1
$\begingroup$

From Statistical Methods for Forecasting by Abraham and Ledolter:

Let $\hat y = X \hat \beta$, where $\hat \beta$ is the least squares estimate $(X^TX)^{-1}X^Ty$.

Then $SSR = \sum(\hat y_t - \bar y)^2 = \sum\hat y_t^2 - n \bar y^2$.

I'm having trouble showing this equality.

I see that expanding and eliminating expressions similar to both sides we're left with showing:

$\sum(y_t - \hat y_t) = 0$, but I can't figure this either.

Anyone have any ideas?

$\endgroup$
1
$\begingroup$

Let's decompose \begin{equation} SSR = \sum\limits_{t=1}^n(\hat y_t - \bar y)^2 = \sum\limits_{t=1}^n \hat y_t^2 - 2\sum\limits_{t=1}^n\hat y_t \bar y + \sum\limits_{t=1}^n \bar y^2 \end{equation} Notice that $\bar y$ does not depend on the summation index $t$ and hence could be extracted outside the sum as \begin{equation} SSR = \sum\limits_{t=1}^n \hat y_t^2 - 2\bar y\sum\limits_{t=1}^n\hat y_t + \bar y^2\sum \limits_{t=1}^n 1 = \sum\limits_{t=1}^n \hat y_t^2 - 2\bar y\sum\limits_{t=1}^n\hat y_t + n\bar y^2 \tag{1} \end{equation} It is true that (As you say) if $\sum(y_t - \hat y_t) = 0$, then we get the desired result. Notice that this happens when $X$ is a square and invertible matrix, because \begin{equation} \hat{\beta} = (X^TX)^{-1}X^Ty =X^{-1}X^{-T} X^{T}y = X^{-1}y \end{equation} where $X^{-T} = (X^{-1})^T$. Now, you can write $\hat{y}$ as \begin{equation} \hat{y} = X\hat{\beta} = XX^{-1}y = y \end{equation} This means that $\sum \hat{y}_t = \sum y = n\bar{y}$. Replacing this in $(1)$, we get \begin{equation} SSR = \sum\limits_{t=1}^n \hat y_t^2 - 2n\bar y^2+ n\bar y^2 = \sum\limits_{t=1}^n \hat y_t^2 - n\bar y^2 \end{equation}

$\endgroup$

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.