Variance of Uniform Distribution

The uniform distribution, also called the rectangular distribution, is a type of probability model in which all outcomes within a given interval are equally likely. The variation of a uniform distribution measures how the values are spread out across the interval.

Mathematically, a uniform distribution is defined by two parameters, a and b

Here,

• a is the lower bound of the interval
• b is the upper bound of the interval

These bounds can either be positive or negative. Based on the nature of the possible outcomes of the random variable, uniform distributions are divided into continuous and discrete.

Continuous Uniform Distribution 

It represents a continuous random variable X that takes any value within a specified range [a, b] with equal probability. The probability density function (PDF) of this distribution is expressed as:

${f\left( x\right) =\begin{cases}\dfrac{1}{b-a},ifa\leq x\leq b\\ 0,otherwise\end{cases}}$

Let us consider a scenario in which a digital timer is used to start a game. The timer is set to begin at a random time within a 30-minute interval. If the timer’s start time is equally likely to occur at any moment within this interval, the start time X is uniformly distributed between a = 0 and b = 30 minutes. This is an example of positive lower and upper bounds.

Formula

The variance of a continuous uniform distribution is obtained using the formula:

Var(X) = ${\dfrac{\left( b-a\right) ^{2}}{12}}$

Derivation

If X is a random variable, the variance of X (Var(X)) is defined as:

Var(X) = E[X²] – (E[X])² …..(i)

Here,

E[X] is the expected value (mean) of X

E[X²] is the expected value of X² 

Calculating E[X] and E[X²] in the closed interval [a, b], we get

E[X] = ${\int ^{b}_{a}xf\left( x\right) dx}$ 

= ${\int ^{b}_{a}\dfrac{x}{b-a}dx}$

= ${\left[ \dfrac{x^{2}}{2\left( b-a\right) }\right] _{a}^{b}}$

= ${\dfrac{b^{2}}{2\left( b-a\right) }-\dfrac{a^{2}}{2\left( b-a\right) }}$

= ${\dfrac{b+a}{2}}$ …..(ii)

E[X²] 

= ${\int ^{b}_{a}\dfrac{x^{2}}{b-a}dx}$

= ${\left[ \dfrac{x^{3}}{3\left( b-a\right) }\right] _{a}^{b}}$

= ${\dfrac{b^{3}}{3\left( b-a\right) }-\dfrac{a^{3}}{3\left( b-a\right) }}$

= ${\dfrac{b^{2}+ab+a^{2}}{3}}$ …..(iii)

Now, substituting the values of (ii) and (iii) in equation (i), we get

Var(X) = ${\dfrac{b^{2}+ab+a^{2}}{3}-\left( \dfrac{b+a}{2}\right) ^{2}}$

= ${\dfrac{b^{2}+ab+a^{2}}{3}-\dfrac{b^{2}+2ab+a^{2}}{4}}$

= ${\dfrac{b^{2}-2ab+a^{2}}{12}}$

= ${\dfrac{\left( b-a\right) ^{2}}{12}}$

Thus, Var(X) = ${\dfrac{\left( b-a\right) ^{2}}{12}}$

Steps To Find

Let us consider a speed camera that monitors vehicles on a stretch of highway where the speed limit is strictly enforced between 5 m/s and 15 m/s. To determine the variance of this uniform distribution, we can follow these steps:

Identifying the Interval

Here, a = 5 (lower bound) and b = 15 (upper bound).

By Using the Formula

Var(X) = ${\dfrac{\left( b-a\right) ^{2}}{12}}$

= ${\dfrac{\left( 15-5\right) ^{2}}{12}}$

= ${\dfrac{\left( 10\right) ^{2}}{12}}$

= ${\dfrac{100}{12}}$

≈ 8.33

Thus, the variance is approximately 8.33

Discrete Uniform Distribution 

It represents a finite number of outcomes where each outcome within the closed interval [a, b] has a probability equal to: 

${f\left( x\right) =\dfrac{1}{n}}$

Here, n = b – a + 1

For example, if a die is rolled, the outcome X can be {1, 2, 3, 4, 5, 6}, each with a probability of ${\dfrac{1}{n}}$ (probability mass function or PMF), which follows a discrete uniform distribution with positive lower and upper bounds.

Formula

The variance of a discrete uniform distribution can be obtained using the formula:

Var(X) = ${\dfrac{n^{2}-1}{12}}$

Here, n = b – a + 1

Derivation

Similar to the continuous uniform distribution, the variance formula for the discrete uniform distribution can also be derived by integrating xf(x) within the interval [a, b]

If X is a random variable, the variance of X (Var(X)) is defined as:

Var(X) = E[X²] – (E[X])² …..(i)

Here,

E[X] is the expected value (mean) of X

E[X²] is the expected value of X² 

Calculating E[X] and E[X²] in the closed interval [a, b], we get

E[X] = ${\sum ^{n}_{i=1}xf\left( x\right)}$

= ${\sum ^{n}_{i=1}\dfrac{x}{n}}$

= ${\dfrac{1}{n}\sum ^{n}_{i=1}x}$

= ${\dfrac{1}{n}\times \dfrac{n\left( n+1\right) }{2}}$

= ${\dfrac{\left( n+1\right) }{2}}$ …..(ii)

E[X²]

= ${\sum ^{n}_{i=1}\dfrac{x^{2}}{n}}$

= ${\dfrac{1}{n}\times \dfrac{n\left( n+1\right) \left( 2n+1\right) }{6}}$

= ${\dfrac{\left( n+1\right) \left( 2n+1\right) }{6}}$ …..(iii)

Now, substituting the values of (ii) and (iii) in equation (i), we get

Var(X) = ${\dfrac{\left( n+1\right) \left( 2n+1\right) }{6}-\left( \dfrac{n+1}{2}\right) ^{2}}$

= ${\dfrac{\left( n+1\right) \left( 2n+1\right) }{6}-\dfrac{\left( n+1\right) ^{2}}{4}}$

= ${\dfrac{2n^{2}+3n+1}{6}-\dfrac{n^{2}+2n+1}{4}}$

= ${\dfrac{2\left( 2n^{2}+3n+1\right) -3\left( n^{2}+2n+1\right) }{12}}$

= ${\dfrac{n^{2}-1}{12}}$

Thus, Var(X) = ${\dfrac{n^{2}-1}{12}}$

Note: When a = 0 and b = 1, the uniform distribution is known as standard uniform distribution. It is used for random number generation with the expected value of ${\dfrac{1}{2}}$ and the variance of ${\dfrac{1}{12}}$

Steps To Find

Let us imagine a fair die is rolled, producing outcomes between 1 and 6. Now, we will calculate the variance of this discrete uniform distribution as follows:

Identifying the Interval and Number of Outcomes

Here, 

a = 1 

b = 6

n = b – a + 1 = 6 – 1 + 1 = 6

By Using the Formula

Var(X) = ${\dfrac{n^{2}-1}{12}}$

= ${\dfrac{6^{2}-1}{12}}$

= ${\dfrac{36-1}{12}}$

= ${\dfrac{35}{12}}$

≈ 2.92

Thus, the variance is approximately 2.92

Solved Examples