Property 1:

Proof: The first and last equations are trivial. We now show the second. By the first equation:

Taking the sum of both sides of the equation over all values of i and then squaring both sides, the desired result follows provided

This follows by substituting ŷi by bxi + ai and simplifying.

Property 2:

Proof: We give the proof for the case where both x and y have mean 0 and standard deviation 1. The general case follows via some tedious algebra.

In the case where and y have mean 0 and standard deviation 1, by Theorem 1 of Method of Least Squares and Property 1 of Method of Least Squares, we know that for all i

Since

it follows that

But

and so n – 1 = SST, which means that

Property 3:

Proof: From Property 2, solving for r2, we have the following by Property 1:

Property 4:

Proof: The first assertion is trivial. The second is a consequence of Property 1and 2 since

Property 5:

a)  The sums of the y values is equal to the sum of the ŷ values; i.e. $\sum_i{y_i}$$\sum_i{\hat{y_i}}$

b)  The mean of the y values and ŷ values are equal; i.e. ȳ = the mean of the ŷi

c)  The sums of the error terms is 0; i.e. $\sum_i{e_i}$ = 0

d)  The correlation coefficient of x with ŷ is sign(b); i.e. rxŷ = sign(rxy)

e)  The correlation coefficient of y with ŷ is the absolute value of the correlation coefficient of x with y; i.e. $r_{y\hat{y}}$ = |$r_{xy}$|

f)  The coefficient of determination of y with ŷ is the same as the correlation coefficient of x with y; i.e. $r_{y\hat{y}}^2$ = $r_{xy}^2$

Proof:

a) By Theorem 1 of Method of Least Squares

and so

b) That the mean of the ŷi is ȳ follows from (a) since the mean of ŷ = $\sum{}$ ŷi/n = $\sum{}$ yi/n = ȳ.

c) This property follows from (a) since

d) First note that by Property 3 of Expectation

and so

Now by Property A of Correlation

Thus, r = 1 if b > 0 and r = -1 if b < 0. If b = 0, r is undefined since there is division by 0. By Property 1 of Method of Least Squares, r = sign(b) = sign(rxy).

e) Using property (b), the correlation of y with ŷ is

By Theorem 1 of Method of Least Squares

for all i, and so

As we saw in the proof of (d),

and so putting all the pieces together, we get

By Property 1 of Method of Least Squaresr = sign(b) = sign(rxy), and so

f) This follows from (e)