       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 is 0, x follows the standard normal
> distribution, x follows the normal distribution with mean 0 and
> standard deviation Sqrt, etc.
>
> It's easy to compute the probability that x > 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 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 may be less than 2" is
>  sloppy, since x 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

```

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