Specific questions are bolded below. I've been unsuccessful in solving the following problem., which is exercise 5.2.3 from Probability and Random Processes by Grimmett and Stirzaker.

Let $X_1, X_2,\ldots$ be independent and identically distributed random variables with the logarithmic mass function $$ f(k) = \frac{(1-p)^k}{k \log(1/p)}, \quad k \geq 1, $$
where $0 < p < 1$. If $N$ is independent of the $X_i$ and has the Poisson distribution with parameter $\mu$, show that $Y=\sum_{i=1}^N X_i$ has a negative binomial distribution.

My strategy is to compute the probability generating function (pgf) for $Y$ and and the pgf for a negative binomial distribution (nbd) and show that they're the same, which would imply that $Y$ is indeed distributed negative binomially. I'm using the fact that random variables $f$ and $g$ have the same distribution iff they have the same pgf. Is it true that random variables $f$ and $g$ have the same distribution iff they have the same pgf? If so, please continue reading my attempted solution.

Since $N$ is independent of each $X_i$, Theorem 5.1.25 of the textbook says that $$G_Y= G_{N} \circ G_{X}.$$

It's easy to compute (and it's done in the textbook) $$G_{N}(s) = \exp(\mu(s-1)).$$ Further $$ \begin{align} G_{X}(s) &= \sum_{k\geq 1}s^{k} \frac{(1-p)^k}{k \log(1/p)} \\ &= \frac{1}{\log(1/p)} \sum_{k\geq 1} \frac{(s(1-p))^k}{k} \\ &= \frac{\log(1/q)}{\log(1/p)} \sum_{k\geq 1} \frac{(1-q)^k}{k\log(1/q)} \qquad (q := 1-s(1-p))\\ &= \frac{\log(1/q)}{\log(1/p)}\qquad \bigg(\text{Since } \frac{(1-q)^k}{k\log(1/q)} \text{ is a pmf.}\bigg) \\ &= \frac{\log q}{\log p} \\ &= \frac{\log(1-s(1-p))}{\log p}. \end{align} $$ Therefore $$ \begin{align} G_{Y}(s) &= G_{N}(G_{X}(s)) \\ &= \exp\bigg(\mu \big(\frac{\log(1-s(1-p))}{\log p} - 1\big)\bigg). \end{align} $$

Now since an nbd with parameters $r$ and $\alpha$ is a sum of $r$ independent geometric random variables (each with parameter $\alpha$), that nbd has pgf given by raising the pgf of each geomtric random variable to the power $r$: $$ \left(\frac{\alpha s}{1-s(1-\alpha)}\right)^{r}. $$ However, setting $$ \exp\bigg(\mu \big(\frac{\log(1-s(1-p))}{\log p} - 1\big)\bigg) = \left(\frac{\alpha s}{1-s(1-\alpha)}\right)^{r} $$ allows me to solve for $\alpha$, but gives something weird, which I doubt is correct. In fact, I suspect that it should be the case that $Y$ is nbd with paramters $\alpha = p$ and $r= \mu$.

Where am I erring?

Is there a better (i.e. simpler) approach to this problem?

As a side question, I noticed that Wikipedia gives the pgf of a negative binomial as $$ \left(\frac{1-\alpha}{1-\alpha s}\right)^{r}, $$ which is apparently different from the one I calculated above (though still not clearly equal to $G_{Y}(s)$ for any values of $\alpha$ and $r$). Why the discrepancy between my calculation of the pgf of an nbd and Wikpedia's entry?

EDIT In the textbook I cited above, a random variable $W_{r}$ is defined to be nbd if it has pmf $$ \mathbb{P}(W_{r} = k) = \binom{k-1}{r-1} \alpha^{r} (1-\alpha)^{k-r}, \qquad k=r,r+1,\ldots. $$ It is then pointed out that $W_{r}$ is the sum of $r$ independent geometric random variables; i.e. $$ W_{r} = Z_{1} + Z_{2} + \cdots + Z_{r}, $$ where each $Z_{i}$ has pmf $$ f_{Z} = \alpha(1-\alpha)^{k-1}, \qquad k=1,2,\ldots. $$

  • $\begingroup$ The pgf of a geometric distribution is $\left(\dfrac{\alpha s}{1-s(1-\alpha)}\right)$, or it is $\left(\dfrac{1-\alpha}{1-\alpha s}\right)^r$, where the $\alpha$ in one case is the same as the $1-\alpha$ in the other case, so those are not actually discrepant. $\endgroup$ Oct 29, 2011 at 23:42

2 Answers 2


The discrepancy results from the fact that you used the geometric distribution supported on the set $\{1,2,3,\ldots\}$ (number of trials needed to get one success, which is $1$ if there's a success on the first trial), whereas the article used the geometric distribution supported on the set $\{0,1,2,3,\ldots\}$ (number of trials before the first success, which is $0$ if there's a success on the first trial). Also, which "negative binomial distribution" are you using? The the distribution of the number of trials needed to get $r$ successes (counting both the successes and the failures among the trials) or the distribution of the number of failures before the $r$th success (or of successes before the $r$th failure, which is how the article phrases it)? The latter distribution is supported on $\{0,1,2,3,\ldots\}$ and the former on $\{r,r+1,r+2,\ldots\}$. The latter is more interesting in some ways because it allows $r$ to be a non-integer and gives you an actually infinitely divisible family of probability distributions.

  • $\begingroup$ Okay, thanks for the explanation of the discrepancy. I'm using the negative binomial distribution as defined (on page 61) in the text I cited: it is the "waiting time for the $r$th success" and is supported on $\{r, r+1, r+2, \ldots\}$. $\endgroup$ Oct 27, 2011 at 23:00
  • $\begingroup$ Good luck with the rest of this. Notice that you can vote on my answer. Or accept it, if it's suitable for that. $\endgroup$ Oct 27, 2011 at 23:43
  • $\begingroup$ I've upvoted your answer, but since you only answered my "side question" I'll not accept it. Thanks again for answering that one though. $\endgroup$ Oct 27, 2011 at 23:52

You have: $$ \exp\left(\mu \left(\frac{\log(1-s(1-p))}{\log p} - 1\right)\right) = \left(\frac{\alpha s}{1-s(1-\alpha)}\right)^r $$

Now suppose $\mu = r\log p$.

Then $$ \begin{align} & \exp\left(\mu \left(\frac{\log(1-s(1-p))}{\log p} - 1\right)\right) = \exp\left(r\log p \left(\frac{\log(1-s(1-p))}{\log p} - 1\right)\right) \\ \\ & = \exp\left(r\log(1-s(1-p)) - r\log p\right) = \exp\left(r \log\left(\frac{1-s(1-p)}{p}\right)\right) \\ \\ & = \left(\frac{1-s(1-p)}{p}\right)^r \end{align} $$ and there you have the pgf of a negative binomial distribution (of the kind supported on ${0,1,2,3,\ldots}$).

So this works with $r=\mu/\log p$.

Two probability distributions supported on $\{0,1,2,3,\ldots\}$ have the same pgf if and only if they are the same distribution. Notice that the power series converge when the argument is less than $1$, so this is an instance of the fact that generally two convergent power series are equal as functions of their argument if and only if they have the same coefficients, i.e. they are the same series. To see this, suppose they have different coefficients, and subtract, and consider what happens in a neighborhood of $0$. (Here you need the fact that convergent power series define continuous functions on the interior of the region of convergence.)


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