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 by
and
- 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
.