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For $k\ge 0$, take $f_k(x)=1_{\{x\ge k+1\}}$ and write $q_k=\mathbb{P}(X\ge k+1)$. Since $f_k^2=f_k$,Its discrete derivative is nonzero only at $k$, soFor a Poisson distribution, $p_{k+1}/p_k=\lambda /(k+1)$ and $q_k\sim p_{k+1}$ as $k\to \infty$. Hencewhere the final estimate follows from Stirling formula. No finite constant $C$ can therefore make the proposed inequality hold for every $f$.

The variance $\mu +{\mu }^2/\theta$ approaches the Poisson distribution variance $\mu$ when $\theta \to \infty$. Equivalently, with $\alpha =1/\theta \ge 0$, the Poisson model is the boundary value $\alpha =0$. The usual Wilks theorem for a likelihood-ratio test assumes that the null parameter is an interior point of a smooth parameter space, so comparing the statistic with an ordinary ${\chi }_1^2$ law is invalid here.

A valid parametric bootstrap proceeds as follows. First fit the null Poisson model and retain its fitted means ${\hat{\mu }}_i$. For each bootstrap repetition $b=1,\dots ,B$, independently simulateusing the original doses, refit both the Poisson and negative-binomial models to that simulated data, and calculateFor the observations calculate the analogous $T_{obs}$. The bootstrap $p$-valueuses the null distribution with its boundary and finite-sample fitting behaviour automatically reproduced. A large $B$ controls the Monte Carlo error of this estimate.

The second fit is a Quasi-Poisson regression. It retains the logarithmic link function and mean modelbut assumes only $\mathrm{Var}(Y_i)=\varphi {\mu }_i$ for an unknown dispersion parameter $\varphi$, rather than a complete Poisson distribution.

Independent photon arrivals have a Poisson distribution: a mean count $\mu$ has variance $\mu$ and standard deviation $\sqrt{\mu }$. Subtraction of independent counting measurements adds their variances.

A Galton-Watson process with a Poisson distribution of offspring of mean $\mu$ has this probability generating function. For $\mu =1+\epsilon >1$ its branching survival probability $r$ satisfies $-\log (1-r)=(1+\epsilon )r$, and $2\epsilon /(1+2\epsilon )\le r\le 2\epsilon$. Indeed $\epsilon ={\sum }_{j\ge 1}{r}^j/(j+1)$ lies between $r/2$ and $r/[2(1-r)]$. These exact inequalities belong to the Poisson branching process, not to every finite binomial branching process.

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.

Given $\Theta$, independent observations $Y_j$ have Poisson distribution with means $m_j\Theta$ for known exposures. A gamma distribution prior of shape $\alpha$ and rate $\beta$ yields a gamma posterior with shape $\alpha +{\sum }_j{Y}_j$ and rate $\beta +{\sum }_j{m}_j$. The likelihood kernel is ${\theta }^{{\sum }_j{Y}_j}{e}^{-\theta {\sum }_j{m}_j}$. Thus the posterior mean depends on total claims and total exposure, not the number of periods alone.

A Poisson generalized linear mixed model gives a count response a conditional Poisson distribution whose logarithmic mean contains both fixed effects and latent random effects.

In a uniform balls-in-bins allocation with $m\to \infty$ and $n/m\to \lambda \in (0,\infty )$, a fixed bin count has convergence in distribution to a Poisson distribution variable of mean $\lambda$ by the Poisson limit theorem. If $N_j$ counts bins containing exactly $j$ balls, symmetry and the linearity of expectation give $\mathbb{E}[N_j]/m=\mathbb{P}(B_1=j)\to e^{-\lambda }{\lambda }^j/j!$ for each fixed $j$. In particular the expected empty fraction tends to $e^{-\lambda }$, and the expected fraction with at least two balls tends to $1-(1+\lambda )e^{-\lambda }$. These are statements about expectations; convergence of the random fractions requires an additional argument.

For three distinct times, the columns $1,t,t^2$ are linearly independent by the Vandermonde determinant. Thus a quadratic Poisson regression for the log mean can reproduce any three positive time-group means. A grouped goodness-of-fit test of that mean form has zero residual degrees of freedom. Replicated observations still allow checks of the Poisson distribution and overdispersion within each time group; saturation of the mean does not establish the response distribution.

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.

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 source counts $S$ and independent background counts $B$ with Poisson distributions, the ideal counting standard deviation is $\sqrt{S+B}$, so the signal-to-noise ratio is $S/\sqrt{S+B}$. Detector noise and uncertainty in the estimated background add further variance. A larger sky aperture raises $B$, while better throughput raises $S$; both affect the faintness limit of a spectrograph.

For every $\lambda ,\mu >0$, an $M/M/\infty$ queue is positive recurrent and has the Poisson distribution with mean $\rho =\lambda /\mu$ as its unique stationary distribution:

The union of independent Poisson point processes with diffuse intensity measures ${\mu }_1,\dots ,{\mu }_m$ is a Poisson point process with intensity measureOn every finite collection of disjoint measurable sets, this follows by adding independent Poisson counts and checking that the resulting counts remain independent.

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 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.

A zeros trick implements a positive likelihood function $L(\theta )$ by adding an observed zero with Poisson distribution mean $K-\log L(\theta )$. If this mean is nonnegative throughout the parameter support, its likelihood is $e^{-K}L(\theta )$ and yields the intended Bayesian posterior. The constant $K$ must be independent of the parameter; arbitrary clipping changes the target.