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Exam
Problem 1: (15 points) Suppose that the classical regression model applies but that the true value of the constant is zero. In order to answer the following questions assume just one independent variable.
- Give the formulae for the two least squares slope estimators (the one with and the one without the constant).
- Calculate their variances.
- Compare the variance of the least squares slope estimator computed without a constant term with that of the estimator computed with an unnecessary constant term.
Solution:
y = β 1 + β 2 x + ε → β 2 =
i( ∑xi^ −^ ¯x)(yi^ −^ y¯) i(xi^ −^ ¯x) 2
y = β˜ 2 x + ε → β˜ 2 =
∑^ i^ xiyi i x
2 i
V ar(β 2 ) =
σ^2 ∑ i(xi^ −^ ¯x) 2
V ar( β˜ 2 ) =
σ^2 ∑ i x 2 i
- The ratio of these two variances is
V ar( β˜ 2 ) V ar(β 2 )
∑^ σ^2 i x^2 i ∑^ σ^2 i(xi−¯x)^2 =
i(xi^ −^ ¯x)
2 ∑ i x 2 ∑^ i (xi − ¯x)^2 =
(x^2 i − 2 xi x¯ + ¯x) =
x^2 i − 2 nx¯¯x + nx¯^2 =
x^2 i − nx¯^2
=
i x
2 i −^ nx¯
2 ∑ i x
2 i
= 1 −
nx¯^2 ∑ i x
2 i
It follows that fitting the constant term when it is unnecessary inflates the variance of the least squares estimator if the mean of the regressor is not zero.
Problem 3: (15 points) The following model is estimated using a balanced panel of five firms over 20 years: Iit = β 1 Fit + β 2 Cit + εit, where the regressors are market value (F ) and capital (C ) and the dependent variable is investment (I ). Suppose that the true error structure of the model is εit = αi + ηit, where α is uncorrelated with the regressors.
- If the model is estimated as a fixed effects model, what will be the statistical properties, in terms of efficiency and consistency, of the esti- mates?
- The estimates for pooled OLS, fixed effects (using dummies) and ran- dom effects models are given in the table below. Use the statistics shown to decide whether the data support a fixed effects or random effects specification. Carefully explain your reasoning.
Dependent Variable is Investment Estimation Constant Market Value Capital (a) OLS -48.030 (-2.236) 0.10509 (9.236) 0.30537 (7.019) (b) Fixed Effects - 0.10598 (6.669) 0.34666 (14.348) (c) Random Effects -61.575 (-0.775) 0.10549 (6.859) 0.34641 (14.350) (t-ratios are shown in brackets) Breush-Pagan LM test for random effects (1 df): 453. Hausman test of fixed vs random effects (2 df): 1.
Solution:
- If the individual effects are strictly uncorrelated with the regressors then a random effects model is the appropriate model. However, if a fixed effect model is estimated the estimates will be consistent but not efficient.
- Breush-Pagan LM test: Test statistic is 453.82, the critical value from the chi-squared table is 3.84, so the null hypothesis that random effects are not needed can be rejected. Hausman Test: Test statistic is 1.27, the critical value from the chi- squared table is 5.99, so the null hypothesis of the random effects model cannot be rejected.
Problem 4: (15 points) Consider the stochastic processes given below. For each process determine what the effects of first differencing the process, i.e. computing yt − yt− 1 , on autocorrelation are, e.g. reduction of the autocorrelation.
- yt = yt− 1 + εt, where εt is normally distributed white noise.
- yt = β 0 + β 1 t + εt, where εt is normally distributed white noise.
- yt = β′xt + εt, where εt = ρεt− 1 + ut and ut is normally distributed white noise. [Hint: Compare the autocorrelation of εt and the autocorrelation of (εt − εt− 1 ).]
Solution:
- ∆yt = yt − yt− 1 = εt, white noise, no more autocorrelation
- ∆yt = yt − yt− 1 = β 1 + εt − εt− 1. This is an MA(1) process with autocorrelation (^) 1+θθ 2 = (^) 1+1^1 = 12.
- ∆yt = yt − yt− 1 = β′(xt − xt− 1 ) + vt, where vt = εt − εt− 1.
V ar(εt) =
σ^2 u 1 − ρ^2
V ar(vt) = V ar(εt − εt− 1 ) = V ar(ρεt− 1 − εt + ut)
= V ar[(ρ − 1)εt− 1 + ut] = (ρ − 1)^2
σ^2 u 1 − ρ^2
2 σ u^2 1 + ρ
Cov[vt, vt− 1 ] = Cov[εt − εt− 1 , εt− 1 − εt− 2 ] = E[εtεt− 1 − ε^2 t− 1 − εtεt− 2 + εt− 1 εt− 2 ]
= ρ
σ^2 u 1 − ρ^2
σ u^2 1 − ρ^2
− ρ^2
σ u^2 1 − ρ^2
σ u^2 1 − ρ^2
σ u^2 (2ρ − 1 − ρ^2 ) 1 − ρ^2
=
σ^2 u(ρ − 1)^2 (ρ − 1)(ρ + 1)
σ u^2 (ρ − 1) ρ + 1 Cov[vt, vt− 1 ] V ar[vt]
ρ − 1 2
