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Write $m=r+s$ and $\beta =\lambda +n$. Conditional independence of the next count given $\theta$ allows us to average its Poisson distribution over the posterior distribution:This Poisson-gamma mixture is the posterior predictive distribution. In the stated negative binomial distribution convention its parameters areIts expected value $m/\beta$ equals the posterior mean of $\theta$, while its variance $m/\beta +m/{\beta }^2$ includes both conditional Poisson variation and uncertainty in $\theta$.

Put $D=1-(1-p)(1-q)=p+q-pq$. By the law of total probability,For $0\le t<1$, termwise differentiation of the geometric series, valid inside its radius of convergence, givesTaking $t=(1-p)(1-q)$ therefore yieldsSince $D-p=(1-p)q$, this is the zero-based geometric distribution with parameterThis illustrates Bernoulli thinning of a geometric count. Here $0<p_{\ast }<1$, as $0<p,q<1$.

For the reverse conditional distribution, Bayes' theorem gives, for $n\ge k$,ThusFor $n<k$ the conditional probability is zero. The conditional distribution of $X-k$ is the failure-count form of a negative binomial distribution, with $k+1$ successes and success parameter $r$: the geometric-binomial posterior is shifted negative binomial.

Let $T_2$ be the time of the second occurrence of $3$, whether or not it is adjacent to the first. On every digit sequence, completion of a consecutive pair requires at least two occurrences of $3$, so $Y⩾T_2$.

The waiting time for the first $3$ has mean $3$. Once that digit has appeared, the fresh independent digits give another waiting time with the same mean for the next $3$. Thus $\mathbb{E}[T_2]=3+3=6$, the mean of the corresponding negative binomial distribution. Monotonicity of expectation now proves $\mathbb{E}[Y]⩾6$, without computing the waiting time for the consecutive pair itself. The argument remains valid even before establishing that $Y$ has finite mean, since these waiting times are nonnegative.

For conditionally independent Poisson observations of mean $\theta$ and a gamma distribution prior of shape $r>0$ and rate $\lambda >0$, the likelihood function contributes ${\theta }^{{\sum }_i{x}_i}{e}^{-n\theta }$. Multiplying by the prior density gives a posterior distribution with shape $r+{\sum }_i{x}_i$ and rate $\lambda +n$. The single-observation update is shape $r+X$ and rate $\lambda +1$.

The posterior mean is the convex combinationThus the prior mean $r/\lambda$ and the sample maximum-likelihood estimate $\bar{x}$ bracket it. Integrating the next conditional Poisson distribution against this posterior distribution gives the Poisson-gamma mixture, a negative binomial distribution; its spread includes uncertainty in $\theta$ as well as the conditional count variation.

If $Y\mid \Lambda \sim \mathrm{Poisson}(\Lambda )$ and $\Lambda$ has a gamma distribution, then $Y$ has a negative binomial distribution. For gamma shape $\nu$ and scale $\alpha$,

A model with $Y\mid \Lambda \sim \mathrm{Poisson}(\Lambda )$ for a nonnegative random intensity. Conditional expectation and total variance give $\mathbb{E}Y=\mathbb{E}\Lambda$ and $\mathrm{Var}Y=\mathbb{E}\Lambda +\mathrm{Var}\Lambda$, explaining overdispersion. A Poisson-gamma mixture gives the negative binomial distribution.

Let $N$ have the positive-integer geometric distribution with success parameter $p\in (0,1)$, and conditionally toss $N$ independent coins with head probability $\theta \in (0,1)$. Exactly one head has conditional probability $n\theta (1-\theta {)}^{n-1}$. Combining this with the prior by Bayes' theorem gives the displayed posterior with $r=(1-p)(1-\theta )$; its normalization uses ${\sum }_{n\ge 1}n{r}^{n-1}=(1-r{)}^{-2}$. For $p=\theta =1/2$, the posterior is $9n/4^{n+1}$. It is also a shifted negative binomial distribution, with $N-1$ equal in law to the number of failures before two successes of probability $1-r$.