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.