# How to derive the sampling distribution of some statistics

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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##?

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