# Poisson distribution: why does time between events follow an exponential distribution?

I was reading an article, and came across the following:

Purchase count follows a Poisson distribution with rate λ. In other words, the timing of these purchases is somewhat random, but the rate (in counts/unit time) is constant. In turn, this implies that the inter-purchase time at the customer level should follow an exponential distribution.

It's been quite a while since I did any statistics so I am struggling with the definitions of a Poisson distribution. What I understand by the "rate is constant" is that if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?

Where I believe I am confused is with the final sentence. Is this saying that the time between a customers purchases would grow exponentially as time goes on? Doesn't this contradict the idea that we have a constant rate of purchase?

• For a full account of this, including definitions, see stats.stackexchange.com/questions/214421. For related posts, see stats.stackexchange.com/… Another useful search (focusing on "memoryless") is stats.stackexchange.com/…. – whuber May 31 at 16:29
• $${}$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $\textbf{Yes}.$ $${}$$ "Is this saying that the time between a customers purchases would grow exponentially as time goes on?" $\textbf{No},$ that's not what it means at all. $\qquad$ – Michael Hardy May 31 at 19:17

## 2 Answers

Let $$X_t$$ be the number of arrivals in the Poisson process with rate $$\lambda$$ between time $$0$$ and time $$t\ge0.$$ Then we have $$\Pr(X_t=x) = \frac{(\lambda t)^x e^{-\lambda t}}{x!} \text{ for } x=0,1,2,3,\ldots$$

Let $$T$$ be the time until the first arrival.

Then the following two events are really both the same event: $$\Big[ X_t=0\Big]. \qquad \Big[ T>t \Big].$$ So they both have the same probability. Thus $$\Pr(T>t) = \Pr(X_t=0) = \frac{(\lambda t)^0 e^{-\lambda t}}{0!} = e^{-\lambda t}.$$ So $$\Pr(T>t) = e^{-\lambda t} \text{ for } t\ge0.$$ That says $$T$$ is exponentially distributed.

• That's very clever. Thanks! – J. Stott Jun 3 at 8:45

Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:$$\lambda e^{-\lambda t}$$ where $$\lambda$$ is Poisson intensity, i.e. average number of events in unit of time, and $$t$$ is the waiting time. The average waiting time is obviously $$1/\lambda$$.