Intro to Econometric Theory Heinz School, Carnegie Mellon University 90-906, Spring 2004-5

Solution #1

1. Consider a sample of people who were asked to report their sex (Everyone answers, and everyone answers either male or female). From their answers, we construct the following matrix:

X = [ 1 M F ]  (1)

In the matrix, each row corresponds to one person’s response; 1 is a column of 1s, M is a column with a 1 in the ith row if the ith person answered male and a zero otherwise, and F is a column with a 1 in the ith row if person i answered female and a zero otherwise.

Is X′X invertible. Why or why not?

No, X′X is not invertible. Notice that M + F = 1. This is a linear dependency in the columns of X. Thus, the rank of X must be less than three. Since the rank of X′X (a 3 × 3 matrix) is the same as the rank of X, its rank must be less than 3. Thus, it is not full rank. Thus it is not invertible.

2. We discussed in class that variance matrixes are always positive semi-definite but not always positive definite. In the case of scalar random variables, a singular variance matrix means a zero variance — that is, that the random variable is not really random at all, but fixed.

Now, let’s think about what a positive semi-definite but not positive definite (hereafter PSDNPD) variance matrix means for the random vector it corresponds to. Let x be a random vector with PSDNPD variance matrix V.

(a) Must there be a fixed non-zero vector α, such that α′V α=0? Why or why not? Yes. If there were no such vector, then V would be positive definite, and we are told that it is not.

(b) Suppose there is such an α, what would the variance of α′x be? Zero. V (α′x) = α′V α = 0.

(c) What does this tell us about α′x?

It is fixed, non-random.

(d) Does this mean that α′x = 0? No, it could just as easily be 8 or 0.73 or any fixed number. The variance of 0 is zero, but the variance of 8 or 0.73 are is zero.

(e) So, what does it mean for a random vector to have a PSD-NPD variance matrix? When a random vector has a PSDNDP variance matrix, this means that there are, in some sense, fewer random variables in the vector than appear to be. Consider a scalar random variable z and a random vector:

x = ( \begin{matrix}z \\ −z \end{matrix} )

As you can see, although x has two elements it does not really contain two random variables, just z. The variance matrix of x is:

V (x) = [ \begin{matrix}σ_{z}^{2} & −σ_{z}^{2} \\ −σ_{z}^{2} & σ_{z}^{2} \end{matrix} ]

This matrix is singular, and:

( 1 1 ) [ \begin{matrix}σ_{z}^{2} & −σ_{z}^{2} \\ −σ_{z}^{2} & σ_{z}^{2} \end{matrix} ] ( \begin{matrix}1 \\ 1 \end{matrix} ) = 0

and: V ((1 1)x) = 0 Another way of saying the same thing is that if V is PSDNPD, then you can always write one of the elements of x as a linear function of the other elements plus a fixed number (this follows from V (α′x) = 0).

3. Suppose we have an estimator of per-capita income in the US, Î, and an estimator of the share of income saved (NOT the average over people of share of income saved but rather the average over dollars of share of income saved!), ŝ. The true values of these parameters are μ_{I} and μ_{s}. These two estimators are derived somehow from a survey of size n, and we have proved the following result for these estimators:

\sqrt{n} (( Î ; ŝ ) − ( \begin{matrix}μ_{I} \\ μ_{s} \end{matrix} )) ∼^{A}N (( \begin{matrix}0 \\ 0 \end{matrix} ) , [ \begin{matrix}400 & 20 \\ 20 & 9 \end{matrix} ])  (2)

Please construct an approximate (why approximate?) 95% confidence interval for the per-capita savings, μ_{I}μ_{s}.

Approximate because we are using asymptotic (or approximate) distributions, not the true distributions of the estimators to construct the CI — this is true even though the problem I gave did not contain a statement about asymptotic distribution since the theorem we will use has only a conclusion for asymptotic distribution. The CI is right if the sample is very large, and (we hope!) pretty close in sample sizes we are likely to encounter.

Anyway, we have a result from class which says:

If \sqrt{n} (x − μ_{x}) ∼^{A}N(0, Σ) then \sqrt{n} (g(x) − g(μ_{x})) ∼^{A}N (0, ▽′_{g}Σ▽_{g}) In this

formula, let’s set x = ( Î ; ŝ ) and g(x) = x_{1}x_{2}. That means that ▽_{g} =

( \begin{matrix}x_{2} \\ x_{1} \end{matrix} ) or ▽_{g} = ( ŝ ; Î )

Now, applying the theorem, we get:

\sqrt{n} (( Î ; ŝ ) − (( μ_{I}μ_{s} ))) = \sqrt{n} ((g(x)) − (g(μ)))
∼^{A} N (0, ( ŝ Î ) [ \begin{matrix}400 & 20 \\ 20 & 9 \end{matrix} ] ( ŝ ; Î ))
= N (0, 400ŝ^{2} + 40ŝÎ+ 9Î^{2})

Let’s call the rather messy quadratic V. Had we been given the values of Î and ŝ, we could have calculated V. Then, our 95% confidence interval would be:

Îŝ ± 1.96V

4. You are interested in the effect of going to the emergency room on a person’s life expectancy. There is a sample of people collected on a particular day, some of whom went to an emergency room and some of whom did not. They are followed over time until they die. From each person there are two items in the dataset: T_{i} the length of time from the day of the survey until death and ER_{i} a variable which equals one if they went to the emergency room and zero if they did not. You wish to run a regression, but the people who conducted the study will not let you have the data, only summary statistics. What should you ask for and how will you use it?

I can run the regression using only \overline{T}, \overline{ER}, σ̂_{T}^{2}, σ̂_{ER}^{2}, and Ĉov(T, ER). Actually, if I only want to regress T on ER, I could do without σ̂_{ER}^{2}. Consider the regression model T = β_{1} + β_{2}ER + ϵ. I will use the sample statistics to calculate the regression coefficients as follows:

β̂_{2,OLS} = \frac{Ĉov(T, ER)}{σ̂_{ER}^{2}}

(3)

β̂_{1,OLS} = \overline{T} − β̂_{2,OLS}\overline{ER}

5. Considering the example in problem 4, consider two linear models:

(a) T_{i} = β_{1} + β_{2}ER_{i} + ϵ_{i}

(b) ER_{i} = β_{3} + β_{4}T_{i} + ϵ_{i}

Let’s call the OLS estimates you would get from regressions run on these equations β̂_{1,OLS}, β̂_{2,OLS}, . . .. What relationship will there be between β̂_{2,OLS}, β̂_{4,OLS}? What sign (±) do you expect for β̂_{2,OLS}, β̂_{4,OLS}?

Just by examining the formula for OLS estimates, we can see that the relationship will be:

β̂_{2,OLS} = β̂_{4,OLS} \frac{σ̂_{T}^{2}}{σ̂_{ER}^{2}}  (4)

Since people who go to the emergency room are probably sick, I expect Cov(T, ER) < 0 and therefore that both β̂_{2,OLS} < 0 and β̂_{4,OLS} < 0.

6. We usually interpret regression models as embodying some causal story (the RHS causing the LHS); however, OLS does not “know” this: it produces estimates regardless of whether our implicit causal story is right. Discuss this fact in the context of the models in question 5 and your expectations about the signs of the relevant coefficients.

Well, the model in question 5a would produce the result that emergency rooms kill people if we were to take its implicit causal story seriously; whereas, the model in question 5b would produce the result that sick people go to the emergency room if we were to take its implicit causal story seriously. Assuming for the moment that we don’t believe that emergency rooms kill people, obviously we can go astray if we don’t think carefully about what causal model we are assuming when we write down and interpret regression models. That is, if we wrote down, estimated, and interpreted the model of question 5a we would be led to the (false) conclusion that emergency rooms kill people; whereas, if we wrote down, estimated, and interpreted the model of question 5b we would be led to the (true) conclusion that sick people go to emergency rooms.

Regressions only use correlation in making their estimates; therefore, they can not tell you about causation. You must already have a causal theory before you run your regression if you want to draw causal conclusions from your results.