# Calculating a Poisson probability from the chacteristic function?

In a previous homework assignment we were given a function that corresponds to an arbitrary angular distribution $A_{FB}=(F-B)/(F+B)=(F-B)/N$, where F = # of events in the forward hemisphere, B = # of events in the backward hemisphere and N = F + B = total # of events. We were told to assume F and B were two independent Poisson distributions, with the total # of events N given by F + B and hence N was variable (i.e. not fixed).

$P_{A_{FB}}=P_{P}(F;v_{F})\cdot P_{P}(B;v_{B})=\frac{(v_{F})^{F}}{F!}e^{-v_{F}}\cdot\frac{(v_{B})^{B}}{B!}e^{-v_{B}}$

Now we want to use the characteristic function $\Phi_{n}(k;v_{n})=e^{-v_{n}(e^{ik}-1)}$ associated with the Poisson distribution to show that the total number of events N = F + B is also Poisson-distributed, (i.e. $P_{P}(N;v_{N})=\frac{(v_{N})^{N}}{N!}e^{-v_{N}}$ , with mean/expected number of events $v_{N}=v_{F}+v_{B}$ and variance $v_{N}=N=F+B$).

My attempt at a solution:

I know that the way this was expected to be solved was to leverage the fact that $\Phi_{N}(k;v_{n})=\Phi_{F+B}(k;v_{n})=\Phi_{F}(k;v_{F})\Phi_{B}(k;v_{B})$.

We then determine: $P_{P}(N;v_{N})=\sum_{k=0}^{\infty}e^{-ikN}\Phi_{F}(k;v_{F})\Phi_{B}(k;v_{B})= \sum_{k=0}^{\infty}e^{-ikF}\Phi_{F}(k;v_{F})e^{-ikB}\Phi_{B}(k;v_{B})=P_{P}(F;v_{F})\cdot P_{P}(B;v_{B})$

and monkey around with the algebra from there to prove the statement in the prompt. However, I wasted quite a bit of time trying to evaluate the infinite sum over k directly and prove it an alternate way:

$P_{P}(N;v_{N})=\sum_{k=0}^{\infty}e^{-ikN}\Phi_{F}(k;v_{F})\Phi_{B}(k;v_{B}) =\sum_{k=0}^{\infty}e^{-ikN} e^{-(v_{F}+v_{B})(e^{ik}-1)}$.

I tried to do this: ${e}^{(v_{F}+v_{B})}=\sum_{k=0}^{\infty}\frac{(v_{F}+v_{B})^{k}}{{k!}}$

and put this back into the expression like so: $P_{P}(N;v_{N})=\sum_{k=0}^{\infty}\frac{(v_{F}+v_{B})^{k}}{{k!}}e^{-ikN}e^{-(v_{F}+v_{B})e^{ik}}$

But after this I could go no farther. Collecting these terms into something like $\frac{(expression)^{k}}{k!}$ to hopefully make it look like another exponent series expansion does not work because of the $e^{e^{ik}}$ term.

-This is the first time I have ever seen any function that contains a double exponential. How does one usually deal with these? There seems to be no decent Taylor series representation, unless $e^{e^{x}}=\sum_{i=0}^{\infty}\frac{(e^{x})^{i}}{i!}$ but I am not sure is this is a useful sum or even allowed.

-Is breaking down an exponent that does not depend on k (the index in the infinite sum) into an infinite series that does depend on k allowed, or do I have to use different indices, such as turning the exponential into an infinite sum over j?

-How can I evaluate the probability I want by using Fourier inversion on the characteristic function with n=N and then solving the infinite sum over k? Every source I have looked up has not had a way to calculate a Poisson probability; it is usually given a priori. Any help?

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The answer to the second question is a definite "no". By using the index of an existing sum instead of introducing an additional sum over a new index, you've replaced the exponential not by a series but by a single term of that series. – joriki Oct 12 '12 at 5:21
What does "and variance $v_N=N=F+B$" mean? The variance isn't random, but the right-hand side is. – joriki Oct 12 '12 at 5:22
@joriki I think that is just meant to say that the variance of N is equal to the sum of means of F and B. Which, since this is a Poisson distribution, also happen to be the variances. That did confuse me a bit as well but I do believe that the aim of this question was to point out that N was 'variable' in our previous working of this problem. Not physically, of course, but in the assumptions of the math. Also, thank you. I thought that was too convenient to be true. Perhaps I should just expand a few terms in the sigma in terms of series, multiply a few terms and try to find a pattern? – user44430 Oct 14 '12 at 6:12
It didn't confuse you "as well" -- I don't think it confuses me; I think it's just wrong :-) You have two different equations for $v_N$; in one of them it's the sum of two random variables and in the other its the sum of their variances; that just doesn't make any sense. If you mean the expectation values and not the variables themselves, then you should write it that way. – joriki Oct 14 '12 at 9:03