deriving-ols-estimator.mdx

title	date	tags	draft	summary
Deriving the OLS Estimator	2020-12-21	next js math ols	false	How to derive the OLS Estimator with matrix notation and a tour of math typesetting using markdown with the help of KaTeX.

Introduction

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Deriving the OLS Estimator

Using matrix notation, let $n$ denote the number of observations and $k$ denote the number of regressors.

The vector of outcome variables $\mathbf{Y}$ is a $n \times 1$ matrix,

\mathbf{Y} = \left[\begin{array}
  {c}
  y_1 \\
  . \\
  . \\
  . \\
  y_n
\end{array}\right]

$$ \mathbf{Y} = \left[\begin{array} {c} y_1 \\ . \\ . \\ . \\ y_n \end{array}\right] $$

The matrix of regressors $\mathbf{X}$ is a $n \times k$ matrix (or each row is a $k \times 1$ vector),

\mathbf{X} = \left[\begin{array}
  {ccccc}
  x_{11} & . & . & . & x_{1k} \\
  . & . & . & . & .  \\
  . & . & . & . & .  \\
  . & . & . & . & .  \\
  x_{n1} & . & . & . & x_{nn}
\end{array}\right] =
\left[\begin{array}
  {c}
  \mathbf{x}'_1 \\
  . \\
  . \\
  . \\
  \mathbf{x}'_n
\end{array}\right]

$$ \mathbf{X} = \left[\begin{array} {ccccc} x_{11} & . & . & . & x_{1k} \\ . & . & . & . & . \\ . & . & . & . & . \\ . & . & . & . & . \\ x_{n1} & . & . & . & x_{nn} \end{array}\right] = \left[\begin{array} {c} \mathbf{x}'_1 \\ . \\ . \\ . \\ \mathbf{x}'_n \end{array}\right] $$

The vector of error terms $\mathbf{U}$ is also a $n \times 1$ matrix.

At times it might be easier to use vector notation. For consistency, I will use the bold small x to denote a vector and capital letters to denote a matrix. Single observations are denoted by the subscript.

Least Squares

Start:
$$y_i = \mathbf{x}'_i \beta + u_i$$

Assumptions:

1. Linearity (given above)
2. $E(\mathbf{U}|\mathbf{X}) = 0$ (conditional independence)
3. rank($\mathbf{X}$) = $k$ (no multi-collinearity i.e. full rank)
4. $Var(\mathbf{U}|\mathbf{X}) = \sigma^2 I_n$ (Homoskedascity)

Aim:
Find $\beta$ that minimises the sum of squared errors:

$$ Q = \sum_{i=1}^{n}{u_i^2} = \sum_{i=1}^{n}{(y_i - \mathbf{x}'_i\beta)^2} = (Y-X\beta)'(Y-X\beta) $$

Solution:
Hints: $Q$ is a $1 \times 1$ scalar, by symmetry $\frac{\partial b'Ab}{\partial b} = 2Ab$.

Take matrix derivative w.r.t $\beta$:

\begin{aligned}
  \min Q           & = \min_{\beta} \mathbf{Y}'\mathbf{Y} - 2\beta'\mathbf{X}'\mathbf{Y} +
  \beta'\mathbf{X}'\mathbf{X}\beta \\
                   & = \min_{\beta} - 2\beta'\mathbf{X}'\mathbf{Y} + \beta'\mathbf{X}'\mathbf{X}\beta \\
  \text{[FOC]}~~~0 & =  - 2\mathbf{X}'\mathbf{Y} + 2\mathbf{X}'\mathbf{X}\hat{\beta}                  \\
  \hat{\beta}      & = (\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}'\mathbf{Y}                              \\
                   & = (\sum^{n} \mathbf{x}_i \mathbf{x}'_i)^{-1} \sum^{n} \mathbf{x}_i y_i
\end{aligned}

$$ \begin{aligned} \min Q & = \min_{\beta} \mathbf{Y}'\mathbf{Y} - 2\beta'\mathbf{X}'\mathbf{Y} + \beta'\mathbf{X}'\mathbf{X}\beta \\ & = \min_{\beta} - 2\beta'\mathbf{X}'\mathbf{Y} + \beta'\mathbf{X}'\mathbf{X}\beta \\ \text{[FOC]}~~~0 & = - 2\mathbf{X}'\mathbf{Y} + 2\mathbf{X}'\mathbf{X}\hat{\beta} \\ \hat{\beta} & = (\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}'\mathbf{Y} \\ & = (\sum^{n} \mathbf{x}_i \mathbf{x}'_i)^{-1} \sum^{n} \mathbf{x}_i y_i \end{aligned} $$

Footnotes

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2. $10 and $20. ↩