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The Definition of Regression Coefficients and Derivation of Estimator Formulas 📂Statistical Analysis

The Definition of Regression Coefficients and Derivation of Estimator Formulas

Definition 1

Y=β0+β1X1+⋯+βpXp+ε Y = \beta_{0} + \beta_{1} X_{1} + \cdots + \beta_{p} X_{p} + \varepsilon In multiple regression analysis, when establishing the above linear model for given pp independent variables X1,⋯ ,XpX_{1} , \cdots , X_{p}, β0,β1,⋯ ,βp\beta_{0} , \beta_{1} , \cdots , \beta_{p} is called the regression coefficient. YY represents the dependent variable, and ε\varepsilon indicates randomly distributed errors.

Formula

[y1y2⋮yn]=[1x11⋯xp11x12⋯xp2⋮⋮⋱⋮1x1n⋯xpn][β0β1⋮βp]+[ε1ε2⋮εn] \begin{bmatrix} y_{1} \\ y_{2} \\ \vdots \\ y_{n} \end{bmatrix} = \begin{bmatrix} 1 & x_{11} & \cdots & x_{p1} \\ 1 & x_{12} & \cdots & x_{p2} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_{1n} & \cdots & x_{pn} \end{bmatrix} \begin{bmatrix} \beta_{0} \\ \beta_{1} \\ \vdots \\ \beta_{p} \end{bmatrix} + \begin{bmatrix} \varepsilon_{1} \\ \varepsilon_{2} \\ \vdots \\ \varepsilon_{n} \end{bmatrix} Given nn data points and denoting it as p<np < n, the linear multiple regression model can be represented in a design matrix as above, and we simply denote it as Y=Xβ+εY = X \beta + \varepsilon. The least squares estimator vector β^\hat{\beta} for β\beta is as follows. β^=[β^0β^1⋮β^p]=(XTX)−1XTY \hat{\beta} = \begin{bmatrix} \hat{\beta}_{0} \\ \hat{\beta}_{1} \\ \vdots \\ \hat{\beta}_{p} \end{bmatrix} = \left( X^{T} X \right)^{-1} X^{T} Y Moreover, since β^\hat{\beta} is the best unbiased estimator of β\beta, it is also known as the Best Linear Unbiased Estimator.

Derivation 2 3

Our goal is ∥ε∥22=∑k=0nεk=[ε0ε1⋯εn][ε0ε1⋮εn]=εTε \left\| \varepsilon \right\|_{2}^{2} = \sum_{k=0}^{n} \varepsilon_{k} = \begin{bmatrix} \varepsilon_{0} & \varepsilon_{1} & \cdots & \varepsilon_{n} \end{bmatrix} \begin{bmatrix} \varepsilon_{0} \\ \varepsilon_{1} \\ \vdots \\ \varepsilon_{n} \end{bmatrix} = \varepsilon^{T} \varepsilon to minimize. Therefore, finding a β\beta that minimizes εTε=(Y−Xβ)T(Y−Xβ)\varepsilon^{T} \varepsilon = \left( Y - X \beta \right)^{T} \left( Y - X \beta \right) follows from ε=Y−Xβ\varepsilon = Y - X \beta.

Gradient of residual sum of squares: f(s):=(y−Xs)TR(y−Xs) f \left( \mathbf{s} \right) := \left( \mathbf{y} - X \mathbf{s} \right)^{T} R \left( \mathbf{y} - X \mathbf{s} \right) Assume that RR is an identity matrix, then the following is obtained. ∂f(s)∂s=−2XT(y−Xs) {{ \partial f \left( \mathbf{s} \right) } \over { \partial \mathbf{s} }} = - 2 X^{T} \left( \mathbf{y} - X \mathbf{s} \right)

By differentiating both sides concerning β\beta ∂∂βεTε=−2XT(Y−Xβ)=−2XT(Y−Xβ)=−2XTY+2XTXβ \begin{align*} {{ \partial } \over { \partial \beta }} \varepsilon^{T} \varepsilon =& - 2 X^{T} \left( Y - X \beta \right) \\ = & - 2 X^{T} \left( Y - X \beta \right) \\ = & - 2 X^{T} Y + 2 X^{T} X \beta \end{align*} results in a zero vector 0\mathbf{0} for β^\hat{\beta} like this. β^=arg min⁡βεTε=(XTX)−1XTY \hat{\beta} = \argmin_{\beta} \varepsilon^{T} \varepsilon = \left( X^{T} X \right)^{-1} X^{T} Y Meanwhile, it can be easily shown that β^\hat{\beta} is an unbiased estimator for β\beta, and since it is derived via the least squares method, there is no unbiased estimator with smaller variance for β\beta, making it the best unbiased estimator.

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If differentiating concerning β\beta in the derivation is not particularly appealing, an alternative approach using matrix algebra is available. In least squares method in matrix algebra, X∗Y=X∗Xβ^ X^{\ast} Y = X^{\ast} X \hat{\beta} the fact that β^\hat{\beta} satisfies being the least squares solution shows that since X∈Rn×pX \in \mathbb{R}^{n \times p} holds, X∗=XTX^{\ast} = X^{T} follows, and consequently, β^=(XTX)−1XTY\hat{\beta} = \left( X^{T} X \right)^{-1} X^{T} Y is derived.

Corollary

If β^\hat{\beta} is the best linear unbiased estimator, then the sum of yky_{k} and the sum of fits y^k=1=β^0+∑j=1pβ^jxj\hat{y}_{k=1} = \hat{\beta}_{0} + \sum_{j=1}^{p} \hat{\beta}_{j} x_{j} are equal: ∑k=1nyk=∑k=1ny^k \sum_{k=1}^{n} y_{k} = \sum_{k=1}^{n} \hat{y}_{k}

Proof

In proving this formula, that β^\hat{\beta} is the best linear unbiased estimator means 0=−2XTY+2XTXβ^  ⟹  0=XT(Y−Xβ^)  ⟹  0=XT[y1−y^1⋮yn−y^n] \begin{align*} & \mathbf{0} = - 2 X^{T} Y + 2 X^{T} X \hat{\beta} \\ \implies & \mathbf{0} = X^{T} \left( Y - X \hat{\beta} \right) \\ \implies & \mathbf{0} = X^{T} \begin{bmatrix} y_{1} - \hat{y}_{1} \\ \vdots \\ y_{n} - \hat{y}_{n} \end{bmatrix} \end{align*} that this holds. Since XX is the design matrix, the first row of XTX^{T} can be regarded as all elements being 11 in a row matrix of ones. Considering only the product of the first row of XTX^{T} with Y−Xβ^Y - X \hat{\beta} yields the following. 0=[1⋯1][y1−y^1⋮yn−y^n]  ⟹  0=(y1−y^1)+⋯+(yn−y^n)  ⟹  0=∑k=1nyk−∑k=1ny^k \begin{align*} & 0 = \begin{bmatrix} 1 & \cdots & 1 \end{bmatrix} \begin{bmatrix} y_{1} - \hat{y}_{1} \\ \vdots \\ y_{n} - \hat{y}_{n} \end{bmatrix} \\ \implies & 0 = \left( y_{1} - \hat{y}_{1} \right) + \cdots + \left( y_{n} - \hat{y}_{n} \right) \\ \implies & 0 = \sum_{k=1}^{n} y_{k} - \sum_{k=1}^{n} \hat{y}_{k} \end{align*} Consequently, the following is obtained. ∑k=1nyk=∑k=1ny^k \sum_{k=1}^{n} y_{k} = \sum_{k=1}^{n} \hat{y}_{k}

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See Also

  1. Hadi. (2006). Regression Analysis by Example(4th Edition): p53. ↩︎

  2. Hadi. (2006). Regression Analysis by Example(4th Edition): p82~84. ↩︎

  3. https://www.stat.purdue.edu/~boli/stat512/lectures/topic3.pdf ↩︎