3
$\begingroup$

Let $X$ be a continuous local martingale with $X_0=0$. Define the exponential local martingale $$\mathcal{E}(X)=e^{X-\frac{1}{2}[X]}.$$ For any $p,q>1$, establish the identity $$\mathcal{E}(X)^p=\mathcal{E}(\sqrt{pq}X)^{1/q}\left(e^{\frac{\sqrt{pq}(\sqrt{pq}-1)}{q-1}X}\right)^\frac{q-1}{q}.$$ Hence prove that $$\mathbb{E}\left[\mathcal{E}(X)_T^p\right]\leq\left(\mathbb{E}\left[e^{\frac{\sqrt{pq}(\sqrt{pq}-1)}{q-1}X_T}\right]\right)^\frac{q-1}{q}$$ for any finite stopping time $T$.

I have managed to obtain the identity fairly easily. However, I am really not sure how to continue. I thought to perhaps use Cauchy-Schwarz to split the expectation into 2, but this gives the wrong exponent because of the squaring. Also, I would like to use the fact that $\mathcal{E}(\sqrt{pq}X)$ is a positive local martingale and thus a supermartingale, but as $T$ is not necessarily bounded I need to show uniform integrability to use the optional stopping theorem, which I do not see why should hold. I therefore feel that my approach is probably wrong. Any advice would be greatly appreciated!

EDIT: Holder's inequality will give the required splitting of integrals. However I am still not sure why I can use the OST.

$\endgroup$
5
  • $\begingroup$ Once you have the identity, apply Hölder inequality. $\endgroup$ May 24 at 11:24
  • $\begingroup$ Thank you, that inequality seems to be the correct approach. How can I apply the OST? $\endgroup$ May 24 at 11:25
  • 1
    $\begingroup$ What is the OST? $\endgroup$ May 24 at 11:36
  • $\begingroup$ Apologies - the optional stopping theorem. $\endgroup$ May 24 at 11:41
  • $\begingroup$ Sorry, I had read only a part of the question. $\endgroup$ May 24 at 11:54

1 Answer 1

4
$\begingroup$

Be careful. Uniform integrability is not sufficient to apply optional stopping theorem to local martingales. Counter-example: if $R$ is a Bessel process of dimension 3 starting at 1, the local martingale $1/R$ is uniformly integrable. Yet, the expectation of $1/R_t$ is a strictly decreasing function of $t$!

An uniform bound is necessary to apply optional stopping theorem to local martingales.

In you situation, let $T_a = \inf\{t \ge 0 : X_t \ge a\}$ for $a \ge 0$. On $[0,T_a \wedge T]$, the local martingale $\mathcal{E}(\sqrt{pq}X)$ remains in $[0,\exp(\sqrt{pq}a)]$, so optional stopping theorem applies to $T_a \wedge T$. Then, apply Fatou's lemma as $a$ goes to infinity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.