Intro to Econometric Theory Heinz School, Carnegie Mellon University 90-906, Spring 2005

Solution #2

1. You may recall that, in the bivariate regression model,

Y = β_{1} + β_{2}X_{2} + ϵ  (1)

the OLS estimate of β_{2} is β̂_{2,OLS} = \frac{Ĉov(Y,X_{2})}{V̂ (X_{2})}. This means that

if the correlation (or covariance) between Y and X_{2} is positive, then β̂_{2} is positive.

Consider now a slightly more complex model:

Y = β_{1}X_{1} + β_{2}X_{2} + ϵ  (2)

Let’s suppose that \overline{Y} = \overline{X_{1}} = \overline{X_{2}} = 0. Please calculate an expression for β̂_{1,OLS} and β̂_{2,OLS} in terms of the variances and covariances of Y, X_{1}, X_{2}. Using your formula, discuss whether or not it can happen that Ĉov(Y, X_{1}) > 0 while β̂_{1,OLS} < 0. If it cannot, why not? If it can, then how? Your goal is both to construct a correct argument and to explain in intuitive English why your result holds.

The formula for OLS says:

( β̂_{1,OLS} ; β̂_{2,OLS} ) = [ \begin{matrix}∑ X_{1}^{2} & ∑ X_{1}X_{2} \\ ∑ X_{2}X_{1} & ∑ X_{2}^{2} \end{matrix} ]^{−1} ( \begin{matrix}∑ Y X_{1} \\ ∑ Y X_{2} \end{matrix} )

= [ \begin{matrix}V̂ (X_{1}) & Ĉov(X_{1}, X_{2}) \\ Ĉov(X_{1}, X_{2}) & V̂ (X_{2}) \end{matrix} ]^{−1} ( \begin{matrix}Ĉov(Y, X_{1}) \\ Ĉov(Y, X_{2}) \end{matrix} )

= \frac{1}{V̂ (X_{1})V̂ (X_{2}) − Ĉov(X_{1}, X_{2})^{2}} ∗

[ \begin{matrix}V̂ (X_{2}) & −Ĉov(X_{1}, X_{2}) \\ −Ĉov(X_{1}, X_{2}) & V̂ (X_{1}) \end{matrix} ] ( \begin{matrix}Ĉov(Y, X_{1}) \\ Ĉov(Y, X_{2}) \end{matrix} )

= \frac{1}{V̂ (X_{1})V̂ (X_{2}) − Ĉov(X_{1}, X_{2})^{2}} ∗

( \begin{matrix}Ĉov(Y, X_{1})V̂ (X_{2}) − Ĉov(Y, X_{2})Ĉov(X_{1}, X_{2}) \\ Ĉov(Y, X_{2})V̂ (X_{1}) − Ĉov(Y, X_{1})Ĉov(X_{1}, X_{2}) \end{matrix} )

Between the first and second equals signs I multiply the (X′X)^{−1} matrix by N and the X′Y matrix by 1/N.

Since X′X is positive definite, its determinant (the fraction above) is positive; this result is in the book, pg 47. So, β̂_{1,OLS} will have the same sign as Ĉov(Y, X_{1})V̂ (X_{2}) − Ĉov(Y, X_{2})Ĉov(X_{1}, X_{2}). By looking at the first term, we can see that one factor tending to make β̂_{1,OLS} positive would be Ĉov(Y, X_{1}) positive, since V̂ (X_{2}) must be positive.

However, it is possible for this factor to be overturned by the second factor in the sum, −Ĉov(Y, X_{2})Ĉov(X_{1}, X_{2}). If, for example, the covariance between Y and X_{2} is also positive and the covariance between X_{1} and X_{2} is positive then it could happen that β̂_{1,OLS} is negative.

So, β̂_{1,OLS} could be negative even though the covariance between Y and X_{1} is positive if there is a strong positive relationship between Y and X_{2} and between X_{1} and X_{2}. This would happen if all three of Y, X_{1}, X_{2} moved together, obviously. This might happen, for example, if increases in X_{2} “drive” both increases in Y and X_{1} but X_{1} itself has a negative influence on Y.

Consider the following example. The covariance (across countries or over time) between life expectancy and the consumption of red meat per capita is positive. Thinking of life expectancy as Y and meat/capita as X_{1} and income as X_{2}, you can see how this kind of situation can come about (assuming for purposes of argument we believe that red meat kills you prematurely). People often call a variable playing the role of X_{2} here a “confounder” because it hides the true relationship between Y and X_{1}.

2. Do problem 9 on page 115 of Greene

The easiest method here is the LR-like method. We will compare the models on pages 2 and 3 of the attached output using the F-statistic:

F − stat = \frac{(SSE_{R} − SSE_{UR})/rank(R)}{SSE_{UR}/(N − K)}
= \frac{(0.15888 − 0.03384)/3}{0.03384/(36 − 5)}
= 38.2

We compare this to the F table with 3 numerator and 31 denominator degrees of freedom. For a significance level of 1%, the table value is about 4.5. So, we reject the null. Prices do affect consumption of gas — hooray for the law of demand.

3. Do problem 10 on page 115 of Greene — in class I said that you need only test the linear restriction γ_{n}c = γ_{u}c

Again, the easiest way to do this is with an LR-like test, so we compare the regressions on pages 4 and 6. Think about how I have imposed the parameter restriction in this model — can this be done for any linear restriction?

Anyway, comparing these models using an F statistic gives:

F − stat = \frac{(SSE_{R} − SSE_{UR})/rank(R)}{SSE_{UR}/(N − K)}
= \frac{(0.0889 − 0.00856)/1}{0.00856/(36 − 10)}
= 1.00

Looking at the F table with 1 numerator and 26 denominator degrees of freedom with a 5% significance level yields a table value of about 7.77. So, we accept the null — there is not enough evidence in the data to reject this null hypothesis.

We could also pretty easily have done this test just using the regression on page 4-5 using a t statistic:

t − stat = \frac{γ̂_{nc} − γ̂_{uc}}{\sqrt{V̂ (γ̂_{nc} − γ̂_{uc})}}

= \frac{0.08313 − (−0.11499)}{\sqrt{0.04301 + 0.0045 − 2(0.0038)}}
= \frac{0.19812}{.200}
= 0.99

Obviously, a t-stat this small will not result in a hypothesis rejection. Notice also that this t-stat is the square root of the F-stat, as theory demands.