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Let $(W_t)$ a Brownian motion (starting at $0$) and $\tau=\inf\{t>0\mid W_t\geq 1\}$. Using reflexion principle in $(*)$, I find that $$\mathbb P\{\tau>t\}=\mathbb P\{\sup_{0\leq s\leq t}W_s<1\}\underset{(*)}{=}\mathbb P\{|W_t|<1\}$$ $$=\sqrt{\frac{2}{\pi}}\int_{0}^1e^{-x^2/2t}dx,$$ and thus $$\mathbb E[\tau]=\int_0^\infty \mathbb P\{\tau>t\}dt=\sqrt{\frac{2}{\pi}}\int_0^\infty \int_0^1 e^{-x^2/2t}dxdt=\infty .$$

I find this result quit non-intuitive, since I would interpret it as : in mean, the BM is smaller than 1, which is - I guess - wrong. So, I was wondering : is my calculation wrong ?

Walace
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  • Maybe it's the notation that's confusing me - but if I take $t$ to be very close to $0$ in your first line, I'm calculating the probability that $\tau$ is greater than some real very close to $0$, yes? Which should be $1$. But it tends to $0$. Conversely for large $t$. – Derek Allums Apr 27 '20 at 09:08
  • I mean, surely (*) is just wrong. – Frank May 01 '20 at 18:56
  • @Frank: Why ? This is just reflexion principle – Walace May 01 '20 at 19:06
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    I don't know enough to comment on the details here, but in a simple (50/50 left/right) one dimensional random walk, the expected time to reach $1$ starting from $0$ is infinite (even though $1/2$ the time you get there in one step). Maybe this is sort of a continuous variation of that? – Ned May 01 '20 at 19:31
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    $\mathbb{E}(\tau)=\infty$ is correct but your interpretation is not. It is possible to show that $\mathbb{P}(\tau<\infty)=1$, i.e the sample paths of Brownian motion will be eventually greater or equal than $1$ at some point. The identity $\mathbb{E}(\tau)=\infty$ tells us that we may have to wait for a very long time until this happens (essentially because Brownian motion may have very long excursion down to $-\infty$) – saz May 03 '20 at 13:37

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