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Suppose $π‘Š$ is a one-dimensional standard Brownian motion defined on some probability space $(\Omega, \mathcal F, P)$ and let $𝑋(𝑑):=\exp\{π‘Š(𝑑)βˆ’\frac{1}{2}π‘‘βˆ’\frac{1}{𝑑+1}\}$ for $𝑑\ge 0$. Note that $𝑋(\infty):=\limsup_{t\to\infty}𝑋(𝑑)=0$ a.s. because $\lim_{t\to\infty}\frac{π‘Š(𝑑)}{𝑑}=0$ a.s.

My question is: How to show that $E[\sup_{0\le t\le \infty}𝑋(𝑑)]=\infty$? Many thanks. (I posed this question in stackexchange earlier today. Not sure if this is the right place to ask this question.)

Many thanks.

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    $\begingroup$ Can we ignore the $1/(t+1)$? $\endgroup$ – Matt F. Feb 19 '20 at 7:26
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    $\begingroup$ The supremum of a Brownian motion with drift -1/2 is distributed as an exponential random variable of mean 1. $\endgroup$ – Timothy Budd Feb 19 '20 at 7:41
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This seems to me a standard exercice in a probability course: Ignore the $1/(t+1)$ term as $X(t)\geq e^{-1}\exp(W(t)-t/2)$. The term $M_t:=\exp(W(t)-t/2)$ is well known to be a martingale so $\mathbb{E}(M_t)=1$ for all $t$. If $\mathbb{E}(\sup_{0\leq t\leq \infty}X_t)<\infty$ then by dominated convergence $$1=\mathbb{E}(M_t)\rightarrow_{t\rightarrow \infty} \mathbb{E}(M_\infty)=0$$

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  • $\begingroup$ Great! Many thanks. $\endgroup$ – epsilon Feb 19 '20 at 17:29

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