Hello again!
In this lecture we are going to discuss the Poisson Distribution and its main characteristics.
For starters, we denote a Poisson distribution with the letters “Po” and a single value
parameter - lambda.
We read the statement below as “Variable “Y” follows a Poisson distribution with
lambda equal to 4”.
Okay!
The Poisson Distribution deals with the frequency with which an event occurs in a specific interval.
Instead of the probability of an event, the Poisson Distribution requires knowing how
often it occurs for a specific period of time or distance.
For example, a firefly might light up 3 times in 10 seconds on average.
We should use a Poisson Distribution if we want to determine the likelihood of it lighting
up 8 times in 20 seconds.
The graph of the Poisson distribution plots the number of instances the event occurs in
a standard interval of time and the probability for each one.
Thus, our graph would always start from 0, since no event can happen a negative amount
of times.
However, there is no cap to the amount of times it could occur over the time interval.
Okay, let us explore an example.
Imagine you created an online course on probability.
Usually, your students ask you around 4 questions per day, but yesterday they asked 7.
Surprised by this sudden spike in interest from your students, you wonder how likely
it was that they asked exactly 7 questions.
In this example, the average questions you anticipate is 4, so lambda equals 4.
The time interval is one entire work day and the singular instance you are interested in
is 7.
Therefore, “y” is 7.
To answer this question, we need to explore the probability function for this type of
distribution.
Alright!
As you already saw, the Poisson Distribution is wildly different from any other we have
gone over so far.
It comes without much surprise that its probability function is much different from anything we
have examined so far.
The formula looks as follows: “p of y, equals, lambda to the power of
y, times the Euler’s number to the power of negative lambda, over y factorial.
Before we plug in the values from our course-creation example, we need to make sure you understand
the entire formula.
Let’s refresh your knowledge of the various parts of this formula.
First, the “e” you see on your screens is known as Euler’s number or Napier’s
constant.
As the second name suggests, it is a fixed value approximately equal to 2.72.
We commonly observe it in physics, mathematics and nature, but for the purposes of this example
you only need to know its value.
Secondly, a number to the power of “negative n”, is the same as dividing 1 by that number
to the power of n.
In this case, “e to the power or negative lambda” is just “1 over, e to the power
of lambda”.
Right!
Going back to our example, the probability of receiving 7 questions is equal to “4,
raised to the 7th degree, multiplied by “E” raised to the negative 4, over 7 factorial,”.
That approximately equals 16384, times 0.183, over 5040, or 0.06.
Therefore, there was only a 6% chance of receiving exactly 7 questions.
So far so good!
Knowing the probability function, we can calculate the expected value.
By definition, the expected value of Y, equals the sum of all the products of a distinct
value in the sample space and its probability.
By plugging in, we get this complicated expression.
Eventually, we get that the expected value is simply lambda.
Similarly, by applying the formulas we already know, the variance also ends up being equal
to lambda.
Both the mean and variance being equal to lambda serves as yet another example of the
elegant statistics these distributions possess and why we can take advantage of them.
Great job, everyone!
Now, if we wish to compute the probability of an interval of a Poisson distribution,
we take the same steps we usually do for discrete distributions.
We find the joint probability of all individual elements within it.