Statistics/Distributions/Uniform

Continuous Uniform Distribution
The (continuous) uniform distribution, as its name suggests, is a distribution with probability densities that are the same at each point in an interval. In casual terms, the uniform distribution shapes like a rectangle.

Mathematically speaking, the probability density function of the uniform distribution is defined as

$$ f\colon[a,b]\to\R $$

$$ f\left(x\right)={1 \over {b-a}} $$

And the cumulative distribution function is:

$$ F\left(x\right)= \begin{cases} 0, & \mbox{if } x \le a \\ {{x-a} \over {b-a}}, & \mbox{if } a < x < b\\ 1, & \mbox{if } x \ge b \end{cases} $$

Mean
We derive the mean as follows.
 * $$\operatorname{E}[X] = \int^\infin_{-\infin}xf(x) dx$$

As the uniform distribution is 0 everywhere but [a, b] we can restrict ourselves that interval
 * $$\operatorname{E}[X] = \int^b_a {1 \over {b-a}} x dx$$
 * $$\operatorname{E}[X] = \left.{1 \over (b-a)}{1 \over 2} x^2 \right|^b_a$$
 * $$\operatorname{E}[X] = {1 \over 2(b-a)}\left[ b^2-a^2 \right]$$
 * $$\operatorname{E}[X] = {b+a \over 2}$$

Variance
We use the following formula for the variance.
 * $$\operatorname{Var}(X) = \operatorname{E}[X^2]-(\operatorname{E}[X])^2$$
 * $$\operatorname{Var}(X) = \left[\int^\infin_{-\infin}f(x) \cdot x^2 dx\right]-\left({b+a \over 2}\right)^2$$
 * $$\operatorname{Var}(X) = \left[\int^b_a {1 \over {b-a}} x^2 dx\right]-{(b+a)^2 \over 4}$$
 * $$\operatorname{Var}(X) = \left. {1 \over {b-a}}{1 \over 3} x^3 \right|^b_a-{(b+a)^2 \over 4}$$
 * $$\operatorname{Var}(X) = {1 \over 3(b-a)}[b^3-a^3] -{(b+a)^2 \over 4}$$
 * $$\operatorname{Var}(X) = {4(b^3-a^3)-3(b+a)^2(b-a) \over 12(b-a)}$$
 * $$\operatorname{Var}(X) = {(b-a)^3 \over 12(b-a)}$$
 * $$\operatorname{Var}(X) = {(b-a)^2 \over 12}$$