3.2 Discrete Uniform Distribution
=================================

3.2.1 Main Idea
---------------

The **discrete uniform distribution** applies when a random variable has a
finite number of possible values and every value is equally likely.

Suppose

.. math::

   X\in\{x_1,x_2,\ldots,x_n\}

and all :math:`n` values are equally likely. Then

.. note::

   **Discrete Uniform Distribution**

   .. math::

      P(X=x_i)=\frac{1}{n},
      \qquad i=1,2,\ldots,n.

The word **uniform** means that the probability is distributed equally over
all possible values.

A discrete uniform PMF therefore has the same probability height at every
allowed value. The possible values remain separate points, but none is more
likely than another.

.. image:: ../_static/figures/discrete_distributions/figure_3_2_1_discrete_uniform_pmf.png
   :width: 68%
   :align: center
   :alt: Discrete uniform probability mass function with equal probability at the consecutive values from a through b.

In the figure, each possible value receives the same probability
:math:`1/(b-a+1)`. This equal-height pattern is the defining graphical feature
of a discrete uniform distribution.


3.2.2 Example 1: Random Digit
-----------------------------

Suppose one digit is selected at random from

.. math::

   \{0,1,2,\ldots,9\}.

Let :math:`X` be the selected digit. There are

.. math::

   n=10

possible values and each is equally likely, so

.. math::

   P(X=x)=\frac{1}{10},
   \qquad x=0,1,\ldots,9.

The PMF therefore has the same height at every possible value.


3.2.3 Consecutive Integer Values
--------------------------------

A common discrete uniform model uses consecutive integers

.. math::

   a,a+1,\ldots,b.

The number of possible values is

.. math::

   n=b-a+1.

Therefore,

.. math::

   P(X=x)
   =
   \frac{1}{n}
   =
   \frac{1}{b-a+1},
   \qquad
   x=a,a+1,\ldots,b.

The formulas for the mean and variance can be derived directly from this
PMF.


3.2.4 Deriving the Mean
-----------------------

For any discrete random variable,

.. math::

   E(X)=\sum_x x\,p(x).

For a discrete uniform random variable on
:math:`\{a,a+1,\ldots,b\}`, every value has probability :math:`1/n`.
Therefore,

.. math::

   E(X)
   =
   \sum_{x=a}^{b}x\left(\frac1n\right)
   =
   \frac1n\sum_{x=a}^{b}x.

To evaluate the sum, recall

.. math::

   1+2+\cdots+m
   =
   \frac{m(m+1)}{2}.

Hence,

.. math::

   \sum_{x=a}^{b}x
   =
   \left(1+2+\cdots+b\right)
   -
   \left(1+2+\cdots+(a-1)\right).

Using the sum-of-integers identity,

.. math::

   \sum_{x=a}^{b}x
   =
   \frac{b(b+1)}{2}
   -
   \frac{(a-1)a}{2}.

Substitute this into the expectation:

.. math::

   E(X)
   =
   \frac1n
   \left[
   \frac{b(b+1)-(a-1)a}{2}
   \right].

Expand the numerator:

.. math::

   b(b+1)-(a-1)a
   =
   b^2+b-a^2+a.

Factor:

.. math::

   b^2+b-a^2+a
   =
   (a+b)(b-a+1).

Because

.. math::

   n=b-a+1,

we obtain

.. math::

   E(X)
   =
   \frac{(a+b)n}{2n}
   =
   \frac{a+b}{2}.

Thus,

.. note::

   **Mean of a Discrete Uniform Distribution**

   .. math::

      \boxed{E(X)=\mu=\frac{a+b}{2}}.

The mean is halfway between the smallest and largest possible values.
This can also be interpreted as a balance point. Because all values carry
equal probability weight, the loading is symmetric around the midpoint.

.. image:: ../_static/figures/discrete_distributions/figure_3_2_2_uniform_mean_balance.png
   :width: 72%
   :align: center
   :alt: Equal probability weights at the values one through six balancing at the midpoint between three and four.

For a fair die, the equal probability weights at :math:`1,2,3,4,5,6`
balance at :math:`3.5`. The expected value does not need to be one of the
possible outcomes; it represents the center of the probability loading.


3.2.5 Deriving the Variance
---------------------------

For a discrete random variable,

.. math::

   \operatorname{Var}(X)
   =
   E(X^2)-[E(X)]^2.

We already know

.. math::

   E(X)=\frac{a+b}{2}.

Therefore, we first derive :math:`E(X^2)`.

Because every value has probability :math:`1/n`,

.. math::

   E(X^2)
   =
   \sum_{x=a}^{b}x^2\left(\frac1n\right)
   =
   \frac1n\sum_{x=a}^{b}x^2.

Recall the sum-of-squares identity

.. math::

   1^2+2^2+\cdots+m^2
   =
   \frac{m(m+1)(2m+1)}{6}.

Therefore,

.. math::

   \sum_{x=a}^{b}x^2
   =
   \left(1^2+2^2+\cdots+b^2\right)
   -
   \left(1^2+2^2+\cdots+(a-1)^2\right).

Applying the identity gives

.. math::

   \sum_{x=a}^{b}x^2
   =
   \frac{b(b+1)(2b+1)}{6}
   -
   \frac{(a-1)a(2a-1)}{6}.

Hence,

.. math::

   E(X^2)
   =
   \frac{1}{6n}
   \left[
   b(b+1)(2b+1)
   -
   (a-1)a(2a-1)
   \right].

Now substitute :math:`E(X^2)` and
:math:`E(X)=(a+b)/2` into the variance formula:

.. math::

   \operatorname{Var}(X)
   =
   \frac{1}{6n}
   \left[
   b(b+1)(2b+1)
   -
   (a-1)a(2a-1)
   \right]
   -
   \left(\frac{a+b}{2}\right)^2.

Using :math:`n=b-a+1` and simplifying the algebra gives

.. math::

   \operatorname{Var}(X)
   =
   \frac{(b-a+1)^2-1}{12}.

Since :math:`n=b-a+1`, this can also be written as

.. math::

   \operatorname{Var}(X)
   =
   \frac{n^2-1}{12}.

Thus,

.. note::

   **Variance and Standard Deviation**

   .. math::

      \boxed{
      \operatorname{Var}(X)
      =
      \frac{(b-a+1)^2-1}{12}
      =
      \frac{n^2-1}{12}
      }

   and

   .. math::

      \boxed{
      \sigma
      =
      \sqrt{\operatorname{Var}(X)}
      }.


3.2.6 Example 2: Rolling a Fair Die
-----------------------------------

Roll one fair six-sided die and let

.. math::

   X=\text{number shown}.

Then

.. math::

   X\in\{1,2,3,4,5,6\},

so

.. math::

   a=1,
   \qquad
   b=6,
   \qquad
   n=b-a+1=6.

Each value has probability

.. math::

   P(X=x)=\frac16.

The PMF is flat because every die face is equally likely.

.. image:: ../_static/figures/discrete_distributions/figure_3_2_3_fair_die_uniform_pmf.png
   :width: 58%
   :align: center
   :alt: Equal-height PMF bars for the six outcomes of a fair die.

The six equal bars make the uniform assumption visible: changing the value of
:math:`x` from 1 to 6 does not change :math:`P(X=x)`.

**Mean**

Using the discrete-uniform mean formula,

.. math::

   E(X)
   =
   \frac{a+b}{2}
   =
   \frac{1+6}{2}
   =
   \frac72
   =
   3.5.

The same result can be checked directly from the PMF:

.. math::

   E(X)
   =
   \sum_{x=1}^{6}x\left(\frac16\right)
   =
   \frac{1+2+3+4+5+6}{6}
   =
   \frac{21}{6}
   =
   3.5.

**Variance**

Using :math:`n=6`,

.. math::

   \operatorname{Var}(X)
   =
   \frac{n^2-1}{12}
   =
   \frac{6^2-1}{12}
   =
   \frac{36-1}{12}
   =
   \frac{35}{12}
   \approx
   2.917.

The same result can also be obtained from
:math:`E(X^2)-[E(X)]^2`:

.. math::

   E(X^2)
   =
   \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6}
   =
   \frac{91}{6}.

Therefore,

.. math::

   \operatorname{Var}(X)
   =
   \frac{91}{6}
   -
   \left(\frac72\right)^2
   =
   \frac{182}{12}
   -
   \frac{147}{12}
   =
   \frac{35}{12}.

**Standard deviation**

.. math::

   \sigma
   =
   \sqrt{\frac{35}{12}}
   \approx
   1.708.

The expected outcome is 3.5. The standard deviation indicates that outcomes
typically vary from this center by about 1.71 units.

For example,

.. math::

   P(X\geq5)
   =
   P(X=5)+P(X=6)
   =
   \frac16+\frac16
   =
   \frac13.


3.2.7 When Should We Use It?
----------------------------

Use a discrete uniform model when:

- there is a finite set of possible values;
- each possible value has the same probability.

Typical examples include:

- one fair die;
- a randomly selected digit;
- a randomly selected numbered tag;
- one randomly selected integer from a specified finite set.

Do not use a discrete uniform distribution merely because the possible values
are equally spaced.

Equal spacing does not imply equal probability.


3.2.8 Quick Practice
--------------------

A box contains five tags labeled

.. math::

   2,3,4,5,6.

One tag is selected at random and each tag is equally likely.
Let :math:`X` be the number on the selected tag.

Find:

(a) :math:`P(X=4)`;

(b) :math:`P(X>4)`;

(c) :math:`E(X)`; and

(d) :math:`\operatorname{Var}(X)`.

**Answer**

There are five equally likely values, so

.. math::

   P(X=4)=\frac15.

For :math:`X>4`, the favorable values are 5 and 6:

.. math::

   P(X>4)=\frac25.

Here,

.. math::

   a=2,
   \qquad
   b=6,
   \qquad
   n=5.

Thus,

.. math::

   E(X)
   =
   \frac{2+6}{2}
   =
   4,

and

.. math::

   \operatorname{Var}(X)
   =
   \frac{5^2-1}{12}
   =
   2.


3.2.9 Common Mistakes
---------------------

A common mistake is to assume that every discrete variable is uniform.

For two coin tosses, the number of heads has possible values
:math:`0,1,2`, but their probabilities are not equal.

Another mistake is to count the integers from :math:`a` through :math:`b`
as :math:`b-a`. Because both endpoints are included, the correct number is

.. math::

   n=b-a+1.

When deriving the variance, do not forget that

.. math::

   \operatorname{Var}(X)
   =
   E(X^2)-[E(X)]^2,

not :math:`E(X^2)-E(X)`.


3.2.10 Summary
--------------

For a discrete uniform random variable on the consecutive integers

.. math::

   a,a+1,\ldots,b,

there are

.. math::

   n=b-a+1

possible values, each with probability

.. math::

   P(X=x)=\frac1n.

The mean is

.. math::

   E(X)=\frac{a+b}{2}.

The variance is

.. math::

   \operatorname{Var}(X)
   =
   \frac{n^2-1}{12}
   =
   \frac{(b-a+1)^2-1}{12}.

The standard deviation is

.. math::

   \sigma
   =
   \sqrt{\operatorname{Var}(X)}.

The key modeling question is:

    Are all possible values equally likely?