[Math] If $M$ is an $L^2$-bounded continuous local martingale, then $M^2-[M]$ is uniformly integrable

martingalesprobability theoryquadratic-variationstochastic-processesuniform-integrability

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 $$\operatorname E\left[[X]_t\right]\le\operatorname E\left[\sup_{s\in[0,\:t]}\left|X_s\right|^2\right]\le 4\operatorname E\left[[X]_t\right]\;\;\;\text{for all }t\ge0\tag2\;.$$

Best Answer

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.

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