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For $\lambda >0$, write the Poisson distribution mass function in exponential dispersion family form:Thus the natural parameter of an exponential family is $\theta =\log \lambda$ and the dispersion parameter is fixed at $1$. The counting-measure support is $0,1,\dots$; there is no independent free dispersion parameter in an ordinary Poisson model. The degenerate boundary $\lambda =0$ is obtained as a limit, rather than a finite natural parameter.

The exponential dispersion family identities give $\mathbb{E}Y=b^{\prime }(\theta )$ and $\mathrm{Var}(Y)=a(\varphi )b^{\prime \prime }(\theta )$. Since both derivatives of $e^{\theta }$ equal $e^{\theta }$,

For $0<p<1$, the Bernoulli distribution has massIn exponential dispersion family notation $\exp \{[y\theta -b(\theta )]/\varphi +c(y,\varphi )\}$, takeThe support is $\{0,1\}$; outside it the probability is zero. The natural parameter of an exponential family is the log odds, and the dispersion parameter is fixed rather than freely estimated. Differentiating the cumulant givesAt $p=0$ or $1$ the degenerate distributions are limiting boundary cases with natural parameter $-\infty$ or $+\infty$.

Use an ideal carrier-sense multiple access scheme on the interference graph. An active station finishes its transmission after an independent unit-rate exponential distribution waiting time. An inactive station has an independent attempt clock with an exponential distribution of rate $e^{{\theta }_r}$; an attempt succeeds only when all its neighbors are inactive. Attempts that are blocked do not change the state. Thus, on the independent sets represented by $S$, the continuous-time Markov chain has transition ratesThe empty independent set can be reached by successive deactivations, and any other independent set can be reached from it by successive activations. Hence this finite chain is an irreducible Markov chain. The weight $a_{\theta }(n)=e^{\theta \cdot n}$ satisfiesBy detailed balance for a continuous-time Markov chain, its unique stationary law isThis is a finite exponential family. Its natural parameter of an exponential family is $\theta$ and its mean parameter of an exponential family is the vector of active fractions.

The throughput region of an interference graph is exactly $\Lambda =\mathrm{conv}S$. To see the possibly less immediate inclusion, suppose a random feasible schedule has mean $v\ge \lambda$. Retain each of its active vertices $r$ independently with probability ${\lambda }_r/v_r$, interpreting zero coordinates as always deleted. A subset of an independent set remains an independent set; the thinned schedule has mean exactly $\lambda$. Thus $\lambda$ lies in the convex hull of $S$. The opposite inclusion follows from the definition using equality. Moreover $0,e_1,\dots ,e_R\in S$, so this convex hull has full dimension.

Now choose $\epsilon >0$ such that $\mu =\lambda +\epsilon 1$ remains in the interior of $\mathrm{conv}S$. Use interior moment matching in a finite exponential family. Consider the convex functionThere is a ball of radius $\delta >0$ around $\mu$ contained in $\mathrm{conv}S$. Maximizing a linear functional over the convex hull is the same as maximizing over its generating set, so, for every $\theta$,The last inequality uses the point $\mu +\delta \theta /\Vert \theta \Vert$ when $\theta \ne 0$. Hence $F$ is a coercive function and has a finite minimizer ${\theta }^{\ast }$. Differentiating the finite normalizing sum givesThis covariance matrix is a positive-definite matrix: a linear functional constant on all states of positive weight must be constant on $0,e_1,\dots ,e_R$, so its coefficients vanish. Thus $F$ is a strictly convex function; its minimizer is unique. At that minimizer the desired strict service margins areFor a backlogged station that transmits at unit speed whenever active, the long-run throughput is ${\mathbb{E}}_{\theta }[n_r]$, by the ergodic theorem for a finite continuous-time Markov chain. Therefore the ideal carrier-sense multiple access scheme can supply a strict service margin for every arrival vector in the interior of the throughput region of an interference graph. The local attempt rates can realize the whole interior capacity region in this sense. Boundary points may require parameters tending to infinity, and the argument gives no uniform delay or mixing-time bound near the boundary.

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.

An exponential dispersion family has densities of the formwhere $\theta$ is the canonical parameter, $\varphi >0$ is the dispersion parameter, and the support does not depend on $\theta$. Its mean and variance are

For the given Gamma distribution, setThengivesThus the gamma laws form the Gamma exponential dispersion family.

Directly integrating, or using the cumulant-generating function of a gamma distribution,soDifferentiation at zero yields

A generalized linear model assumes independent responses from one exponential dispersion family,where ${\sigma }_i^2=a_i{\sigma }^2$, ${\mu }_i=\mathbb{E}[Y_i]=b^{\prime }({\theta }_i)$, and a link function relates the mean to the linear predictor by $g(\mu )=X\beta$. It follows from normalization of the density after shifting the natural parameter thatIndependence therefore gives the joint moment-generating functionwhenever every shifted natural parameter lies in its natural-parameter domain.

Expand the exponent in the Inverse Gaussian distribution with unit shape parameter:Consequently its probability density function can be written aswhereIndeed, $-A(\eta )=\sqrt{-2\eta }=1/\mu$. This is the canonical form of a one-parameter exponential family, and its natural parameter is therefore

Here the natural statistic is $T(X)=X$. The exponential-family derivative identities giveandThus

A one-parameter exponential family has densitieswith support independent of the natural parameter $\theta$. In the present canonical setting the natural statistic is $T(y)=y$.

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

A one-parameter exponential family with natural statistic $Y$ has density or probability mass functionwhere $\theta$ is the natural parameter of an exponential family and $K$ is the cumulant function of an exponential family. Its mean parameter is

The exponential-family deviance from ${\theta }_1$ to ${\theta }_2$ is twice the Kullback-Leibler divergence:The carrier $h(Y)$ cancels from the likelihood ratio, so, with ${\mu }_1={\mathbb{E}}_{{\theta }_1}Y$,

For the Poisson distribution with mean $\mu$,ThereforeWriting ${\mu }_2={\mu }_1+\delta$ and applying the Taylor series of $\log (1+\delta /{\mu }_1)$ givesso the second-order approximation is