Brownian motion exiting a corridor


Link to the double barrier problem:


If T = inf {t > 0 : B_t > a or B_t < -b} then 0 = B(0) = E B(t) = a p_a - b (1 - p_a), where p_a it the probability of exiting at a, hence p_a = b/(a + b). However this does not give us the distribution of T.


jw945 wrote Sep 22, 2013 at 4:03 PM

Do we need to write this function as the corridor is symmetric or not? as the function might be different is upper level and lower level is different in absolute value.

keithalewis wrote Sep 23, 2013 at 1:27 AM

Let's stick with symmetric corridors for now. Even the closed form solution for that case will have to be inverted with 1-d root finding.

It is easy to generate stopping times for one sided barriers (even including drift) so it might be worthwhile figuring out how to use that to simulate 2 sided barriers. We know the probability which barrier will be hit, so we could generate that first, and look at the hitting time for that barrier. Unfortunately that includes paths that hit the other barrier first, so a more subtle analysis will be required.

Let T = T_a,b be the double sided (non-symmetric) hitting time. Then P(T < t) = P(T < t, B_T = a) + P(T < t, BT =b) = P(T < t) [ P(B_T = a | T < t) + P(B_T = b | T < t) ].

jw945 wrote Sep 23, 2013 at 5:20 PM

should this be 4P(Bt>a) - 4P(Bt>3a)?

4*(Bt>3a) is the probability of brownian motion hit a and -a both within time t..

keithalewis wrote Sep 23, 2013 at 8:36 PM

Ignore my equation above. It should be
P(T < t) = P(T < t, B_T = a) + P(T < t, B_T = -b) 
= P(B_T = a)P(T < t | B_T = a) + P(B_T = -b)P(T < t | b_T = -b). 
We know P(B_T = a) and P(B_T = -b). It is not clear how to compute P(T < t | B_T = a) and the similar term involving -b.

The link has the formula. It involves theta functions. Maybe we should try for an approximation.

See Reflection Principle for the reflections principle. You can find more things here.

For the symmetric barrier we can use the fact B_t^2 - t is a martingale (why?) and the optional sampling theorem, 0 = EB^_T - T = a^2 - ET, so ET = a^2.

If u(t, x) satisfies u_t + (1/2) u_xx = 0, then u(t, B_t) is a martingale. (E.g., u(t, x) = x^2 - t).

Show u(t, x) = x^4 - 6 x^2 t + 3 t^2 is a solution and use that to find E[T^2].

More generally, show
u(t,x) = sum_0^k (-1/2)^j p^(2j)(x) t^j/j!
is a solution for any polynomial of degree 2k. This can be used to find all moments of T by taking p(x) = x^{2k}.

If we can find a tractable distribution matching the first couple of moments we can just use that.

wrote Sep 24, 2013 at 2:06 AM

wrote Sep 27, 2013 at 8:29 PM