1 September 2004 Statistical Inference in the Classical Linear Regression Model A. Introduction In this section, we will summarize the properties of estimators in the classical linear regression model previously developed, make additional distributional assumptions, and develop further properties associated with the added assumptions. Before presenting the results, it will be useful to summarize the structure of the model, and some of the algebraic and statistical results presented elsewhere. B. Statement of the classical linear regression model The classical linear regression model can be written in a variety of forms. Using summation notation we write it as yt = $1 + $2 xt2 + $3xt3 + ... + gt t
E(gt xt1, xt2, ...xtk) = 0 t
Var(gt xt1,...,xtk) = F2 t
E(gt gs) = 0 t
xti is a known constant
No xi is a linear combination of the other x's gt - N(0,F2)
We can also write it in matrix notation as follows
The ordinary least squares estimator of $ in the model is given by
The fitted value of y
and the estimated vectors of residuals ( e ) in the model are defined by
2 The variance of , (F2) is usually estimated using the estimated residuals as (4)
C. The fundamental matrices of linear regression 1. M - the residual creation matrix The residuals from the least squares regression can be expressed as
a. b. c. d. e.
The matrix MX is symmetric and idempotent. MX X = 0. e = MX , eNe = yNMX y. eNe = ,NMX ,.
2. P - the projection matrix Consider a representation of the predicted value of y
a. P is symmetric and idempotent. b. PX X = X c. PX MX = 0 3. An – The deviation transformation matrix Consider the matrix An below which transforms a vector or matrix to deviations from the mean.
An is symmetric and idempotent.
An MX = MX (first column of X is a column of ones) Proof: First write An in a different fashion noting that the vector of ones we called j, is the same as the first column of the X matrix in a regression with a constant term.
Now consider the product of An and MX (9)
From previous results, MX X = 0n×k, which implies that X`MX = 0k×n. This then implies that . Given that this product is a row of zeroes, it is clear that the entire second term vanishes. This then implies (10) D. Some results on traces of matrices The trace of a square matrix is the sum of the diagonal elements and is denoted tr A or tr (A). We will state without proof some properties of the trace operator. a. trace (In ) = n b. tr(kA) = k tr(A) c. trace (A + B) = trace (A) + trace (B)
d. tr(AB) = tr(BA) if both AB and BA are defined e. trace (ABC) = trace (CAB) = trace (BCA) The results in part e hold as along as the matrices involved are conformable, though the products may be different dimensions. We will also use Theorem 17 from the lecture on characteristic roots and vectors. A proof of this theorem is given there. Theorem 17: Let A be a square symmetric idempotent matrix of order n and rank r. Then the trace of A is equal to the rank of A, i.e., tr(A) = r(A). E. Some theorems on quadratic forms and normal variables (stated without proof) 1. Quadratic Form Theorem 1: If y ~N(:y, Ey ), then z = Cy ~ N( :z = C:y; Ez = C Ey CN ) where C is a matrix of constants. 2. Quadratic Form Theorem 2 Let the nx1 vector y~N(0, I ), then yNy - P2(n). 3. Quadratic Form Theorem 3: If y-N(0,F2I) and M is a symmetric idempotent matrix of rank m then (11) Corollary: If the nx1 vector y~N(0,I) and the nxn matrix A is idempotent and of rank m, then y'Ay - P2(m). 4. Quadratic Form Theorem 4: If y~N(0,F2I), M is a symmetric idempotent matrix of order n, and L is a kxn matrix, then Ly and yNMy are independently distributed if LM = 0. 5. Quadratic Form Theorem 5: Let the nx1 vector y~N(0,I), let A be an nxn idempotent matrix of rank m, let B be an nxn idempotent matrix of rank s, and suppose BA = 0. Then yNAy and yNBy are independently distributed P2 variables. 6. Quadratic Form Theorem 6 (Craig’s Theorem) If y~N(:, S) where S is positive definite, then q1 = yNAy and q2 = yNBy are independently distributed iff ASB = 0. 7. Quadratic Form Theorem 7 If y is a nx1 random variable and y-N(:,G) then (y - :)NG-1(y - :) P2(n). 8. Quadratic Form Theorem 8: Let y - N(0, I). Let M be a nonrandom idempotent matrix of dimension nxn (rank(M)=r # n). Let A be a nonrandom matrix such that AM = 0. Let t1 = My and let t2 = Ay. Then t1 and t2 are independent random vectors. F. Some finite sample properties of the ordinary least squares estimator in the classical linear regression model can be derived without specific assumptions about the exact distribution of the error term 1. Unbiasedness of Given the properties of the model, we can show that
is unbiased as follows if X is a nonstochastic
6 matrix of full rank
2. Variance of y We know that yt depends on the constants xt and $, and on the stochastic error, ,t. We write this as (13) This implies that (14) Furthermore with E(gt gs) = 0 t
s, i.e., the covariance between yt and yt+s is zero, implying that (15) 3. Variance of We can determine the variance of by writing it out and then using the information we have on the variance of y and the formula for the variance of any quadratic form.
is the best linear unbiased estimator of We can show that is the best linear unbiased estimator of by showing that any other linear unbiased estimator has a variance which is larger that the variance of by a positive definite matrix. The least squares estimator is given by (17) Consider another linear unbiased estimator given by form of . We can determine the restrictions on G for
. Linearity is imposed by the linear to be unbiased by writing it out as
The variance of
is similar to the variance of (19)
Now let D = G - C = G - (XNX)-1XN, so that G = D +C. Now rewrite the variance of
Now substitute in equation 20 for GX = Ik and XNGN = Ik.
The variance of
is thus the variance of
plus a matrix that can be shown to be positive definite.
5. Unbiasedness of s2 Given the properties of the model, we can show that s2 is an unbiased estimator of F2. First write eNe as a function of ,.
Now take the expected value of eNe, use the property of the trace operator that tr (ABC) = tr (BCA), and then simplify
We find the trace of M using the properties on sums, products, and identity matrices.
9 6. Covariance of
Given the properties of the model, we can show that the covariance of both and e as functions of , from equations 2 and 5.
and e is zero. First write
Remember that has an expected valued of expected valued of zero as follows
because it is unbiased. We can show that e has an
We then have (27)
Now compute the covariance directly
10 G. Distribution of
given normally distributed errors
1. introduction Now make the assumption that gt - N(0,F2) or
. Given that
then y is also distributed normally because we are simply adding a constant vector to the random vector ,. The error vector , is not transformed in forming y. Given E(,) = 0, E(y) = X$, and Var(y) = F2I, we then have (30) 2. exact distribution of We can write
as a linear function of the normal random variable y from equation 2 as follows (31)
We can find its distribution by applying Quadratic Form Theorem 1. From this theorem and . Substituting we obtain
Therefore we have (33) We can also show this by viewing
directly as a function of , and then applying the theorem.
H. Distribution of s2 Consider the quantity (35)
This can be written
The random variable (g/F) is a standard normal variable with mean zero and variance I. The matrix MX is symmetric and idempotent. By Theorem 3 on quadratic forms, this ratio is distributed as a P2 variable with (n-k) degrees of freedom, that is (37)
where we found the trace of MX in equation 24. Given that
, we can use information on the properties of chi-squared random
variables to find the variance of s2. First remember that the variance of a P2 variable is equal to twice its degrees of freedom, i.e., Var (P2 (