What is the distribution of the sample variance of the Skellam distribution? I want to estimate the parameter  $\mu$ using the difference between two Poisson distributions with the same parameter, i.e. a Skellam distribution with $\mu_1=\mu_2 = \mu$. 
I can calculate the variance of the Skellam distribution as the average variance over multiple samples, however, depending on sample size and some good/bad luck, I have quite a bit of error in my estimate (red crosses in the figure below); the variance should be $2\mu$. If I knew what distribution I was looking for, I could fit that instead to hopefully get a more robust estimate.
Why am I doing this if I appear to know $\mu$? In my application, I only know $\hat{\mu}=\beta_1\mu+\beta_0$, so I need a second estimate of $\mu$ to find out the betas. 

 A: Some of the comments have pointed out that there may be better ways to solve your problem that do not involve finding the distribution of the sample variance of the Skellam distribution.  Here I will answer the title question irrespective of those other issues.

The exact distribution of the sample variance from a Skellam distribution is complicated, since it is a quadratic function of Poisson random variables.  Using the large-sample approximation with the chi-squared distribution (see e.g., O'Neill 2004) you can get the approximate distribution:
$$S_n^2 \sim 2 \mu \cdot \frac{\text{Chi-Sq}(DF_n)}{DF_n} \quad \quad \quad DF_n = \frac{2n \mu^2}{3 \mu^2 + (n-3) / (n-1)}.$$
This should get you a reasonable approximation to the distribution of the sample variance, so long as $n$ is not too small.
A: I'm not sure what the problem is: sample variance is a random variable. I would guess it is distributed as a (re-scaled) Chi-square for this distribution. For a fixed number of samples, the standard error of the variance statistic will be proportional to the population variance you are trying to estimate. Thus I would expect these histograms to get 'wider' as the population variance increases.  
