Random arrivals

A Geiger counter clicks at random, twice a second on average. Guess what random looks like, then watch ten minutes of clicks.

What random looks like

Before the clip you picked a drawing. Real random arrivals are the one with clumps and gaps. On the timeline the ticks bunch together in places and leave long silences in others, though the counter never changed.

Most people expect random events to be spread out evenly, so a cluster or a long gap looks like a sign that something changed. It is not. Evenly spaced ticks would be the surprising picture: they would mean each tick knows where the last one was.

The clicks here have two properties:

• A steady rate. On average $\lambda$ clicks arrive per second, here $\lambda=2$, the same at every moment.
• No memory. What happens in one stretch of time tells nothing about any other stretch that does not overlap it.

Events with these two properties form a Poisson process. The letter $\lambda$ is a rate here, as in Continuous random variables; in the linear algebra course it is an eigenvalue.

Counts in a window

How many clicks land in a window of $t$ seconds? Cut the window into $n$ tiny slices, so thin that two clicks never share one. Each slice holds a click with chance $\lambda t/n$, independently of the others. The count is then binomial, $\mathrm{Bin}(n,\lambda t/n)$, as in Discrete random variables.

Now let the slices get thinner and thinner. The number of tries grows and the chance per try shrinks, while their product stays $\lambda t$. The binomial probabilities settle on the Poisson law with mean $\lambda t$:

I write $X\sim\mathrm{Poisson}(\lambda t)$. Its mean and its variance are both $\lambda t$. In the picture the windows are 3 seconds long, so the counts have mean 6 and variance 6.

Over the whole 600 seconds the expected number of clicks is 1200, give or take about $\sqrt{1200}=34.6$. The clip’s run gave 1270, about two of those give-or-takes above, and its 200 windows held 6.35 clicks on average with a variance of 6.02: mean and variance nearly equal, as they are for every Poisson count.

Gaps: the exponential law

How long until the next click? A gap longer than $t$ means no click in a window of length $t$, and the Poisson law gives that chance directly:

So the gap $T$ follows the exponential law $\mathrm{Exp}(\lambda)$, with density $\lambda e^{-\lambda t}$ and mean $1/\lambda$: half a second here. That is the curve under the second histogram. Short gaps are the most common, which is why ticks bunch.

The exponential law has no memory, as Continuous random variables showed for light bulbs:

Merging and splitting streams

Two independent Poisson streams, at rates $\lambda_1$ and $\lambda_2$, merge into one Poisson stream at rate $\lambda_1+\lambda_2$: each tiny slice now has two independent chances of a click.

Keep each click of a stream with chance $p$, independently, and throw the rest away. The kept clicks form a Poisson stream at rate $p\lambda$, and, less obviously, the kept and the dropped clicks are independent of each other. This is called thinning.

The waiting-time paradox

Buses arrive as a Poisson process, on average every 10 minutes. You walk to the stop at a random moment. How long do you wait, on average?

By the missing memory, your wait from the moment you arrive is exponential with the same mean, 10 minutes, not 5. That seems to clash with gaps of 10 minutes on average. The answer is that you are more likely to arrive during a long gap than a short one, because long gaps cover more of the clock. The gap you land in has mean 20 minutes, twice the typical gap, and you wait for the part of it still to come.

If the buses ran exactly every 10 minutes, your average wait would be 5. Random arrivals double it.

Worked example

1. Clicks in a window. At 2 per second, a 3-second window has mean count 6.

Even the most likely counts, 5 and 6, which are equally likely, each turn up in only about one window in six.

2. Gaps. A gap longer than 1 second has chance $e^{-2}=0.1353$. Half the gaps are shorter than $\ln2/2=0.3466$ s, though the mean gap is 0.5 s. After a second of silence, the chance of another full second of silence is still $0.1353$.

3. Many rare chances. A thousand independent tries, each succeeding with chance 0.003, give a binomial count; the Poisson law with mean 3 is close to it:

4. Merging and thinning. Streams of 2 and 3 clicks per second merge into one of 5 per second. Keeping each click of the merged stream with chance 0.4 leaves 2 per second. Keeping each click of the 2-per-second stream with chance 0.4 leaves 0.8 per second.

5. Buses. With Poisson buses every 10 minutes on average, a rider arriving at random waits 10 minutes on average; with buses every 6 minutes, 6. With evenly spaced buses every 10 minutes, the average wait is 5.

6. Counting photons. A camera pixel collects on average 10 000 photons in an exposure. The count is Poisson, so it wobbles with standard deviation $\sqrt{10\,000}=100$. The ratio of signal to this noise is $10\,000/100=100$, or 40 dB. Four times the light only doubles it: the noise grows as the square root.

Try it yourself. At 2 per second, the mean count in 5 seconds is 10. The chance of no click in 1 second is $e^{-2}=0.135$. After 10 seconds without a click, the rate is still 2 per second.

Where you’ll meet this

Traffic. Calls reaching an exchange, requests reaching a web server and packets reaching a router are modelled as Poisson streams. Queues are built from them, and the model predicts how long requests wait.

Radioactivity and light. Decays of a radioactive sample, and photons reaching a sensor, are Poisson. The photon count’s square-root wobble is shot noise, which limits every camera in dim light.

Failures. A part that fails at random, not by wearing out, has exponential lifetimes, and its failures in a fleet form a Poisson stream.

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