Let $X \thicksim Negbin(r,p)$ where $(0\lt p \lt 1) $

I want to derive skewness and kurtosis of $X$ by getting the Cgf of X.

First, since Followance of Negative Binomial equals to the distribution r-th repetition of $Geo(p)$, such as $X \equiv Z_1+Z_2+...+Z_r$ where $Z_i \thicksim Geo(p), \forall i\in\{1,2,3,..r\} $

then since Mgf of $Z_i = pe^t(1-e^t(1-p))^{-1}$, Mgf of $X = (pe^t(1-e^t(1-p))^{-1})^r$

Then Cgf of $X$,

$C(t) = \log(pe^t(1-e^t(1-p))^{-1})^r\\ = r[\log p + t - \log(1-e^t(1-p))]$

then since $-\log(1-A) = A +1/2A^2+1/3A^3 ..$,

$C(t) = r[\log p + t + \sum_{i=1}^\infty\dfrac{[e^t(1-p)]^i}{i}]$

then I need to change above identity into the form of

$\sum_{i=1}^\infty \dfrac{c_i(0)t^i}{i!}$ so that I could easily assume the characteristic of distribution from the coefficients, such as mean/variance/skewness/kurtosis etc.

But I can't figure out how could I rearrange the $\sum_{i=1}^\infty\dfrac{[e^t(1-p)]^i}{i}$ into a form of $t^i$ not $(e^t)^i$.

Any hints?

  • 1
    $\begingroup$ This should probably have a self-study tag (see the tag wiki) -- fortunately you seem to already be following its guidelines. $\endgroup$
    – Glen_b
    Jul 9, 2017 at 8:02

1 Answer 1


This won't be the easy way to do it, but if you can't see any neat way to get anywhere, this would seem amenable to simply expanding each series in the last term and rearranging/collecting terms (as long as the conditions hold for doing that -- so you need your various series to be absolutely convergent in some radius of convergence that at least includes a neighborhood of $t=0$).

That is, noting that $(e^t)^k=e^{kt}$, you could expand each of the exponential series and collect terms in constants, in $t$, in $t^2/2!$, and so on up to fourth order. Each term in $t^j/j!$ will have a series, and each such series is fairly easy to recognize (assuming I did it right).

Of course if you just rely on additivity of the cumulants you could proceed from the geometric (which doesn't save much, admittedly, just the multiplication by $r$)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.