Re: Cumulative probability that random walk variable

*To*: mathgroup at smc.vnet.net*Subject*: [mg104876] Re: [mg104834] Cumulative probability that random walk variable*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Thu, 12 Nov 2009 06:04:56 -0500 (EST)*References*: <200911110928.EAA29352@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

For this, we need a probability density for t itself, not just for x[t]. Assuming the uniform distribution, the problem goes like this: Clear[xCDF] xCDF[x_, t_] = CDF[NormalDistribution[0, Sqrt@t], x] 1/2 (1 + Erf[x/(Sqrt[2] Sqrt[t])]) Clear[tPDF] tPDF[t_] = Boole[0 <= t <= 5]/5 1/5 Boole[0 <= t <= 5] xCDF[x_] = Integrate[tPDF[t] (1 - xCDF[x, t]), {t, 0, 5}, Assumptions -> {0 < x <= 5}] (E^(-(x^2/ 10)) (-Sqrt[10] x + 5 E^(x^2/10) Sqrt[\[Pi]] Erfc[x/Sqrt[10]] + E^(x^2/10) Sqrt[\[Pi]] x^2 Erfc[x/Sqrt[10]]))/(10 Sqrt[\[Pi]]) Now specifically for x == 2: xCDF[2] N@% (-2 Sqrt[10] + 9 E^(2/5) Sqrt[\[Pi]] Erfc[Sqrt[2/5]])/(10 E^( 2/5) Sqrt[\[Pi]]) 0.0947972 If the distribution of t is complicated, you'll need NIntegrate, rather than Integrate. If the distribution of t is discrete, you'll need Sum. Bobby On Wed, 11 Nov 2009 03:28:02 -0600, Kelly Jones <kelly.terry.jones at gmail.com> 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. > -- DrMajorBob at yahoo.com

**References**:**Cumulative probability that random walk variable exceeds given value***From:*Kelly Jones <kelly.terry.jones@gmail.com>