The mean of a continuous uniform random variable defined over the support \(a<x<b\) is:

\(\mu=E(X)=\dfrac{a+b}{2}\)

Proof

The variance of a continuous uniform random variable defined over the support \(a<x<b\) is:

\(\sigma^2=Var(X)=\dfrac{(b-a)^2}{12}\)

Proof

Because we just found the mean \(\mu=E(X)\) of a continuous random variable, it will probably be easiest to use the shortcut formula:

\(\sigma^2=E(X^2)-\mu^2\)

to find the variance. Let's start by finding \(E(X^2)\):

Now, using the shortcut formula and what we now know about \(E(X^2)\) and \(E(X)\), we have:

\(\sigma^2=E(X^2)-\mu^2=\dfrac{b^2+ab+a^2}{3}-\left(\dfrac{b+a}{2}\right)^2\)

Simplifying a bit:

\(\sigma^2=\dfrac{b^2+ab+a^2}{3}-\dfrac{b^2+2ab+a^2}{4}\)

and getting a common denominator:

\(\sigma^2=\dfrac{4b^2+4ab+4a^2-3b^2-6ab-3a^2}{12}\)

Simplifying a bit more:

\(\sigma^2=\dfrac{b^2-2ab+a^2}{12}\)

and, finally, we have:

\(\sigma^2=\dfrac{(b-a)^2}{12}\)

as was to be proved.

The moment generating function of a continuous uniform random variable defined over the support \(a < x < b\) is:

\(M(t)=\dfrac{e^{tb}-e^{ta}}{t(b-a)}\)

Proof