Answer to Question #267312 in Statistics and Probability for John

Question

Compute the mean and variance of random variable whose distribution is uniform on the interval[a,b].

(a) It is not enough to simplify state values. You must give the details of the computation

Answer 1

Probability density function:

$f(x)=\begin{cases} \frac{1}{b-a} &\text{for } a\le x\le b \\ 0 &\text{for } x<a\ or\ x>b \end{cases}$

mean:

$E(X)=\int^b_axf(x)dx=\int^b_a\frac{x}{b-a}dx=\frac{x^2}{2(b-a)}|^b_a=\frac{b^2-a^2}{2(b-a)}=\frac{b+a}{2}$

variance:

$V(X)=E(X^2)-(E(X))^2$

$E(X^2)=\int^b_ax^2f(x)dx=\int^b_a\frac{x^2}{b-a}dx=\frac{x^3}{3(b-a)}|^b_a=\frac{b^3-a^3}{3(b-a)}=\frac{a^2+ab+b^2}{3}$

$(E(X))^2=\frac{(b+a)^2}{4}$

$V(X)=\frac{a^2+ab+b^2}{3}-\frac{(b+a)^2}{4}=\frac{4(a^2+ab+b^2)-3(a^2-2ab-b^2)}{12}=\frac{a^2-2ab+b^2}{12}=\frac{(a-b)^2}{12}$