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

#color(white)(aaaa)##p##color(white)(aaaa)##1/n##color(white)(aaaa)##1/n##color(white)(aaaa)##1/n##color(white)(aaaa)##..........##color(white)(aaaaaa)##1/n##color(white)(aaaaa)##1/n#

#color(white)(aaaa)##Xp##color(white)(aaa)##1/n##color(white)(aaaa)##2/n##color(white)(aaaa)##3/n##color(white)(aaaa)##..........##color(white)(aaaa)##(n-1)/n##color(white)(aaaa)##n/n#

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