3
$\begingroup$

I'm working on this identity $$\sum_{i-1}^n (y_i - \hat {\beta_0} - \hat {\beta_1}x_i)^2 = \sum_{i=1}^n y_i^2 - \hat {\beta_0}\sum_{i=1}^n y_i - \hat {\beta_1} \sum_{i=1}^n x_iy_i$$

I have these relationships to work with:

$$\hat {\beta_1} = \frac { n\sum_{i_1}^n x_iy_i - \left ( \sum_{i=1}^n x_i \right ) \left ( \sum_{i=1}^n y_i \right )}{ n \left ( \sum_{i=1}^n x_i^2 \right ) -\left (\sum_{i=1}^n x_i \right )^2 }$$

$$ \hat {\beta_0}= \overline {y} - \hat {\beta_1} \overline {x}$$

A little manipulation also shows

$$\hat {\beta_1}= \frac { \sum_{i=1}^n ( x_i - \overline {x}) y_i}{ \left ( \sum_{i=1}^n x_i^2 \right ) -n \overline {x}^2 }$$

My strategy has to substitute $\hat {\beta_1}$ out of the equation, but I just keep getting $\overline {y}$ in too many terms and not any terms with $y_i$. The formula for $\hat {\beta_1}$ is too complicated to consider working with. What am I missing?

$\endgroup$
  • $\begingroup$ By selecting suitable units in which to measure the $x_i$ and the $y_i$ you can make the $x_i$ sum to $0$, the $y_i$ sum to $0$, the $x_i^2$ sum to $n$, and the $y_i^2$ sum to $n$. The formulas for $\hat\beta_0$ and $\hat\beta_1$ will greatly simplify with no loss in generality. $\endgroup$ – whuber May 14 '14 at 20:47
4
$\begingroup$

OK, I will do some parts and leave the rest for you to do it yourself. I dropped the index of summations for simplicity. Start from expanding the L.H.S to have $$L.H.S=\sum y_i^2+\sum(\hat{\beta_0}+\hat{\beta_1}x_i)^2-2\hat{\beta_0}\sum y_i-2\hat{\beta_1}\sum (y_ix_i)$$ which is $$L.H.S=\Big[\sum y_i^2-\hat{\beta_0}\sum y_i-\hat{\beta_1}\sum (y_ix_i)\Big]+\sum(\hat{\beta_0}+\hat{\beta_1}x_i)^2-\hat{\beta_0}\sum y_i-\hat{\beta_1}.\sum (y_ix_i).$$ Now what we have inside the bracket is actually the R.H.S. So you need to show that the rest is zero i.e. $$\sum(\hat{\beta_0}+\hat{\beta_1}x_i)^2-\hat{\beta_0}\sum y_i-\hat{\beta_1}\sum (y_ix_i)=0.$$ Now to show this, I will give you some hints. You need to do them correctly and step by step.

  1. Replace $\hat{\beta_0}$ by $\bar{y}-\hat{\beta_1}\bar{x}$ to re-write it all based on $\hat{\beta_1}$.
  2. Expand the terms and simplify (some terms will be canceled out).
  3. Now use two facts:
    (1): $\hat{\beta_1}=\dfrac{S_{xy}}{S_{xx}},$ where $S_{xy}=\sum(x_i-\bar{x})(y_i-\bar{y})$ and $S_{xx}=\sum(x_i-\bar{x})^2$ and
    (2): $S_{xy}=\sum (x_iy_i)-\dfrac{\sum x_i .\sum y_i}{n}$
    to write everything in terms of $S_{xy}$ and $S_{xx}$.
  4. Simplify to show that it is zero.
$\endgroup$
8
$\begingroup$

You are trying to reprove the Pythagorean Theorem. Understanding the connection provides a powerful intuition for understanding ordinary least squares regression and (incidentally) makes short work of the proof.


Let $y$ represent the vector $(y_1, y_2, \ldots, y_n)$, $\mathbf{1}$ the $n$-vector $(1,1,\ldots,1)$, and $x$ the vector $(x_1,x_2, \ldots, x_n)$. Denote by $\hat{y}$ the vector $\hat\beta_0\mathbf{1} + \hat\beta_1 x.$ In this notation the identity (after combining the last two sums) is

$$ ||y-\hat y||^2 = \sum_{i-1}^n (y_i - \hat {\beta_0} - \hat {\beta_1}x_i)^2 = \sum_{i=1}^n y_i^2 - \sum_{i=1}^n\left(\hat {\beta_0} - \hat {\beta_1} x_i\right)y_i = ||y||^2 - \hat y \cdot y.$$

Because $\hat\beta_0$ and $\hat\beta_1$ are formulas for the least-squares coefficients, by definition $y-\hat{y}$ minimizes the squared distance from $y$ to the line generated by $\hat{y}$. Because $y$ and $\hat{y}$ span a space of at most two dimensions, understanding their relationship is a matter of planar Euclidean geometry which is faithfully illustrated with a simple diagram:

Figure

The formulae for $\hat\beta_0$ and $\hat\beta_1$ are usually derived by demonstrating the geometrically obvious fact that $y-\hat y$ must be perpendicular to $\hat y$. In terms of vector operations, this means their dot product is zero:

$$\hat y \cdot (y - \hat y) = 0.$$

Expanding this dot product shows it is equivalent to the key relationship

$$||\hat y||^2 = \hat y \cdot \hat y = \hat y \cdot y.$$

Consider the Pythagorean Theorem. In this right triangle it asserts that the square of one leg equals the square of the hypotenuse minus the square of the other leg:

$$||y-\hat y||^2 = ||y||^2 - ||\hat y||^2.$$

The key relationship provides another expression for the last term, yielding

$$||y-\hat y||^2 = ||y||^2 - \hat y \cdot y,$$

which is the desired identity, QED.

$\endgroup$
  • $\begingroup$ I having trouble following this derivation. How do you know $y$ and $\hat{y}$ span only 2 dims? How do you know that $\hat{y}$ and $y - \hat{y}$ are perpendicular/dot product=0? $\endgroup$ – Mitch Jan 4 '17 at 21:42
  • 1
    $\begingroup$ @Mitch (1) Two vectors and their common origin are three points. Assuming they are distinct, Euclid states they determine a plane. The graphic here is fully general; the only case it does not accurately depict is when $y$ and $\hat y$ are collinear: they span just one (or zero) dimension. (2) The perpendicularity follows from the least squares criterion: the squared residual length is minimized when the dot product is zero. This is geometrically clear: if the angle were not ninety degrees, you could shorten the distance--and thereby reduce the squared distance--by changing the projection. $\endgroup$ – whuber Jan 4 '17 at 21:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.