Let $J_n = \sum_{i=1}^n S_i$ be a renewal process with gamma distributed interval times $S_i \sim Gamma(\alpha,\beta)$.

Consider the value $J_k$ of the first event when the process passes a certain point $t_{boundary}$.

What is the distribution of $t = J_k - t_{boundary}$? More specifically what is the distribution in the limit $t_{boundary} \to \infty$?

Example computation/simulation of the process and the distribution for $S_i \sim Gamma(\alpha = 5,\beta = 20)$ and $t_{boundary} = 10$


alpha = 5
beta = 20

sim = function(x = 10) {
  t = 0
  ### keep adding gamma distributions
  ### untill t is above some boundary point
  while(t<x) {
    dt = rgamma(1,alpha,beta)
    t = t+dt
  return(t-x) ### return the difference with the boundary point

time = replicate(10^4,sim())  
hist(time, freq = 0, breaks = seq(0,max(time)+0.2,0.05))

1 Answer 1


We can argue that the interval length of the last interval $S_k$ is distributed as $f(t) \sim Gamma(\alpha+1,\beta)$. That is, the $\alpha$ parameter is one more than for the unconditional distribution of interval times $S$. See for a motivation: Intuitive explanation of paradoxical interval times distribution

If the interval had size $S_k$, we can imagine the point $t_{boundary}$ to be uniform distributed within that interval. Then the distribution of $t$ is like a compound distribution

$$t \sim U(0,S_k) \\ S_k \sim Gamma(\alpha+1,\beta)$$

This is related to the integral

$$g(t) = \int_t^\infty \frac{1}{s} f(s,\alpha+1,\beta) ds$$ and we can express this in terms of the cumulative distribution of a Gamma distribution

$$g(t) = \frac{\beta}{\alpha} (1-F(t,\alpha,\beta))$$

Here is a code that adds this distribution to the histogram plot

ts = seq(0,max(time),0.01)

histogram with distribution added

More generally the solution is

$$g(t) = \frac{(1-F(t,\alpha,\beta))}{E[S]}$$

as mentioned in How to calculate average length of adherence to vegetarianism when we only have survey data about current vegetarians?


Not the answer you're looking for? Browse other questions tagged or ask your own question.