Show that $\sup\limits_t\left|X_t^n\right|\to0$ in probability iff $\lim\limits_{t\to\infty}[X^n]_t\to0$ in probability when $n\to\infty$

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge 0}$ be a filtration of $\mathcal A$

Let $X^n$ be a real-valued continuous local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ with $X_0^n=0$ for $n\in\mathbb N$. I want to show that$^1$ $$\sup_{t\ge0}\left|X_t^n\right|\xrightarrow{n\to\infty}0\text{ in probability}\Leftrightarrow\lim_{t\to\infty}[X^n]_t\xrightarrow{n\to\infty}0\text{ in probability }\tag3\;.$$

How can we prove $(3)$?

For the first direction "$\Rightarrow$": Let $\varepsilon>0$. By assumption, $$\operatorname P\left[\sup_{t\ge0}\left|X^n_t\right|>\varepsilon\right]\xrightarrow{n\to\infty}0\tag4\;.$$ Let $$Y^n:=(X^n)^2-[X^n]$$ and $$\tau_n:=\inf\left\{t\ge0:|X^n_t|>\varepsilon\right\}$$ for $n\in\mathbb N$. By $(2)$, we obtain $$\operatorname E\left[[X^n]_t^{\tau_n}\right]\le\varepsilon^2\;\;\;\text{for all }t\ge0\tag5\;,$$ but I don't know how I need to proceed.

(By definition, the limit as $t\to\infty$ on the right-hand side of $(3)$ is equal to the supremum over $t\ge0$. Am I missing something or isn't it possible that this supremum is $\infty$?)

---

$^1$ 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 $$\operatorname E\left[[X]_t\right]\le\operatorname E\left[\sup_{s\in[0,\:t]}\left|X_s\right|^2\right]\le 4\operatorname E\left[[X]_t\right]\;\;\;\text{for all }t\ge0\tag2\;.$$

Answer 1

Let $X^n=\{X^n_t,t\ge 0\}$ be locally continuous martingale and $X^{n\ast }_t =\sup_{s\le t}|X_s^n|$. To prove the following equavalence $$ \text{pr-}\lim_{n\to\infty}X^{n\ast}_T=0 \quad\Longleftrightarrow \text{pr-}\lim_{n\to\infty}[X^n]_T=0, \qquad \text{for all stopping time $T$,}\tag{1}$$ where $\text{pr-}\lim_{n\to\infty}$ denotes the convergence limit in probability, the key points is following relations, $$ \mathsf{E}(X^n_T)^2\le \mathsf{E}[X^n]_T\le \mathsf{E}[(X^{n\ast }_T)^2], \qquad \text{for all bounded stopping time $T$}\tag{2}$$ From (2) using Lenglart's results we could got following relations(cf. J. Jacod and A. N. Shiryaev, Limit Theorems for Stochastic Processes, 2nd Ed.(2003), Springer Verlag, Ch1, 3.30, P.35.), \begin{align} \mathsf{P}(X^{n\ast}_T\ge \varepsilon)&\le \frac{\eta}{\varepsilon^2}+ \mathsf{P}([X^n]_T\ge \eta), \qquad \text{for all stopping time $T$,} \tag{3}\\ \mathsf{P}([X]_T\ge \varepsilon)&\le \frac{\eta}{\varepsilon}+\mathsf{P}(X^{n\ast}_T)^2\ge \eta),\qquad \text{for all stopping time $T$}. \tag{4} \end{align} Now it is easy to deduce (1) from (3),(4).