For $B_t$ Brownian Motion with drift $\mu<0$, I need to prove that the max value, $X = \max_{0 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.
Max of Brownian motion with drift is finite almost surely
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statistics
probability-theory
stochastic-processes
brownian-motion
2 Answers
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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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0What 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. – 2012-07-11
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0I 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.. – 2012-07-11
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0Yes it never reaches + infinity. But the statement that it stays finite almost surely is not the correct way to express the result. – 2012-07-11
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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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0Yes, 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? – 2012-07-11