42 Uniform Distribution

42.1 Probability Density Function

A random variable \(X\) is said to have a Uniform Distribution with parameters \(a\) and \(b\) if its pdf is

\[f(x)=\left\{ \begin{array}{ll} \frac{1}{b-a}, & a\leq x \leq b\\ 0 & elsewhere \end{array} \right. \]

42.2 Cumulative Density Function

\[\begin{aligned} F(x) &= \int\limits_{a}^{x}\frac{1}{b-a}dt \\ &= \frac{t}{b-a}|_{a}^{x} \\ &= \frac{x}{b-a}-\frac{a}{b-a} \\ &= \frac{x-a}{b-a} \end{aligned}\]

\[F(x)=\left\{ \begin{array}{lll} 0 & x<a\\ \frac{x-a}{b-a},& a\leq x\leq b\\ 1 & elsewhere \end{array}\right. \]

The figures on the left and right display the Uniform probability and cumulative distirubtion functions, respectively, for $a=0, b=5$.

Figure 42.1: The figures on the left and right display the Uniform probability and cumulative distirubtion functions, respectively, for \(a=0, b=5\).

42.3 Expected Values

\[\begin{aligned} E(X) &= \int\limits_{a}^{b}x\frac{1}{b-a}dx \\ &= \frac{1}{b-a}\int\limits_{a}^{b}x\ dx \\ &= \frac{1}{b-a}\cdot \Big[\frac{x^2}{2}\Big]_a^b \\ &= \frac{1}{b-a}\cdot\Big[\frac{b^2}{2}-\frac{a^2}{2}\Big] \\ &= \frac{1}{b-a}\cdot \frac{b^2-a^2}{2} \\ &= \frac{b^2-a^2}{2(b-a)} \\ &= \frac{(b-a)(b+a)}{2(b-a)} \\ &= \frac{b+a}{2} \\ \\ \\ E(X^2) &= \int\limits_{a}^{b}x^2\frac{1}{b-a}dx \\ &= \frac{1}{b-a}\int\limits_{a}^{b}x^2\ \frac{1}{b-a}dx \\ &= \frac{1}{b-a}\Big[\frac{x^3}{3}\Big]_a^b \\ &= \frac{1}{b-a}\Big[\frac{b^3-a^3}{3}\Big] \\ &= \frac{1}{b-a}\Big[\frac{(b-a)(b^2+ab+a^2)}{3}\Big] \\ &= \frac{(b-a)(b^2+ab+a^2)}{3(b-a)} \\ &= \frac{(b^2+ab+a^2)}{3} \\ \\ \\ \mu &= E(X) \\ &= \frac{b+a}{1} \\ \\ \\ \sigma^2 &= E(X^2) - E(X)^2 \\ &= \frac{b^2+ab+a^2}{3} - \frac{(b-a)^2}{4} \\ &= \frac{4(b^2+ab+a^2)-3(b+a)^2}{12} \\ &= \frac{4(b^2+ab+a^2-3(b^2+2ab+a^2)}{12} \\ &= \frac{4b^2+4ab+4a^2-3b^2-6ab-3a^2)}{12} \\ &= \frac{4b^2-3b^2+4ab-6ab+4a^2-3a^2}{12} \\ &= \frac{b^2-2a+a^2}{12}=\frac{(b-a)^2}{12} \end{aligned}\]

42.4 Moment Generating Function

\[\begin{aligned} M_X(t) &= E(e^{tX})=\int\limits_{a}^{b}e^{tx}\frac{1}{b-a}dx \\ &= \frac{1}{b-a}\int\limits_{a}^{b}e^{tx}dx \\ &= \frac{1}{b-a}\Big[\frac{e^{tb}-e^{ta}}{t}\Big] \\ &= \frac{e^{t(b-a)}}{t(b-a)} \end{aligned}\]

\(M_X^{(k)}(0)\) will lead to an undefined operation (division by 0). Thus, in the case of the Uniform distribution, we are unable to use the method of moments to identify parameter values.

42.5 Theorems for the Uniform Distribution

42.6 Validity of the Distribution

\[\int\limits_{a}^{b}\frac{1}{b-a} = 1\]

Proof:

\[\begin{aligned} \int\limits_{a}^{b}\frac{1}{b-a} &= \frac{x}{b-a}\Big|_a^b \\ &= \frac{b}{b-a}-\frac{a}{b-a} \\ &= \frac{b-a}{b-a} \\ &= 1 \end{aligned}\]