6
$\begingroup$

I'm reading the The Elements of Statistical Learning and come across the following:

enter image description here

Can anyone explain how they moved from 3.3 to 3.6 or even to 3.4 ?

$\endgroup$
1

1 Answer 1

14
$\begingroup$

Our loss function is $RSS(\beta) = (y - X\beta)^T(y -X\beta)$. Expanding this and using the fact that $(u - v)^T = u^T - v^T$, we have $$ RSS(\beta) = y^Ty - y^TX\beta - \beta^TX^Ty + \beta^T X^T X \beta. $$ Noting that $y^TX\beta$ is a scalar, and for any scalar $r \in \mathbb R$ we have $r = r^T$ we have $y^T X \beta = (y^T X \beta)^T = \beta^T X^T y$ so all together $$ RSS(\beta) = y^T y - 2 \beta^T X^T y + \beta^T X^T X \beta. $$

Now we'll differentiate with respect to $\beta$: $$ \frac{\partial RSS}{\partial \beta} = \frac{\partial}{\partial \beta} y^T y - 2 \frac{\partial}{\partial \beta} \beta^T X^T y + \frac{\partial}{\partial \beta} \beta^T X^T X \beta $$ $$ = 0 - 2 X^T y + 2 X^T X \beta. $$ If you haven't seen derivatives with respect to a vector before, the Matrix Cookbook is a popular reference.

We want to find the minimum of $RSS$ so we'll set the derivative equal to $0$. This leads us to $$ \frac{\partial RSS}{\partial \beta} \stackrel{\text{set}}= 0 \implies -2X^T y + 2X^T X \beta = 0 $$ $$ \implies X^T y - X^T X \beta = 0 \implies X^T(y - X \beta) = 0. $$

Now we do use the assumption that $X$ is full column rank, which means we know $X^T X$ is positive definite and therefore invertible. This means $$ X^Ty = X^T X \beta \implies (X^T X)^{-1}X^T y = \hat \beta $$ where we achieved this by left-multiplying by $(X^T X)^{-1}$.

$\endgroup$
1
  • 1
    $\begingroup$ See this answer for a proof that $X$ is full column rank iff $X^T X$ is invertible. Also, I think it should be noted that if $X$ is not full column rank, we can remove columns to make it full column rank, as explained here. $\endgroup$ Aug 25, 2018 at 6:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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