The mean of a running maximum

Suppose $$π$$ is a one-dimensional standard Brownian motion defined on some probability space $$(\Omega, \mathcal F, P)$$ and let $$π(π‘):=\exp\{π(π‘)β\frac{1}{2}π‘β\frac{1}{π‘+1}\}$$ for $$π‘\ge 0$$. Note that $$π(\infty):=\limsup_{t\to\infty}π(π‘)=0$$ a.s. because $$\lim_{t\to\infty}\frac{π(π‘)}{π‘}=0$$ a.s.

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

Many thanks.

• Can we ignore the $1/(t+1)$? – Matt F. Feb 19 '20 at 7:26
• The supremum of a Brownian motion with drift -1/2 is distributed as an exponential random variable of mean 1. – Timothy Budd Feb 19 '20 at 7:41

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 $$1=\mathbb{E}(M_t)\rightarrow_{t\rightarrow \infty} \mathbb{E}(M_\infty)=0$$