above inequality. Do we have $\{[M]_t\}$ is increasing process? I found a theorem states that if $M$ is $L^2$ continuous martingale, then we have $\{[M]_t\}$ is increasing, but $M$ is just local martingale by our assumption.

My second question is I don't have goo idea to bound the second term of right hand side of $(\star)$. Please give me some suggestion. Thanks.

Answer 1

WLOG $M_0=0$ (otherwise replace $M_t$ by $M_t-M_0$). By the very definition of the quadratic variation,

$$N_t := M_{t \wedge T_n}^2-[M]_{t \wedge T_n}$$

is a martingale. In particular,

$$\mathbb{E}(N_t) = \mathbb{E}(N_0) = 0,$$

i.e.

$$\mathbb{E}(M_{t \wedge T_n}^2) = \mathbb{E}([M]_{t \wedge T_n}).$$

(In your calculation this corresponds to the fact that the second term in $(\ast)$ vanishes since the stochastic integral with respect to a martingale is again a martingale.) Since the quadratic variation is increasing, this implies

$$\mathbb{E}(M_{t \wedge T_n}^2) \leq \mathbb{E}([M]_{\infty}) < \infty.$$

Hence, by Fatou's lemma,

$$\mathbb{E}(M_t^2) = \mathbb{E} \left( \lim_{n \to \infty} M_{t \wedge T_n}^2 \right) \leq \mathbb{E}([M]_{\infty}).$$

Finally, Doob's maximal inequality shows

$$\mathbb{E} \left( \sup_{t \leq T} M_t^2 \right) \leq 4 \mathbb{E}([M]_{\infty}) \qquad \text{for all} \, \, T \geq 0.$$

This proves the boundedness in $L^2$. Clearly, this implies boundedness in $L^1$, and therefore it follows from

$$\mathbb{E}(M_{t \wedge T_n} \mid \mathcal{F}_s) = M_{s \wedge T_n}, \qquad s \leq t,$$

the continuity of the sample paths and the dominated convergence theorem that $(M_t)_{t \geq 0}$ is a martingale.