Continuous Uniform Distribution

The continuous uniform distribution represents a probability distribution where any interval of a specified length within the range of possible values has an equal probability of occurring, and outside this range, the probability is zero.

A continuous uniform random variable is denoted as followed:

\[ X \sim \text{Uniform}(a,b) \]

where

• \(a\) represents the lower bound

• \(b\) represents the upper bound

Probability Distribution Function (PDF)

The pdf for a continuous random distribution is given as followed:

\[\begin{split} f_X(x) = \begin{cases} \frac{1}{b-a}, & \text{if } a < x < b \\ 0, & \text{otherwise} \end{cases} \end{split}\]

Expectation

The expected value of a continuous uniform can be derived as such:

\[\begin{split} \begin{align} EX &= \int_a^b x f_X(x) dx \\ &= \int_a^b \frac{x}{b-a} dx \\ &= \frac{1}{b-a} \int_a^b {x} dx \\ &= \frac{1}{b-a} \left[ \frac{x^2}{2} \right]_a^b \\ &= \frac{1}{b-a} \left[ \frac{b^2 - a^2}{2} \right] \\ &= \frac{b^2 - a^2}{2(b-a)} \\ &= \frac{(b-a)(b+a)}{2(b-a)} \\ &= \frac{b+a}{2} \\ \end{align} \end{split}\]

Variance

To find the variance, we’ll use the following equation:

\[ Var(X) = EX^2 - (EX)^2 \]

First, we’ll find \(EX^2\):

\[\begin{split} \begin{align} EX^2 &= \int_a^b x^2 f_X(x) dx \\ &= \int_a^b \frac{x^2}{b-a} dx \\ &= \frac{1}{b-a} \int_a^b x^2 dx \\ &= \frac{1}{b-a} \left[ \frac{x^3}{3} \right]_a^b \\ &= \frac{b^3 - a^3}{3(b-a)} \\ &= \frac{(b-a)(b^2 + ab + a^2)}{3(b-a)} \\ &= \frac{b^2 + ab + a^2}{3} \\ \end{align} \end{split}\]

Now, we can find \(Var(X)\):

\[\begin{split} \begin{align} Var(X) &= EX^2 - (EX)^2 \\ &= \frac{b^2 + ab + a^2}{3} - \left( \frac{b+a}{2} \right)^2 \\ &= \frac{b^2 + ab + a^2}{3} - \left( \frac{b^2 + 2ab + a^2}{4} \right) \\ &= \frac{4b^2 - 3b^2 + 4ab -6ab + 4a^2 - 3a^2}{12} \\ &= \frac{b^2 - 2ab + a^2}{12} \\ &= \frac{(b-a)^2}{12} \\ \end{align} \end{split}\]