How to update Poisson conjugate prior with observations of arrival time instead of counts?

Question

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

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

Answer 1

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 $t = \sum_{i=1}^n a_i$.