Here is a nice application of series which appeared in my Calculus II course.

In class we defined a *probability density function (pdf)* to be a function such that

- for all , and
- .

We then discussed how to describe a pdf using two important quantities:

- the
*mean*, which is a measure of the center and is given byand

- the
*standard deviation*, which is a measure of how far from typical measurements tend to be (or how “spread out” the pdf is) and is given by

In all of this we are assuming that any real number is a possible value for (though we can restrict this to intervals by defining for otherwise).

** Discrete distributions **

Suppose that we are measuring some variable, but the only possible values for the variable are integers. We want to repeat what we did earlier, defining the mean and standard deviation of a probability distribution.

Let’s call the possible measurement values and let be the probability of measuring a value of . We require two properties of :

- We must have for each and
- the total probability must add up to one:

As before we define the mean to be the sum of each possible value, multiplied by the probability of that value:

and the standard deviation as the “distance from being all average”:

These formulas are particular interesting when for an infinite number of integers , as we need to deal with convergence issues.

** Example **

Suppose that we have and that the probabilities of measuring each value is given by

We know from geometric series that the total probability is because and thus

We can therefore try to compute the mean for this probability distribution, which would be given by

It is easy to see (do it!) that this series converges, but what does it converge to?

In fact, it is possible to compute the mean exactly. To do this, we first write

We then define function by

so that .

One can check that converges absolutely for . Furthermore,

Therefore .

We now consider the standard deviation, given by

We can split this in to three sums, the second and third of which can be computed using what we know already:

and

(In fact, there is a general phenomena going on here — see below.)

For the first sum in the expression for , we repeat the trick from before: Let

Thus we conclude that

and hence

** Exercises **

- Suppose is the pdf given for by
Find and .

- Let be any discrete probability distribution. Use the fact that to deduce that
What is the integration version of this identity?

- Suppose that is given, for , by
Find the mean and the deviation using the methods of the example.

- Let be given by
for . Explain why is a valid probability distribution, then find the mean and deviation .