Proving Equivalence of Arithmetic Mean and Least Square Criterion

[kreil]

Homework Statement

Show that the calculation of an arithmetic mean for a given set of values y1,y2...yn is equivalent to fitting the line y=y_av to the data under the least square criterion

Homework Equations

distance between the data points and the best fit curve:
d(y_n,f(x))= \Sigma_{i=1}^{n}(y_i-f(x_i,a_1,a_2,...,a_n))^2

arithmetic mean of a set of values y_n:
y_av= \frac{1}{n} \Sigma_{i=1}^{n}y_i

The Attempt at a Solution

I need help getting started please!

 

[Dick]

In your distance formula, just put f=y_av. Now how would you minimize d with respect to y_av?

 

[kreil]

set the partial derivative of d equal to zero?

 

[Dick]

set the partial derivative equal to zero?

Absolutely. What do you get?

 

[kreil]

\frac{d(R^2)}{dy_{av}}= -2 \Sigma_{i=1}^n(y_i-y_{av})=0

 

[Dick]

What's your next move?

 

[kreil]

i was thinking maybe distribute the sum, but i don't know how the -2 helps, since I am trying to get the formula in the form of the arithmetic mean

 

[Dick]

-2*x=0 means x=0, doesn't it?

 

[kreil]

how is this:

\frac{d(R^2)}{dy_{av}}= -2 \Sigma_{i=1}^n(y_i-y_{av})=0 \implies \Sigma_{i=1}^n(y_i-y_{av})=0 \implies \Sigma_{i=1}^ny_i=\Sigma_{i=1}^ny_{av}=y_{av}

 

[Dick]

The sum from 1 to n of y_av is NOT = y_av.

 

[kreil]

oh ok that sum is n y_av, so the sum of 1/ny_i is y_av correct?

 

[Dick]

That's what you are trying to prove, right?

 

[kreil]

right! thanks a lot man