Let $X$ be the number of failures before the first success in a sequence of Bernoulli trials with probability of success $\theta$. Then $P_{\theta}[X = k] = (1-\theta)^{k}\theta$, $k = 1,2,\ldots$ This is called the geometric distribution ($\mathcal{G}(\theta))$.\

(a) Show that the family of geometric distributions is a one-parameter exponential family with $T(x) = x$.

(b) Deduce that if $X_{1},X_{2},\ldots, X_{n}$ is a sample from $\mathcal{G}(\theta)$, then the distributions of $\sum_{i=1}^{n}X_{i}$ form a one-parameter exponential family.

(c) Show that $\sum_{i=1}^{n}X_{i}$ in part (b) has a negative binomial distribution with parameters $(n,\theta)$ defined by \begin{align*} P_{\theta}\left[\sum_{i=}^{n}X_{i} = k\right] = {{n+k-1}\choose{k}}(1-\theta)^{k}\theta^{n} \end{align*}

where $k = 0,1,2,\ldots$


So far, I am fine with the first. As to the one-parameter exponential family, the author considers the following expression \begin{align*} p(x,\theta) = h(x)\exp(\eta(\theta)T(x) - B(\theta)) \end{align*}

Could someone help me with the second and third questions? Thanks in advance!

  • 3
    $\begingroup$ there are many ways to show this result, one way that is somewhat simple to do (if you assume each trial is independent) is to write out the generating/characteristic functions for each trial. Multiplication of these should bring you to the corresponding negative binomial generating/characteristic function. $\endgroup$ – Lucas Roberts May 13 at 22:49
  • 1
    $\begingroup$ The approach suggested by @Lucas is a fine one. An intuitively appealing one refers to the contexts in which these distributions typically arise: namely, the number of independent trials needed to observe $k$ failures obviously equals the sum of the numbers of trials needed to observe the first failure, then the second, ..., and finally the $k^\text{th}.$ $\endgroup$ – whuber May 13 at 23:15

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