2
$\begingroup$

Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\ge0}$ be a filtration of $\mathcal A$
  • $M$ be a real-valued continuous $L^2(\operatorname P)$-bounded local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ with $M_0=0$

How can we show that$^1$ $M^2-[M]$ is a uniformly integrable $\mathcal F$-martingale?

I've tried the following:

  • Let $$[M]_\infty:=\lim_{t\to\infty}[M]_t$$ and $(\tau_n)_{n\in\mathbb N}$ be a localizing sequence for $M^2-[M]$
  • Then, $$\operatorname E\left[\left(M^2-[M]\right)^{\tau_n}_t\right]=0\;\;\;\text{for all }t\ge0\tag3$$ for all $n\in\mathbb N$ and hence \begin{equation} \begin{split} \operatorname E\left[[M]_\infty\right]&=\lim_{n\to\infty}\operatorname E\left[[M]_n^{\tau_n}\right]\\ &=\lim_{n\to\infty}\operatorname E\left[M^2_{\tau_n\:\wedge\:n}-\left(M^2-[M]\right)_n^{\tau_n}\right]\\ &=\lim_{n\to\infty}\operatorname E\left[M^2_{\tau_n\:\wedge\:n}\right] \end{split}\tag4 \end{equation} by Lebesgue's monotone convergence theorem

Now, the idea is to show that $$\operatorname E\left[[M]_\infty\right]\le\sup_{t\ge 0}\operatorname E\left[M^2_t\right]<\infty\tag5$$ and $$\sup_{t\ge 0}M_t^2\in\mathcal L^1(\operatorname P)\tag6\;.$$ Then we could conclude that $$\left|M^2-[M]\right|\le\sup_{t\ge 0}M_t^2+[M]_\infty=:X\in\mathcal L^1(\operatorname P)\tag7\;.$$ By $(7)$, $M^2-[M]$ is an $\mathcal F$-martingale.

So, how can we show $(5)$ and $(6)$. And how can we show that $M^2-[M]$ is uniformly integrable?


$^1$ If $X$ is a real-valued continuous local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$, there is a real-valued $\mathcal F$-adapted stochastic process $[X]$ on $(\Omega,\mathcal A,\operatorname P)$, unique up to indistinguishability, with

  1. $[X]_0=0$
  2. $[X]$ is continuous
  3. $[X]$ is of locally bounded variation
  4. $X^2-[X]$ is a local $\mathcal F$-martingale
  5. $[X]$ is nondecreasing

If $(\sigma_n)_{n\in\mathbb N}$ is a localizing sequence for $X$, then $(\sigma_n\wedge n)_{n\in\mathbb N}$ is a localizing sequence for both $X$ and $X^2-[X]$. If $X$ is an $\mathcal F$-martingale, then $X^2-[X]$ is an $\mathcal F$-martingale. If $\tau$ is an $\mathcal F$-stopping time on $(\Omega,\mathcal A)$, then $$[X^\tau]=[X]^\tau\tag1\;.$$ If $X_0=0$, then $$\operatorname E\left[[X]_t\right]\le\operatorname E\left[\sup_{s\in[0,\:t]}\left|X_s\right|^2\right]\le 4\operatorname E\left[[X]_t\right]\;\;\;\text{for all }t\ge0\tag2\;.$$

$\endgroup$
0

1 Answer 1

1
$\begingroup$

Let $X_t:=M^2_t-[M]_t$, $t\ge 0$. Because $M$ is $L^2$-bounded, you have that $\lim_{t\to\infty}M_t=M_\infty$ exists, both a.s. and in $L^2$. Therefore $M^2_t$ converges in $L^1$ to $M_\infty^2$. Also, $[M]_t$ converges monotonically to the integrable r.v. $[M]_\infty$; thus $[M]_t\to[M]_\infty$ in $L^1$ as well. It follows that the martingale $(X_t)$ converges in $L^1$, and therefore $(X_t)$ is uniformly integrable.

$\endgroup$
5
  • $\begingroup$ How do you conclude the a.s. and $L^2$ convergence of $M_t$ as $t\to\infty$? It's clear to me that by $L^2$-boundedness $M$ is uniformly integrable and hence $M_t$ converges in $L^1$ as $t\to\infty$. $\endgroup$
    – 0xbadf00d
    Commented May 24, 2017 at 22:59
  • 1
    $\begingroup$ The a.s. convergence comes from the martingale convergence theorem. By the uniform integrablility that you note, $M_t=E[M_\infty|\mathcal F_t]$, so $0\le E[(M_t-M_\infty)^2]= E[M_\infty^2]-E[M_t^2]$ for all $t>0$. On the other hand, by Fatou, $E[M_\infty^2]\le\liminf_{t\to\infty} E[M_t^2]=\lim_{t\to\infty}E[M_t^2]$. It follows that $\lim_{t\to\infty} E[M_t^2]=E[M_\infty^2]$, and so $M_t\to M_\infty$ in $L^2$. Consequently, $M_t^2\to M_\infty^2$ in $L^1$. $\endgroup$ Commented May 25, 2017 at 17:37
  • $\begingroup$ I only know the martingale convergence theorem for strict martingales, but $M$ is a local martingale. I don't see that the assumptions imply that $M$ is already strict. That would be the case, if $\sup_{t\ge0}\left|M_t\right|\in L^1(\operatorname P)$, but I don't think that $L^2$-boundedness implies this for local martingales. I've asked for that here. $\endgroup$
    – 0xbadf00d
    Commented Jun 8, 2017 at 12:03
  • $\begingroup$ Ack. I seem to have read past local in the description of $M$. $\endgroup$ Commented Jun 8, 2017 at 17:03
  • $\begingroup$ There is another thing: In order to apply the monotone convergence theorem in $(4)$, we need $[M]_t^{τ_n}\in L^1(\text P)$ for all $n\in\mathbb N$ and $t\ge0$. How do we see that this is the case? My attempt is $$\text E\left[[M]_t^{τ_n}\right]\stackrel{(a)}\le\text E[[M]_t]\stackrel{(b)}\le\text E\left[\sup_{s\in[0,\:t]}|M_s|^2\right]\stackrel{(c)}\le 4\text E[|M_t|^2]\stackrel{(d)}<∞\;.$$ $(a)$ holds, since $[M]$ is increasing. $(b)$ holds by Lemma 2 here. $(c)$ holds by Doob's inequality. And $(d)$ holds, since $M$ is square-integrable. Do you agree? $\endgroup$
    – 0xbadf00d
    Commented Jun 8, 2017 at 20:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .