Linear regression: *Why* can you partition sums of squares?

This post refers to a bivariate linear regression model, $Y_i = \beta_0 + \beta_1x_i$ . I have always taken the partitioning of total sum of squares (SSTO) into sum of squares for error (SSE) and sum of squares for the model (SSR) on faith, but once I started really thinking about it, I don't understand why it works...

The part I do understand:

$y_i$: An observed value of y

$\bar{y}$: The mean of all observed $y_i$s

$\hat{y}_i$: The fitted/predicted value of y for a given observation's x

$y_i - \hat{y}_i$: Residual/error (if squared and added up for all observations this is SSE)

$\hat{y}_i - \bar{y}$: How much the model fitted value differs from the mean (if squared and added up for all observations this is SSR)

$y_i - \bar{y}$: How much an observed value differs from the mean (if suared and added up for all observations, this is SSTO).

I can understand why, for a single observation, without squaring anything, $(y_i - \bar{y}) = (\hat{y}_i - \bar{y}) + (y_i - \hat{y}_i)$. And I can understand why, if you want to add things up over all observations, you have to square them or they'll add up to 0.

The part I don't understand is why $(y_i - \bar{y})^2 = (\hat{y}_i - \bar{y})^2 + (y_i - \hat{y}_i)^2$ (eg. SSTO = SSR + SSE). It seems to be that if you have a situation where $A = B + C$, then $A^2 = B^2 + 2BC + C^2$, not $A^2 = B^2 + C^2$. Why isn't that the case here?

• You left out the summation in your last paragraph. SST = SSR + SSE is a sum over $i$, but your equality you wrote immediately before it is not actually true without the summation sign there. Jan 26 '17 at 15:08
• In your last paragraph, you want (i.e. SSTO = SSR + SSE) not (e.g. SSTO = SSR + SSE). "e.g." is an abbreviation for the Latin phrase "exempli gratia," or "for example" in English. "i.e." is an abbreviation for "id est" and can be read in English as "that is." Jan 27 '17 at 21:58

It seems to be that if you have a situation where $A = B + C$, then $A^2 = B^2 + 2BC + C^2$, not $A^2 = B^2 + C^2$. Why isn't that the case here?

Conceptually, the idea is that $BC = 0$ because $B$ and $C$ are orthogonal (i.e. are perpendicular).

In the context of linear regression here, the residuals $\epsilon_i = y_i - \hat{y}_i$ are orthogonal to the demeaned forecast $\hat{y}_i - \bar{y}$. The forecast from linear regression creates an orthogonal decomposition of $\mathbf{y}$ in a similar sense as $(3,4) = (3,0) + (0,4)$ is an orthogonal decomposition.

Linear Algebra version:

Let:

$$\mathbf{z} = \begin{bmatrix} y_1 - \bar{y} \\ y_2 - \bar{y}\\ \ldots \\ y_n - \bar{y} \end{bmatrix} \quad \quad \mathbf{\hat{z}} = \begin{bmatrix} \hat{y}_1 - \bar{y} \\ \hat{y}_2 - \bar{y} \\ \ldots \\ \hat{y}_n - \bar{y} \end{bmatrix} \quad \quad \boldsymbol{\epsilon} = \begin{bmatrix} y_1 - \hat{y}_1 \\ y_2 - \hat{y}_2 \\ \ldots \\ y_n - \hat{y}_n \end{bmatrix} = \mathbf{z} - \hat{\mathbf{z}}$$

Linear regression (with a constant included) decomposes $\mathbf{z}$ into the sum of two vectors: a forecast $\hat{\mathbf{z}}$ and a residual $\boldsymbol{\epsilon}$

$$\mathbf{z} = \hat{\mathbf{z}} + \boldsymbol{\epsilon}$$

Let $\langle .,. \rangle$ denote the dot product. (More generally, $\langle X,Y \rangle$ can be the inner product $E[XY]$.)

\begin{align*} \langle \mathbf{z} , \mathbf{z} \rangle &= \langle \hat{\mathbf{z}} + \boldsymbol{\epsilon}, \hat{\mathbf{z}} + \boldsymbol{\epsilon} \rangle \\ &= \langle \hat{\mathbf{z}}, \hat{\mathbf{z}} \rangle + 2 \langle \hat{\mathbf{z}},\boldsymbol{\epsilon} \rangle + \langle \boldsymbol{\epsilon},\boldsymbol{\epsilon} \rangle \\ &= \langle \hat{\mathbf{z}}, \hat{\mathbf{z}} \rangle + \langle \boldsymbol{\epsilon},\boldsymbol{\epsilon} \rangle \end{align*}

Where the last line follows from the fact that $\langle \hat{\mathbf{z}},\boldsymbol{\epsilon} \rangle = 0$ (i.e. that $\hat{\mathbf{z}}$ and $\boldsymbol{\epsilon} = \mathbf{z}- \hat{\mathbf{z}}$ are orthogonal). You can prove $\hat{\mathbf{z}}$ and $\boldsymbol{\epsilon}$ are orthogonal based upon how the ordinary least squares regression constructs $\hat{\mathbf{z}}$.

$\hat{\mathbf{z}}$ is the linear projection of $\mathbf{z}$ onto the subspace defined by the linear span of the regressors $\mathbf{x}_1$, $\mathbf{x}_2$, etc.... The residual $\boldsymbol{\epsilon}$ is orthogonal to that entire subspace hence $\hat{\mathbf{z}}$ (which lies in the span of $\mathbf{x}_1$, $\mathbf{x}_2$, etc...) is orthogonal to $\boldsymbol{\epsilon}$.

Note that as I defined $\langle .,.\rangle$ as the dot product, $\langle \mathbf{z} , \mathbf{z} \rangle = \langle \hat{\mathbf{z}}, \hat{\mathbf{z}} \rangle + \langle \boldsymbol{\epsilon},\boldsymbol{\epsilon} \rangle$ is simply another way of writing $\sum_i (y_i - \bar{y})^2 = \sum_i (\hat{y}_i - \bar{y})^2 + \sum_i (y_i - \hat{y}_i)^2$ (i.e. SSTO = SSR + SSE)

The whole point is showing that certain vectors are orthogonal and then use Pythagorean theorem.

Let us consider multivariate linear regression $Y = X\beta + \epsilon$. We know that the OLS estimator is $\hat{\beta} = (X^tX)^{-1}X^tY$. Now consider the estimate

$\hat{Y} = X\hat{\beta} = X(X^tX)^{-1}X^tY = HY$ (H matrix is also called the "hat" matrix)

where $H$ is an orthogonal projection matrix of Y onto $S(X)$. Now we have

$Y - \hat{Y} = Y - HY = (I - H)Y$

where $(I-H)$ is a projection matrix onto orthogonal complement of $S(X)$ which is $S^{\bot}(X)$. Thus we know that $Y-\hat{Y}$ and $\hat{Y}$ are orthogonal.

Now consider a submodel $Y = X_0\beta_0 + \epsilon$

where $X = [X_0 | X_1 ]$ and similarily we have the OLS estimator and estimate $\hat{\beta_0}$ and $\hat{Y_0}$ with projection matrix $H_0$ onto $S(X_0)$. Similarily we have that $Y - \hat{Y_0}$ and $\hat{Y_0}$ are orthogonal. And now

$\hat{Y} - \hat{Y_0} = HY - H_0Y = HY - H_0HY = (I - H_0)HY$

where again $(I-H_0)$ is an orthogonal projection matrix on complement of $S(X_0)$ which is $S^{\bot}(X_0)$. Thus we have orthogonality of $\hat{Y} - \hat{Y_0}$ and $\hat{Y_0}$. So in the end we have

$||Y - \hat{Y}||^2 = ||Y||^2 - ||\hat{Y}||^2 = ||Y - \hat{Y_0}||^2 + ||\hat{Y_0}||^2 - ||\hat{Y} - \hat{Y_0}||^2 - ||\hat{Y_0}||^2$

and finally $||Y - \hat{Y_0}||^2 = ||Y - \hat{Y}||^2 + ||\hat{Y} - \hat{Y_0}||^2$

Lastly, the mean $\bar{Y}$ is simply the $\hat{Y_0}$ when considering the null model $Y = \beta_0 + e$.

• Thank you for your answer! What is S() (as in S(X) in your post)? Jan 27 '17 at 18:21
• $S(X)$ is a subspace generated by the columns of matrix $X$ Jan 27 '17 at 19:26