Let $M = \{M_t\}_{t\ge0}$ be the exponential martingale of Brownian motion $W= \{W_t\}_{t\ge0}$, that is,
$$
M_t = \mathcal E(W)_t = \mathrm{exp} \left( W_t – \frac{t}{2}\right).
$$
Question: Is $M$ uniformly absolutely continuous, that is, for every $\epsilon >0$ there exists $\delta >0$ such that, for every measurable $A$,
$$
\mathbf P(A)<\delta \Rightarrow \sup_{t\ge0}E(M_t;A)<\epsilon\ ?
$$
What I know is the uniformly absolute continuity is implied the uniform integrability, but the exponential martingale $M$ is only uniformly integrable on finite time interval but not on $[0,\infty)$.
Any comments or hints will be appreciated. TIA…
EDIT: If not, can we guarantee a slightly weaker result that
$$
\mathbf P(A)<\delta \Rightarrow \limsup_{t\ge0}E(M_t;A)<\epsilon\ ?
$$
Best Answer
Since $W_t$ has the same distribution as $\sqrt N$, where $N$ is standard normal, it follows that $$ \mathbb E\left[M_t\right]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left(\sqrt{t}x-\frac{x^2}2-\frac t2\right)dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left(-\frac 12\left(x-\sqrt t\right)^2\right)dx=1 $$ and $M_t\geqslant 0$ hence $(M_t)_{t\geqslant 0}$ is bounded in $\mathbb L^1$. If the claim in the question (or the version with the $\limsup$) was true, we would get uniform integrability of $\left(M_t\right)$, but it has been shown that it was not the case in a linked thread.