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.