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Re: Cumulative probability that random walk variable exceeds given value

  • To: mathgroup at smc.vnet.net
  • Subject: [mg104887] Re: Cumulative probability that random walk variable exceeds given value
  • From: Bill Rowe <readnews at sbcglobal.net>
  • Date: Thu, 12 Nov 2009 06:07:08 -0500 (EST)

On 11/11/09 at 4:28 AM, kelly.terry.jones at gmail.com (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).

It is easy? It certainly is simple to compute the probability a
randomly selected value from the distribution x[5] is greater
than 2. But since x[5] is a distribution not a variable, I don't
see how you even begin to make a comparison between x[5] and 2.

I suspect what you have in mind here is

1-CDF[NormalDistribution[0, Sqrt[5]], 2]

as being the desired probability.

>How do I compute the probability that x[t] > 2 for 0 <= t <= 5.

You don't without more information. For each value of t you have
a different distribution. You have no way to say something about
the probability of a value randomly selected from one of these
distributions is greater than some given value until you say
something about the probability of choosing say distribution x[2].

>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.

This is a reasonably well understood problem. You might start
with looking up "random walk" on Wikipedia.

>% 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.

You really aren't being that clear.



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