exponential distribution - get some meaning

Question

I read that a continuous random variable having an exponential distribution can be used to model inter-event arrival times in a Poisson process. Examples included the times when asteroids hit the earth, when earthquakes happen, when calls are received at a call center, etc. In all these examples, the expected value of the number of events per unit of time, lambda, is known and is constant over time. Moreover, each event's occurrence is independent of previous events' occurrences. And the exponential variable that models inter-event arrivals has the same lambda parameter as the Poisson variable that models the number of events.

And now, my problem. It don't get the connection between the intuition behind the exp distrib and its pdf. It seems obvious that the more time it passes by without an earthquake happening, the more likely it is than an earthquake will happen. Assume my understanding of lambda is correct, i.e., lambda = the rate at which the event happens, e.g., 5 earthquakes per minute on some remote, angry planet. On the pdf graphic, the prob of generating a value between 0 and 1 is greater than the prob of choosing a value between 4 and 5 for instance. How is this graphic related to the fact that on average we need to have 5 earthquakes per minute?

Answer 1

Just restating my comment as an answer:

You are confusing "long interarrival times are rare" with "the more time passes by, the more likely it is an earthquake will happen". The first is true for an exponential distribution if your mean interarrival time is short. The second one is never true for an exponential distribution, as Did and the others already have noted.

I find it an interesting question though because this is something you will come across relatively often in newspapers. Statements like "a big eruption of Yellowstone park is overdue" or "the last big asteroid crash dated from ..., considering the mean rate is ..., we are overdue for a crash.".

They create a false sense of impending doom. What they are really stating is that we are experiencing an unusually long interarrival time, but insofar as the process is Poisson, the probability of the event happening in the next days, months or years is still independent of how long it has been since the last event.

Add to that the fact that these processes are only approximately Poisson or even that the rates are variable and you realize that one should really relativize all this sensationalist press. For instance, asteroid crashes become ever rarer as planetary orbits get gradually cleared of asteroids through the solar system's history. There are just less and less asteroids to crash. So the rate of the process is not constant.

Answer 2

It seems obvious that the more time it passes by without an earthquake happening, the more likely it is than an earthquake will happen.

As explained in the comments, to use an exponential distribution is to assume the opposite: that whatever time passed by without an earthquake happening, still as likely as ever it is than an earthquake will happen.

So, if one thinks this rate should not be constant, how to model the time $X$ until the first earthquake happens? The hypothesis is that, for every $t\geqslant0$, when $s\to0^+$, $$ P[X\leqslant t+s\mid X\geqslant t]=r(t)s+o(s), $$ for some function $r$. Thus, the function $\bar F:t\mapsto P[X\geqslant t]$ is such that $$ \bar F(t+s)=\bar F(t)\cdot(1-r(t)s+o(s)), $$ for every $t\geqslant0$, that is, $$ \bar F'(t)=-r(t)\bar F(t). $$ Since $\bar F(0)=1$ by definition, one gets $$ \bar F(t)=\exp\left(-\int_0^tr(u)\,\mathrm du\right), $$ and finally, $X$ has PDF $$ f(t)=\exp\left(-\int_0^tr(u)\,\mathrm du\right)r(t)\,\mathbf 1_{t\geqslant0}. $$ One sees that every density $f$ can be written like that, since one recovers the instantaneous rate $r$ through the formula $$ r(t)=\frac{f(t)}{\bar F(t)}=\frac{f(t)}{1-F(t)}. $$

Answer 3

The key property of the exponential distribution is that it is memoryless, so that translation of the PDF in time $x\to x-c$ is just scaling of the PDF $f\to e^{\lambda c} f$ plus cutting it off at zero. Probabilistically this says $P(X>s+t|X>s)=P(X>t)$. That is, it doesn't matter how long it was since the last event, the probability that it takes a certain time from now is the same as it would be in any other situation. For somebody who has been waiting the whole time it might seem very unlikely that it will take 20 minutes, but given that it doesn't happen in the first quarter an hour, their new estimate of the likelihood of the total wait (not the wait stating at quarter an hour) being at least 20 minutes has gone up - it is now the same they thought the probability of it's taking over 5 minutes was at the start. Thus you have to be careful what you are asking.

Your feeling that the likelihood of an earthquake goes up over time is indeed contrary to the principle of the exponential dist. as Did said. However, one thing is true: e.g. let $p$ be the prob. of the next earthquake being between 15-20 minutes at the start, say at midnight; let $X$ be the time past midnight. Let $q$ be the the probability it happened in that same time given that you know it doesn't happen in the first 10 minutes. Then $q>p$. But let $Y$ be the time taken for the first earthquake after $00:10$. The probability that $Y$ is between 5 and 10 is unchanged by the knowledge that there was nothing between midnight and 10 past midnight. It was always $q$.

As to the mean, well, there's no reason why earlier times being more likely can prevent the mean being anything in particular - if there are large times with some probability then it can all balance out.

Answer 4

The link between the Poisson point process and the exponential is probably the key to understanding this.

Let's say that asteroids hit your planet as a Poisson point process rate $\lambda$ per minute. This means that if $X_i$ is the number of asteroids that arrive in the $i$th minute then the sequence $X_i$ are independent Poisson distributed random variables with mean $\lambda$. Furthermore the exact arrival times of the $X_i$ asteroids in minute $i$ are distributed uniformly in the interval $(i-1,i)$.

The "magic" of the Poisson distribution then means that in any period of $T$ minutes, the number of asteroids that arrive will be a Poisson distribution mean $\lambda T$. Even if $T$ is not an integer.

So, what is the probability that the first asteroid lands in the $4$th minute? That's two events, firstly at least one asteroid lands in the $4$th minute, and secondly, that no asteroid arrives in the first three minutes.

That's reasonably easy to work out. The probability that at least one asteroid lands in the $4$th minute is $1-e^{-\lambda}$ and the probabilty that no asteroid lands in the first three minutes. is $e^{-3\lambda}$. So the probability that the first asteroid lands in the $4th$ minute is

$$(1-e^{-\lambda})e^{-3\lambda} = e^{-3\lambda}-e^{-4\lambda} = \frac 1\lambda\int_3^4e^{-\lambda t}dt.$$

The maths works out the same for any interval. But the intuition is hopefully a bit clearer as to why the pdf is decreasing. If the first asteroid arives at time $t$ it's because no asteroids have arrived in the first $t$ minutes.

As the (infinitesimal) probability that an asteroid arrives at time $t$ doesn't depend on $t$, the probability density that the first asteroid arrives at time $t$ is directly proportional to the probabilty that no asteroids arrive in the interval $(0,t)$ which must be decreasing.