# How to find the closed form formula for $\hat{\beta}$ while using ordinary least squares estimation?

According to Wikipedia's article on Linear Regression:

Given a data set $\{y_i,x_{i1},\ldots,x_{ip}\}_{i=1}^{i=n}$ of $n$ statistical units, a regression model assumes that the relationship between the dependent variable $y_i$ and the $p-\text{vector}$ or regressors $x_i$ is linear. This relationship is modelled through a disturbance term or error variable $\varepsilon_i$-an unobserved random variable that adds random noise to the to the linear relationship between the dependent variables and regressors. The model takes the form $y_i=\beta_0(1)+\beta_1x_{i1}+\cdots+\beta_px_{ip}+\epsilon_i=x_i^T \beta + \varepsilon_i$

These equations can be written in vector form as $$y=\mathbf{X\beta+\epsilon}$$

For the Ordinary Least Square estimation they say that the closed form expression for the estimated value of the unknown parameter $\beta$ is

$$\hat{\mathbf{\beta}}=(\mathbf{X^{T}X})^{-1}\mathbf{X}^{T}y$$

I'm not sure how they get this formula for $\hat{\beta}$. It would be very nice if someone can explain me the derivation.

• @RobertIsrael No clue how Schwarz–Christoffel mapping is related to this.
– user400242
Commented Dec 21, 2017 at 19:50
• Sorry, wrong link. This is the standard linear algebra formula for Linear least squares Commented Dec 21, 2017 at 19:58
• I believe you have a typo. It's $\mathbf{X}^{T}\mathbf{X}$, not $\mathbf{X}\mathbf{X}^{T}$. Commented Dec 21, 2017 at 20:01
• Of the three answers posted so far, mine is the only one that doesn't use differentiation. Commented Dec 21, 2017 at 20:54

I'm going to show this using partial differentiation.

Consider the assumed linear model $$y_i = \mathbf{x}_i^{T}\boldsymbol\beta + \epsilon_i$$ where $y_i, \epsilon_i \in \mathbb{R}$ and $\mathbf{x}_i=\begin{bmatrix} x_{i0} \\ x_{i1} \\ \vdots \\ x_{ip} \end{bmatrix}, \boldsymbol\beta = \begin{bmatrix} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_p \end{bmatrix} \in \mathbb{R}^{p+1}$ for $i = 1, \dots, n$, with $x_{i0} = 1$.

Our aim is to solve for $\hat{\boldsymbol\beta}$ by minimizing the residual sum of squares, or minimizing $$\text{RSS}(\boldsymbol\beta) = \sum_{i=1}^{n}(y_i-\mathbf{x}_i^{T}\boldsymbol\beta)^2\text{.}$$ To compute this sum, consider the vector of residuals $$\mathbf{e}=\begin{bmatrix} y_1 - \mathbf{x}_1^{T}\boldsymbol\beta \\ y_2 - \mathbf{x}_2^{T}\boldsymbol\beta \\ \vdots \\ y_n - \mathbf{x}_n^{T}\boldsymbol\beta \end{bmatrix}$$ Then $\text{RSS}(\boldsymbol\beta) = \mathbf{e}^{T}\mathbf{e}$. Our next step is to find the "partial derivatives" of $\text{RSS}(\boldsymbol\beta)$.

To do this, note that for $k = 1, \dots, p$, $$\dfrac{\partial \text{RSS}}{\partial \beta_k}=\dfrac{\partial}{\partial\beta_k}\left\{\sum_{i=1}^{n}\left[y_i- \sum_{j=0}^{p}\beta_jx_{ij}\right]^2 \right\}=-2\sum_{i=1}^{n}x_{ik}\left(y_i - \sum_{j=0}^{p}\beta_jx_{ij}\right)\text{.}$$ "Stacking" these, we obtain \begin{align} \dfrac{\partial \text{RSS}}{\partial \boldsymbol\beta}&=\begin{bmatrix} \dfrac{\partial \text{RSS}}{\partial \beta_0} \\ \dfrac{\partial \text{RSS}}{\partial \beta_1} \\ \vdots \\ \dfrac{\partial \text{RSS}}{\partial \beta_p} \end{bmatrix} \\ &= \begin{bmatrix} -2\sum_{i=1}^{n}x_{i0}\left(y_i - \sum_{j=0}^{p}\beta_jx_{ij}\right) \\ -2\sum_{i=1}^{n}x_{i1}\left(y_i - \sum_{j=0}^{p}\beta_jx_{ij}\right) \\ \vdots \\ -2\sum_{i=1}^{n}x_{ip}\left(y_i - \sum_{j=0}^{p}\beta_jx_{ij}\right) \end{bmatrix} \\ &= -2\begin{bmatrix} \sum_{i=0}^{n}x_{i0}(\mathbf{y}-\mathbf{x}_1^{T}\boldsymbol\beta)\\ \sum_{i=0}^{n}x_{i1}(\mathbf{y}-\mathbf{x}_1^{T}\boldsymbol\beta) \\ \vdots \\ \sum_{i=0}^{n}x_{ip}(\mathbf{y}-\mathbf{x}_1^{T}\boldsymbol\beta) \end{bmatrix} \\ &= -2\left(\begin{bmatrix} \sum_{i=0}^{n}x_{i0}\mathbf{y}\\ \sum_{i=0}^{n}x_{i1}\mathbf{y} \\ \vdots \\ \sum_{i=0}^{n}x_{ip}\mathbf{y} \end{bmatrix} - \begin{bmatrix} \sum_{i=0}^{n}x_{i0}\mathbf{x}_1^{T}\boldsymbol\beta)\\ \sum_{i=0}^{n}x_{i1}\mathbf{x}_1^{T}\boldsymbol\beta) \\ \vdots \\ \sum_{i=0}^{n}x_{ip}\mathbf{x}_1^{T}\boldsymbol\beta) \end{bmatrix}\right)\\ &= -2(\mathbf{X}^{T}\mathbf{y}-\mathbf{X}^{T}\mathbf{X}\boldsymbol\beta)\text{.} \end{align} where $$\mathbf{X} = \begin{bmatrix} \mathbf{x}_1^{T} \\ \mathbf{x}_2^{T} \\ \vdots \\ \mathbf{x}_n^{T} \end{bmatrix}\text{.}$$ Setting $\dfrac{\partial \text{RSS}}{\partial \boldsymbol\beta} = \mathbf{0}$, we obtain $$\mathbf{X}^{T}\mathbf{X}\boldsymbol\beta = \mathbf{X}^{T}\mathbf{y}$$ and assuming $\mathbf{X}^{T}\mathbf{X}$ is invertible, $$\hat{\boldsymbol\beta} = (\mathbf{X}^{T}\mathbf{X})^{-1}\mathbf{X}^{T}\mathbf{y}$$

– user400242
Commented Dec 21, 2017 at 20:36
• @Blue Residual Sum of Squares Commented Dec 21, 2017 at 20:36

Essentially we're trying to minimize $\epsilon ^T \epsilon$ with respect to $\beta$.

So we're trying to get

\begin{align} \frac{ d \epsilon ^ T \epsilon }{ d \beta } &= 0\\ \\ \frac{ d ( y - X\beta ) ^ T ( y - X\beta) }{ d \beta } &= 0\\ \\ \frac{d}{d\beta}(-2y^T(X\beta) + \beta^TX^TX\beta) &= 0 \\ \\ 2(X^TX)\beta &= 2 X^Ty \\ \\ \hat \beta &= (X^TX)^{-1} X^Ty \end{align} As required, assuming $X^TX$ is invertible.

Proposition: The value of $\beta$ that minimizes $\|Y-X\beta\|^2$ is the one that makes $X\beta$ equal to the orthogonal projection onto the column space of $X.$

Proof: The desired value is called $\widehat\beta.$ Let $X\beta_P$ be the orthogonal projection of $Y$ onto the column space of $X.$ Then for any value of $\beta$ we have \begin{align} \|Y - X\widehat\beta_P\|^2 & = \|(Y-X\beta_P) + (X\beta_P - X\widehat\beta)\|^2 \\[10pt] & = \|Y-X\beta_P\|^2 + (Y-X\beta_P)^T(X\beta_P-X\widehat\beta) + \| X(\beta_P-\widehat\beta) \|^2. \end{align} The middle term is $0$ because it's the dot product of something orthogonal to the column space and something in the column space. We therefore have $$\|Y-X\beta_P\|^2 + \|X(\beta_P-\widehat\beta)\|^2.$$ The value of $\widehat\beta$ that minimizes this is the one that makes the makes the second term $0,$ and that is $\widehat\beta=\beta_P. \quad \blacksquare$

Proposition: Suppose $X$ has linearly independent columns (but not rows, since $X$ has many more rows than columns). Then the orthogonal projection onto the column space of $X$ is $$y \mapsto X(X^TX)^{-1}X^T y.$$ Proof: Suppose first that $y$ is in the column space of $X.$ Then $y = X\alpha$ (for some $\alpha$). So $$X(X^TX)^{-1}X^Ty = X(X^TX)^{-1}X^T (X\alpha) = X\alpha = y.$$ Next suppose $y$ is orthogonal to the column space of $X.$ Then $X^Ty=0$ so $X(X^TX)^{-1}X^Ty = 0. \qquad \blacksquare$

Therefore, for the desired value $\widehat\beta,$ we have $$X\widehat\beta = X(X^TX)^{-1}X^T y. \tag 1$$ Since $X$ has more rows than columns, it does not have a inverse matrix, but since its columns are linearly independent, it has a left inverse (more than one of them, in fact) and one such left inverse is $(X^TX)^{-1}X^T,$ as you can see by multiplying. So multiply both sides of $(1)$ by that left inverse, and you've got it.