Analytical solution to linear-regression coefficient estimates

Question

I'm trying to understand matrix notation, and working with vectors and matrices.

Right now I'd like to understand how the vector of coefficient estimates $\hat{\beta}$ in multiple regression is computed.

The basic equation seems to be

$$ \frac{d}{d\boldsymbol{\beta}} (\boldsymbol{y}-\boldsymbol{X\beta})'(\boldsymbol{y}-\boldsymbol{X\beta}) = 0 \>. $$

Now how would I solve for a vector $\beta$ here?

Edit: Wait, I'm stuck. I'm here now and don't know how to continue:

$ \frac{d}{d{\beta}} \left( \left(\begin{smallmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{smallmatrix}\right) - \left(\begin{smallmatrix} 1 & x_{11} & x_{12} & \dots & x_{1p} \\ 1 & x_{21} & x_{22} & \dots & x_{2p} \\ \vdots & & & & \vdots \\ 1 & x_{n1} & x_{n2} & \dots & x_{np} \\ \end{smallmatrix}\right) \left(\begin{smallmatrix} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_p \end{smallmatrix}\right) \right) ' \left( \left(\begin{smallmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{smallmatrix}\right) - \left(\begin{smallmatrix} 1 & x_{11} & x_{12} & \dots & x_{1p} \\ 1 & x_{21} & x_{22} & \dots & x_{2p} \\ \vdots & & & & \vdots \\ 1 & x_{n1} & x_{n2} & \dots & x_{np} \\ \end{smallmatrix}\right) \left(\begin{smallmatrix} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_p \end{smallmatrix}\right) \right) $

$ \frac{d}{d{\beta}} \sum_{i=1}^n \left( y_i - \begin{pmatrix} 1 & x_{i1} & x_{i2} & \dots & x_{ip} \end{pmatrix} \begin{pmatrix} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_p \end{pmatrix} \right)^2$

With $x_{i0} = 1$ for all $i$ being the intercept:

$ \frac{d}{d{\beta}} \sum_{i=1}^n \left( y_i - \sum_{k=0}^p x_{ik} \beta_k \right)^2 $

Can you point me in the right direction?

Answer 1

We have

$\frac{d}{d\beta} (y - X \beta)' (y - X\beta) = -2 X' (y - X \beta)$.

It can be shown by writing the equation explicitly with components. For example, write $(\beta_{1}, \ldots, \beta_{p})'$ instead of $\beta$. Then take derivatives with respect to $\beta_{1}$, $\beta_{2}$, ..., $\beta_{p}$ and stack everything to get the answer. For a quick and easy illustration, you can start with $p = 2$.

With experience one develops general rules, some of which are given, e.g., in that document.

Edit to guide for the added part of the question

With $p = 2$, we have

$(y - X \beta)'(y - X \beta) = (y_1 - x_{11} \beta_1 - x_{12} \beta_2)^2 + (y_2 - x_{21}\beta_1 - x_{22} \beta_2)^2$

The derivative with respect to $\beta_1$ is

$-2x_{11}(y_1 - x_{11} \beta_1 - x_{12} \beta_2)-2x_{21}(y_2 - x_{21}\beta_1 - x_{22} \beta_2)$

Similarly, the derivative with respect to $\beta_2$ is

$-2x_{12}(y_1 - x_{11} \beta_1 - x_{12} \beta_2)-2x_{22}(y_2 - x_{21}\beta_1 - x_{22} \beta_2)$

Hence, the derivative with respect to $\beta = (\beta_1, \beta_2)'$ is

$ \left( \begin{array}{c} -2x_{11}(y_1 - x_{11} \beta_1 - x_{12} \beta_2)-2x_{21}(y_2 - x_{21}\beta_1 - x_{22} \beta_2) \\ -2x_{12}(y_1 - x_{11} \beta_1 - x_{12} \beta_2)-2x_{22}(y_2 - x_{21}\beta_1 - x_{22} \beta_2) \end{array} \right) $

Now, observe you can rewrite the last expression as

$-2\left( \begin{array}{cc} x_{11} & x_{21} \\ x_{12} & x_{22} \end{array} \right)\left( \begin{array}{c} y_{1} - x_{11}\beta_{1} - x_{12}\beta_2 \\ y_{2} - x_{21}\beta_{1} - x_{22}\beta_2 \end{array} \right) = -2 X' (y - X \beta)$

Of course, everything is done in the same way for a larger $p$.

Answer 2

You can also use formulas from Matrix cookbook. We have

$$(y-X\beta)'(y-X\beta)=y'y-\beta'X'y-y'X\beta+\beta'X'X\beta$$

Now take derivatives of each term. You might want to notice that $\beta'X'y=y'X\beta$. The derivative of term $y'y$ with respect to $\beta$ is zero. The remaining term

$$\beta'X'X\beta-2y'X\beta$$

is of form of function

$$f(x)=x'Ax+b'x,$$

in formula (88) in the book in page 11, with $x=\beta$, $A=X'X$ and $b=-2X'y$. The derivative is given in the formula (89):

$$\frac{\partial f}{\partial x}=(A+A')x+b$$

so

$$\frac{\partial}{\partial \beta}(y-X\beta)'(y-X\beta)=(X'X+(X'X)')\beta-2X'y$$

Now since $(X'X)'=X'X$ we get the desired solution:

$$X'X\beta=X'y$$

Answer 3

Here is a technique for minimizing the sum of squares in regression that actually has applications to more general settings and which I find useful.

Let's try to avoid vector-matrix calculus altogether.

Suppose we are interested in minimizing $$ \newcommand{\err}{\mathcal{E}}\newcommand{\my}{\mathbf{y}}\newcommand{\mX}{\mathbf{X}}\newcommand{\bhat}{\hat{\beta}}\newcommand{\reals}{\mathbb{R}} \err = (\my - \mX \beta)^T (\my - \mX \beta) = \|\my - \mX \beta\|_2^2 \> , $$ where $\my \in \reals^n$, $\mX \in \reals^{n\times p}$ and $\beta \in \reals^p$. We assume for simplicity that $p \leq n$ and $\mathrm{rank}(\mX) = p$.

For any $\bhat \in \reals^p$, we get $$ \err = \|\my - \mX \bhat + \mX \bhat - \mX \beta\|_2^2 = \|\my - \mX \bhat\|_2^2 + \|\mX(\beta-\bhat)\|_2^2 - 2(\beta - \bhat)^T \mX^T (\my - \mX \bhat) \>. $$

If we can choose (find!) a vector $\bhat$ such that the last term on the right-hand side is zero for every $\beta$, then we would be done, since that would imply that $\min_\beta \err \geq \|\my - \mX \bhat\|_2^2$.

But, $(\beta - \bhat)^T \mX^T (\my - \mX \bhat) = 0$ for all $\beta$ if and only if $\mX^T (\my - \mX \bhat) = 0$ and this last equation is true if and only if $\mX^T \mX \bhat = \mX^T \my$. So $\err$ is minimized by taking $\bhat = (\mX^T \mX)^{-1} \mX^T \my$.

---

While this may seem like a "trick" to avoid calculus, it actually has wider application and there is some interesting geometry at play.

One example where this technique makes a derivation much simpler than any matrix-vector calculus approach is when we generalize to the matrix case. Let $\newcommand{\mY}{\mathbf{Y}}\newcommand{\mB}{\mathbf{B}}\mY \in \reals^{n \times p}$, $\mX \in \reals^{n \times q}$ and $\mB \in \reals^{q \times p}$. Suppose we wish to minimize $$ \err = \mathrm{tr}( (\mY - \mX \mB) \Sigma^{-1} (\mY - \mX \mB)^T ) $$ over the entire matrix $\mB$ of parameters. Here $\Sigma$ is a covariance matrix.

An entirely analogous approach to the above quickly establishes that the minimum of $\err$ is attained by taking $$ \hat{\mB} = (\mX^T \mX)^{-1} \mX^T \mY \>. $$ That is, in a regression setting where the response is a vector with covariance $\Sigma$ and the observations are independent, then the OLS estimate is attained by doing $p$ separate linear regressions on the components of the response.

Answer 4

One way which may help you understand is to not use matrix algebra, and differentiate with each respect to each component, and then "store" the results in a column vector. So we have:

$$\frac{\partial}{\partial \beta_{k}}\sum_{i=1}^{N}\left(Y_{i}-\sum_{j=1}^{p}X_{ij}\beta_{j}\right)^{2}=0$$

Now you have $p$ of these equations, one for each beta. This is a simple application of the chain rule:

$$\sum_{i=1}^{N}2\left(Y_{i}-\sum_{j=1}^{p}X_{ij}\beta_{j}\right)^{1}\left(\frac{\partial}{\partial \beta_{k}}\left[Y_{i}-\sum_{j=1}^{p}X_{ij}\beta_{j}\right]\right)=0$$ $$-2\sum_{i=1}^{N}X_{ik}\left(Y_{i}-\sum_{j=1}^{p}X_{ij}\beta_{j}\right)=0$$

Now we can re-write the sum inside the bracket as $\sum_{j=1}^{p}X_{ij}\beta_{j}=\bf{x}_{i}^{T}\boldsymbol{\beta}$ So you get:

$$\sum_{i=1}^{N}X_{ik}Y_{i}-\sum_{i=1}^{N}X_{ik}\bf{x}_{i}^{T}\boldsymbol{\beta}=0$$

Now we have $p$ of these equations, and we will "stack them" in a column vector. Notice how $X_{ik}$ is the only term which depends on $k$, so we can stack this into the vector $\bf{x}_{i}$ and we get:

$$\sum_{i=1}^{N}\bf{x}_{i}\rm{Y}_{i}=\sum_{i=1}^{N}\bf{x}_{i}\bf{x}_{i}^{T}\boldsymbol{\beta}$$

Now we can take the beta outside the sum (but must stay on RHS of sum), and then take the invervse:

$$\left(\sum_{i=1}^{N}\bf{x}_{i}\bf{x}_{i}^{T}\right)^{-1}\sum_{i=1}^{N}\bf{x}_{i}\rm{Y}_{i}=\boldsymbol{\beta}$$