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Re: Cumulative probability that random walk variable exceeds

  • To: mathgroup at smc.vnet.net
  • Subject: [mg104863] Re: [mg104834] Cumulative probability that random walk variable exceeds
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Thu, 12 Nov 2009 06:02:27 -0500 (EST)
  • References: <200911110928.EAA29352@smc.vnet.net>

Kelly Jones wrote:
> 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.

Should be something like integrating 1-CDF[...,2] from 0 to 5, then 
divide by length of integration range (which of course is just 5).

In[9]:= Integrate[f[t, 2], {t, 0, 5}]/5
Out[9]= 1/5*(-(Sqrt[(10/Pi)]/E^(2/5)) - 9/2*(-1 + Erf[Sqrt[2/5]]))

In[8]:= N[%]
Out[8]= 0.0947972

This assumes your t is uniformly distributed. It also assumes I know 
what I'm talking (writing) about, which in this instance is iffy at best.

Daniel Lichtblau
Wolfram Research


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