Poisson é exponencial como Gamma-Poisson é o quê?

Uma distribuição Poisson pode medir eventos por unidade de tempo, e o parâmetro é $\lambda$ . A distribuição exponencial mede o tempo até o próximo evento, com o parâmetro $\frac{1}{\lambda}$ . Pode-se converter uma distribuição na outra, dependendo se é mais fácil modelar eventos ou horários.

Agora, um gamma-poisson é um poisson "esticado" com uma variação maior. Uma distribuição Weibull é uma exponencial "esticada" com uma variação maior. Mas esses dois podem ser facilmente convertidos um no outro, da mesma maneira que Poisson pode ser convertido em exponencial?

Ou existe alguma outra distribuição que seja mais apropriada para uso em combinação com a distribuição gama-poisson?

A gama-poisson também é conhecida como distribuição binomial negativa, ou NBD.

— zbicyclist
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Respostas:

Este é um problema bastante direto. Embora exista uma conexão entre as distribuições de Poisson e Binomial Negativo, acho que isso não ajuda em nada sua pergunta específica, pois incentiva as pessoas a pensar em processos binomiais negativos. Basicamente, você tem uma série de processos de Poisson:

Y_{i} (t_{i}) | λ_{i} \sim P o i s s o n (λ_{i} t_{i})

$Y_i(t_i)|\lambda_i\sim Poisson(\lambda_i t_i)$

Onde é o processo e é o tempo que você observá-lo, e $Y_i$ $t_i$ $i$ denota os indivíduos. E você está dizendo que esses processos são "semelhantes", vinculando as taxas por uma distribuição:

λ_{i} \sim G a m m a (α, β)

$\lambda_i\sim Gamma(\alpha,\beta)$

Ao fazer a integração / mixagem sobre , você tem: $\lambda_i$

Y_{i} (t_{i}) | α β \sim N e g B i n (α, p_{i}) w h e r e p_{i} = \frac{t_{i}}{t_{i} + β}

$Y_i(t_i)|\alpha\beta\sim NegBin(\alpha,p_i)\;\;\; where \;\;p_i=\frac{t_i}{t_i+\beta}$

Isso tem um pmf de:

P r (Y_{i} (t_{i}) = y_{i} | α β) = \frac{Γ (α + y_{i})}{Γ (α) y_{i}!} p_{i}^{y_{i}} (1 - p_{i})^{α}

$Pr(Y_i(t_i)=y_i|\alpha\beta) = \frac{\Gamma(\alpha+y_i)}{\Gamma(\alpha)y_i!}p_i^{y_i}(1-p_i)^\alpha$

Para obter a distribuição do tempo de espera, observamos que:

P r (T_{i} \leq t_{i} | α β) = 1 - P r (T_{i} > t_{i} | α β) = 1 - P r (Y_{i} (t_{i}) = 0 | α β)

$Pr(T_i\leq t_i|\alpha\beta)=1-Pr(T_i> t_i|\alpha\beta)=1-Pr(Y_i(t_i)=0|\alpha\beta)$

= 1 - (1 - p_{i})^{α} = 1 - {(1 + \frac{t_{i}}{β})}^{- α}

$=1-(1-p_i)^\alpha=1-\left(1+\frac{t_i}{\beta}\right)^{-\alpha}$

Diferencie isso e você terá o PDF:

p_{T_{i}} (t_{i} | α β) = \frac{α}{β} {(1 + \frac{t_{i}}{β})}^{- (α + 1)}

$p_{T_i}(t_i|\alpha\beta)=\frac{\alpha}{\beta}\left(1+\frac{t_i}{\beta}\right)^{-(\alpha+1)}$

Este é um membro das distribuições generalizadas de Pareto, tipo II. Eu usaria isso como sua distribuição do tempo de espera.

Para ver a conexão com a distribuição Poisson, observe que , de modo que, se definirmos $\frac{\alpha}{\beta}=E(\lambda_i|\alpha\beta)$ e, em seguida, pegue o limite, obtemos: $\beta=\frac{\alpha}{\lambda}$ $\alpha\to\infty$

lim_{α \to \infty} \frac{α}{β} {(1 + \frac{t_{i}}{β})}^{- (α + 1)} = lim_{α \to \infty} λ {(1 + \frac{λ t_{i}}{α})}^{- (α + 1)} = λ \exp (- λ t_{i})

$\lim_{\alpha\to\infty}\frac{\alpha}{\beta}\left(1+\frac{t_i}{\beta}\right)^{-(\alpha+1)}=\lim_{\alpha\to\infty}\lambda\left(1+\frac{\lambda t_i}{\alpha}\right)^{-(\alpha+1)}=\lambda\exp(-\lambda t_i)$

Isso significa que você pode interpretar como um parâmetro de sobre-dispersão. $\frac{1}{\alpha}$

— probabilityislogic
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You can also note that the waiting time distribution is, roughly speaking, an exponential distribution with a Gamma random rate parameter and strictly speaking this is a Beta distribution of the second kind, as for any Gamma distribution with a Gamma random rate parameter.

— Stéphane Laurent

Using @probabilityislogic as a basis, I found the following article providing more detail on the relationship between NBD and Pareto: Gupta, Sunil and Donald G. Morrison. Estimating Heterogeneith in Consumers' Purchase Rates. Marketing Science, 1991, 10(3), 264-269. Thanks to all who helped me answer this question.

— zbicyclist

+1, I guess this nice analytical form may no longer exist for

P o i s s o n (λ_{i} t_{i} + c)

$Poisson(\lambda_i t_i + c)$ , where

c

$c$ is a constant.

— Randel

@randel - you could get a "nice-ish" form by noting this rv is the sum of two independent rvs...

Z_{i} = Y_{i} + X_{i}

$Z_i=Y_i+X_i$ where

Y_{i}

$Y_i$ is the same as above and

X_{i} \sim p o i s s o n (c)

$X_i\sim poisson (c)$ . As

X_{i}

$X_i$ doesn't depend on

λ_{i}

$\lambda_i$ or

Y_{i}

$Y_i$ the pdf of

Z_{i}

$Z_i$ is the convolution of the above negative binomial pdf and a poisson pdf. To get the waiting time distribution just multiply

P r (Y_{i} = 0)

$Pr(Y_i=0)$ in the above answer by

P r (X_{i} = 0) = e^{- c}

$Pr(X_i=0)=e^{-c}$ . You then get waiting time cdf of

1 - e^{- c} {(1 + \frac{t_{i}}{β})}^{- α}

$1-e^{-c}\left(1+\frac{t_i}{\beta}\right)^{-\alpha}$ and pdf of

e^{- c} \frac{α}{β} {(1 + \frac{t_{i}}{β})}^{- (α + 1)}

$e^{-c}\frac {\alpha}{\beta}\left(1+\frac{t_i}{\beta}\right)^{-(\alpha+1)}$ .

— probabilityislogic

This won't work in terms of the mixing distribution, because you need

λ_{i} < c t_{i}^{- 1}

$\lambda_i <ct_i^{-1}$ (else the poisson mean is negative). The gamma mixing distribution would need to be truncated (I also assumed that

c > 0

$c>0$ in my previous answer). This would mean no nb distribution.

— probabilityislogic

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 time $t$ 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.

Choose $p \lt 1$ .
Let $X_1, X_2, X_3, ...$ be IID with logarithmic distributions, so $P(x_i = k) = \frac{-1}{\log(1-p)} \frac{p^k}{k}.$
Let $N$ be a Poisson process with constant rate $-\log(1-p)$ , so $N(t) = \text{Pois}(-t \log(1-p)).$
Let $NBP$ be the process so that

N B P (t) = \sum_{i = 1}^{N (t)} X_{i} .

$NBP(t) = \sum_{i=1}^{N(t)} X_i.$

$NBP$ is a pure jump process with logarithmically distributed jumps. The gaps between jumps follow an exponential distribution with rate $-\log(1-p).$

I don't think it is obvious from this description that $NBP(t)$ 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.

— Douglas Zare
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No, any compound Poisson process has an exponential waiting time. This means you add

Pois (λ t)

$\text{Pois}(\lambda t)$ IID random variables with some distribution.

— Douglas Zare

No, that is not what is meant by a compound Poisson process. en.wikipedia.org/wiki/Compound_Poisson_process " The jumps arrive randomly according to a Poisson process and the size of the jumps is also random, with a specified probability distribution." I did not say IID Poisson variables. You take the

N

$N$ th partial sum of IID logarithmic random variables where

N

$N$ is the value of a Poisson process.

— Douglas Zare

If you multiply a Poisson process by

2

$2$ , this is not a Poisson process and the waiting times remain exponential.

— Douglas Zare

let us continue this discussion in chat

— Douglas Zare

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.

— Meadowlark Bradsher
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cause of the overdispersion: This is consumer purchase data. Individual consumers are poisson, each with a rate of purchase lambda. But not every consumer has the same lambda -- that's the cause of the overdispersion. The lambda purchasing rates are considered to be distributed as gamma. This is a common model (traces back to A.S.C. Ehrenberg), but I haven't found anything in his writing that answers this question.

— zbicyclist