Re: Second Opinion

*To*: mathgroup at smc.vnet.net*Subject*: [mg26377] Re: [mg26373] Second Opinion*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Sat, 16 Dec 2000 02:40:09 -0500 (EST)*References*: <200012130741.CAA17254@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

John Lai wrote: > > Hello all, > I tried to calculate Poisson Distribution in a backdoor way and used > mathematica to model it. I could not get what I wanted. I don't think it > is mathematica problem and more than likely my method is flawed. So I toss > this out to see if some of you may spot my error. > > Poisson Distribution,P(n) =1-Summation [exp(-n)*(n^x)]/Factorial(x) where x > goes from 0 to N-1 > > For given n and N, P(n) can be determined easily. However, I want to > determine N if P(n) and n are specified and I do not want to get access to > Poisson lookup table. My idea is to calculate P(n) with a series of n and N > (essentially generating the tables). Plot a surface curve whose variables > are n, P(n) and N. The idea was once this surface is obtained, with x-axis > as n, y-axis as P(n) and z-axis as N, then for a given n and P(n) I can > obtain N. > > I wrote a C program to generate P(n) and use mathematica to plot this > surface. I have 14 sets of n and in each set of n, I have 139 variables > (i.e. N runs from 1 to 140 ), so there are 139 corresponding values of P(n) > for each n. When I tried to use the function Fit to estimate this surface, > it took about ½ hr for my 500MHz desktop to calculate! And the resultant > expression is huge! > > Then, I cut down the dimension of my data set. For each n, I generated 10 > values of N and repeated the process again. However, no matter what > combination of polynomial I used (x,x^-1,Exp(-x),Exp(-x^2),Exp(-x-y).), the > resulting equation of the surface is meaningless. It doesn't look right (at > least I expected it to resemble some sort of Poisson or even Gaussian shape) > and substituting P(n) and n back, I got garbage. I have enclosed a .nb file > for reference. [Contact the author to obtain this file - moderator] > > So after all this, does it mean that my scheme of calculating Poisson > Distribution is fundamentally wrong? > Any suggestions are appreciated and thanks in advance. > > John Lai I do not claim to have understood much of the above, but I think you may be trying to replicate the Quantile function. In[23]:= <<Statistics`; In[24]:= Quantile[PoissonDistribution[1.8], .4] Out[24]= 1 In[25]:= Quantile[PoissonDistribution[1.8], .1] Out[25]= 0 In[26]:= Quantile[PoissonDistribution[1.8], .8] Out[26]= 3 Daniel Lichtblau Wolfram Research

**References**:**Second Opinion***From:*"John Lai" <john.lai@worldnet.att.net>