The tutor offers proof of a formula he recalls from Stats.

In my university Stats courses, a formula referred to often was

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

in which

V(X)=population variance

E(X^2)=expected value of X^2

E(X)=expected value of X

Here’s how I believe one might prove it:

By definition,

V(X)=Σ_i=0ⁿ(x_i-μ)^2/n

By expansion,

(x_i-μ)^2 = x_i^2-2x_iμ+μ^2

Therefore,

V(X)=Σ_i=0ⁿ(x_i^2-2x_iμ+μ^2)/n

which further equals, by taking separate sums,

V(X)=Σ_i=0ⁿx_i^2/n-2Σ_i=0ⁿx_iμ/n+Σ_i=0ⁿμ^2/n

Now, using summation rules, we can rewrite the above as

V(X)=Σ_i=0ⁿx_i^2/n-2μΣ_i=0ⁿx_i/n+nμ^2/n

Of course, by definition, E(X)=μ

Furthermore, since we are calculating for the entire population,

E(X^2)=Σ_i=0ⁿx_i^2/n

and also

E(X)=Σ_i=0ⁿx_i/n

Substituting the above definitions into our expanded, then simplified, formula, we arrive at

V(X)=E(X^2)-2E(X)*E(X)+(E(X))^2

and then

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

then finally

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

I’ll be verifying this variance formula with an actual list of values in a coming post.

HTH:)

Source:

Ross, Sheldon A. A First Course in Probability, 3rd ed. New York: Macmillan, 1988.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.