Uniform Distributions

Definition For real numbers $a < b$, a random variable $X$ has the continuous uniform distribution on $(a, b)$, written $X \sim \operatorname{Uniform}(a, b)$, if its density is

Changing $f$ at the two endpoints changes no probability, so the intervals $(a, b)$ and $[a, b]$ describe the same distribution. The CDF is $F(x) = 0$ for $x < a$, $F(x) = 1$ for $x > b$, and for $a \le x \le b$

For $a \le c < d \le b$,

and the answer is the same with either endpoint included, since a uniform distribution has no atoms. An event that reaches outside $(a, b)$ is first cut down to its part inside. Because a constant density must have total area $1$, the interval has to be bounded, and no uniform distribution exists on the whole real line or on $[0, \infty)$.

Definition Let $X$ have CDF $F$ and let $0 < p < 1$. The $p$-quantile of $X$ is the smallest number $q$ with $F(q) \ge p$. When $F$ is continuous, $F(q) = p$ at that number, and $q$ is the only solution of $F(q) = p$ unless $F$ stays at the level $p$ over an interval. The $100p$th percentile is another name for the $p$-quantile, so the $90$th percentile is the $0.9$-quantile. For $X \sim \operatorname{Uniform}(a, b)$, solving $(q - a)/(b - a) = p$ gives

Definition A median of $X$ is a number $m$ with $P(X \le m) \ge \tfrac12$ and $P(X \ge m) \ge \tfrac12$. The $0.5$-quantile is always a median. When $F$ is continuous, the condition becomes $F(m) = \tfrac12$. A median need not be unique, since every point of an interval on which $F$ stays at $\tfrac12$ is a median. For $X \sim \operatorname{Uniform}(a, b)$, the median is unique and equals the midpoint $(a + b)/2$.

Definition If $X \sim \operatorname{Uniform}(a, b)$, then

and $\operatorname{SD}(X) = (b - a)/\sqrt{12}$. The mean comes from

For the variance, let $d = (b - a)/2$ be the half-width. The centered variable $X - E[X]$ has density $1/(2d)$ on $(-d, d)$, so

Definition Let $X \sim \operatorname{Uniform}(a, b)$ and $a \le c < d \le b$. The event $\{c < X < d\}$ has probability $(d - c)/(b - a) > 0$, and given this event, $X$ is uniform on $(c, d)$. For $c \le s < t \le d$, the common factor $1/(b - a)$ cancels in the ratio of probabilities, so

More generally, conditioning a variable with density $f$ on an interval $I$ with $P(X \in I) > 0$ gives the conditional density $f(x)/P(X \in I)$ for $x$ in $I$ and $0$ elsewhere. The conditional law is uniform here only because $f$ is constant on the interval.

Definition Let $X$ have density $f_X$, and let $Y = \alpha + cX$ for constants $\alpha$ and $c > 0$, as in a change from minutes to seconds or from degrees Celsius to degrees Fahrenheit. Then $Y$ has density

The reason is the substitution $x = (y - \alpha)/c$ with $dx = dy/c$, which matches each integral of $f_X$ over an interval of $x$ values with an integral of $f_Y$ over the corresponding interval of $y$ values. Probabilities of the same event are unchanged, and when the moments of $X$ are finite, $E[Y] = \alpha + cE[X]$, $\operatorname{SD}(Y) = c\operatorname{SD}(X)$ and $\operatorname{Var}(Y) = c^2 \operatorname{Var}(X)$. In particular, if $X \sim \operatorname{Uniform}(a, b)$, then $Y \sim \operatorname{Uniform}(\alpha + ca, \alpha + cb)$.