Rectangular distributionAQA A-Level Further Maths: Revision notes

A continuous random variable $X$ has a rectangular distribution (also called uniform) on $[a,b]$ when every value in the interval is equally likely, so intervals of equal width have equal probability. The pdf is $f(x)=\frac{1}{b-a}\quad(a\le x\le b),\qquad f(x)=0\ \text{otherwise}.$ The graph is a rectangle of width $b-a$ and height $\frac{1}{b-a}$, so the total area is $1$.

Use a rectangular distribution when:

• the variable is continuous and takes values only within a fixed interval,
• there is no reason to favour any part of the interval, so any two sub-intervals of equal width are equally likely. Typical examples are the waiting time for something that arrives at a random moment in a fixed cycle, and rounding errors. It is not suitable if values cluster (for example commuters who time their arrival to the timetable) or if the interval is not fixed.

Probability is area, so for $a\le c<d\le b$: $\mathrm{P}(c<X<d)=\int_c^d\frac{1}{b-a}\,dx=\frac{d-c}{b-a}.$ Worked example: $X$ rectangular on $[2,10]$. Then $f(x)=\frac18$ and $\mathrm{P}(3<X<5.5)=\frac{2.5}{8}=\frac{5}{16}$. For a probability such as $\mathrm{P}(|E|>0.4)$ on $[-0.5,0.5]$, add the two tails: $0.1+0.1=0.2$.

$\mathrm{E}(X)=\int_a^b x\cdot\frac{1}{b-a}\,dx=\left[\frac{x^2}{2(b-a)}\right]_a^b=\frac{b^2-a^2}{2(b-a)}=\frac{a+b}{2}.$ This is the midpoint of the interval, as symmetry suggests. The factorisation $b^2-a^2=(b-a)(b+a)$ is the key step.

First $\mathrm{E}(X^2)=\int_a^b\frac{x^2}{b-a}\,dx=\frac{b^3-a^3}{3(b-a)}=\frac{a^2+ab+b^2}{3},$ using $b^3-a^3=(b-a)(a^2+ab+b^2)$. Then $\mathrm{Var}(X)=\frac{a^2+ab+b^2}{3}-\frac{(a+b)^2}{4}=\frac{(b-a)^2}{12},\qquad \sigma=\frac{b-a}{\sqrt{12}}.$ Example: trains every 15 minutes gives $\mathrm{Var}=\frac{225}{12}=18.75$ and $\sigma=4.33$ minutes.