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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.

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.

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.

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.

Keep baseline covariates fixed and put $q=1-\pi$, ${\mu }_j=e^{{\beta }_0+{\beta }^T{x}_i+{\alpha }_j}$, and $r=1/\tau$ for $\tau >0$. Integration of a Poisson count over the Gamma component givesThis negative binomial distribution has mean ${\mu }_j$ and variance ${\mu }_j+\tau {\mu }_j^2$. Therefore the unconditional one-term distribution is the zero-inflated negative binomial distributionIn particular its zero probability is $\pi +q(1+\tau {\mu }_j{)}^{-1/\tau }$, not simply $\pi$. As $\tau \downarrow 0$ this becomes a Zero-inflated Poisson distribution.

The mixing effect has $\mathbb{E}{b}_i=q$, $\mathbb{E}{b}_i^2=q(1+\tau )$ and $\mathrm{Var}(b_i)=q(\tau +\pi )$. The law of total variance and law of total covariance now giveEquivalently, for $\mu =({\mu }_1,{\mu }_2,{\mu }_3{)}^T$, $\mathbb{E}{Y}_i=q\mu$ and $\mathrm{Cov}(Y_i)=\mathrm{diag}(q\mu )+q(\tau +\pi )\mu {\mu }^T$. This also supplies raw second moments by adding $(\mathbb{E}{Y}_i)(\mathbb{E}{Y}_i{)}^T$. Dependence arises from both shared heterogeneity and shared structural zeros, even when the positive random effect is constant.

The printed moment variance matches the hierarchical variance at the same parameter values, but its printed off-diagonal covariance is $\tau {\mu }_j{\mu }_k$. Equality would require $\tau =q(\tau +\pi )$, or $\pi (\tau -q)=0$. It is therefore false to claim that the two approaches consistently estimate all the same parameters in general.

The precise distinction is between consistency of mean ratios and identification of structural parameters. In the hierarchical model, with $m_j=q{\mu }_j$, both the quadratic variance coefficient and the covariance coefficient in terms of the marginal mean areFor the moment parameterization, writing its parameters as ${\pi }_Q,{\tau }_Q,q_Q=1-{\pi }_Q$, these coefficients are $A_Q=({\pi }_Q+{\tau }_Q)/q_Q$ and $B_Q={\tau }_Q/q_Q^2$. Its intercept is identifiable from the mean only as ${\gamma }_{0Q}={\beta }_{0Q}+\log q_Q$. Matching the true first two moments demands $A_Q=B_Q=\kappa$. One solution always isFor $\kappa >1$ there is also a solution $q_Q=1/\kappa$, ${\pi }_Q=1-1/\kappa$, ${\tau }_Q=1/\kappa$, with the corresponding shifted intercept. Indeed eliminating ${\tau }_Q$ gives ${\pi }_Q({\tau }_Q-q_Q)=0$. This is moment aliasing in a shared zero-inflated count model: matching the moments does not identify the actual structural-zero probability.

For a concrete counterexample take $\pi =0.2,\tau =0.3$. Then $q=0.8$ and $\kappa =0.625$. The only admissible moment-matching choice is ${\pi }_Q=0,{\tau }_Q=0.625$, with an intercept shifted by $\log 0.8$; it cannot recover the true structural-zero proportion 0.2. Correct marginal means can give consistent term and baseline-covariate slopes with sandwich standard errors; they do not justify consistent estimation of the intended $\pi ,\tau$, or latent intercept. Nor does a mean-only estimating equation identify $\pi$ at all. Full-distribution inference provides information beyond these first two moments.

Conditioning on the intensity in the Poisson mixture givesThus $N$ has the negative binomial distribution with two successes and success probability $p$, counting failures; explicitly $\mathbb{P}(N=j)=(j+1)p^2{q}^j$ for $j\ge 0$. The probability generating function is finite for real $z<1/q$.

The claim-size moment-generating function is $M_X(t)=(1-\mu t{)}^{-1}$. Substituting into the aggregate moment-generating function and using $p+q=1$ givesPut $L(t)=p/(p-\mu t)$, the moment-generating function of an exponential distribution with rate $p/\mu$. The identitythen yieldsThe gamma-mixed Poisson aggregate with exponential claims has three nonnegative mixture weights summing to one. By uniqueness of the moment-generating function near zero, the aggregate distribution isHere ${\delta }_0$ is the Dirac measure at zero. In particular $\mathbb{P}(S=0)=p^2$, consistently with $\mathbb{P}(N=0)$. The positive components have respective expected values $\mu /p$ and $2\mu /p$; their mixture distribution accounts for the possibility of no aggregate payout.

A feasible per-flow allocation satisfies $x_r>0$ for each active route and ${\sum }_r{A}_{jr}{n}_r{x}_r\le C_j$. It has proportional fairness if every feasible competitor obeysBy the first-order inequality for the strictly concave logarithm, this is equivalent to maximizing weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$ over the feasible allocations. Inactive-route rates are immaterial; set them to zero as a convention.

Write $a=n_{\{1\}}$, $b=n_{\{2\}}$, $c=n_{\{1,2\}}$, and $N=a+b+c$. The capacity constraints are $a{x}_1+c{x}_0\le 1$ and $b{x}_2+c{x}_0\le 1$. For $c>0$, put $q=c{x}_0$. Each active local route uses its remaining capacity $1-q$. Omitting constants, the utility isIts maximum has $q=c/N$; when $a+b=0$, the maximum is the endpoint $q=1$. Hence for active routesThese formulas also give $x_1=1/a$ and $x_2=1/b$ when $c=0$. When $N=0$ there is no allocated service. Equivalently, the aggregate route rates for $N>0$ areThis is the proportionally fair allocation on a two-resource linear network for a two-resource linear flow network.

For independent Poisson processes of documents and independent document sizes with exponential distribution of rate ${\mu }_r$, the memorylessness of the exponential distribution makes the population a continuous-time Markov chain. Its transition intensities areThe departure intensity sums the $n_r$ independent completion hazards ${\mu }_r{x}_r$. There is no departure on an inactive route. Set ${\rho }_r={\nu }_r/{\mu }_r$, with ${\mu }_r>0$.

To prove the proposed stationary distribution, define the balance function of a flow-level networkFor an active local route, $\Psi (n-e_r)/\Psi (n)=(a+b)/N$, while for the through route this ratio is $c/N$. Thus ${\Lambda }_r(n)=\Psi (n-e_r)/\Psi (n)$. The weightssatisfy detailed balance on every arrival-departure pair:Each total jump rate is bounded by ${\sum }_r({\nu }_r+{\mu }_r)$, so the chain is a nonexplosive Markov chain; normalized detailed-balance weights therefore give its stationary distribution, the stationary law of a two-resource linear flow network.

Use the negative binomial series to sum first over $c$:The remaining two geometric series giveAll summands are nonnegative. Consequently the sum is finite exactly under the stability conditions for a two-resource linear flow networkThese say that the offered load on each unit-capacity resource is strictly below one. Equality is insufficient: at least one geometric series then diverges. If all arrival rates are positive, the chain is irreducible and the normalizable law is its unique equilibrium law.

Finally put $q_i={\rho }_i/(1-{\rho }_0)$, $i=1,2$. Summing over $c$ yieldsThus the two local-route counts are independent, each with geometric distribution on the nonnegative integers. The conditional through-route count has negative binomial distributionFor positive traffic rates this depends on the local counts, so all three route counts are not independent. More quantitatively, $\mathrm{Cov}(a,c)={\rho }_0\mathrm{Var}(a)/(1-{\rho }_0)>0$, and likewise for $b$. If zero arrival rates are allowed, the exceptional independent cases are ${\rho }_0=0$, or ${\rho }_1={\rho }_2=0$, when the relevant counts are deterministic zero. This distinction is local-count independence with through-flow dependence.

For fixed known $r>0$, take the natural parameter of an exponential familyThe negative binomial distribution has log-likelihoodThus this is a negative binomial exponential family, with dispersion parameter $\varphi =1$ in the convention of part (a). The carrier contains the Gamma function and depends on the fixed size $r$, which need not be an integer. Differentiating the cumulant function of an exponential family givesThereforeThe size $r$ controls overdispersion relative to a Poisson distribution; it is different from the unit dispersion parameter used by this fixed-size generalized linear model. Another common convention applies to the scaled response $Z=Y/r$: then $b(\theta )=-\log (1-e^{\theta })$ and $\varphi =1/r$. Keeping track of the response scale prevents confusing these two representations. At $\lambda =0$ the law is concentrated at zero and is obtained as a boundary limit, rather than by a finite natural parameter of an exponential family.

Let $I_0=\{i:y_i=0\}$ and $I_{+}=\{i:y_i>0\}$. Under the proposed zero-inflated negative binomial model, independence gives the likelihood functionThe positive-count factor uses the ordinary, untruncated negative binomial distribution. With $f_i=f_{\mathrm{NB}}(y_i;r,{\lambda }_i)$, the log-likelihood isFor latent indicators $a_i$, the complete-data log-likelihood iswhere $a_i=1$ is allowed only when $y_i=0$. This separation is what makes the expectation-maximization algorithm convenient.

For the process to stop on toss $n$, the last toss must be the $k$th head, while the first $n-1$ tosses must contain exactly $k-1$ heads. Hence the negative binomial distribution gives

For $X=x$, the last flip must be the $r$th head. Among the preceding $x+r-1$ flips, exactly $x$ are tails and $r-1$ are heads. There are $\left(\genfrac{}{}{0ex}{}{x+r-1}{x}\right)$ choices for their positions, and every resulting sequence has probability $(1-p{)}^x{p}^r$. The negative binomial stopping argument therefore givesso $X$ has the failures-before-the-$r$th-success negative binomial distribution.

To put the mass function into exponential family form, setSince $p=1-e^{\theta }$,whereThus the natural parameter of an exponential family is $\theta$, and the Fisher-Neyman factorization theorem shows that $T(X)=X$ is a sufficient statistic. This is the negative binomial exponential family.

The exponential-family derivative identities now giveand