Approximation of skellam distribution by a Gaussian one

[sabbagh80]

Hi, everybody

Let n_1 ~ Poisson (\lambda_1) and n_2 ~ Poisson (\lambda_2).
Now define n=n_1-n_2. We know n has "Skellam distribution" with mean \lambda_1-\lambda_2 and variance \lambda_1+\lambda_2, which is not easy to deal with.
I want to find the Pr(n \geq 0). Is it possible to find a good approximation for the above probability by employing an approximated "Gaussian distribution"? If "Gaussian" is not a good candidate, which distribution can I replace it with?

 

[pmsrw3]

If at least one of the lambdas is large, the Gaussian with the same mean and variance will be a good approximation.

 

[sabbagh80]

But it is not always the case. I want to deal with the more general cases.

 

[pmsrw3]

Yes, I know. But for small lambda, I don't think there's any simpler approximation. Of course, you could in that case just truncate the distributions. If the mean is small, the probably of large n is vanishingly small, so it won't introduce much inaccuracy to leave them out.

 

[bpet]

To approximate one distribution with another use maximum likelihood, i.e. maximize
E[\log(f(X;t))
wrt the parameter vector t, where f is the pdf or pmf of the approximating distribution. E.g. solving for the normal distribution we get \mu=E[X] and \sigma^2=E[X^2]-E[X]^2.

 

[sabbagh80]

To approximate one distribution with another use maximum likelihood, i.e. maximize
E[\log(f(X;t))]
wrt the parameter vector t, where f is the pdf or pmf of the approximating distribution. E.g. solving for the normal distribution we get \mu=E[X] and \sigma^2=E[X^2]-E[X]^2.

Could you please explain it in more details.