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OMD
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Nice Brownian Motion exercise (not really a teaser)

February 25th, 2005, 4:37 am

Covariance of stopping time and value at the stopped time:This is really cool. I am really enjoying this stuff right now I must say!Let B_t be Brownian motion. Let A and B be greater than zero.Let T be the stopping time given by T = inf{t : B_t = A or B_t = -B}Through elementary techniques (though creative) or through a really nice first use of martingales and the canonical theorems that people prove about them in conjunction with stopping times, it can be shown that:P(B_T = A) = B/(A + B)E(T) = ABThis alone constitutes a good exercise if you have not done this, but what really made my jaw drop in fascination was a beautiful technique shown in Steele's stochastic calculus book for calculating Cov(B_T,T).Obviously this comes down to calculating E(T * B_T) since E(B_T) = 0.It was a really nice application of Ito's lemma, a small PDE result that follows immediately from Ito's lemma, the fact that Ito integrals are local martingales, the fact that a bounded local martingale is a martingale, along with Doob's stopping time theorem for martingales. Without looking this up, can you calculuate E(T * B_T) and show work?Sorry! I just had to share this because I thought it was really neat!
Last edited by OMD on February 25th, 2005, 11:00 pm, edited 1 time in total.
 
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greenleaf
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Joined: March 10th, 2004, 6:02 pm

Nice Brownian Motion exercise (not really a teaser)

February 25th, 2005, 3:18 pm

Steele's writing is a masterpiece, I have no doubt about that.
 
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karakfa
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Nice Brownian Motion exercise (not really a teaser)

February 26th, 2005, 2:17 am

We've discussed elementary solutions to the same problem here
 
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OMD
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Nice Brownian Motion exercise (not really a teaser)

February 26th, 2005, 4:59 am

QuoteOriginally posted by: karakfaWe've discussed elementary solutions to the same problem hereI looked there and the link discusses E(B_T) and E(T), but not the question I was asking - namely E(T*B_T).The answer is E(T*B_T) = AB(A-B)/3, but the derivation is really cool.
Last edited by OMD on February 25th, 2005, 11:00 pm, edited 1 time in total.
 
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allstar
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Joined: July 9th, 2004, 1:23 pm

Nice Brownian Motion exercise (not really a teaser)

March 5th, 2005, 5:34 am

I think the function (B_t)^3 will do the work.
 
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OMD
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Nice Brownian Motion exercise (not really a teaser)

March 5th, 2005, 8:10 pm

QuoteOriginally posted by: allstarI think the function (B_t)^3 will do the work.How do you use that to relate it to T*B_T?
 
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OMD
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Nice Brownian Motion exercise (not really a teaser)

March 15th, 2005, 6:10 am

QuoteOriginally posted by: OMDCovariance of stopping time and value at the stopped time:This is really cool. I am really enjoying this stuff right now I must say!Let B_t be Brownian motion. Let A and B be greater than zero.Let T be the stopping time given by T = inf{t : B_t = A or B_t = -B}Through elementary techniques (though creative) or through a really nice first use of martingales and the canonical theorems that people prove about them in conjunction with stopping times, it can be shown that:P(B_T = A) = B/(A + B)E(T) = ABThis alone constitutes a good exercise if you have not done this, but what really made my jaw drop in fascination was a beautiful technique shown in Steele's stochastic calculus book for calculating Cov(B_T,T).Obviously this comes down to calculating E(T * B_T) since E(B_T) = 0.It was a really nice application of Ito's lemma, a small PDE result that follows immediately from Ito's lemma, the fact that Ito integrals are local martingales, the fact that a bounded local martingale is a martingale, along with Doob's stopping time theorem for martingales. Without looking this up, can you calculuate E(T * B_T) and show work?Sorry! I just had to share this because I thought it was really neat!I guess I will just throw up the solution:Consider f(t,x) where we replace x with B_t when we use Ito's formula.We would like f(t,x) = tx + g(t,x) where f(t,B_t) is a martingale.Note that Ito integrals are local martingales.By Ito, we have:edited - to say that forgot to put the (1/2) in the last integral above! D'OH!To make f(t,B_t) a local martingale, we just need to kill off the integrals with respect to ds.Therefore we need to find f(t,x) = tx + g(t,x) such that f_t + (1/2)f_xx = 0.This can easily be done by taking f(t,x) = tx - (x^3)/3Therefore we have M_t = tB_t - ((B_t)^3)/3 is a local martingale.When we introduce the stopping time T discussed in the problem, we know by a variation of Doob's stopping time theorem that if we let u = min(T,t), then M_u is also a local martingale. Furthermore, M_u is both bounded and continuous, so it is indeed an honest martingale.Therefore, E(M_u) = M_0 = 0 for all u.Note that M_u is dominated by the integrable function max(A,B).Therefore, by the dominated convergence theorem, we get lim_{u---> \infty} E(M_u) = E(M_T) = 0 by the martingale property.So, E(M_T) = E(T*B_T) - E((B_T)^3)/3 = 0So, E(T*B_T) = E((B_T)^3)/3 = P(B_T = A)*(A^3)/3 + P(B_T = -B)*(-B^3)/3 and since P(A) = B/(A+B) and P(B) = 1-P(A), we getE(T*B_T) = (B(A^3)/3 - A(B^3)/3)/(A+B) = AB(A-B)/3.QED - very nice I think!For some reason, I could only use the equation feature for posting once - it would not work the subsequent times Itried to use them.
Last edited by OMD on March 14th, 2005, 11:00 pm, edited 1 time in total.
 
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OMD
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Nice Brownian Motion exercise (not really a teaser)

March 15th, 2005, 7:37 am

Here is one that was an exercise as opposed to an example - similar principles at work and really nice result that is well known, but proven by the unexpected methods of probability - although one should not really be too surprised since probability is just a special branch of real analysis:Let h(x,y) be a harmonic function - i.e twice differentiable continuously where h_xx + h_yy = 0 defined over a domain containing a disc of radius r around (x_0,y_0).Using the theory of martingales in a method similar to the previous methods below, prove the well known fact that:where the integral is done on the circle we will call C of radius r about (x_0,y_0)Proof:Let B^1_t and B^2_t be independent Brownian motions with (B^1_t, B^2_t) starting at the point (x_0,y_0) in R^2.Consider h(B^1_t, B^2_t). Since B^1_t and B^2_t are independent, h_t = 0, and h is harmonic, we know by Ito's formula thath(B^1_t, B^2_t) is a local martingale.Now let T be the stopping time defined by T = inf{t : h(B^1_t, B^2_t) is on circle C}Let u = min{t,T}Then M_u := h(B^1_u,B^2_u) is a bounded continuous local martingale which implies that it is an honest martingale.We know that E(M_u) = M_0 = h(x_0, y_0).By the dominated convergence theorem, we know that:lim_{u --> \infty} E(M_u) = E(M_T) = h(x_0, y_0), but we also know thatedited - to say that the above integral is goofed a little - the integral is not from 0 to t! Of course, it is over the circle of radius r centered at (x_0,y_0). D'OH!QED - real nice as well!Note that by the Cauchy-Riemann equations that the real and imaginary parts of a complex valued function (with domain the complexes as well of course) are harmonic, so this extends nicely to holomorphic functions as well. Then this can be used to prove Liouville's theorem that if an entire function has a maximum, then it is constant. Reason:If f: C --> C is entire with a max at w and f(w) = M, then the integral on any circle of radius r centered at w divided by 2 * pi * r must be less than or equal to f(w) since f is always less than or equal to f(w) over the whole circle, but by the previous considerations, it must equal f(w) on the whole circle of radius r (except at a measure zero - but by continuity of f, it is = M on the whole circle). Since r was arbitrary f(z) = M for all complex z.Of course this is true if w is only a local maximum since we would be able to find a disc of small radius where f must be constant, but by the properties of holomorphic functions, this means that f is constant over the entire domain where it is defined - remember domains are connected.Martingales are real nice!
Last edited by OMD on March 14th, 2005, 11:00 pm, edited 1 time in total.