Super-Poissonian distribution

In probability theory it is common to say a distribution, D, is a sub-distribution of another distribution E if D 's moment-generating function, is bounded by E 's up to a constant.

In other words

:$$

E_{X\sim D}[\exp(t X)] \le E_{X\sim E}[\exp(C t X)].

for some C > 0.{{Cite book |last=Vershynin |first=Roman |url=https://books.google.com/books?id=TahxDwAAQBAJ&dq=high+dimensional+probability&pg=PR11 |title=High-Dimensional Probability: An Introduction with Applications in Data Science |date=2018-09-27 |publisher=Cambridge University Press |isbn=978-1-108-24454-1 |language=en}}

This implies that if $X\_1$ and $X\_2$ are both from a sub-E distribution, then so is $X\_1+X\_2$.

A distribution is strictly sub- if C ≤ 1.

From this definition a distribution, D, is sub-Poissonian if

:$$

E_{X\sim D}[\exp(t X)]

\le E_{X\sim \text{Poisson}(\lambda)}[\exp(t X)]

= \exp(\lambda(e^t-1)),

for all t > 0.{{Cite journal |last=Ahle |first=Thomas D. |date=2022-03-01 |title=Sharp and simple bounds for the raw moments of the binomial and Poisson distributions |url=https://www.sciencedirect.com/science/article/pii/S0167715221002662 |journal=Statistics & Probability Letters |language=en |volume=182 |pages=109306 |doi=10.1016/j.spl.2021.109306 |arxiv=2103.17027 |issn=0167-7152}}

An example of a sub-Poissonian distribution is the Bernoulli distribution, since

:$E[\backslash exp(t\; X)]\; =\; (1-p)+p\; e^t\; \backslash le\; \backslash exp(p(e^t-1)).$

Because sub-Poissonianism is preserved by sums, we get that the binomial distribution is also sub-Poissonian.

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Category:Poisson point processes

Category:Types of probability distributions

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