An interesting problem about Brownian motions

 Consider a standard 1 dimensional Brownian motion B_t, starting at 0. We would like to show that for any c > 0, the chance that |B_t| does not exceed c for t < 1 is always positive. This is equivalent to saying that the stopping time tau = inf {t: B_t > c} has unbounded support on RR.
  The idea is to first establish the result for a particular value c. Here we can take c = 1 and use second moment of tau or c =2 and use exponential moment of tau. We present both approaches. It illustrates the power of Martingale method in the continuous setting.

  Recall that B_t^2 – t is a Martingale. One way to see it is by Ito’s formula. d (B_t^2 – t) = 2 B_t dB_t + dt – dt  = 2B_t dB_t, which shows it has no drift term, and almost surely bounded by continuity of B_t.
  Thus by optional stopping theorem, Etau = E B_tau^2 = 1 since tau has finite mean, otherwise we would already have tau > c with positive probability for any c!
  Now we could use another Brownian Martingale to establish that tau is not identically 1, by showing that var tau > 0. But it suffices to have the result for some c < 1. This is easily accomplished by a contradiction argument. (To illustrate how uninitiated I am, I don’t know how to use standard inequalities, e.g., Markov, to get this)

  Alternatively one could use the Martingale exp(B_t – t/2), and obtain P(tau < c) = P(-tau/2 > -c/2) le E exp(B_tau – tau/2) / exp(B_tau – c/2) = 1/exp(1-c/2) = exp(c/2 – 1).
Thus one could take c = 2 -epsilon to get P(tau ge c) > 0, or c =2 if one can further control second moment of tau.

  So in summary we have established P(tau > c ) > 0 for some c > 0. Next Let c’ be the first time P(tau > c’) = 0 for the sake of contradiction. Then at time c’ – epsilon, there is a positive probability that B_t has not exited the interval [-1,1], and conditioned on this event, there is positive probability that B_{c’ -epsilon} in (a-epsilon’,a+epsilon’), a small interval centered at a > 0 that’s strictly contained in (-1,1), with a + epsilon’ < 1. Further condition on  that, we can consider the event that B_{c’-epsilon) exits the interval (-1,a+epsilon’) first through -1. This has positive probability again by exploiting the Martingale B_t -t. Hence with positive probability the brownian motion passes through 0 after c’ – epsilon without exiting [-1,1]. Let tau’ be the first time it does that. Then we can restart the Brownian motion from 0 at time tau’, using the strong Markov property. If we define tau” to be the conditional stopping time that it exits [-1.1], after tau’, then P(tau” > c’ – epsilon) > 0, by the same reasoning as above. Thus multiplying all the positive probability together, we obtain that P(tau > 2(c’ -epsilon)) > 0. Iterating we see that tau has infinite support.
   

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About aquazorcarson

math PhD at Stanford, studying probability
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