Prove that a martingale is uniformly integrable.

Question

Let's consider the following stochastic process: $$ M_t = e^{\theta X_t - \psi(\theta)t}, $$ where: $\quad X_t = (r-\frac{\sigma^2}{2})t + \sigma W_t \quad$ and $\quad \psi(\theta) = (r-\frac{\sigma^2}{2})\theta + \frac{\sigma^2}{2}\theta^2$.

Moreover, $W_t$ is a Wiener process.

It is not so difficult to show that this process is a martingale and I know how to do it.

But how to show that it is uniformly integrable?

I know the definition of uniform integrability. It states that a process $\{X_t\}$ is uniformly integrable if $$ \lim_{a \rightarrow \infty} \sup_{t} \mathbb{E}[|X_t|, |X_t|>a] = 0. $$

But I do not know how to apply it to check uniform integrability for process $M_t$, because it seems to be too theoretical.

Is there maybe another method to check it?

Answer 1

$\sup_t E(M_t)^{2} <\infty$ is a sufficient condition for uniform integrability.

Answer 2

The martingale $M_t$ is not uniformly integrable.

If it were, then by standard martingale facts, it would converge a.s. and in $L^1$ to some $M_\infty$. (An $L^1$-bounded martingale always converges a.s., and if it's uniformly integrable, then the convergence is also $L^1$ by Vitali.) In particular we would have $E[M_\infty] = \lim E[M_t] = 1$.

However, this martingale converges a.s. to zero, and that's a contradiction.

One way to see $M_t \to 0$ is to note, after simplifying, that $$M_t = \exp\left(\theta \sigma W_t - \frac{\theta^2 \sigma^2 t}{2}\right).$$ By the strong law of large numbers, $W_t / t \to 0$ a.s., and hence $$\frac{1}{t} \left(\theta \sigma W_t - \frac{\theta^2 \sigma^2 t}{2}\right) \to -\frac{\theta^2 \sigma^2}{2} < 0$$ which implies $$\theta \sigma W_t - \frac{\theta^2 \sigma^2 t}{2} \to -\infty.$$