In my university Stats courses, a formula referred to often was<\/p>\n
V(X)=E(X^2)-(E(X))^2<\/p>\n
in which<\/p>\n
V(X)=population variance<\/p>\n
E(X^2)=expected value of X^2<\/p>\n
E(X)=expected value of X<\/p>\n
Here’s how I believe one might prove it:<\/p>\n
By definition,<\/p>\n
V(X)=Σi=0<\/sub>n<\/sup>(xi<\/sub>-μ)^2\/n<\/p>\n By expansion,<\/p>\n (xi<\/sub>-μ)^2 = xi<\/sub>^2-2xi<\/sub>μ+μ^2<\/p>\n Therefore,<\/p>\n V(X)=Σi=0<\/sub>n<\/sup>(xi<\/sub>^2-2xi<\/sub>μ+μ^2)\/n<\/p>\n which further equals, by taking separate sums,<\/p>\n V(X)=Σi=0<\/sub>n<\/sup>xi<\/sub>^2\/n-2Σi=0<\/sub>n<\/sup>xi<\/sub>μ\/n+Σi=0<\/sub>n<\/sup>μ^2\/n<\/p>\n Now, using summation rules, we can rewrite the above as<\/p>\n V(X)=Σi=0<\/sub>n<\/sup>xi<\/sub>^2\/n-2μΣi=0<\/sub>n<\/sup>xi<\/sub>\/n+nμ^2\/n<\/p>\n Of course, by definition, E(X)=μ<\/p>\n Furthermore, since we are calculating for the entire population,<\/p>\n E(X^2)=Σi=0<\/sub>n<\/sup>xi<\/sub>^2\/n<\/p>\n and also<\/p>\n E(X)=Σi=0<\/sub>n<\/sup>xi<\/sub>\/n<\/p>\n Substituting the above definitions into our expanded, then simplified, formula, we arrive at<\/p>\n V(X)=E(X^2)-2E(X)*E(X)+(E(X))^2<\/p>\n and then<\/p>\n V(X)=E(X^2)-2(E(X))^2+(E(X))^2<\/p>\n then finally<\/p>\n V(X)=E(X^2)-(E(X))^2<\/p>\n I’ll be verifying this variance formula with an actual list of values in a coming post.<\/p>\n HTH:)<\/p>\n Source:<\/p>\n Ross, Sheldon A. A First Course in Probability<\/u>, 3rd ed. New York: Macmillan, 1988.<\/p>\n