How to prove that this stochastic process converges in mean square as $t \to \infty$

Question

I have that $\{ X_t ; t=0,1...\}$ is a martingale with finite second moments (meaning that $E(X_t^2) < \infty$). Assume that
$E(X_t^2) \to C \quad $ as $t \to \infty$
where $C < \infty$ is a constant. Prove that $\{ X_t;t=0,1,..\}$ converges in mean square as $t \to \infty$. I dont know where to start. Can somebody help me?

Answer 1

Since $\mathbb{E}[X_t^2] \rightarrow C < \infty$, we have that $\sup_{t} \mathbb{E}[X_t^2] < \infty$ so $(X_t)$ is uniformly integrable (this is discrete time, but I'm not certain that this exact argument goes through in continuous time without some additional assumptions). Doob's martingale convergence theorem then guarantees that there exists $X_\infty$ such that $(X_t) \rightarrow X_\infty$ almost surely, and since $(X_t)$ is uniformly integrable this implies $(X_t) \rightarrow X_\infty$ in $L^2$. I'll also prove that for completeness: Fix $\varepsilon > 0$. Then

\begin{align*} \mathbb{E}[|X_t - X_\infty|^2] &= \mathbb{E}[|X_t - X_\infty|^21_{|X_t - X_\infty| > \varepsilon}] + \mathbb{E}[|X_t - X_\infty|^21_{|X_t - X_\infty| \le \varepsilon}] \\ &\le 2(\mathbb{E}[|X_t|^21_{|X_t - X_\infty| > \varepsilon}] + \mathbb{E}[ |X_\infty|^21_{|X_t - X_\infty| > \varepsilon}]) + \varepsilon^2. \end{align*}

Since $(X_t) \rightarrow X_\infty$ almost surely, we can choose $T > 0$ to make $\sup_{t \ge T} \mathbb{P}(|X_t-X_\infty| > \varepsilon)$ arbitrarily small, so by uniform integrability we can choose $T> 0$ such that $\mathbb{E}[|X_t|^21_{|X_t - X_\infty| > \varepsilon}] + \mathbb{E}[ |X_\infty|^21_{|X_t - X_\infty| > \varepsilon}] \le \frac{\varepsilon}{2}$ for all $t \ge T$. Thus we conclude

\begin{align*} \lim_{t \rightarrow \infty}\mathbb{E}[|X_t - X_\infty|^2] &\le \lim_{t \rightarrow \infty}2(\mathbb{E}[|X_t|^21_{|X_t - X_\infty| > \varepsilon}] + \mathbb{E}[ |X_\infty|^21_{|X_t - X_\infty| > \varepsilon}]) + \varepsilon^2 \\ &\le \varepsilon + \varepsilon^2 \end{align*}

and since $\varepsilon$ was arbitrary, we conclude $\lim_{t \rightarrow \infty}\mathbb{E}[|X_t - X_\infty|^2] = 0.$