hw 1 answers

Economics 120C Name: _________________________ Professor Yongil Jeon Winter 2009 Student ID#: _________________________

Answer Key to Homework #1 – Winter 2009, Econ 120C

(Midterm Exam, Summer 2008-Session 2, Econ 120C) Answer all questions on separate paper. This problem set should be handed in to Professor Jeon at the beginning of your class on Wednesday, February 4th, 2009. Problem sets may not be handed in once solutions have been distributed. Please write down your name and PID clearly. Good luck!

1) (3 points) The GLS assumptions include all of the following, with the exception of

a. the Xi are fixed in repeated samples. b. Xi and ui have nonzero finite fourth moments. c. E(UU′|X) = Ω(X), where Ω(X) is n×n matrix-valued that can depend on X. d. E(U|X) = 0n. Answer: a

2) (3 points) The assumption that X has full column rank implies that

a. the number of observations equals the number of regressors. b. binary variables are absent from the list of regressors. c. there is no perfect multicollinearity. d. none of the regressors appear in natural logarithm form.

Answer: c

3) (3 points) The OLS estimator

a. has the multivariate normal asymptotic distribution in large samples. b. is t-distributed. c. has the multivariate normal distribution regardless of the sample size. d. is F-distributed.

Answer: a

4) (3 points) The leading example of sampling schemes in econometrics that do not result in independent observations is

a. cross-sectional data. b. experimental data. c. the Current Population Survey.

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2 Answer to hw#1, ECON 120C, Winter 2009

d. when the data are sampled over time for the same entity.

Answer: d

5) (3 points) Finite-sample distributions of the OLS estimator and t-statistics are complicated, unless

a. the regressors are all normally distributed. b. the regression errors are homoskedastic and normally distributed,

conditional on 1,..., nX X . c. the Gauss-Markov Theorem applies. d. the regressor is also endogenous.

Answer: b

6) (10 points) Suppose that Y and X are related by the regression Y=1.0 +2.0X+u. A

researcher has observations on Y and X, where 0 20,X≤ ≤ where the conditional variance is var( | ) 1i iu X x= = for 0 10,x≤ ≤ and var( | ) 16i iu X x= = for 10 20x< ≤ . Instead of using WLS, the researcher decides to compute the OLS estimator using only the observations for which 0 10,x≤ ≤ then using only the observations for which 10x > , then average the two OLS of estimators. Is this more efficient than WLS?

Answer: The Gauss-Markov theorem implies that the WLS is the best linear conditionally unbiased estimator, and thus that the averaged estimator cannot be better than WLS.

7) (20 points) A researcher, using a sample of 2,868 individuals, is investigating how the probability of a respondent obtaining a bachelor’s degree from a four-year college is related to the respondent’s score on ASVABC. 26.7 percent of the respondents earned bachelor’s degrees. ASVABC ranged from 22 to 65, with mean value 50.2, and most scores were in the range 40 to 60. Defining a variable BACH to be equal to 1 if the respondent has a bachelor’s degree (or higher degree) and 0 otherwise, the researcher fitted the OLS regression (standard errors in parentheses):

CHAB ˆ = –0.864 + 0.023ASVABC, R2 = 0.21 (0.042) (0.001)

She also fitted the following logit regression:

Z = –11.103 + 0.189 ASVABC (0.487) (0.009)

where Z is the variable in the logit function. Using this regression, she plotted the probability and marginal effect functions shown in the diagram.


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(1) (10 points) Give an interpretation of the OLS regression and explain why OLS is not a satisfactory estimation method for this kind of model.

(2) (10 points) Sketch the probability and marginal effect diagrams for the OLS

regression and compare them with those for the logit regression. (In your discussion, make use of the information in the first paragraph of this question.)

Answer: (1) The slope coefficient of the linear probability model indicates that the

probability of earning a bachelor’s degree rises by 2.3 percent for every additional point on the ASVABC score. While this may be realistic for a range of values of ASVABC, it is not for very low ones. Very few of those with scores in the low end of the spectrum earned bachelor’s degrees and variations in the ASVABC score would be unlikely to have an effect on the probability. The intercept literally indicates that an individual with a 0 score would have a minus 86.4 percent probability of earning a bachelor’s degree. Given the way that ASVABC was constructed, a score of 0 was in fact impossible. However the linear probability model predicts nonsense negative probabilities for all those with scores less of 37 or less, of whom there were many in the sample. The linear probability model also suffers from the problem that the standard errors and t and F tests are invalid because the disturbance term does not have a normal distribution and it is heteroskedastic. Its distribution is not even continuous, consisting of only two possible values for each value of ASVABC.

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ASVABC

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(2) The counterpart to the cumulative probability curve in the figure is a straight line using the regression result. At the ASVABC mean it predicts that there is a 29% chance of the respondent graduating from college, considerably more than the logit figure, but for a score of 65 it predicts a probability of only 63%. It is particularly unsatisfactory for low ASVABC scores. It predicts a –36% chance of graduating for respondents with the lowest score of 22, and negative probabilities for all scores lower than 38. The OLS counterpart to the marginal probability curve is a horizontal straight line at 0.023, showing that the marginal effect is under estimated for ASVABC scores above 50 and overestimated below that figure.

(8) (10 points) Consider the simple regression model Yi = β0 + β1Xi + ui where Xi > 0 for all i, and the conditional variance is 4var( | )i i iu X Xθ= where θ is a known constant with θ > 0.

(a) (4 points) Write the weighted regression as iY% = β0 0iX% + β1 1iX% + iu% . How would

you construct iY% , 0iX% and 1iX% ?

Answer: 2i

ii

YYX

=% , 0 2

1i

i

XX

=% , and 1 2

1ii

i i

XXX X

= =% .

(b) (4 points) Prove that the variance of iu% is homoskedastic.

Answer: 4

2 4 4

var( | )var( | ) var |i i i ii i i

i i i

u u X Xu X XX X X

θ θ⎛ ⎞

= = = =⎜ ⎟⎝ ⎠

% , which is constant.

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(c) (2 points) When interpreting the regression results, which of the two equations should you use, the original or the modified model?

Answer: The modified model is simply used to obtain estimates of the original

model. The modified model should therefore not be used for interpretation.

9) (5 points) Consider estimating a consumption function from a large cross-section sample of households. Assume that households at lower income levels do not have as much discretion for consumption variation as households with high income levels. After all, if you live below the poverty line, then almost all of your income is spent on necessities, and there is little room to save. On the other hand, if your annual income was $1 million, you could save quite a bit if you were a frugal person, or spend it all, if you prefer. Sketch what the scatterplot between consumption and income would look like in such a situation. What functional form do you think could approximate the conditional variance var( | )iu Income ?

Answer: See the accompanying figure. var( | )iu Income could be a b Income+ × or 2a b Income+ × . Hence there would be heteroskedasticity.

10) (15 points) Consider the following probit regression

Pr( 1| )Y X Φ= = (8.9 – 0.14×X)

Calculate the change in probability for X increasing by 10 for X = 40 and X = 60.

Why is there such a large difference in the change in probabilities?

Note that . di normal(-0.9) .18406013; . di normal(0.5) .69146246 . di normal(1.9) .97128344; . di normal(3.3) .99951658

where “normal” in Stata returns the cumulative standard normal distribution Answer: Pr(Y=1|X=40) = 0.999; Pr(Y=1|X=50) = 0.971; Pr(Y=1|X=60) = 0.691;

Pr(Y=1|X=70) = 0.184. The large differences happen as a result of the non-linearity of the function, and the points at which they are calculated.


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11) (10 points) Consider the model 1i i iY X uβ= + , where | |i i iu c X e= and all of the X’s

and e’s are each i.i.d, distributed as N(0,1) and N(0, 2eσ ) and independent of each other.

(a) (5 points) Which of the (multiple) linear regression assumptions are satisfied

here? Prove your assertions. Answer: The extended least squares assumptions are:

1. E( iu |Xi) =E( | |i ic X e |Xi) = 0 (conditional mean zero) – this holds here since the X’s and e’s are i.i.d; (note that even if Xi is not normal but ei is normal and independent of Xi, it is still true)

2. (Xi, Yi), i = 1,…, n are independent and identically distributed (i.i.d.) draws from their joint distribution - this applies here;

3. (Xi, ui) have nonzero finite fourth moments – this follows from the normal distribution, which has moments of all orders.

4. var(ui|Xi) = 2uσ (homoskedasticity) – this fails since var(ui|Xi) =

2 2i ec X σ ; and

5. The conditional distribution of ui given Xi is normal (normal errors) – this holds since iX and iu are both normal.

(b) (5 points) Would an OLS estimator of 1β be efficient here?

Answer: Since the model is heteroskedastic, WLS offers efficiency gains.

12. (10 points) Consider the linear probability model 0 1i i iY X uβ β= + + , where

0 1Pr( 1| )i i iY X Xβ β= = + . Show that ui is heteroskedastic


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Answer: Since Yi is binary variable, we know

E(Yi |Xi) = 1 × Pr(Yi = 1 |Xi) + 0 × Pr(Yi = 0 |Xi) = Pr(Yi = 1 |Xi) = β0 + β1Xi. Thus

0 1 0 1( | ) [ ( )| ] ( | ) ( ) 0i i i i i i i iE u X E Y X X E Y X Xβ β β β= − + = − + =

We have

= = − +

= + − +0 1 0 1

var( | ) Pr( 1| )[1 Pr( 1| )]( )[1 ( )].

i i i i i i

i i

Y X Y X Y XX Xβ β β β

Thus

= − +

= = + − +0 1

0 1 0 1

var( | ) var[ ( ) | ]var( | ) ( )[1 ( )].

i i i i i i

i i i i

u X Y X XY X X X

β ββ β β β

var(ui |Xi) depends on the value of Xi, so ui is heteroskedastic.

13 (5 points) Why are the coefficients of probit and logit models estimated by maximum likelihood instead of OLS? Answer: OLS cannot be used because the regression function is not a linear function of the regression coefficients (the coefficients appear inside the nonlinear functions). The maximum likelihood estimator is efficient and can handle regression functions that are nonlinear in the parameters.


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