Find the variance of $U[a,b]$ from $U[0,1]$

Question

I know the variance of uniform distribution $U[a,b]$ is $(b-a)^2/12$, but I'm having trouble to prove it from $U[0,1]$.

i.e. let $X = a+(b-a)U[0,1], Var(X) = \mathbb{E}(X^2)+\mathbb{E}(X)^2 = \mathbb{E}(a^2+(b-a)^2U^2+2a(b-a)U)-(a+b)^2/4 = a^2+a(b-a)+(b-a)^2\mathbb{E}(U^2)-(a+b)^2/4 = \frac{-3(a^2+b^2)+4b-4a+10ab}{12}$

where $\mathbb{E}(U[0,1]^2)=\frac{1}{3(b-a)}$

what did I do wrong?

Answer 1

You calculated it wrong :( Please try again.

For reference,

$Var (X)$

$= \mathbb{E}(a^2 + (b-a)^2U^2 - 2a(b-a)U) - \mathbb{E}(a + (b-a)U)^2$

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

$= a^2 + \frac{4a^2+4b^2-8ab}{12} + ab-a^2-\frac{3a^2+3b^2+6ab}{12}$

$= (12ab + 4a^2+4b^2-8ab - 3a^2-3b^2-6ab)/12$

$=( a^2+b^2+(12-8-6)ab)/12$

$=( a^2+b^2-2ab)/12$

$= (b-a)^2/12.$

Q.E.D.

Answer 2

By "from" they mean if $X\sim U(a,\,b)$ then $X=a+(b-a)Y$ with $Y\sim U(0.\,1)$, so $\operatorname{Var}X=(b-a)^2\operatorname{Var}Y=(b-a)^2/12$.

Edit, since @BenjaminWang has suggested proving $\operatorname{Var}Y=\tfrac{1}{12}$: since $\Bbb EY^n=\int_0^1y^ndy=\tfrac{1}{n+1}$ for $n>-1$, $\operatorname{Var}Y=\tfrac13-(\tfrac12)^2=\tfrac{4-3}{12}=\tfrac{1}{12}$.