Statistics: Proof of V(X)=E(X²)-[E(X)]²
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)=Σi=0n(xi-μ)^2/n
By expansion,
(xi-μ)^2 = xi^2-2xiμ+μ^2
Therefore,
V(X)=Σi=0n(xi^2-2xiμ+μ^2)/n
which further equals, by taking separate sums,
V(X)=Σi=0nxi^2/n-2Σi=0nxiμ/n+Σi=0nμ^2/n
Now, using summation rules, we can rewrite the above as
V(X)=Σi=0nxi^2/n-2μΣi=0nxi/n+nμ^2/n
Of course, by definition, E(X)=μ
Furthermore, since we are calculating for the entire population,
E(X^2)=Σi=0nxi^2/n
and also
E(X)=Σi=0nxi/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.
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