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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?

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    $\begingroup$ I think it's not uniformly integrable. Let's take $r=0$, $\theta = 1$, $\sigma = 1$. Then $M_t = e^{W_t - t/2}$. By SLLN or other methods, $W_t - t/2 \to -\infty$ a.s., so $M_t \to 0$ a.s. If $M_t$ were uniformly integrable, then Vitali would imply $E[M_t] \to E[0] = 0$, which is not the case since it's a martingale and $E[M_t] = 1$ for all $t$. Or am I mistaken? $\endgroup$ – Nate Eldredge Apr 7 '18 at 16:08
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    $\begingroup$ If this were a UI martingale, then you could use it to apply a drift to BM and find a probability measure under which the original BM with drift is also BM on the entire positive real line (not just on a bounded time interval) by Girsanov's theorem. That would lead to some contradictory results about the distribution of hitting times of BM. Just mentioning this in case that is why you wanted your martingale to be UI. $\endgroup$ – Calculon Apr 7 '18 at 19:03
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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 $$\theta \sigma W_t - \frac{\theta^2 \sigma^2 t}{2} \to -\infty.$$

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

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  • $\begingroup$ Thank you four your response. I did the following things: $\mathbb{E}[M_t]^2 = \mathbb{E}[e^{\theta X_t - \psi(\theta) t}]^2 = \mathbb{E}[e^{2\theta X_t - 2\psi(\theta) t}] = e^{\sigma^2 \theta^2 t}$. But taking supremum over this function I will obtain infinity. Does it mean, that it is not uniformly integrable? $\endgroup$ – MathMen Apr 7 '18 at 12:30
  • $\begingroup$ No, the converse part does not hold. I approve Nate Eldredge's answer. Sorry for my useless comment. $\endgroup$ – Kavi Rama Murthy Apr 9 '18 at 0:36

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