Cooper City High School
Statistics

Is the coefficient of Determination the same as the coefficient of Correlation squared?

...Yes,..it is. We will prove this now....Let's start !...

We define the coefficient of correlation ( r ) as :

(1)

Where:

S_x is the standard deviation for X, and S_y is the standard deviation for Y.

n is the number of events.

We can distribute the product (X_i - X_bar)(Y_i-Y_bar) = X_i Y_i - X_i Y_bar - X_bar Y_i + X_barY_bar   

So,  Σ[X_iY_i - X_i Y_bar - X_bar Y_i + X_barY_bar] = ΣX_iY_i - NX_barY_bar - NX_barY_bar + NX_barY_bar,

Simplifying ,  

(2)       Σ[X_iY_i - X_i Y_bar - X_bar Y_i + X_barY_bar] = ΣX_i Y_i - NX_barY_bar,

So, formula (1) can be rewritten as :

                        (3)  r = (1/(N-1))[ (ΣX_iY_i - NX_barY_bar)/(S_x S_y)]

or,          

                        (3a)    [ΣX_iY_i - NX_barY_bar] = (N-1)(S_x S_y)r

Now, forget for a moment about this result and let’s work with the Least-Square Regression Line(LSRL).

The condition to get the least-squares regression is that the sum (LS) should be a minimum:

                                    (4)   (LS) =  Σ (Y_i – (Y_hat)_i)²

where (Y_hat)_i is the value of Y we obtain using the LSRL expression:

                                    (5) (Y_hat)_i  = a + b X_i

The first condition is that the derivative of the function (LS), respect to the parameter a should be ZERO(0). From this condition, we get the result :

(6) Y_bar =  a + bX_bar    or equivalently,

(7)  a = Y_bar - bX_bar

The second condition is that the derivative of the function (SL), respect to the parameter b should be ZERO(0). From this condition, we get the result :

(8) ΣX_iY_i =  a N X_bar + b N (X_bar)²

Now, if we plug in the value of a given in the formula (7) into the formula (8), and solving for b, we get:

(9) b = (ΣX_iY_i - NY_barX_bar) / (N-1) S_x²

and, from here, we get :

(10) (ΣX_iY_i - NY_barX_bar) = b(N-1) S_x²

Now, let’s go back to the formula (3a), using the formula (10):

(11)    (N-1)(S_x S_y)r = b(N-1) S_x²

and, if we simplify, and solve for r :

(12)    r  =  b S_x/S_y

Next, let’s work with the expression of (LS):

                                    (13) (LS) =  Σ (Y_i – (Y_hat)_i)²

Using (5), this formula, can be rewritten as :

                                    (14) (LS) =  Σ (Y_i – a + b X_i)²

if we distribute the product, and use the expression (7) for  a , we get the result:

                                    (15) (LS) = (N-1) S_y² – b² (N-1)S_x² ,

Finally, let’s work with the definition of the coefficient of determination (R²).

        (16) R² =  [ Σ (Y_i-Y_bar)² - Σ (Y_i – (Y_hat)_i)²]/Σ (Y_i-Y_bar)²            ,

Because of the definition of the standard deviation,

                                    (17) S_y² = Σ (Y_i-Y_bar)² /(N-1),

so, we can transform (17) into:

                                    (18)            Σ (Y_i-Y_bar)² =  (N-1) S_y²

Next, using the definition of (LS) (13), and the expression (15), my obtain:

             (19)   R² =  {(N-1) S_y² – [(N-1) S_y² - b (N-1)S_x² ] } / (N-1) S_y² ,

Simplifying,

                                    (20) R² =   b² S_x² / S_y²

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