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Suppose a random variable $N \sim \operatorname{Pois}(\mu)$, with Gamma conjugate prior such that $\mu \sim \operatorname{Gamma}(\alpha, \beta)$. Then given a sequence of $n$ observations of counts $\mathbf{k} = (k_1, \dots, k_n)$ in the next $n$ units of time, then the way to derive the posterior distribution of $\mu|\mathbf{k}$ is by updating the prior so that:

$$\mu|\mathbf{k} \sim \operatorname{Gamma}(\alpha + \sum_{i=1}^n k_i, \beta + n)$$

Suppose instead of knowing the counts $\mathbf{k}$, I only know that the next success was observed in $t$ units of time. Is it then appropriate to update the prior so that:

$$\mu|\text{time to success} = t \sim \operatorname{Gamma}(\alpha + 1, \beta + t)$$

If this is appropriate, why does this hold and how does it relate to the previous update, and if it is not appropriate, how do you incorporate one observation of time of next success?

For example, I can see that when the time of next success is a small fraction of the unit time, and let $t_1, \dots, t_{k_1}$ denote the inter-arrival times of the first $k_1$ successes, then

$$\operatorname{Gamma}(\alpha + k_i, \beta + 1) \approx \operatorname{Gamma}(\alpha + k_i, \beta + \sum_{i=1}^{k_1} t_i)$$

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I think your confusion comes with the Poisson distribution and what $n$ is. On the linked wikipedia page you have $k_i \stackrel{ind}{\sim} Po(\mu)$ for $i=1,\ldots,n$. The relationship I think you are looking for is below.

A Poisson process, i.e. $N(t)\sim Po(\mu t)$, has exponentially distributed interarrival times (times between arrivals), i.e. $$ A_i \stackrel{ind}{\sim} Exp(\mu) $$ where $E[A_i] = 1/\mu$ where this is the same $\mu$ from the Poisson process. Assume the same prior on $\mu$, namely $$ \mu \sim Ga(\alpha,\beta). $$ Suppose you observe this system until the $n$th arrival which occurs at time $t = \sum_{i=1}^n a_i$. Then the posterior is $$ \mu|n,t \sim Ga(\alpha+n,\beta+t) $$ which can be derived either using the count: $$ p(\mu|n,t) \propto (\mu t)^n e^{-\mu t} \mu^{\alpha-1} e^{-\beta \mu} \propto \mu^{\alpha+n-1} e^{-(\beta+t) \mu} $$ or from the interarrival times $$ p(\mu|a_1,\ldots,a_n) \propto \left[ \prod_{i=1}^n \mu e^{-\mu a_i} \right] \mu^{\alpha-1} e^{-\beta \mu} \propto \mu^{\alpha+n-1} e^{-(\beta+t) \mu} $$ since $t = \sum_{i=1}^n a_i$.

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  • $\begingroup$ Thanks for your answer. However, my understanding is that the number of counts is always measured in terms of a fixed interval, e.g. from the wikipedia article: "overflow floods occur once every 100 years on average". Let's say we observed $k_i$ floods in the next 100 years. It is almost surely guaranteed that the time to see these floods is not exactly 100 years. In this case, doesn't the update no longer agree as the sum of the inter-arrival times is less than 100? $\endgroup$ – Alex Feb 8 '17 at 23:09
  • $\begingroup$ I think what you are trying to say is that my first formula to update the prior using the observed counts is incorrect, instead of adding $n$ to $\beta$, I should add the fractional number of unit time intervals that tightly contain all the observed successes. In this case, I can see that the update rules agree. Or, in the last remaining bit of time $s$ in the unit interval where no succeses are recorded, update the prior using # success = 0 in time $s$. I suspect the second interpretation is the one that fully incorporates all information available to us. $\endgroup$ – Alex Feb 8 '17 at 23:13
  • $\begingroup$ You can let $t$ be the total time (including the additional time after the last observation) and use the posterior formulas provided. This is trivially true for the Poisson. For the exponential, you need to add a censored data term due the fact that you know the next observation must occur later than that additional amount of time. But the resulting posterior is exactly the same. $\endgroup$ – jaradniemi Feb 9 '17 at 17:07

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