From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Transformed Poisson distribution equals ...
Date: 7 Jan 1999 01:09:47 GMT
Newsgroups: sci.math

In article <3693C25F.41C6@gwdg.de>, Daniel Weiss  <dweiss1@gwdg.de> wrote:

>Given a random variable A with Poisson distribution (mean a,
>std. dev. sqrt(a)), and given B=log(A), what can be said about B:

It's undefined when A = 0 (which occurs with positive probability in
any Poisson distribution).

>- Is it still Poisson distributed? In that case, what mean
>  and standard deviation? I tried evaluating the sum
>  E(B)=Sum(log(i)*p(i)) instead of E(A)=Sum(i*p(i)) (with p(i) the
>  Poisson probability for i events), but didn't succeed.

Undefined, because log(0) is undefined.  Or -infinity if you wish.

>The reason for my question: I am counting photons from a coherent source
>(Poisson distributed), and am using the negative logarithm of the
>photon count to determine absorption in the probe. I am interested in
>the statistical properties of the derived variable 'absorption'.

So your "a" is large enough that A=0 is extremely unlikely, and to avoid
the unpleasantness at 0 you might as well define, say, 
B = { log(A) if A > 0
    { 0      if A = 0

I doubt that you'll find exact "closed-form" formulas for the mean or other
moments, but for large a you can find asymptotic formulas.  Write 
B = log(a) + log(1 + (A-a)/a) 
  = log(a) + (A-a)/a - (A-a)^2/(2a^2) + (A-a)^3/(3a^3) - ...
so the mean is
   E[B] = log(a) + E[A-a]/a - E[(A-a)^2]/(2a^2) + E[(A-a)^3]/(3a^3) - ...
        = log(a)  - 1/(2 a) + 1/(3a^2) + O(a^(-3))

Similarly
B^2 = log(a)^2 + 2 log(a)(A-a)/a + ... (I'm too lazy to write out so many 
terms) from which you can get the variance and standard deviation of B.

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2