1) Suppose that b is the least squares coecient vector of the regression of y on

Question

1) Suppose that b is the least squares coecient vector of the regression of y on X and that c is any other K 1 vector. Prove that the dierence between the two sums of squared residuals can be written as:

(y - Xc)'(y - Xc) - (y - Xb)'(y - Xb) = (c - b)'X'X(c - b)

Prove also that this diference is positive.

2) Observations with missing data for at least one explanatory variable are common in applied econometrics, but there is no good way of dealing with them. Two possible strategies are (i) to drop the observations with missing data; and (ii) to replace the missing values with the mean of the available values.
Suppose that your original regression model includes only a constant and one explanatory variable, x, on the right-hand side of the equation.

(a) Derive expressions for the OLS estimators of the constant and the coecient of x in case (i).
(b) Derive expressions for the OLS estimators of the constant and the coecient of x in case (ii).
(c) Do strategies (i) and (ii) yield the same results for the two coecients?

Explanation / Answer

solution(1):

First we have to Prove that:

the difference in the two sums of squared residuals is

(y Xc)’ (y Xc) (y Xb)’ (y Xb) = (c b)’X’X(c b).

Write c as b+(c-b).

Then,

the sum of the squared residuals based on c is

(y Xc)’ (y Xc) = [y X(b + (c b))]’ [y X(b + (c b))]

= [(y Xb) + X(c b)]’ [(y Xb) + X(c b)]

= (y Xb)’ (y Xb) + (c b)’X’X(c b) + 2(c b)’X’ (y Xb).

But,

the third term is zero,

as 2(c b)’X’ (y Xb) = 2(c b)’X’ (y Xb) = 0.

Therefore,

(y Xc)’ (y Xc) = (y Xb)’ (y Xb) + (c b)’X’X(c b)

(y Xc)’ (y Xc) (y Xb)’ (y Xb)= (c b)’X’X(c b). …(1)

Hence proved.

Second we have to prove that:

(y Xc)’ (y Xc) (y Xb)’ (y Xb) this difference is positive

Let d= X(c b)

Therefore d’= X’(c b)’

This put in equation (1), we get

(y Xc)’ (y Xc) (y Xb)’ (y Xb)=d’d

so it is necessaril y positive.

This confirms what we knew at the outset, least squares is least squares.

That is,

(y Xc)’ (y Xc) (y Xb)’ (y Xb) this difference is positive

Hence proved.

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