# Inequality on expectation of exponential martingale

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.

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

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 $$t$$!
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.