A How to derive the sampling distribution of some statistics

[Ad VanderVen]

TL;DR Summary

Given a geometric Erlang distribution, how can I derive the sampling distribution of the coefficient of variation and the skewness for a sample of size $n$.

Assume that $T$ has an Erlang distribution:
$$\displaystyle f \left(t \, | \, k \right)=\frac{\lambda ^{k }~t ^{k -1}~e^{-\lambda ~t }}{\left(k -1\right)!}$$
and $K$ has a geometric distribution
$$\displaystyle P \left( K=k \right) \, = \, \left( 1-p \right) ^{k-1}p$$
Then the compound distribution has the following form.
$$\displaystyle g \left(t \right)= \sum _{k=1}^{\infty} f \left(t \, | \, k \right)~P \left(K =k \right)=\frac{\lambda ~p }{e^{\lambda ~t ~p }}$$
with expectation:
$$\displaystyle \mu_{{1}}\, = \,{\frac {1}{\lambda\,p}}$$
variance:
$$\displaystyle \mu_{{2}}\, = \,{\frac {1}{{\lambda}^{2}{p}^{2}}}$$
and third central moment:
$$\displaystyle \mu_{{3}}\, = \, {\frac {2}{{\lambda}^{3}{p}^{3}}}$$
The coefficient of variation $c_v$ is given by:
$$\displaystyle {\it c_v}\, = \,{\frac { \sqrt{\mu_{{2}}}}{\mu_{{1}}}}=1$$
and the skewness $\tilde{\mu}_3$ by:
$$\displaystyle {\it \tilde{\mu}_3}\, = \,{\frac {\mu_{{3}}}{{\mu_{{2}}}^{3/2}}}=2$$
Is it possible to derive a formula for the sampling distribution of the coefficient of variation and the skewness with a sample size $n$?

 

[mathman]

https://en.wikipedia.org/wiki/Exponential_distribution
Should answer your question.

 

[Ad VanderVen]

https://en.wikipedia.org/wiki/Exponential_distribution
Should answer your question.

I am now aware that $g(t)$ is in fact an exponential distribution with rate parameter $\lambda \, p$. But the Wikipedia site on the exponential distribution makes no mention of sampling distributions.

 

[mathman]

Section on statistical parameters? I am not sure what you want.