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For $B_t$ Brownian Motion with drift $\mu<0$, I need to prove that the max value, $X = \max_{0<t<\infty}B_t$ is finite almost surely, ie $P(X<\infty)=1$.

Now, I know that because the mean is negative, it will go more and more negative, and it is also a supermartingale. But I don't know how to prove almost surely...

Appreciate any hints.

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2 Answers 2

up vote 4 down vote accepted

Hint: Try the strong law of large numbers. What does it say about $\lim_{t \to \infty} B_t/t$? What does this say about the sign of $B_t$ for large $t$?

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What I use to call "law of large numbers" is for discrete-time processes, and its application to continuous time continuous processes is not totally obvious. I agree that it works, as well as some refinements (LIL), but there may be some non-trivial things to show, depending on what Ido is expected to know. –  D. Thomine Jul 11 '12 at 17:47
    
I actually did think about using the SLLN here, by saying that the BM is the limit of some increasingly-dense random walks. In this case this is clear. I don't know if I can present it as this limit, however.. –  yoki Jul 11 '12 at 17:50
    
Yes it never reaches + infinity. But the statement that it stays finite almost surely is not the correct way to express the result. –  Michael Chernick Jul 11 '12 at 18:08

Actually that is wrong in the sense that it is not finite in the limit. Brownian Motion with a negative drift will wander off to -∞ almost surely. It will tend to go down in a linear fashion with the slope equal to the drift parameter.

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Yes, this is more than what I need to prove, I just need to prove that the maximal value is less than positive infinity. How do I do it? –  yoki Jul 11 '12 at 17:34

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