How do I prove that sum_(i=1)^n(x_i-mu)^2=sum_(i=1)^n(x_i)^2-nmu^2?

1 Answer
Geoff K.
Aug 7, 2018

Start with:

sum_(i=1)^n (x_i-mu)^2

Within the sum, distribute the square:

=sum_(i=1)^n (x_i-mu)(x_i-mu)

=sum_(i=1)^n (x_i^2-2x_i mu + mu^2)

Distribute the sum notation to each term:

=sum_(i=1)^n x_i^2- sum_(i=1)^n 2x_i mu + sum_(i=1)^n mu^2

Factor out the 2mu from the middle sum (since 2mu doesn't depend on i) and notice that the last term is just underset (n" times") underbrace(mu^2 + mu^2 + ... + mu^2):

=sum_(i=1)^n x_i^2- 2musum_(i=1)^n x_i + nmu^2

Using mu = 1/n sum_(i=1)^n x_i, rearrange to get sum_(i=1)^n x_i = nmu, then sub this in:

=sum_(i=1)^n x_i^2- 2mu(nmu) + nmu^2

=sum_(i=1)^n x_i^2- 2nmu^2 + nmu^2

Combine like terms:

=sum_(i=1)^n x_i^2- nmu^2

Done!

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