How do you calculate the Prob that a standard BM stays positive in some interval [t1, t2], for t2>t1>0?i.e. Prob(Bt>0) for t in [t1, t2].

I should first ask: do you know how to answer this question when t_1 = 0, and what the answer is? If you don't, then you should read up on the "reflection principle" (discussed elsewhere in this forum, or in any textbook on BM)

hmmm....and if I do? in my problem t1 is > 0, anyway.

Once you understand how to do t_1 = 0 (which, as I mentioned, is usually explained to the student in textbooks), it's not that hard to use similar reasoning to extend to t_1 > 0---although not exactly the same reasoning, which is why it's important to understand, and not just memorize, the t_1 = 0 answer!

could you kindly mention the names of the textbooks that has a proof of the formula for t1=0?I promise not to memorise the answer! LOL

Isn't a Brownian motion supposed to take both positive and negative values on any interval [0,t]? If so the probability is 0 for any t>0, whatever small it is.The problem seems different for me if t1>0 car then the question is:- Is Bt1 > 0?- If yes, will it remain positive until Bt2 (which is possible even if the interval has a positive length) ?

Like I mentioned, just look up "reflection principle" in this forum or Google, and you should find some explanations of the answer. If you're looking for something on paper, I'm sure it's also somewhere in Durrett's probability textbook, and probably somewhere in Billingsley's textbook.Jungix does make a reasonable point, so the question I should be asking is: Let a < 0. What is the probability that standard Brownian motion started at 0 does not dip below level a from time t_1=0 to time t_2?

Well, my thinking is: Prob that Bt >0 over the time interval [t1, t2] is equal to 1/2(1-q), where q is the probability that Bt hits zero within time interval [t1, t2]. Now I could use conditioning on Bt1 and the formula for the distribution of the first-passage time. This however requires not-so-obvious calculation of integrals. Therefore, I wonder if there is any other solution to this problem. And no, this problem is not a simple application of a reflection principle. If it is, please prove me wrong

Ooh, I think you almost had me! I finally sat down and used the reflection principle, which is similar to your reasoning, and I get the probability This looks messy (and I was stumped here for a bit!), but on the other hand, a little closer inspection makes it look like polar coordinates would be useful (because we have u^2 and v^2 in the exponents). The tricky thing here is how to redefine the intervals. If you draw the region 0 < v < infinity and -infinity < u < -(t_1/(t_2-t_1))v in the plane (with v on the horizontal axis, u on the vertical axis), you get a "pizza slice" of the plane bounded in the fourth quadrant by the lines v = 0 and u = -(t_1)/(t_2-t_1) * v, so it follows that 0 < r < infinity, and -pi/2 < theta < arctan(-t_1/(t_2-t_1)). Thus, the probability becomeswhich I think should be the answer.It seems like you have to condition on the value of B_{t_1} at some point, so I'd be surprised if there's a way that doesn't involve calculation of some integrals.

cool. looks like the right answer. sorry for making you calculating integrals ...but you made it sound in the begining that it's almost an elementary problem, that's why I had a doubt. thanks!

Yeah...as you probably suspected, I had thought about how to do the problem when I said that it was elementary, but it wasn't until I finally sat down and did it that I noticed that it looked a bit messier than I anticipated!

QuoteOriginally posted by: gentinexOoh, I think you almost had me! I finally sat down and used the reflection principle, which is similar to your reasoning, and I get the probability This looks messy (and I was stumped here for a bit!), but on the other hand, a little closer inspection makes it look like polar coordinates would be useful (because we have u^2 and v^2 in the exponents). The tricky thing here is how to redefine the intervals. If you draw the region 0 < v < infinity and -infinity < u < -(t_1/(t_2-t_1))v in the plane (with v on the horizontal axis, u on the vertical axis), you get a "pizza slice" of the plane bounded in the fourth quadrant by the lines v = 0 and u = -(t_1)/(t_2-t_1) * v, so it follows that 0 < r < infinity, and -pi/2 < theta < arctan(-t_1/(t_2-t_1)). Thus, the probability becomeswhich I think should be the answer.It seems like you have to condition on the value of B_{t_1} at some point, so I'd be surprised if there's a way that doesn't involve calculation of some integrals.At the risk of sounding a little pedantic, I think there should be a square root inside the Arctan

Edit: double post.

i go into this in great detail in Concepts etc.

good to know! will have a look.P.S. sometimes it's useful not just to have a book, but actually read it LOL