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Deriving the OLS Estimator

Introduction

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Deriving the OLS Estimator

Using matrix notation, let nn denote the number of observations and kk denote the number of regressors.

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

\mathbf{Y} = \left[\begin{array}
    {c}
    y_1 \\
    . \\
    . \\
    . \\
    y_n
\end{array}\right]
Y=[y1...yn]\mathbf{Y} = \left[\begin{array} {c} y_1 \\ . \\ . \\ . \\ y_n \end{array}\right]

The matrix of regressors X\mathbf{X} is a n×kn \times k matrix (or each row is a k×1k \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]
X=[x11...x1k...............xn1...xnn]=[x1...xn]\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 U\mathbf{U} is also a n×1n \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: yi=xiβ+uiy_i = \mathbf{x}'_i \beta + u_i

Assumptions:

  1. Linearity (given above)
  2. E(UX)=0E(\mathbf{U}|\mathbf{X}) = 0 (conditional independence)
  3. rank(X\mathbf{X}) = kk (no multi-collinearity i.e. full rank)
  4. Var(UX)=σ2InVar(\mathbf{U}|\mathbf{X}) = \sigma^2 I_n (Homoskedascity)

Aim: Find β\beta that minimises sum of squared errors:

Q=i=1nui2=i=1n(yixiβ)2=(YXβ)(YXβ)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: QQ is a 1×11 \times 1 scalar, by symmetry bAbb=2Ab\frac{\partial b'Ab}{\partial b} = 2Ab.

Take matrix derivative w.r.t β\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}
minQ=minβYY2βXY+βXXβ=minβ2βXY+βXXβ[FOC]   0=2XY+2XXβ^β^=(XX)1XY=(nxixi)1nxiyi\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}

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