This is related to my previous question: How to update Poisson conjugate prior with observations of arrival time instead of counts?

Using the same notation, suppose $N \sim \operatorname{Pois}(\mu)$ is a Poisson distributed random variable, with interarrival times between observations distributed accordingly as $A \sim \operatorname{Exp}(\mu)$. Let $\mu$ be distributed according to the conjugate prior $\mu \sim \operatorname{Gamma}(\alpha, \beta)$.

Suppose I start waiting for the next observation of my Poisson process, and I have waited a long time $t$ without observing any, that is, in time $t$, there are $n = 0$ observations. Is it possible to derive a posterior distribution for $\mu|n = 0, t$? Would it be:

$$\mu|n = 0, t \sim \operatorname{Gamma}(\alpha, \beta + t)$$

I am not sure how this is consistent with Wikipedia article on updating the prior of an exponential distribution, as that explicitly relies on at least one observation of an interarrival time.


1 Answer 1


Not only is it possible to update your prior for the observed waiting time, it is necessary to do this in order to faithfully capture the observed data. What you are dealing with is a censored observation where the time until the event is known to be greater than the waiting time $t$. Letting $n(t)$ be the number of arrivals at time $t$, you have $n(t) \sim \text{Pois}(\mu t)$.

The sampling distribution for your observation is:

$$p(n(t) = 0| \mu ) = \text{Pois}(0 | \mu t) = \exp (- \mu t).$$

Hence, your posterior distribution is:

$$\begin{equation} \begin{aligned} p(\mu | n(t) = 0) &\propto \exp(-\mu t) \cdot \mu^{\alpha -1} \exp(-\beta \mu) \\ &= \mu^{\alpha -1} \exp(- (\beta + t) \mu) \\ &\propto \text{Ga}(\mu |\alpha, \beta + t ). \end{aligned} \end{equation}$$

So yes, you can derive the posterior, and your proposed posterior distribution is correct.


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