An Introduction to the Poisson Distribution - YouTube

Let's look at an introduction to the Poisson distribution,

an important discrete probability distribution.

Suppose we are counting the number of occurrences of an event

in a given unit of time or distance or area or volume.

For example, we might be counting up the number of car accidents in a day,

in a city like Toronto perhaps.

Or the number of dandelions in a square metre plot of land.

Then the number of events is going to be a random variable

that may or may not have the Poisson distribution,

depending on the specifics of the situation.

But a Poisson random variable is a count of the number of occurrences of an event.

I'm going to phrase the following in terms of time,

but the same ideas hold if we are discussing distance or area or volume etc.

Suppose events are occurring independently.

In other words, knowing when one event happens

gives absolutely no information about when another event will occur.

And the probability that an event occurs in a given length of time does not change through time.

In other words, the theoretical rate at which the events are occurring

does not change through time.

A little more loosely we might say that the events are occurring randomly and independently.

If these conditions hold, then the random variable X,

which represents the number of events in a fixed unit of time,

has the Poisson distribution.

Here's the probability mass function for the Poisson distribution,

what we'll use to calculate probabilities.

The probability the random variable X takes on the value little x,

which you'll sometimes see written as p(x),

is equal to lambda^x times e^(-lambda) over x!

e, like pi, is an important mathematical constant, the base of natural logarithms.

It is approximately 2.71828, but it is an irrational number

that has infinite non-repeating decimal places.

We've discussed factorials previously, but as a specific example of this x!,

5! would be 5 times 4 times 3 times 2 times 1,

and that would be 120.

It's not a probability distribution until we say what values X can take on.

Here the random variable is a count of the number of events in a given unit of time,

and so it can take on any non-negative whole number value.

So this is the probability mass function that we use to calculate probabilities

for any value of x that's 0,1, 2, off to infinity.

There is no upper bound on the value that X can take on.

But depending on the situation the probabilities eventually

get tiny for large values of X.

The mean of the Poisson distribution is lambda.

So mu, the mean of the random variable X, is equal to lambda.

So we could have used mu as our parameter, and some sources do that.

But we often use lambda for the Poisson distribution.

The variance of the Poisson distribution,

which we'll label as sigma squared, is also equal to lambda.

For the Poisson distribution, the mean and the variance are equal.

Let's look at an example.

Plutonium-239 is an isotope of plutonium that is used in nuclear weapons and reactors.

One nanogram, or 1 billionth of a gram, of plutonium 239

will have an average of 2.3 radioactive decays per second.

And the number of decays in a given period will follow,

to a very close approximation, a Poisson distribution.

Here we'd like to know: what is the probability that in a

randomly selected two second period there are exactly 3 radioactive decays?

We'll let the random variable X

represent the number of decays in a two second period.

Lambda is the mean number of decays in that period.

So here we have an average of 2.3 radioactive decays per second,

but we're talking about a two second period

and so in that period, the mean number of occurrences,

which is going to equal lambda, is going to be 2.3 times 2, which is 4.6.

And so X has a Poisson distribution with lambda equal to 4.6.

We want to find the probability that the random variable X takes on the value 3.

And the Poisson probability mass function

is lambda^x times e to the minus lambda, over x!

And here that's going to be

4.6, lambda, raised to the third power

times e to the -4.6 divided by 3!

If we worked that out on a calculator or computer

we'd see that's equal to 0.163, when rounded to three decimal places.

So that is the probability of getting exactly three radioactive decays in a two second period.

If we were to calculate the probabilities for the

different possible values of X and plot them, we'd get this.

This is the probability distribution of the random variable X in this spot,

a Poisson distribution with lambda equal to 4.6.

The number we calculated, the probability that X takes on the value 3, is here.

That's what we just calculated to be 0.163.

We can see here that X takes on the possible values 0, 1, 2, on up.

I've truncated the plot over here at 15,

since the possible values go off to infinity,

but the probabilities start getting very very small.

But for a Poisson distribution there is no upper bound

on the values the random variable X can take on.

For this distribution, the mean mu is equal to lambda, and that's 4.6 here.

The variance is also equal to lambda, so that's also equal to 4.6

And if we wanted the standard deviation sigma, we'd simply take the square root of 4.6.

If we look closely we can see that there's a hint of right-skewness in this distribution.

The Poisson distribution has some right skewness,

but it depends on the value of lambda.

When lambda is large, the distribution will be close to symmetric,

when lambda is close to 0, the right skewness can be pretty strong.

Suppose we wanted a different probability,

the probability there are no more than three radioactive decays.

Here that's the red bit on the plot.

We'd need to work out the probability of 0, of 1, of 2, and of 3

using the Poisson probability mass function,

and add them together. So let's go ahead and do that.

Here we need to find the probability that the random variable X

takes on a value less than or equal to 3,

which is the sum up the probabilities of 0, 1, 2, and 3.

We put these values of x

into the Poisson probability mass function with a lambda of 4.6,

and when rounded to three decimal places,

these probabilities work to these 4 values,

and they sum to 0.326.

Working out probabilities like this can be a bit of a pain if there are a lot of values,

so we often rely on software to carry out the calculations.

There is an important relationship that sometimes helps us determine

whether a random variable has a Poisson distribution.

The binomial distribution tends toward the Poisson distribution

as n tends to infinity, p tends to 0 and np stays constant.

For us, at the moment, the important bit is that the Poisson distribution

with lambda equal to np from the binomial distribution,

closely approximates the binomial distribution if n is large and p is small.

In fact, this is why the radioactive decays of plutonium has a Poisson distribution.

Even for a tiny bit a plutonium, there are a very large number of atoms,

and each one has a tiny probability of experiencing a radioactive decay in a two second period.

So in the example we just worked through,

it was in its underlying nature a binomial problem

with a very large n and a very small p.

And that's why the number of radioactive decays

is very well approximated by the Poisson distribution.

I have videos that explore this relationship in greater detail.

Like many models in probability and statistics,

the Poisson distribution is typically used as an approximation to the true underlying reality.

In most situations where we use the Poisson,

we know that the Poisson distribution doesn't fit the scenario precisely

but we use it as an approximation. Possibly a very good approximation.

But it can be difficult to determine whether

a random variable has a Poisson distribution to a reasonable approximation.

So I'm going to look at a few examples and discuss some considerations in another video.