A brownian motion W(0) = 0. What is the probability of (W(1)>0 && W(2)<0) ?I understand that it comes down to an integral But it does not seem to be obvious way to get a numerical value out of it. Does anybody have an idea, or may be a suggestion of another approach?

The answer is 1/8. You have not written well the integrals: The limit of the innermost should be from minus infinity to minus x, for example. At a recent interview, it was pointed out to me that there is a very simple solution that requires no calculations and no integrations. Can you find it out?

Function have a rotation symmetry around 0. So, when you integrate the function over a domaine x>0, x<z (x-y=z, y<0) you get 1/8 of the integral over the whole plane.Another way to see the result is to see that events W(1)>0, W(2)-W(1)<0 and abs(W(1))<abs(W(2)-W(1)) are independent, with probability of each equal to 1/2.

Hi Vassil,check it once again, the integrals are correct. The secondterm stands for the transitional probability from x to y,and y should be integrated from -inf to 0. I thinkErge got the point, really good solution.

QuoteOriginally posted by: ErgeFunction have a rotation symmetry around 0. So, when you integrate the function over a domaine x>0, x<z (x-y=z, y<0) you get 1/8 of the integral over the whole plane.Another way to see the result is to see that events W(1)>0, W(2)-W(1)<0 and abs(W(1))<abs(W(2)-W(1)) are independent, with probability of each equal to 1/2.Hi Erge,thanks a lot! The first one is excellent, and the second one is obvious as well.Good job!

Can you clarify how you got this integral please? I dont quite get it.To me this problem looks like that

*Sorry, just noticed that Erge had posted this method*We know W(1) and W(2)-W(1) are i.i.d and hence we can't "distinguish" them. So P( |W(2)-W(1)| > |W(1)| ) = 1/2.Thus, P( W(1)>0, W(2)<0 ) = P( W(1)>0, W(2)-W(1) < 0, |W(2)-W(1)| > |W(1)|)For Brownian motion, signs and lengths are independent, so,P( W(1)>0, W(2)<0 ) = P( W(1)>0) P(W(2)-W(1) < 0) P(|W(2)-W(1)| > |W(1)|)=1/2*1/2*1/2=1/8

QuoteOriginally posted by: ronwiseCan you clarify how you got this integral please? I dont quite get it.To me this problem looks like thatW_1 and W_2 are not independent. SoHowever the BM increments are independent and so the following is true

Thanks a lot!

I understand the other two terms. Can some one explain to me where this term P(|W(2)-W(1)| > |W(1)|)come from ?ThanksL.

QuoteOriginally posted by: VassiliThe answer is 1/8. You have not written well the integrals: The limit of the innermost should be from minus infinity to minus x, for example. At a recent interview, it was pointed out to me that there is a very simple solution that requires no calculations and no integrations. Can you find it out? The way Vassili suggests is also the way I'd approach it.Let x = w1, and z = w2 - w1. The set of interest is 0 < x < infinity, -infinity < z < -x, where the integral over z is the inner integral. That integral gives us We now rewrite the integral asAnd this gives us 1/8. Seems pretty straightforward

QuoteThus, P( W(1)>0, W(2)<0 ) = P( W(1)>0, W(2)-W(1) < 0, |W(2)-W(1)| > |W(1)|)where does it come from?