$M$ is a continuous local martingale. Prove the consequences

Question

$M$ is a continuous local martingale. Show that:
a) If $M_0=0$ and $\mathbb{E} (< M >) _t <\infty$ for $t\ge 0$ then $M \in \mathcal M^{2,c}$
b) There is a sequence of stopping moments $\tau_n \to T$ such that $M^{\tau_n} \mathbb 1_{\{\tau_n>0\}}$ is a bounded martingale
c) For any stopping moment $\tau$, the process $M^{\tau}$ is a local martingale
$\mathcal M^{2,c}$ is family of continuous martingales $\in L^2$

My thoughts:
a) Assume that $\mathbb{E} (< M >) _t\le A$ then: $$E[M_t^2] = E\sum (M_{t_{i+1}}-M_{t_i})^2 \leq A E[\sup_i|M_{t_{i+1}}-M_{t_i}|].$$ b) Probably optional stopping theorem is helpful

Unfortunately, I don't know what to do next... Can anyone help me? I spent several hours on this task, but I don't understand how to get to the solution.

Answer 1

$M$ is a continuous local martingale. Show that:

a) If $M_0=0$ and $\mathbb{E} (< M >) _t <\infty$ for $t\ge 0$ then $M \in \mathcal M^{2,c}$

We will show that finite quadratic variation implies that M is L2-bounded martingale. We follow Revuz-Yor (1.23) Proposition.

L2-bounded.

Since $M-[M]$ is a local martingale, we get a sequence of stopping times ${T_n},n\ge1$ that diverge $T_n\to\infty$ and $\mathbb{E}[M^2_{t\wedge T_n}1_{T_{n}>0}-[M]_{t\wedge T_n}]$ is bounded. Using that $ E[<M>_{\infty}]<\infty$ we bound by:

$$\mathbb{E}[M^2_{t\wedge T_n}1_{T_{n}>0}]\leq E[<M>_{\infty}]=E[M_{0}^{2}]=:K$$

and by Fatou's lemma

$$\mathbb{E}[M^2_{t}]\leq\lim_{n}\mathbb{E}[M^2_{t\wedge T_n}]\leq K.$$

So this gives L2-bounded.

Martingale

This also gives that $M_{t\wedge T_n}1_{T_{n}>0}$ is uniformly integrable and so we can apply Vitali convergence theorem for the conditional relation

$$E[M_{t\wedge T_n}1_{T_{n}>0}|\mathcal{F}_{s}]=M_{s\wedge T_n}1_{T_{n}>0}$$

to get martingale

$$E[M_{t}|\mathcal{F}_{s}]=M_{s}.$$

see also $L^1$-bounded quadratic variation of a continuous local martingale $\implies$ it is a true martingale, $L^2$-bounded and Is continuous L2 bounded local martingale a true martingale?.

b) There is a sequence of stopping moments $\tau_n \to T$ such that $M^{\tau_n} \mathbb 1_{\{\tau_n>0\}}$ is a bounded martingale

We fix $$ D_n = \left\{\frac{k}{2^n}, k= 0,1,2,\ldots\right\}\subset D_{n+1}\subset \cdots $$

and so we define the approximation $\tau_n$ $$ \tau_n(\omega) = \inf\left\{t\in D_n; t\ge T(\omega)\right\}. $$ Then $T_n\ge T_{n+1}\to T $ and $T_n$ is a stopping time. For $0\le s\le t$, we similarly define $$ t_n = \inf\left\{u\in D_n; u\ge t\right\} \ge t_{n+1}\ge\cdots $$ and $$ s_n = \inf\left\{u\in D_n; u\ge s\right\} \ge s_{n+1}\ge\cdots $$ It follows from the discrete time optional sampling theorem that $$ \mathbb{E}\left[M_{t_n\wedge T_n}|\mathcal{F}_{s_m}\right] = M_{s_m \wedge T_n} $$ for any integers $m\ge n$.

c) For any stopping moment $\tau$, the process $M^{\tau}$ is a local martingale

This is shown here $\tau$ be a stopping time, $(M_t)_{t\ge 0}$ a martingale. Is $(M_{\tau\wedge t})_{t\ge 0}$ a martingale?

indeed the main idea is to use the optional stopping theorem but for a dyadic approximation. Letting $m\to\infty$, we have $s_m \to s$ decreasing and $\mathcal{F}_{s_m}\to\mathcal{F}_s$. By Lévy's Downward theorem and the continuity of process $M$, we have $$ \mathbb{E}\left[M_{t_n\wedge \tau_n}|\mathcal{F}_{s}\right] = M_{s_m \wedge \tau_n} $$ Observe that $(M_{t_n\wedge \tau_n},\mathcal{F}_{t_n\wedge \tau_n})$ is a backward martingale, whence it is uniformly integrable. Letting $n\to\infty$, we arrive at $$ \mathbb{E}\left[M_{t\wedge T}|\mathcal{F}_{s}\right] = M_{s\wedge T}. $$