Linear Regression Models – Gaurav Sharma's Blog

Suppose that a scalar ${y_t}$ is related to a ${(k \times 1)}$ vector, ${x_t}$ and a disturbance term ${u_t}$ according to the regression model.

$\displaystyle y_t = x_t^T \beta + u_t \ \ \ \ \ (1)$

In this article, we will study the estimation and hypothesis testing of ${\beta}$ when ${x_t}$ is deterministic and ${u_t}$ is i.i.d. Gaussian.

1. The Algebra of Linear Regression

Given a sample of T values of ${y_t}$ and the vector ${x_t}$ , the ordinary least squares (OLS) estimate of ${\beta}$ , denoted as ${b}$ , is the value of ${\beta}$ which minimizes the residual sum of squares (RSS).

$\displaystyle RSS = \sum_{t=1}^T (y_t - x_t^T b)^2 \ \ \ \ \ (2)$

The OLS estimate of ${\beta}$ , b, is given by:

$\displaystyle b = \bigg(\frac{1}{T}\sum_{t=1}^T x_tx_t^T \bigg)^{\!-1} \!\!\cdot\, \frac{1}{T}\sum_{t=1}^n x_ty_t. \ \ \ \ \ (3)$

The model is written in matrix notation as:

$\displaystyle y = X\beta + u. \ \ \ \ \ (4)$

$\displaystyle \bold{y}= \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_T \end{bmatrix},\quad \bold{x}= \begin{bmatrix} x_1^T \\ x_2^T \\ \vdots \\ x_T^T \end{bmatrix},\quad \bold{u}= \begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_T \end{bmatrix}. \ \ \ \ \ (5)$

where ${y}$ is a ${T \times 1}$ vector, ${X}$ is a ${T \times k}$ matrix, ${\beta}$ is a ${k \times 1}$ vector and ${u}$ is a ${T \times 1}$ vector.

Thus,

$\displaystyle b = (X^TX)^{-1}X^Ty. \ \ \ \ \ (6)$

Similarly,

$\displaystyle \hat u = y - Xb = y - X(X^TX)^{-1}X^Ty = [I_T - X(X^TX)^{-1}X^T]y = M_Xy. \ \ \ \ \ (7)$

where ${M_X}$ is defined as:

$\displaystyle M_X = [I_T - X(X^TX)^{-1}X^T]. \ \ \ \ \ (8)$

${M_X}$ is a projection matrix. Hence it is symmetric and idempotent.

$\displaystyle M_X = M_X^T. \ \ \ \ \ (9)$

$\displaystyle M_XM_X = M_X. \ \ \ \ \ (10)$

Since ${M_X}$ is the projection matrix for the space orthogonal to ${X}$ ,

$\displaystyle M_X^TX = M_XX = 0. \ \ \ \ \ (11)$

Thus, we can verify that the sample residuals are orthogonal to ${X}$ .

$\displaystyle u^TX = y^TM_X^TX = 0. \ \ \ \ \ (12)$

The sample residual is constructed from the sample estimate of ${\beta}$ which is ${b}$ . The population residual is a hypothetical construct based on the true population value of ${\beta}$ .

$\displaystyle u_t = y_t - x_t^T \beta. \ \ \ \ \ (13)$

$\displaystyle \hat u_t = y_t - x_t^T b. \ \ \ \ \ (14)$

$\displaystyle \hat u = y - Xb = [I_T - X(X^TX)^{-1}X^T]y = M_Xy = M_XX b + u. \ \ \ \ \ (15)$

$\displaystyle b = (X^TX)^{-1}X^Ty = (X^TX)^{-1}X^T(X\beta + u) = \beta + (X^TX)^{-1}X^Tu. \ \ \ \ \ (16)$

The fit of OLS is described in terms of ${R_u^2}$ , which is defined as the ratio of squares of the fitted values ( ${x_t^Tb)}$ to the observed values of ${y}$ .

$\displaystyle R_u^2 = \frac{\sum_{t=1}^T b^Tx_tx_t^Tb}{\sum_{t=1}^T y_t^2} = \frac{b^TX^TXb}{y^Ty}. \ \ \ \ \ (17)$

2. Assumptions on X and u

We shall assume that

(a) X will be deterministic

(b) ${u_t}$ is i.i.d with mean 0 and variance ${\sigma^2}$ .

2.1. Properties of Estimated b Under Above Assumptions

Since,

$\displaystyle b = (X^TX)^{-1}X^Ty = (X^TX)^{-1}X^T(X\beta + u) = \beta + (X^TX)^{-1}X^Tu. \ \ \ \ \ (18)$

Taking expectations of both sides, we have,

$\displaystyle \mathop{\mathbb E}(b) = \beta + (X^TX)^{-1}X^T\mathop{\mathbb E}(u) = \beta. \ \ \ \ \ (19)$

And the variance covariance matrix is given by,

$\displaystyle \mathop{\mathbb E}[(b - \beta)(b - \beta)^T] = \mathop{\mathbb E}[((X^TX)^{-1}X^Tu)((X^TX)^{-1}X^Tu)^T] = \sigma^2(X^TX)^{-1}. \ \ \ \ \ (20)$

Thus b is unbiased and is a linear function of y.

2.2. Distribution of Estimated b Under Above Assumptions

As u is Gaussian,

$\displaystyle b = \beta + (X^TX)^{-1}X^Tu. \ \ \ \ \ (21)$

implies that b is also Gaussian.

$\displaystyle b \sim N(\beta, \sigma^2(X^TX)^{-1}). \ \ \ \ \ (22)$

2.3. Properties of Estimated Sample Variance Under Above Assumptions

The OLS estimate of variance of u, ${\sigma^2}$ is given by:

$\displaystyle s^2 = RSS / (T - k) = {\hat u}^T\hat u / (T - k) = u^TM_X^TM_Xu / (T - k) = u^TM_Xu / (T - k). \ \ \ \ \ (23)$

Since ${M_X}$ is a projection matrix and is symmetric and idempotent, it can be written as:

$\displaystyle M_X = P\Lambda P^T. \ \ \ \ \ (24)$

where

$\displaystyle P P^T = I_T. \ \ \ \ \ (25)$

and ${\Lambda}$ is a diagonal matrix with eigenvalues of ${M_X}$ on the diagonal.

Since,

$\displaystyle M_XX = 0. \ \ \ \ \ (26)$

that is, since the two spaces that they represent are orthogonal to each other, it follows that:

$\displaystyle M_Xv = 0. \ \ \ \ \ (27)$

whenever v is a column of X. Since we assume X to be of full rank, there are k such vectors and their eigenvalue is the right hand side, which is 0.

Also since

$\displaystyle M_X = I_T - X(X^TX)^{-1}X^T. \ \ \ \ \ (28)$

Thus, it follows that

$\displaystyle M_Xv = v. \ \ \ \ \ (29)$

whenever v is orthogonal to X. Since there are (T – k) such vectors, ${M_X}$ has (T – k) eigenvectors with eigenvalue 1.

Thus ${\Lambda}$ has k zeroes and (T – k) 1s on the diagonal.

$\displaystyle u^TM_Xu = u^TP\Lambda P^Tu. \ \ \ \ \ (30)$

Let

$\displaystyle w = P^Tu \ \ \ \ \ (31)$

Then,

$\displaystyle u^TM_Xu = u^TP\Lambda P^Tu = w^T\Lambda w = w_1^2 \lambda_1 + w_2^2 \lambda_2 + \dots + w_T^2 \lambda_T. \ \ \ \ \ (32)$

$\displaystyle u^TM_Xu = w_1^2 \lambda_1 + w_2^2 \lambda_2 + \dots + w_{T - k}^2 \lambda_{T - k}. \ \ \ \ \ (33)$

As these ${\lambda}$ s are all unity, we have:

$\displaystyle u^TM_Xu = w_1^2 + w_2^2 + \dots + w_{T - k}^2 . \ \ \ \ \ (34)$

Also,

$\displaystyle \mathop{\mathbb E}(w^Tw) = \mathop{\mathbb E}(P^Tu u^T P) = \sigma^2I_T. \ \ \ \ \ (35)$

Thus, elements of w are uncorrelated with each other, have mean 0 and variance ${\sigma^2}$ .

Since each w has expectation of ${\sigma^2}$ ,

$\displaystyle \mathop{\mathbb E}(u^TM_Xu) = (T - k)\sigma^2. \ \ \ \ \ (36)$

Hence,

$\displaystyle \mathop{\mathbb E}(s^2) = \sigma^2. \ \ \ \ \ (37)$

2.4. Distribution of Estimated Sample Variance Under Above Assumptions

Since

$\displaystyle w = P^Tu \ \ \ \ \ (38)$

when u is Gaussian, w is also Gaussian.

Then,

$\displaystyle u^TM_Xu = w_1^2 \lambda_1 + w_2^2 \lambda_2 + \dots + w_{T - k}^2 \lambda_{T - k}. \ \ \ \ \ (39)$

implies that ${u^TM_Xu}$ is the sum of squares of (T – k) independent ${N(0, \sigma^2)}$ random variables.

Thus,

$\displaystyle RSS^2 / \sigma^2 = u^TM_Xu / \sigma^2 \sim \chi^2 ( T - k). \ \ \ \ \ (40)$

Also, b and ${\hat u}$ are uncorrelated, since,

$\displaystyle \mathop{\mathbb E}[\hat u(b - \beta)^T] = \mathop{\mathbb E}[M_Xu u^T X (X^TX)^{-1} = 0. \ \ \ \ \ (41)$

Since b and ${\hat u}$ are independent, b and ${s^2}$ are also independent.

2.5. t Tests about ${\beta}$ Under Above Assumptions

We wish to test the hypothesis that the ith element of ${\beta}$ , ${\beta_i}$ , is some particular value ${\beta_i^0}$ .

The t-statistic for testing this null hypothesis is

$\displaystyle t = \frac{b_i - \beta_i^0}{\hat \sigma_{b_i}} = \frac{b_i - \beta_i^0}{s (\xi^{ii})^2} \ \ \ \ \ (42)$

where ${ \xi^{ii}}$ denotes the ith column and ith row element of ${(X^TX)^{-1}}$ and ${\hat \sigma_{b_i}}$ is the standard error of the OLS estimate of the ith coefficient.

Under the null hypothesis,

$\displaystyle b_i \sim N(\beta_i^0, \sigma^2 \xi^{ii}). \ \ \ \ \ (43)$

Thus,

$\displaystyle \frac{b_i - \beta_i^0}{\sqrt{\sigma^2 \xi^{ii}}} \sim N(0, 1). \ \ \ \ \ (44)$

Thus,

$\displaystyle t = \frac{{(b_i - \beta_i^0)} / {\sqrt{\sigma^2 \xi^{ii}}}}{\sqrt{s^2 / \sigma^2 }} \ \ \ \ \ (45)$

Thus the numerator is N(0, 1) and the denominator is the square root of a chi-square distribution with (T – k) degrees of freedom. This gives a t-distribution to the variable on the left side.

2.6. F Tests about ${\beta}$ Under Above Assumptions

To generalize what we did for t tests, consider that we have a matrix ${R}$ that represents the restrictions we want to impose on ${\beta}$ , that is ${R\beta}$ gives a vector of the hypothesis that we want to test. Thus,

$\displaystyle H_0 \colon R\beta = r \ \ \ \ \ (46)$

Since,

$\displaystyle b \sim N(\beta, \sigma^2(X^TX)^{-1}). \ \ \ \ \ (47)$

Thus, under ${H_0}$ ,

$\displaystyle Rb \sim N(r, \sigma^2R(X^TX)^{-1}R^T). \ \ \ \ \ (48)$

Theorem 1 If ${z}$ is a ${(n \times 1)}$ vector with ${z \sim N(0, \Sigma^2)}$ and non singular ${\Sigma}$ , then ${z^T\Sigma^{-1} z \sim \chi^2(n)}$ .

Applying the above theorem to the ${Rb - r}$ vector, we have,

$\displaystyle (Rb - r)^T (\sigma^2R(X^TX)^{-1}R^T)^{-1}(Rb - r) \sim \chi^2 (m). \ \ \ \ \ (49)$

Now consider,

$\displaystyle F = (Rb - r)^T (s^2R(X^TX)^{-1}R^T)^{-1}(Rb - r) / m. \ \ \ \ \ (50)$

where sigma has been replaced with the sample estimate s.

Thus,

$\displaystyle F = \frac{[(Rb - r)^T (\sigma^2R(X^TX)^{-1}R^T)^{-1}(Rb - r)] / m}{[RSS / (T - k)]/ \sigma^2} \ \ \ \ \ (51)$

In the above, the numerator is a ${\chi^2(m)}$ distribution divided by its degree of freedom and the denominator is a ${\chi^2(T - k)}$ distribution divided by its degree of freedom. Since b and ${\hat u}$ are independent, the numerator and denominator are also independent of each other.

Hence, the variable on the left hand side has an exact ${F(m , T - k)}$ distribution under ${H_0}$ .

Category: Linear Regression Models

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