Suppose X is a uniform discrete random variable with possible values X=1,2,....,n. Can you show that the variance of X is (n^2-1)/12? Hint: 1^2 + 2^2 + 3^2+ .... + n^2=(n(n+1)(2n+1))/6

1 Answer
May 6, 2017

See proof below

Explanation:

We build a chart

color(white)(aaaa)Xcolor(white)(aaaa)1color(white)(aaaa)2color(white)(aaaaaa)3color(white)(aaaa)..........color(white)(aaaa)(n-1)color(white)(aaaa)n

color(white)(aaaa)pcolor(white)(aaaa)1/ncolor(white)(aaaa)1/ncolor(white)(aaaa)1/ncolor(white)(aaaa)..........color(white)(aaaaaa)1/ncolor(white)(aaaaa)1/n

color(white)(aaaa)Xpcolor(white)(aaa)1/ncolor(white)(aaaa)2/ncolor(white)(aaaa)3/ncolor(white)(aaaa)..........color(white)(aaaa)(n-1)/ncolor(white)(aaaa)n/n

color(white)(aaaa)X^2pcolor(white)(aa)1^2/ncolor(white)(aaaa)2^2/ncolor(white)(aaaa)3^2/ncolor(white)(aaaa)..........color(white)(a)(n-1)^2/ncolor(white)(aaaa)n^2/n

The Expected value is

E(X)=sum_1^nXp=1/n(1+2+3+...+(n-1)+n)

=1/n*n/2(n+1)

=1/2(n+1)

sum_1^nX^2p=1/n(1^2+2^2+3^2+........+(n-1)^2+n^2)

=1/n*n/6(n+1)(2n+1)

=1/6(n+1)(2n+1)

The variance is

var(X)=(sum_1^nX^2p)-(E(X))^2

=1/6(n+1)(2n+1)-1/4(n+1)^2

=1/2(n+1)(1/3(2n+1)-1/2(n+1))

=1/2(n+1)*1/6(4n+2-3n-3)

=1/12(n+1)(n-1)

=1/12(n^2-1)

QED

I hope that this is clearer.