Inequality on expectation of exponential martingale

Question

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.

Answer 1

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.