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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 sample coefficient of variation and the sample skewness?

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  • $\begingroup$ I am interested in the derivation of the formula for the sampling distribution of the sample coefficient of variation as an estimator of $c_v$ and the sample skewness as an estimator of $\tilde{\mu}_3$. $\endgroup$
    – AdVen
    Jun 23 at 13:56
  • $\begingroup$ Are you aware there are various different formulas for both those things? They don't differ enough to worry about here, but it's always good to be specific. $\endgroup$
    – whuber
    Jun 23 at 16:14

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