Jest to dość prosty problem. Chociaż istnieje związek między rozkładami Poissona i ujemnych dwumianów, tak naprawdę uważam, że nie jest to pomocne w przypadku konkretnego pytania, ponieważ zachęca ludzi do myślenia o negatywnych procesach dwumianowych. Zasadniczo masz szereg procesów Poissona:
Yi(ti)|λi∼Poisson(λiti)
Gdzie jest procesem, a t i jest czas obserwować go, a ja oznacza fizycznych. Mówisz, że te procesy są „podobne”, wiążąc stawki razem poprzez rozkład:Yitii
λi∼Gamma(α,β)
Wykonując integrację / miksowanie nad , masz:λi
Yi(ti)|αβ∼NegBin(α,pi)wherepi=titi+β
Ma to pmf:
Pr(Yi(ti)=yi|αβ)=Γ(α+yi)Γ(α)yi!pyii(1−pi)α
Aby uzyskać rozkład czasu oczekiwania, zauważamy, że:
= 1 - ( 1 - p i ) α = 1 - ( 1 +
Pr(Ti≤ti|αβ)=1−Pr(Ti>ti|αβ)=1−Pr(Yi(ti)=0|αβ)
=1−(1−pi)α=1−(1+tiβ)−α
Zróżnicuj to i masz plik PDF:
pTi(ti|αβ)=αβ(1+tiβ)−(α+1)
Jest to członek uogólnionych dystrybucji Pareto, typ II. Użyłbym tego jako twojego rozkładu czasu oczekiwania.
Aby zobaczyć związek z rozkładem Poissona, zwróć uwagę, że , więc jeśli ustawimyβ=ααβ=E(λi|αβ) a następnie weź granicęα→∞otrzymujemy:β=αλα→∞
limα→∞αβ(1+tiβ)−(α+1)=limα→∞λ(1+λtiα)−(α+1)=λexp(−λti)
This means that you can interpret 1α as an over-dispersion parameter.
One possibility: Poisson is to Exponential as Negative-Binomial is to ... Exponential!
There is a pure-jump increasing Lévy process called the Negative Binomial Process such that at timet the value has a negative binomial distribution. Unlike the Poisson process, the jumps are not almost surely 1 . Instead, they follow a logarithmic distribution. By the law of total variance, some of the variance comes from the number of jumps (scaled by the average size of the jumps), and some of the variance comes from the sizes of the jumps, and you can use this to check that it is overdispersed.
There may be other useful descriptions. See "Framing the negative binomial distribution for DNA sequencing."
Let me be more explicit about how the Negative Binomial Process described above can be constructed.
Choosep<1 .
LetX1,X2,X3,... be IID with logarithmic distributions, so P(xi=k)=−1log(1−p)pkk.
LetN be a Poisson process with constant rate −log(1−p) , so N(t)=Pois(−tlog(1−p)).
LetNBP be the process so that
I don't think it is obvious from this description thatNBP(t) has a negative binomial NB(t,p) distribution, but there is a short proof using probability generating functions on Wikipedia, and Fisher also proved this when he introduced the logarithmic distribution to analyze the relative frequencies of species.
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I am not able to comment yet so I apologize is this isn't a definitive solution.
You are asking for the appropriate distribution to use with an NB but appropriate isn't entirely defined. If an appropriate distribution means appropriate for explaining data and you are starting with an overdispersed Poisson then you may have to look further into the cause of the overdispersion. The NB doesn't distinguish between a Poisson with heterogeneous means or a positive occurrence dependence (that one event occurring increases the probability of another occurring). In continuous time there is also duration dependence, eg positive duration dependence means the passage of time increases the probability of an occurrence. It was also shown that negative duration dependence asymptotically causes an overdispersed Poisson[1]. This adds to the list of what might be the appropriate waiting time model.
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