The Discrete Uniform Distribution, introduction to probability lesson

A fair die is modeled by giving each face from $1$ to $6$ the same probability, and a raffle that draws one ticket uniformly from tickets numbered $1$ to $500$ gives each number the same chance. In both models the probability of an event comes from counting values. The average result and the typical distance from it still take some work. This lesson defines equal mass on a finite set of values, derives the mean and variance for consecutive integers from $a$ to $b$ , and shows which parts of those formulas fail when the values are unevenly spaced.

Key terms

Discrete uniform distribution

Definition Let $S$ be a finite nonempty set of $N$ distinct real numbers. A random variable $X$ has the discrete uniform distribution on $S$ if its PMF is

p_X(x) = \begin{cases} \tfrac{1}{N}, & x \in S, \\[0.3em] 0, & \text{otherwise}. \end{cases}

Then $S$ is the support of $X$ , and for every set $A$ of real numbers

P(X \in A) = \frac{\lvert A \cap S \rvert}{N},

where $\lvert A \cap S \rvert$ is the number of support values that lie in $A$ . The support has to be finite. No distribution on the positive integers gives every positive integer the same probability, because a common mass $c > 0$ makes the total infinite and the mass $c = 0$ makes the total $0$ .

In plain words Every listed value gets the same share of probability, and anything not listed gets none. The equal shares belong to the values of $X$ , not to physical objects. A die with faces $1$ , $1$ , $2$ , $3$ , $4$ and $5$ has six equally likely faces, but the number it shows equals $1$ with probability $\tfrac{1}{3}$ , so that number is not uniform on $\{1, 2, 3, 4, 5\}$ .

Example A selection device picks one of the labels $2$ , $3$ , $4$ , $5$ and $6$ , each with probability $\tfrac{1}{5}$ , and $X$ is the label. The event $X \ge 4$ contains the three values $4$ , $5$ and $6$ , so $P(X \ge 4) = \tfrac{3}{5}$ . The event $X > 4$ contains only $5$ and $6$ , so $P(X > 4) = \tfrac{2}{5}$ . Equal masses turn every event into a counting problem, but the wording of the event still decides which values are counted.

Number of support values

Definition For integers $a \le b$ , the set of consecutive integers $\{a, a + 1, \dots, b\}$ contains

N = b - a + 1

values, since both endpoints are included. A discrete uniform variable on this set gives each value the mass $\tfrac{1}{N}$ , and an event that contains exactly $m$ of the support values has probability $\tfrac{m}{N}$ .

In plain words Count the values, not the gaps between them. Going from $a$ to $b$ takes $b - a$ steps of size $1$ , and there is one more value than there are steps, for the same reason that a fence with four gaps needs five posts.

Example The labels from $2$ to $6$ give $N = 6 - 2 + 1 = 5$ values, with $4$ gaps between neighbors. The integers from $-3$ to $3$ give $N = 3 - (-3) + 1 = 7$ values, so each has mass $\tfrac{1}{7}$ . Using $b - a = 6$ instead would give each of the seven values the mass $\tfrac{1}{6}$ , and the masses would add to $\tfrac{7}{6}$ .

Mean by symmetry

Definition For a discrete uniform variable on a finite set $S$ with $N$ values, the mean is the plain average of the support values,

E[X] = \frac{1}{N} \sum_{x \in S} x.

If $S$ is symmetric about a number $c$ , meaning that $x \in S$ exactly when $2c - x \in S$ , then $E[X] = c$ . The reason is that $X$ and $2c - X$ have the same distribution, so $E[X] = E[2c - X] = 2c - E[X]$ . A set of consecutive integers from $a$ to $b$ is symmetric about its midpoint, so

E[X] = \frac{a + b}{2}.

In plain words When the support is symmetric, pair the smallest value with the largest, the second smallest with the second largest, and so on. Each pair balances at the same midpoint, and with equal masses no pair pulls harder than another. When $N$ is odd, the unpaired middle value is the midpoint itself. The mean need not be a possible value of $X$ .

Example For the labels $2$ to $6$ , the pair $2$ and $6$ and the pair $3$ and $5$ both balance at $4$ , and the middle value is $4$ . So $E[X] = 4$ , in agreement with $(2 + 6)/2$ and with the direct average $20/5$ . Now let $X$ be uniform on $\{2, 4, 7, 11\}$ . This is still a discrete uniform distribution, with each mass $\tfrac{1}{4}$ , but its support is not symmetric. Its mean is $24/4 = 6$ , while the midpoint of the endpoints is $6.5$ .

Sum-of-squares identity

Definition For every positive integer $N$ ,

\begin{aligned} &\sum_{k=1}^{N} k^2 \\[0.4em] &\quad = \frac{N(N + 1)(2N + 1)}{6}. \end{aligned}

One proof adds the identities $(k + 1)^3 - k^3 = 3k^2 + 3k + 1$ for $k = 1, \dots, N$ . The left sides telescope to $(N + 1)^3 - 1$ . The right sides add to $3\sum_{k=1}^{N} k^2 + \tfrac{3}{2}N(N + 1) + N$ , using the sum of the first $N$ positive integers. Solving the resulting equation for the sum of squares gives the formula.

In plain words This is an algebra fact about finite sums, not a probability assumption. Probability enters when the squares are averaged with equal weights. For $K$ uniform on $\{1, 2, \dots, N\}$ the identity gives the second moment

E[K^2] = \frac{(N + 1)(2N + 1)}{6}.

Example For $N = 5$ the squares $1$ , $4$ , $9$ , $16$ and $25$ add to $55$ , and the formula gives $5 \cdot 6 \cdot 11 / 6 = 55$ . A variable $K$ uniform on $\{1, 2, 3, 4, 5\}$ therefore has $E[K^2] = 55/5 = 11$ . With $E[K] = 3$ , its variance is $11 - 3^2 = 2$ .

Variance of consecutive integers

Definition Let $a \le b$ be integers, let $N = b - a + 1$ , and let $X$ be uniform on $\{a, a + 1, \dots, b\}$ . Then

\operatorname{Var}(X) = \frac{N^2 - 1}{12},

and $\operatorname{SD}(X) = \sqrt{(N^2 - 1)/12}$ . To derive this, first take $K$ uniform on $\{1, 2, \dots, N\}$ , with $E[K] = (N + 1)/2$ and $E[K^2]$ from the sum-of-squares identity. Then the variance of $K$ is

\begin{aligned} &E[K^2] - (E[K])^2 \\[0.4em] &\quad = \frac{(N + 1)(2N + 1)}{6} \\[0.4em] &\qquad - \frac{(N + 1)^2}{4} \\[0.4em] &\quad = \frac{N^2 - 1}{12}. \end{aligned}

Over the common denominator $12$ the numerator is $(N + 1)\bigl(2(2N + 1) - 3(N + 1)\bigr) = (N + 1)(N - 1)$ . The variable $X$ has the same distribution as $K + (a - 1)$ , and adding a constant leaves the variance unchanged. The formula assumes consecutive integers with equal masses. For $N = 1$ it gives $0$ , as it should for a variable with one possible value.

In plain words The spread depends only on how many consecutive values there are, not on where the block of values sits on the number line. Sliding the block moves the mean and leaves the variance alone.

Example For the labels $2$ to $6$ , $N = 5$ and the formula gives $(25 - 1)/12 = 2$ . The direct check uses the squared deviations from the mean $4$ , which are $4$ , $1$ , $0$ , $1$ and $4$ . Their average is $10/5 = 2$ , so $\operatorname{Var}(X) = 2$ squared label units and $\operatorname{SD}(X) = \sqrt{2} \approx 1.41$ label units. Equal probabilities do not mean zero spread.

Evenly spaced support

Definition A finite set is evenly spaced with gap $d > 0$ if it has the form $\{c, c + d, \dots, c + (N - 1)d\}$ . If $X$ is uniform on such a set, then $J = (X - c)/d$ is uniform on $\{0, 1, \dots, N - 1\}$ , because the map is one-to-one, and $X = c + dJ$ . Hence

\begin{gathered} E[X] = c + \frac{(N - 1)d}{2}, \\[0.4em] \operatorname{Var}(X) = \frac{d^2 (N^2 - 1)}{12}. \end{gathered}

The mean is the midpoint of the smallest and largest values, and $\operatorname{SD}(X) = d\sqrt{(N^2 - 1)/12}$ . For a support that is not evenly spaced, the mean and variance are computed directly from the support values.

In plain words Equal spacing makes the variable a rescaled copy of consecutive integers, so the formulas carry over with the gap as the unit of length. The variance scales by $d^2$ and the standard deviation by $d$ . Equal masses survive any one-to-one relabeling, but a function that merges values need not keep them equal. If $X$ is uniform on $\{-1, 0, 1\}$ , then $X^2$ equals $1$ with probability $\tfrac{2}{3}$ , so it is not uniform on $\{0, 1\}$ .

Example The set $\{1, 3, 5, 7\}$ has $c = 1$ , $d = 2$ and $N = 4$ , so $E[X] = 1 + 3 = 4$ and $\operatorname{Var}(X) = 4 \cdot 15/12 = 5$ . Directly, the squared deviations $9$ , $1$ , $1$ and $9$ average to $5$ . The set $\{2, 4, 7, 11\}$ has gaps $2$ , $3$ and $4$ , so the shortcut formulas do not apply. From its mean $6$ the deviations are $-4$ , $-2$ , $1$ and $5$ , and

\begin{aligned} \operatorname{Var}(X) &= \frac{16 + 4 + 1 + 25}{4} \\[0.4em] &= \frac{23}{2}. \end{aligned}

The consecutive-integer formula with $N = 4$ would give $\tfrac{5}{4}$ , far too small.

Check yourself

Let $X$ be an integer chosen uniformly from $-3$ through $3$ , both endpoints included. How many support values are there, what is the mass of each, and what is $P(\lvert X \rvert \le 1)$ ?

Show the answer

The support is $-3$ , $-2$ , $-1$ , $0$ , $1$ , $2$ and $3$ , so $N = 3 - (-3) + 1 = 7$ and each value has mass $\tfrac{1}{7}$ . The event $\lvert X \rvert \le 1$ contains the three values $-1$ , $0$ and $1$ , so its probability is $\tfrac{3}{7}$ .

For the same $X$ , find $E[X]$ and $\operatorname{Var}(X)$ , once with the consecutive-integer formulas and once directly.

Show the answer

The formula gives $E[X] = (a + b)/2 = (-3 + 3)/2 = 0$ . Directly, the seven values add to $0$ , so their average is $0$ , as the symmetry of the support about $0$ also shows. The variance formula gives $(7^2 - 1)/12 = 48/12 = 4$ . Directly, the squared deviations from $0$ are $9$ , $4$ , $1$ , $0$ , $1$ , $4$ and $9$ . They add to $28$ , and $28/7 = 4$ .

With the same $X$ , let $Y = X^2$ . A student says that $Y$ is uniform on $\{0, 1, 4, 9\}$ and concludes that $E[Y] = 14/4 = 3.5$ . Find the correct PMF of $Y$ and the correct $E[Y]$ .

Show the answer

The value $0$ comes only from $X = 0$ , while each of $1$ , $4$ and $9$ comes from two values of $X$ . So $P(Y = 0) = \tfrac{1}{7}$ and $P(Y = 1) = P(Y = 4) = P(Y = 9) = \tfrac{2}{7}$ , which is not uniform. Then

\begin{aligned} E[Y] &= \frac{0 + 2 + 8 + 18}{7} \\[0.4em] &= 4. \end{aligned}

This agrees with the previous answer, as it must. Because $E[X] = 0$ , the shortcut $\operatorname{Var}(X) = E[X^2] - (E[X])^2$ gives $E[Y] = E[X^2] = \operatorname{Var}(X) = 4$ .

The Discrete Uniform Distribution

Key terms

Discrete uniform distribution

Number of support values

Mean by symmetry

Sum-of-squares identity

Variance of consecutive integers

Evenly spaced support

Check yourself

How this lesson connects

Builds on