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If host-specific equilibrium mean burden has a gamma distribution, conditional Poisson distributions produce a Poisson-gamma mixture, hence a negative binomial distribution. With mean $m$ and shape $k$, its zero-burden probability is $(1+m/k{)}^{-k}$. Decreasing $k$ increases aggregation: a smaller infected-host prevalence can accompany the same mean burden and more heavily infected individuals.

The number of insurance claims in a specified exposure period. Its probability generating function combines with the moment-generating function of an independent claim size through the random-sum transform identity. A Poisson distribution, geometric distribution, or negative binomial distribution provides a possible count model, with different dispersion and zero-claim probabilities.

A compound mixed Poisson distribution is a random sum of independent claims whose count has a Poisson mixture law: conditional on a nonnegative random intensity $\Lambda$, $N$ has Poisson distribution with parameter $\Lambda$. The claim sizes are independent and identically distributed random variables, independent of the intensity and count mechanism. If $M_X$ is their moment-generating function, then $M_S(t)=\mathbb{E}\exp (\Lambda (M_X(t)-1))$ wherever finite. A gamma distribution intensity of shape $r$ and rate $\alpha$ gives a negative binomial distribution count with shape $r$ and success probability $\alpha /(\alpha +1)$. Independent intensities with common gamma rate add their shapes by additivity of independent gamma distributions with a common rate.

For independent rate-$\lambda$ exponential distribution claims and a zero-based negative binomial distribution count of shape $k$ and parameter $p$, the random-sum transform identity gives $M_S(t)=[p(\lambda -t)/(p\lambda -t){]}^k$. This equals the transform of a compound binomial distribution with count $\mathrm{Bin}(k,1-p)$ and rate-$p\lambda$ claim sizes. Conditional on its positive count $n$, the total has Erlang distribution, whose survival function is the inner exponential-polynomial sum. Conditioning yields the displayed finite mixture distribution formula for $x>0$. It replaces an infinite mixture and includes the zero-total atom $p^k$ through the omitted $n=0$ term.

For known size $\theta >0$, the negative binomial distribution is an exponential family with natural parameter $\eta =\log [\mu /(\theta +\mu )]<0$ and cumulant $b_{\theta }(\eta )=-\theta \log (1-e^{\eta })$. A generalized linear model can use a logarithmic link function for its mean, even though that link is not canonical. Jointly estimating the size changes the family and variance function, requiring additional updates beyond a single fixed-family GLM fit; this is what MASS::glm.nb does.

Let $\mathbb{P}(X=n)=p(1-p{)}^n$ for $n\ge 0$ and $Y\mid X=n\sim \mathrm{Binomial}(n,q)$, with $0<p,q<1$. Write $D=p+q-pq$. Summing the joint probability mass function with the differentiated geometric series identity ${\sum }_{n=k}^{\infty }\left(\genfrac{}{}{0ex}{}{n}{k}\right)t^{n-k}=(1-t{)}^{-k-1}$ gives $\mathbb{P}(Y=k)=(p/D)((1-p)q/D{)}^k$. By Bayes' theorem,Thus $X-k$, given $Y=k$, has the failure-count negative binomial distribution for $k+1$ successes and success parameter $D$. Its conditional expectation is $k+(k+1)(1-D)/D$ for $X$ itself. The shift by $k$, the number of retained successes, must be retained when identifying the posterior distribution.

In the stationary law of a two-resource linear flow network, the local counts are independent geometric variables with parameters $q_i={\rho }_i/(1-{\rho }_0)$. Conditional on local counts $a,b$, the through count has a negative binomial distribution with mean ${\rho }_0(a+b+1)/(1-{\rho }_0)$. With positive loads, it is correlated with each local count, even though those two local counts are independent. Degenerate zero-load cases must be distinguished from the positive-load assertion.

Suppose independent Bernoulli trials give a player a win with probability $p$, and the match ends when either player reaches $r$ wins. The player's match-win probability isFor a win with $n$ losses, the final trial must be a win. The preceding $r+n-1$ trials contain exactly $r-1$ wins and $n$ losses; the binomial coefficient counts their arrangements. Because $n<r$, the opponent cannot already have won, and because the player has only $r-1$ previous wins, the match has not ended earlier. Summing the disjoint terminal events gives the formula, a truncated negative binomial distribution sum. The constant game-win probability must describe independent games, or equivalently remain the same conditional on every previous history.

A mixture of Poisson distributions over a nonnegative random intensity. With exposure $n>0$ and prior density $\pi (\theta )$, its probability mass function is $p_n(s)={\int }_0^{\infty }{e}^{-n\theta }(n\theta {)}^s\pi (\theta )d\theta /s!$. A gamma mixing distribution gives a negative binomial distribution. Counts sharing the same latent intensity are conditionally independent, but generally dependent after mixing.

With known size $\theta >0$, the mean-dependent log mass of a negative binomial distribution is $y\log \mu -(y+\theta )\log (\mu +\theta )$. Its derivative is $\theta (y-\mu )/[\mu (\mu +\theta )]$, so the saturated mean is $y$, including its boundary value zero. Subtracting the fitted log mass from that maximum gives the displayed deviance. This formula needs no fitted-intercept cancellation. As $\theta \to \infty$, its second term tends to $-2(y_i-e_i)$ and it becomes the full Poisson deviance.

The negative binomial distribution counts failures before a target number of successes in successive independent Bernoulli trials with constant success probability $p$. In this stopping model, $r$ is a positive integer and $0<p\le 1$. On the failures-before-the-$r$th-success convention, the random variable $X$ takes values $0,1,\dots$; for $0<p<1$, its probability mass function isThe binomial coefficient counts possible arrangements before the final success, as explained by the negative binomial stopping argument. At $p=1$, every trial succeeds: $X=0$ with probability one, so its expected value and variance are both zero. This handles the endpoint without needing to interpret $0^0$. At $p=0$, the target is never reached and there is no proper finite waiting-count law.

For a worked example, toss a fair coin until the second head, calling heads $S$ and tails $F$. Thus $r=2$ and $p=1/2$. Exactly two failures means that the second success is on trial four. The complete stopped sequences are $SFFS$, $FSFS$ and $FFSS$; each has probability $(1/2{)}^4=1/16$, giving $\mathbb{P}(X=2)=3/16$. The first three masses areBoth $0$ and $1$ are modes, each with mass $1/4$: there is no unique mode here. Later masses decrease, since $\mathbb{P}(X=x+1)/\mathbb{P}(X=x)=(x+2)/(2x+2)<1$ for $x\ge 1$.

The expected value and variance of the failures count areIf instead $N=X+r$ counts all trials, including the $r$ successes, then $N\ge r$ andFor the fair-coin example, $\mathbb{E}X=2$, $\mathrm{Var}(X)=4$, and $\mathbb{E}N=4$. Shifting the count changes its expected value but preserves its variance. When $r=1$, $X$ has the failures-count geometric distribution, while $N=X+1$ has the all-trials geometric distribution. The two conventions and the moments are treated in Kyle Siegrist, The Negative Binomial Distribution, Random Services.

The negative binomial distribution is useful when a procedure runs until it collects a target number of successes: for example, obtaining $r$ usable samples from independent attempts with a stable acceptance probability. A binomial distribution instead counts successes in a fixed number of trials. The attractive connection is that these two experimental questions can be answered from the same trial sequence, using the negative binomial stopping argument. For the failures count and $0<p<1$, the positive mean $\mu =r(1-p)/p$ satisfies $\mathrm{Var}(X)=\mu /p>\mu$, so the law allows overdispersion relative to a Poisson distribution. At $p=1$, both quantities vanish and the inequality is not strict. The separate Poisson-gamma mixture explains another way this count law arises; its noninteger shape parameters do not represent a literal target number of successes.

When each observation counts untagged captures before the $r$th tagged capture with immediate replacement, its negative binomial distribution has success probability $p=k/N$. For $n$ independent runs, the log-likelihood is $nr\log p+({\sum }_i{x}_i)\log (1-p)$ plus a constant. Differentiation gives the displayed estimator and expected Fisher information. For an interior true parameter, its asymptotic variance is $p^2(1-p)/(nr)$. A confidence interval $(L,U)$ for $p$ transforms to $(k/U,k/L)$ for the population size, with endpoint order reversed.

For the negative binomial distribution with mean $\mu >0$ and size $\theta >0$, the one-observation score function in the mean is $u_{\mu }=\theta (y-\mu )/[\mu (\mu +\theta )]$. Its size derivative is $(y-\mu )/(\mu +\theta {)}^2$, whose expectation is zero. Thus the expected Fisher information matrix is diagonal in $(\mu ,\theta )$. Regular interior maximum-likelihood estimators have zero asymptotic cross covariance, without necessarily being independent in finite samples.

Negative binomial regression models overdispersed counts with a negative binomial distribution, commonly using a logarithmic link for the mean.

For a nonnegative integer $L$ and $|z|<1$, this specialization of the binomial series sums the normalizing coefficients of a negative binomial distribution.

For the negative binomial distribution with positive integer target $r$ and success probability $0<p<1$, the $r$th success occurs after exactly $x$ failures only if the final trial is a success, while the preceding $x+r-1$ trials contain $x$ failures and $r-1$ successes. Choosing the failure locations gives the binomial coefficient $\left(\genfrac{}{}{0ex}{}{x+r-1}{x}\right)$. Each such stopped sequence has probability $(1-p{)}^x{p}^r$ by independence, and the sequences describe mutually exclusive events, so their probabilities add. This proves the probability mass function. Counting arrangements among all $x+r$ positions would wrongly include sequences that already reached the target earlier. The same argument is used for three heads in MIT 18.05, spring 2022, problem set 9 solutions, problem 2(b).

A short route to normalization and moments is to split the failures count into successive waiting blocks:where $F_i$ counts failures after success $i-1$ and before success $i$, with $F_1$ starting at the first trial. Fresh independent Bernoulli trials make the blocks independent, each with the failures-count geometric distribution. Each block has total mass $p{\sum }_{j=0}^{\infty }(1-p{)}^j=1$ by the geometric series, so their sum also defines a normalized probability distribution. The linearity of expectation and variance additivity for independent random variables give $\mathbb{E}X=r(1-p)/p$ and $\mathrm{Var}(X)=r(1-p)/p^2$. This makes the moments transparent without a long calculation of sums. For $p=1$, every block is zero.

Let $p+q=1$ and let the original independent count have negative binomial distribution with probability mass function $\left(\genfrac{}{}{0ex}{}{n+k-1}{k}\right)p^n{q}^k$ on $k\ge 0$. With unit-rate exponential distribution severities, its moment-generating function is $[p+qp/(p-t){]}^n$ for $t<p$. Thus it also has a random sum of independent claims representation with independent binomial distribution count $B$ and rate-$p$ exponential distribution steps. For $n=2$ the law is $p^2{\delta }_0+2pq\mathrm{Exp}(p)+q^2\mathrm{Erlang}(2,p)$. The hurdle decomposition of a positive random sum therefore has zero mass $p^2$ and positive density $p^2{e}^{-px}(2q+q^{2}x)/(1-p^2)$.

The nonnegative integer laws whose probabilities satisfy this recurrence for all $n\ge 1$. Their probability generating function solves $(1-az)G^{\prime }(z)=(a+b)G(z)$. This is the $(a,b,0)$ count class used in the Panjer aggregate algorithm; the count recurrence itself should be distinguished from that aggregate recursion. It includes the Poisson distribution, binomial distribution, and negative binomial distribution with suitable parameters.

A microparasite typically multiplies within an infected host on a much shorter time scale than between-host transmission. A useful first model classifies hosts by infection status. A macroparasite, such as an adult parasitic worm, instead requires a count of parasites per host: two infected hosts can carry very different within-host parasite burdens. These are modelling distinctions rather than a rule determined only by an organism's physical size.

For a microparasite with no lasting immunity, a homogeneous SIS model isThe basic reproduction number is $\beta /\gamma$, and the prevalence $I/N$ describes the infected-host fraction. A SIR model adds an immune class, and a SEIR model adds a latent class. Contact structure, age, susceptibility, recovery rate and infectiousness can require several such classes. Then a next-generation matrix or operator, rather than a single averaged contact rate, sets the invasion threshold. Stochastic epidemic models are relevant to early extinction and finite populations; network or spatial models represent nonuniform contact opportunities.

For a simple macroparasite burden model, let $p_j$ be the fraction of hosts carrying $j$ adult worms. Assume acquisitions at rate $\lambda =\beta L$, independent adult-worm losses at rate $\mu j$, and a number $L$ of infective stages in a well-mixed environmental reservoir. The immigration-death worm-burden model hasThis birth-death process needs the burden distribution or its probability generating function, not just an infected/not-infected partition. Its mean $m={\sum }_{j}j{p}_j$ satisfies $\dot{m}=\beta L-\mu m$. With host number $N$ fixed and each adult producing infective stages at rate $\sigma$, one possible environmental closure isFor this asexual, unsaturated transmission approximation, one worm's basic reproduction number is $(\sigma /\mu )\beta N/(d_L+\beta N)$. Growth above the invasion threshold requires additional density regulation if a finite endemic burden is sought. Sex-dependent mating would change this closure.

With constant external exposure $\lambda$, the equilibrium burden follows a Poisson distribution with mean $m=\lambda /\mu$, giving infected-host prevalence $1-e^{-m}$. If exposure rates vary across hosts as a gamma distribution, the Poisson-gamma mixture gives a negative binomial distribution with mean $m$ and aggregation parameter $k$:The same mean burden can therefore coexist with very different prevalence and highly infected subgroups. Aggregation of macroparasite burdens affects treatment coverage, burden-dependent host mortality, immunity and parasite fecundity. For sexually reproducing worms, hosts need both sexes for fertile output; a mean alone may not determine it. With independent equally likely sexes and Poisson burden, the probability of both sexes is $(1-e^{-m/2}{)}^2$, illustrating mating-limited macroparasite transmission and its low-burden bottleneck.

Both microparasite and macroparasite models need heterogeneous exposure and host responses. The special extra difficulty for macroparasites is translating a distributed worm burden, and possibly worm sex or developmental stage, into transmission and host harm. Conversely, complicated within-host microparasite processes can also require more than a binary host state when the simple fast-within-host approximation fails.

Condition on the claim count. The empty random sum of independent claims is zero, and independence gives $\mathbb{E}[e^{tS}\mid N=n]=M_X(t{)}^n$. Therefore the random-sum transform identity iswherever the composition is finite. This follows directly from the law of total expectation, rather than replacing the random count by its mean.

Put $q=1-p$. For the specified negative binomial distribution, the negative binomial series gives $G_N(z)=[p/(1-qz){]}^k$. A rate-$\lambda$ exponential distribution has moment-generating function $M_X(t)=\lambda /(\lambda -t)$. HenceThe last expression is exactly the moment-generating function of a compound binomial distribution with independent count and severitiesThe moment-generating function exists in a neighborhood of zero, so its uniqueness identifies the distributions. In particular, the new claim sizes have mean $1/(p\lambda )$, not $1/\lambda$, and both aggregate distributions have an atom $p^k$ at zero. This is the negative-binomial sum of exponential claims representation with the severity rate explicitly retained.

Conditional on $N=n\ge 1$, the aggregate has Erlang distribution with density ${\lambda }^n{s}^{n-1}{e}^{-\lambda s}/(n-1)!$ on $s>0$. Repeated integration by parts of its tail givesIndeed, after substituting $v=\lambda s$, the integral ${\int }_{\lambda x}^{\infty }{e}^{-v}{v}^{n-1}dv/(n-1)!$ reduces recursively to the corresponding integral with exponent $n-2$, terminating at $e^{-\lambda x}$. Applying the law of total probability yields the original infinite mixture distribution tailApplying the same argument to the compound binomial distribution gives instead the finite Erlang-mixture tail for a negative-binomial exponential aggregate:The two tails are equal. The binomial representation replaces an infinite sum by a finite exact calculation, eliminating the need to choose a claim-count truncation and control its omitted tail. The inner sums can be evaluated recursively, using $a_{j+1}=a_j(p\lambda x)/(j+1)$ for their exponential-weighted terms.