**Celebrities die 2.7218 at a time**

The claim that celebrities die in threes is usually dismissed as the result of the human propensity to see patterns where there are none. But celebrities don’t die at regularly spaced intervals either. It would be very weird if a celebrity predictably died on the 14th of every month. And once you deviate from a regularly spaced pattern, some amount of clustering is inevitable. Can we make this more precise?

Rather than trying to define exactly what constitutes a celebrity,
I’ll simply assume that they die at a fixed rate and that they do so
independently of each other (The Day the Music
Died
notwithstanding). It follows that celebrity deaths is a Poisson
process with intensity

As an example, suppose we define celebrityhood in such a way that
twelve celebrities die each year on average. Then

What does it mean for celebrities to die

Here is a diagram of 10 years worth of randomly generated deaths with 12 deaths per year and clusters as defined above highlighted:

**Average cluster size**

Suppose a celebrity has just died after a longer than average
wait. This death will start a new cluster, and we want to figure out
what the size of it is.

In a Poisson process the waiting time between two events is
exponentially distributed with parameter

The cluster size will be 1 when the waiting time for the next death is
larger than or equal to the average (which is

$\text{P}(C = 1) = \text{P}(W > 1/\lambda)$

The probability that the cluster will have size 2 is the same as the probability that the next waiting time is shorter than average and the next one after that is longer:

$\text{P}(C = 2) = \text{P}(W \le 1/\lambda)\cdot \text{P}(W > 1/\lambda)$

For size three, it’s the probability that the next two waiting times are shorter and the third one longer:

$\text{P}(C = 3) = \text{P}(W \le 1/\lambda)^2\cdot \text{P}(W > 1/\lambda)$

In general, the probability that the next cluster will be size

$\text{P}(C = k) = \text{P}(W \le 1/\lambda)^{k - 1}\cdot \text{P}(W > 1/\lambda)$

So what’s the average size of a Celebrity Death Cluster? The expected
value of

$\displaystyle \text{E}[C] = \sum_{k=1}^\infty k \cdot \text{P}(C = k) = \sum_{k=1}^\infty k\cdot \text{P}(W \le 1/\lambda)^{k - 1}\cdot \text{P}(W > 1/\lambda)$

Plugging in the distribution function for the exponential distribution, we get:

$ \begin{align*} \text{E}[C] &= \sum_{k=1}^\infty k \cdot (1 - e^{- \lambda \cdot (1/\lambda) })^{k - 1} \cdot (1 - (1 - e^{- \lambda \cdot (1 / \lambda)}))\\ &= \sum_{k=1}^\infty k \cdot (1 - e^{- 1})^{k - 1} \cdot e^{-1} \end{align*} $

It’s not hard to show that this infinite series has sum