#### 26 Jun 2013 ssp»(Master)

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 $\lambda$ where $\lambda$ is the number of deaths that occur in some fixed time period.

As an example, suppose we define celebrityhood in such a way that twelve celebrities die each year on average. Then $\lambda = 12/\text{year}$, and because the time between events in a Poisson process is exponentially distributed with parameter $\lambda$, the average time between two deaths is $1/\lambda$ = 1/12th year, or one month.

What does it mean for celebrities to die $n$ at a time? We will simply say that two celebrities die together if the period between their deaths is shorter than expected. If the celebrity death rate is 12/year, then two celebrities died together if their deaths were less than one month apart. Similarly, three celebrities died together if the period between death 1 and death 2 and the period between death 2 and death 3 were both shorter than a month. In general, $k$ celebrities died together if the $k - 1$ periods between their deaths were all shorter than expected.

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 $\lambda$, so it can be modelled with a stochastic variable $W \sim Exp(\lambda)$. The cluster size itself is modelled with another stochastic variable, $C$, whose distribution is derived as follows.

The cluster size will be 1 when the waiting time for the next death is larger than or equal to the average (which is $1/\lambda$ for the exponential distribution):

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

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 $k$ is:

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

So what’s the average size of a Celebrity Death Cluster? The expected value of $C$ is given by:

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

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 $e$ (Hint: Use the fact that $k x^{k - 1}$ is the derivative of $x^k$), so on average, celebrities die 2.7218 at a time.

Syndicated 2013-06-26 00:00:00 from Søren Sandmann Pedersen