Re: Cumulative probability that random walk variable exceeds given
- To: mathgroup at smc.vnet.net
- Subject: [mg104849] Re: Cumulative probability that random walk variable exceeds given
- From: Roland Franzius <roland.franzius at uos.de>
- Date: Thu, 12 Nov 2009 05:59:34 -0500 (EST)
- References: <hde054$sn0$1@smc.vnet.net>
Kelly Jones schrieb:
> How can I use Mathematica to solve this problem?
>
> Let x[t] be a normally-distributed random variable with mean 0 and
> standard deviation Sqrt[t].
>
> In other words, x[0] is 0, x[1] follows the standard normal
> distribution, x[2] follows the normal distribution with mean 0 and
> standard deviation Sqrt[2], etc.
>
> It's easy to compute the probability that x[5] > 2 (for example).
>
> How do I compute the probability that x[t] > 2 for 0 <= t <= 5.
>
> In other words, the probablity that x[t] surpassed 2 at some point
> between t=0 and t=5, even though x[5] may be less than 2 itself. Notes:
>
> % My goal: predicting whether a continuous random walk will exceed a
> given value in a given period of time.
>
> % I realize that saying things like "x[5] may be less than 2" is
> sloppy, since x[5] is a distribution, not a value. Hopefully, my
> meaning is clear.
>
> % I tried doing this by adding/integrating probabilities like this
> (psuedo-code):
>
> P(x[t] > 2 for 0 <= t <= 5) = Integral[P(x[t] > 2),{t,0,5}]
>
> but this overcounts if x[t] > 2 for multiple values of t.
>
From symmetry and the strong Markov property it follows that starting
at the random time T_first(X=x) of the first hit on the boundary at x >
0 the random path for later times
X(t) - X(T_first(X=x)), t>T_first(X=x)
ist normal distributed with startpoint x. So you get from the so called
mirror principle equal weights for points X_t >x and X_t < x with a
visit in [x,oo) for any s<t
Pr[{X(s)> x at least once in 0<s<t} ]
= Pr[T_first(x)< t]
=Pr[X(t) > x] + Pr[{X(s)< x && X_s =x at least once in 0<s<t} ]
= 2 Pr[ X_t > x ]
= 1-Erf[x/(Sqrt[2t]]
From this you get the probability for a first hit on x=2 in 0<t<5
Erf[2/Sqrt[2*2]]-Erf[2/Sqrt[2*5]]
--
Roland Franzius