Poisson distribution ($\mathrm{Pois}(\lambda )$)

The Poisson distribution with mean lambda assigns mass $e^{-\lambda }{\lambda }^k/k!$.

A Poisson distribution of parameter $a$ conditioned to lie in $\{0,\dots ,C\}$. It is the occupancy law of a unit-resource loss network. Its mean is the carried load of an Erlang loss resource, $a[1-E(C,a)]$. Differentiating the mean with respect to $\log a$ gives its variance, which is positive for $a>0$ and $C\ge 1$; this proves strict monotonicity of the carried load.

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.

For a uniform prior distribution on $(l,u)$ with $0<l<u$, conditionally independent Poisson distribution observations have sufficient count $s={\sum }_i{x}_i$. Their Bayesian posterior density is proportional to $t^s{e}^{-nt}$ on that bounded interval. The displayed posterior mean is the Bayes estimator under squared error loss, and is generally not the affine Bühlmann credibility estimate.

When the random intensity is $\alpha +Y$ with $Y$ having an exponential distribution of rate $\beta$, the mixed count is the sum of independent Poisson distribution and support-zero geometric distribution counts. The Poisson parameter is $\alpha$ and the geometric success probability is $\beta /(\beta +1)$. At $\alpha =0$ the law is geometric.

For the shifted-exponential Poisson mixture, the displayed relation holds for $n\ge 2$, initialized by $p_0=e^{-\alpha }\beta /(\beta +1)$ and $p_1=(\alpha +1/(\beta +1))p_0$. It follows by differentiating the probability generating function and comparing coefficients. Nonnegativity follows from the independent Poisson-geometric convolution of independent random variables representation.

A Poisson distribution conditioned on a positive count has probability mass function $\mathbb{P}(Y=y)={\theta }^y/[y!(e^{\theta }-1)]$ for $y\ge 1$ and $\theta >0$.

For the zero-truncated Poisson distribution, $21_{\{Y\text{ even}\}}$ is the unique integrable nonrandomized unbiased estimator of $p=1-e^{-\theta }$. Indeed unbiasedness gives ${\sum }_{y\ge 1}t(y){\theta }^y/y!=e^{\theta }+e^{-\theta }-2$, and power series coefficients force the parity values. Its variance is $2p-p^2$; clipping it to $1_{\{Y\text{ even}\}}$ gives mean squared error $p/2$, strictly smaller for $0<p<1$.

The sum of independent Poisson random variables is Poisson, with parameter equal to the sum of their parameters. This follows by multiplying their probability generating functions $\exp ({\lambda }_i(z-1))$.

For Bernoulli random variables $X_1,\dots ,X_n$, which need not be independent, put $p_i=\mathbb{P}(X_i=1)$, $S={\sum }_i{X}_i$, and $\lambda ={\sum }_i{p}_i$. ThenThe dependence penalty is the total correlation. The proof compares the joint Bernoulli law with the product of Poisson laws of means $p_i$, then applies the data processing inequality for relative entropy to addition.

A Poisson observation model represents a nonnegative count as conditionally Poisson with a mean supplied by an underlying process or regression model.

A Poisson observation is an integer-valued count modeled by a Poisson distribution of specified intensity, possibly conditionally on a latent field. Independent Poisson observations motivate Poisson data fidelity through their log-likelihood.

A zero-inflated Poisson distribution is a mixture of a point mass at zero and a Poisson distribution. If the Poisson component has weight $\pi$ and mean $\lambda$, then its mean is $\pi \lambda$ and its variance is $\pi \lambda +\pi (1-\pi ){\lambda }^2$.

A zero-inflated Poisson regression specifies a structural-zero probability $\pi$ and a susceptible-component Poisson distribution mean $\mu$. In this parametrization,The mean is $(1-\pi )\mu$. Logistic predictors for $\pi$ and logarithmic predictors for $\mu$ allow covariates to affect component membership and count intensity differently. A susceptible individual can still generate a zero count; susceptibility is not equivalent to an observed positive response.

In a zero-inflated Poisson regression, Bayes theorem separates structural zeros from susceptible zero counts. A positive count rules out the structural-zero class; a zero raises its posterior probability according to the displayed formula, but usually does not determine class membership with certainty. This same probability supplies the E-step of the EM algorithm for zero-inflated Poisson regression.

A logarithmic count-component contrast $b$ multiplies the susceptible mean by $e^b$, while a zero-logit contrast $g$ multiplies the structural-zero odds by $e^g$. If the baseline zero predictor is $\eta$, the marginal mean ratio isThus the count ratio alone is not a population-average effect when a covariate changes both components. With interaction terms, the relevant count contrast must first include the interactions for the stated reference group.

The expectation-maximization algorithm introduces $Z_i=1$ for a structural zero. Its E-step gives ${\tau }_i=0$ at positive counts and ${\tau }_i={\pi }_i/[{\pi }_i+(1-{\pi }_i)e^{-{\mu }_i}]$ at zero. The M-step maximizes a logistic log-likelihood with fractional responses ${\tau }_i$ and a Poisson log-likelihood with weights $1-{\tau }_i$. With only a binary predictor, write $n_j$ for group size, $S_j={\sum }_{x_i=j}{\tau }_i$ and $C_j={\sum }_{x_i=j}{Y}_i$; the explicit updates are ${\pi }_j=S_j/n_j$ and ${\mu }_j=C_j/(n_j-S_j)$. Log and logit group contrasts recover the regression coefficients, with boundary values interpreted through limits.

If $N_{\lambda }\sim \mathrm{Pois}(\lambda )$ and $\lambda \to \infty$, thenIndeed, its characteristic function is

If $X_{\lambda }$ has a Poisson distribution, then $\mathbb{P}(X_{\lambda }\le \lfloor \lambda \rfloor )\to 1/2$ as $\lambda \to \infty$. The Poisson central limit theorem standardizes by mean $\lambda$ and variance $\lambda$; the cutoff $(\lfloor \lambda \rfloor -\lambda )/\sqrt{\lambda }$ tends to zero. Squeezing between fixed nearby cutoffs and using continuity of the standard normal distribution function gives the limit.

Independent Poisson counts conditioned on their sum have a multinomial distribution.

For each group $g$, let independent cell counts have Poisson means ${\mu }_{gj}$. Conditional on their total $n_g$, the cells have a multinomial distribution with probabilitiesIncluding a free group main effect in the Poisson log-linear model makes its profiled likelihood for the remaining parameters proportional to the corresponding multinomial likelihood. The two fits therefore give the same fitted proportions and likelihood-ratio comparisons.

For positive group totals, the improper prior $du/u$ on an independent Poisson baseline yields the multinomial likelihood kernel after integration. This prior is flat in $\log u$. A proper finite-variance log-rate prior generally introduces an additional coefficient-dependent factor.

A mean-zero normal distribution prior of variance $s^2$ on the log baseline multiplies the multinomial kernel by the displayed correction, up to a constant independent of coefficients. The factor is not constant in $S$, so the posterior equivalence is approximate. Dominated convergence makes it tend to one as $s^2$ tends to infinity.

For large $\lambda$, a Poisson random variable of mean $\lambda$ is approximately normal with mean and variance $\lambda$.