Uniform Random Variable

\textbf{Definition} We say a continuous random variable $\mathbb{X}$ is a uniformly distributed in [a, b], denoted $\mathbb{X} \sim uniform(a, b)$ if $$\mathbb{X} \sim f(x) = \begin{cases} \frac{1}{b-a} & \text{if } x \in [a, b] \\ 0 & \text{otherwise} \end{cases}$$

Property If $\mathbb{X} \sim uniform(a, b)$, then $\mu = \frac{a+b}{2}, \sigma^2 = \frac{(b-a)^2}{12}$

Proof 1 By definition $\mu = \int_a^b \frac{x}{b-a} \mathrm{d}x = \frac{a+b}{2}$ $\sigma^2 = \int_a^b \frac{x^2}{b-a} \mathrm{d}x - (\frac{a+b}{2})^2 = \frac{b^2+ab+a^2}{3} - \frac{a^2+2ab+b^2}{4} = \frac{(b-a)^2}{12}$

Proof 2 By moment generating function

Since $D_t f(x)e^{xt} = D_t \frac{e^{xt}}{b-a} = \frac{xe^{xt}}{b-a}$ is continuous on $\mathbb{R}^2$, and $M_X(t) = \int_a^b \frac{e^{xt}}{b-a} \mathrm{d}x = \begin{cases} \frac{e^{bt}-e^{at}}{t(b-a)} & \text{ if } t \neq 0 \\ 1 & \text{ if } t = 0 \end{cases}$

exists for all $t$, all conditions that garantee the validity of m.f.g method can be met. so $\mu = M_X'(0) = \lim_{t\to 0} \frac{1}{t}( \frac{e^{bt}-e^{at}}{t(b-a)} - 1 ) = \lim_{t\to 0} \frac{e^{bt}-e^{at}-t(b-a)}{t^2(b-a)}$ $\text{ (apply L'hospital twice) }= \lim_{t\to 0} \frac{b^2e^{bt}-a^2e^{at}}{2(b-a)} = \frac{a+b}{2}$ And $M_X''(0) = \lim_{t\to 0} \frac{1}{t}( \frac{t(b-a)(be^{bt}-ae^{at})-(e^{bt}-e^{at})(b-a)}{t^2(b-a)^2} - \frac{a+b}{2} )$ $= \lim_{t\to 0} \frac{2t(b-a)(be^{bt}-ae^{at})-2(e^{bt}-e^{at})(b-a) - (a+b)t(b-a)^2)}{2t^3(b-a)^2}$ $\text{ (apply L'hospital 3 times ) }= \frac{4(b-a)(b^3-a^3)}{12(b-a)^2} = \frac{a^2 + ab + b^2}{3}$

So $\sigma^2 = \frac{a^2 + ab + b^2}{3} - (\frac{a+b}{2})^2 = \frac{(b-a)^2}{12}$