## Models for Panel Data

1. The following is a panel of data on investment (y) and profit (x) for n=3 firms over 7=10 periods. i=1 i=2 i=3

y |
x |
y |
x |
y |
x | ||||||

t= 1 |
13. |
32 |
12 |
.85 |
20. |
30 |
22. |
93 |
8. |
85 |
8.65 |

t= 2 |
26. |
30 |
25 |
.69 |
17. |
47 |
17. |
96 |
19. |
60 |
16.55 |

t= 3 |
2. |
62 |
5 |
.48 |
9. |
31 |
9. |
16 |
3. |
87 |
1.47 |

t= 4 |
14. |
94 |
13 |
.79 |
18. |
01 |
18. |
73 |
24. |
19 |
24.91 |

t= 5 |
15. |
80 |
15 |
.41 |
7. |
63 |
11. |
31 |
3. |
99 |
5.01 |

t= 6 |
12. |
20 |
12 |
.59 |
19. |
84 |
21. |
15 |
5. |
73 |
8.34 |

t= 7 |
14. |
93 |
16 |
.64 |
13. |
76 |
16. |
13 |
26. |
68 |
22.70 |

t= 8 |
29. |
82 |
26 |
.45 |
10. |
00 |
11. |
61 |
11. |
49 |
8.36 |

t= 9 |
20. |
32 |
19 |
.64 |
19. |
51 |
19. |
55 |
18. |
49 |
15.44 |

t=10 |
4. |
77 |
5 |
.43 |
18. |
32 |
17. |
06 |
20. |
84 |
17.87 |

(a) Pool the data and compute the least squares regression coefficients of the model yit =a+ P'x,j + sii.

(b) Estimate the fixed effects model of (13-2), then test the hypothesis that the constant term is the same for all three firms.

(c) Estimate the random effects model of (13-18), then carry out the Lagrange multiplier test of the hypothesis that the classical model without the common effect applies.

(d) Carry out Hausman's specification test for the random versus the fixed model.

The pooled least squares estimator is y = -.747476 + 1.058959x, e'e = 120.6687 (.95595) (.058656)

The fixed effects regression can be computed just by including the three dummy variables since the sample sizes are quite small. The results are y = - 1.4684/1 - 2.8362/2 + -12166/3 + 1.102192x e'e = 79.183.

The F statistic for testing the hypothesis that the constant terms are all the same is

F[26,2] = [(120.6687 - 79.183)/2]/[79.183/26] = 6.811. The critical value from the F table is 19.458, so the hypothesis is not rejected.

In order to estimate the random effects model, we need some additional parameter estimates. The group means are y x

Group 1 15.502 14.962 Group 2 15.415 16.559 Group 3 14.373 12.930 In the group means regression using these three observations, we obtain y. = 10.665 + .29909 x. with e«'e« = .19747.

There is only one degree of freedom, so this is the candidate for estimation of cte2/7 + ct2. In the least squares dummy variable (fixed effects) regression, we have an estimate of cte2 of 79.183/26 = 3.045. Therefore, our

estimate of ct2 is ctu = .19747/1 - 3.045/10 = -.6703. Obviously, this won't do. Before abandoning the random effects model, we consider an alternative consistent estimator of the constant and slope, the pooled ordinary least squares estimator. Using the group means above, we find

2 ?=i [ y,. - (-.747476) - 1.058959 x ,f = 3.9273. One ought to proceed with some caution at this point, but it is difficult to place much faith in the group means regression with but a single degree of freedom, so this is probably a preferable estimator in any event. (The true model underlying these data -- using a random number generator -- has a slope, p of 1.000 and a true constant of zero. Of course, this would not be known to the analyst in a real world situation.) Continuing, we

now use ct 1= 3.9273 - 3.045/10 = 3.6227 as the estimator. (The true value of p = ct„2/(ct„2+ct62) is .5.) This leads to 6 = 1 - [3.04551/2/(10(3.6227) + 3.045)1/2] = .721524. Finally, the FGLS estimator computed according to (16-48) is y = -1.3415(.786) + 1.0987 (.028998)x.

For the LM test, we return to the pooled ordinary least squares regression. The necessary quantities are e'e = 120.6687, St e1t = -.55314, St e2t = -13.72824, St e3t = 14.28138. Therefore,

LM = {[3(10)]/[2(9)]}{[(-.55314)2 + (13.72824)2 + (14.28138)2]/120.687 - 1}2 = 8.4683 The statistic has one degree of freedom. The critical value from the chi-squared distribution is 3.84, so the hypothesis of no random effect is rejected. Finally, for the Hausman test, we compare the FGLS and least squares dummy variable estimators. The statistic is x2 = [(1.0987 - 1.058959)2]/[(.058656)2 - (.05060)2] = 1.794373. This is relatively small and argues (once again) in favor of the random effects model.

2. Suppose that the model of (13-2) is formulated with an overall constant term and n-1 dummy variables (dropping, say, the last one). Investigate the effect that this has on the set of dummy variable coefficients and on the least squares estimates of the slopes.

There is no effect on the coefficients of the other variables. For the dummy variable coefficients, with the full set of n dummy variables, each coefficient is y i * = mean residual for the ith group in the regression of y on the xs omitting the dummy variables. (We use the partitioned regression results of Chapter 6.) If an overall constant term and n-1 dummy variables (say the last n-1) are used, instead, the coefficient on the ith dummy variable is simply yi* - y i* while the constant term is still y 1* For a full proof of these results, see the solution to Exercise 5 of Chapter 8 earlier in this book.

3. Use the data in Section 13.9.7 (these are the Grunfeld data) to fit the random and fixed effects models. There are five firms and 20 years of data for each. Use the F, LM, and/or Hausman statistics to determine which model, the fixed or random effects model, is preferable for these data.

The regression model is Iit = p1i + p2Fit + p3Cit + pit. We first fit the model by pooled OLS, ignoring the specific firm effect.

+-----------------------------------------------------------------------+

OLS Without Group Dummy Variables Ordinary least squares regression

Dependent variable is I Mean = 248.95700, S.D. = 267.8654

Model size: Observations = 100, Parameters = 3, Deg.Fr. = 97

Residuals: Sum of squares= 0.157088E+07 Std.Dev. = Fit: R-squared = 0.77886, Adjusted R-squared =

Diagnostic: Log-L = -624.9928, Restricted(B=0) Log-L = Panel Data Analysis of I [ONE way]

Unconditional ANOVA (No regressors) Source Variation Deg. Free. Mean Square

Between 0.487817E+07 4. 0.121954E+07

Residual 0.222527E+07 95. 23423.9

Total 0.710344E+07 99. 71751.9

127.25831 0.77430 0.00000 -700.4398

Variable Coefficient Standard Error t-ratio P[*T*$t] Mean of X

F 0.10509

C 0.30537

Constant -48.030

0.11378E-01 9.236 0.00000 1922. 0.43508E-01 7.019 0.00000 311.1 21.480 -2.236 0.02764

The least squares regression with firm specific effects is

+-----------------------------------------------------------------------+

Least Squares with Group Dummy Variables

Ordinary least squares regression Weighting variable = ONE Dependent variable is I Mean = 248.95700, S.D. = 267.8654

Model size: Observations = 100, Parameters = 7, Deg.Fr. = 93

Residuals: Sum of squares= 444288. Std.Dev. = 69.11797

| Fit: R-squared = 0.93745, Adjusted R-squared = 0.93342 |

| Model test: F[ 6, 93] = 232.32, Prob value = 0.00000 |

Diagnostic: Log-L = -561.8468, Restricted(fi=0) Log-L = -700.4398 +-----------------------------------------------------------------------+

Variable Coefficient Standard Error t-ratio P[*T*$t] Mean of X

F 0.10598 0.15891E-01 6.669 0.00000 1922.

### C 0.34666 0.24161E-01 14.348 0.00000 311.1

To estimate the variance components for the random effects model, we also computed the group means regression. The sum of squared residuals from the LSDV estimator is 444,288. The sum of squares from the group means regression is 22382.1. The estimate of aS2 is 444,288/93 = 4777.29. The estimate of ct„2 is 22,382.1/2 - (1/20)4777.29 = 10,952.2. The model is then reestimated by FGLS using these estimates:

+--------------------------------------------------+

Estimates: Var[e] = 0.477729E+04

Corr[v(i,t),v(i,s)] = 0.696284 Lagrange Multiplier Test vs. Model (3) = 453.82 ( 1 df, prob value = 0.000000)

Fixed vs. Random Effects (Hausman) = 3.14

( 2 df, prob value = 0.208081) +--------------------------------------------------+

Variable Coefficient Standard Error z=b/s.e. P[*Z*$z] Mean of X

F 0.10489 0.14711E-01 7.130 0.00000 1922.

C 0.34602 0.24112E-01 14.350 0.00000 311.1

### Constant -60.291 54.167 -1.113 0.26568

The F and LM statistics are not useful for comparing the fixed and random effects models. The Hausman statistic can be used. The value appears above. Since the Hausman statistic is small (only 3.14 with two degrees of freedom), we conclude that the GLS estimator is consistent. The statistic would be large if the two estimates were significantly different. Since they are not, we conclude that the evidence favors the random effects model.

4. Derive the log-likelihood function for the model in (13-18) assuming that ,it and ui are normally distributed. [Hints: Write the log-likelihood function as lnL = E "=1 lnLi where lnL, is the log-likelihood function for the T observations in group i. These T observations are joint normally distributed with covariance matrix given in (14-20).] The log-likelihood is the sum of the logs of the joint normal densities of the n sets of T observations,

Sit + U = yit - a - p'x,, This will involve the inverse and determinant of Q. Use (2-66) to prove that

Q-1 = (1/ae2){I - [a„2/(ae2 + Ta „2)]ii']> To find the determinant, use the product of the characteristic roots. Note first that

|aI + aii'| = (ae2)|I + (a„2/ae2)ii'|. The roots are determined by [I + (a„2/ae2)ii']c = Xc or (a„2/ae2)ii'c = (X-1)c. Any vector whose elements sum to zero is a solution. There are T-1 such independent vectors, so T-1 characteristic roots are (X-1)=0 or X=1. Premultiply the expression by i' to obtain the remaining characteristic root. (Remember to add 1 to the result.) Now, collect terms to obtain the log-likelihood. The ith group of T observations,

e,1 + ui |
y,1 |
-P'xn " | ||

XiP = |
e, 2 + u, |
= |
y, 2 |
-P' x, 2 |

e,T + u, _ |
_ y,T |
-P' x,T _ |

, is normally distributed with mean vector 0 and the covariance matrix given in (14-20). We have included the constant term in Xi. The joint density of these T observations is, therefore, L; = fw,) = (2TC)"T/2|Q[1/2exp[(-1/2)w/Q-1wi]. The log of the joint density is lnLi = -(T/2)ln(2n) - (1/2)ln|Q| - (1/2)w/Q-1w, and, finally, for the full sample, lnL = SjlnLj. Consider the log-determinant first. As suggested above, we write Q = ae2[I + (CT„2/CTe2)ii']. Then, |Q| = (cte2)t|I +(CT„2/CTe2)ii'| or ln|Q| = 7lnae2 + ln|I + (CT„2/CTe2)ii'|. The determinant of a matrix equals the product of its characteristic roots, so the log determinant equals the sum of the logs of the roots. The characteristic roots of the matrix above remain to be determined. As shown in the exercise, T-1 of the T roots equal 1. Therefore, the logs of these roots are zero, so the log-determinant equals the log of the remaining root. It remains only to find the other characteristic root. Premultiply the result

(CT„2/CTe2)ii'c = (X-1)c by i' to obtain (CTu2/CTe2)i'ii'c = (X-1)i'c. Now, i'i = T. Divide both sides of the equation by i'c -- remember, we now seek the characteristic root which corresponds to the characteristic vector whose elements do not sum to zero -- and obtain T(a„2/ae2) = X - 1 or X = 1 + T(ct„2/ct62). ln|Q| = 7lnae2 + ln[1 + T(ct„2/ct62)].

Therefore, By writing we obtain

We now turn to the exponential term. The inverse matrix is given in the exercise, so it remains only to multiply it out. Thus, w,'Q-1wi = w/w,/CTe2 - (wi'i)2/[CTe2 + Tct„2]

Since w, = y, - X,P w/Q-1w, = (y, - X,P)' (y, - X,P)/CTe2 - [i' (y, - X,P)]2/[ cte2 + Tctu2].

The first term is the usual sum of squared deviations. The numerator in the second can be written as

[i'(yi - Xip)]2 = [T( y. - P' X £)]2 Collecting terms, lnLi = -(T/2)ln(2n) - [(T-1)/2]lnCTe2 - ^(y, - Xp)' (y, - XiP)/ct62 - ^[T(y. - P' X £)]2/[cte2 + Tctu2]. Finally, the log-likelihood for the full sample is the sum of these expressions over the i=1 to n groups.

5. Unbalanced design for random effects. Suppose that the random effects model of Section 13.4 is to be estimated with a panel in which the groups have different numbers of observations. Let Ti be the number of observations in group i.

(a) Show that the pooled least squares estimator in (13-11) is unbiased and consistent in spite of this complication.

(b) Show that the estimator in (13-29) based on the pooled least squares estimator of P (or, for that matter, any consistent estimator of P) is a consistent estimator of cts2.

The model in (13-11) is a generalized regression model. As we saw in Chapter 10, OLS is consistent in the GR model. The unequal group sizes here does not have any effect on the result. The residual

using any estimators of a and P is eit = yti - a - P' xit and ei = yi - a - P' x,. . Thus the estimator in (1329) is [1/(nT-n-K)]SISt (e,t -^)2 = [1/(nT-n-K)]SISt [y -yL) - P' (xIi -X, )]2. The probability limit is the

same as the probability limit of the statistic which results when P is replaced with its probability limit. If P is a consistent estimator of P, then the estimator converges to plim[1/(nT-n-K)]SiSt [(yit -yi ) - P'(xit - x,. )]2. But,

(y,t - y, ) - P'(x,t - x;. ) = eit - e,. So, our estimator has the same limiting behavior as ict^ = [1/(nT-n-K)]SiEt

(eit - e,. )2. Write this as ctJ: = (1/n)S[Et(eit - e, )2]/[(T-1) - K/n]. The expected value of sum of squared deviations in the brackets is (T-1)cte2 Each term in the outer sum has the same expectation, so the exact expectation is 1/n times n times this expectation, or

E[ ol ] = [(7-1) Oe2]/[(7-1) - Kin] This obviously converges to oe2as The exact variance of the estimator depends upon what we assume about the fourth moment of eit. If we assume only that the fourth moment of eit is finite, then the variance of each term in the inner sum is of the form

[7/(7 - 1 - K/n)][f/7 + ^2/7 2 + ^3/7 3] = f If ^ is finite, then the variance of the entire expression is 4>/n which converges to 0. This completes the proof. To summarize the argument, we have shown that the limiting behavior of the statistic in (13-27) based on any consistent estimator of p is the same as that of a statistic which converges in mean square to oe2 if the fourth moment of e is finite.

6. What are the probability limits of (1/n)LM, where LM is defined in (13-31) under the null hypotheses that ou2 = 0 and under the alternative that ou2 ^ 0?

To find plim(1/n)LM = plim [7/(2(7-1))]{[SI(SeIt)2MYIYeIt2] - 1}2 we can concentrate on the sums inside the curled brackets. First, Y^Ee^)2 = n72{(1/n)EI[(1/7)Eeit]2} and Y^e*2 = n7(1/(n7))YIYeIt2 The ratio equals p^^/EYe2] = 7{(1/n)S,[(1/7)SeIi]2}/{(1/(n7))SISeIi2}. Using the argument used in Exercise 8 to establish consistency of the variance estimator, the limiting behavior of this statistic is the same as that which is computed using the true disturbances since the OLS coefficient estimator is consistent. Using

the true disturbances, the numerator may be written (1/n)Y ,[(1/7) E^e J2 = (1/n)Z,-si. Since E[ e,.] = 0, plim(1/n)S,-62 = Var[ e,. ] = oe27 + ou2The denominator is simply the usual variance estimator, so plim( 1/(n7)) Y Ete ,t2 = Var[e it] = Oe2+ o„2Therefore, inserting these results in the expression for LM, we find that plim (1/n)LM = [7/(2(7-1))]{[7(oe27 + ou2)]/[oe2+ ou2] - 1}2. Under the null hypothesis that o2 = 0, this equals 0. By expanding the inner term then collecting terms, we find that under the alternative hypothesis that o2 is not equal to 0, plim (1/n)LM = [7(7-1)/2][ a„2/(CTe2+CT„2)]2. Within group i, Corr2[e it,e ¿J = p2 = ou /(ou + oe) so plim (1/n)LM = [7(7-1)/2](p2)2. It is worth noting what is obtained if we do not divide the LM statistic by n at the outset. Under the null hypothesis, the limiting distribution of LM is chi-squared with one degree of freedom. This is a random variable with mean 1 and variance 2, so the statistic, itself, does not converge to a constant; it converges to a random variable. Under the alternative, the LM statistic has mean and variance of order n (as we see above) and hence, explodes. It is this latter attribute which makes the test a consistent one. As the sample size increases, the power of the LM test must go to 1.

7. A two way fixed effects model: Suppose the fixed effects model is modified to include a time specific dummy variable as well as an individual specific variable. Then, yit = a, + yt + P'xit + eit. At every observation, the individual- and time-specific dummy variables sum to one, so there are some redundant coefficients. The discussion in Section 13.3.3 shows one way to remove the redundancy. Another useful way to do this is to include an overall constant and to drop one of the time specific and one of the time-dummy variables. The model is, thus, yit = S + (ai - aj) + (yt - yj) + p'xit + eit. (Note that the respective time or individual specific variable is zero when t or i equals one.) Ordinary least squares estimates of p can be obtained by regression of yit - yi - yt + y on xit - xi. - x.i + x Then, (arai) and (yt-yi) are estimated using the expressions in (13-17) while d = y - b' x .

Using the following data, estimate the full set of coefficients for the least squares dummy variable using the expressions in (13-17) while d = y - b' x .

Using the following data, estimate the full set of coefficients for the least squares dummy variable

t=1 |
t=2 |
t=3 |
t=4 |
t=5 |
t=6 |
t=7 |
t=8 |
t=9 |
t=10 | |

y |
21.7 |
10.9 |
33.5 |
22.0 |
17.6 |
16.1 |
19.0 |
18.1 |
14.9 |
23.2 |

Xl |
26.4 |
l7.3 |
23.8 |
17.6 |
26.2 |
21.1 |
17.5 |
22.9 |
22.9 |
14.9 |

X2 |
5.79 |
2.60 |
8.36 |
5.50 |
5.26 |
1.03 |
3.11 |
4.87 |
3.79 |
7.24 |

y |
21.8 |
21.0 |
33.8 |
18.0 |
12.2 |
30.0 |
21.7 |
24.9 |
21.9 |
23.6 |

Xl |
19.6 |
22.8 |
27.8 |
14.0 |
11.4 |
16.0 |
28.8 |
16.8 |
11.8 |
18.6 |

X2 |
3.36 |
1.59 |
6.19 |
3.75 |
1.59 |
9.87 |
1.31 |
5.42 |
6.32 |
5.35 |

y |
25.2 |
41.9 |
31.3 |
27.8 |
13.2 |
27.9 |
33.3 |
20.5 |
16.7 |
20.7 |

Xl |
13.4 |
29.7 |
21.6 |
25.1 |
14.1 |
24.1 |
10.5 |
22.1 |
17.0 |
20.5 |

X2 |
9.57 |
9.62 |
6.61 |
7.24 |
1.64 |
5.99 |
9.00 |
1.75 |
1.74 |
1.82 |

y |
15.3 |
25.9 |
21.9 |
15.5 |
16.7 |
26.1 |
34.8 |
22.6 |
29.0 |
37.1 |

Xi |
14.2 |
18.0 |
29.9 |
14.1 |
18.4 |
20.1 |
27.6 |
27.4 |
28.5 |
28.6 |

X2 |
4.09 |
9.56 |
2.18 |
5.43 |
6.33 |
8.27 |
9.16 |
5.24 |
7.92 |
9.63 |

Test the hypotheses that (1) the "period" effects are all zero, (2) the "group" effects are all zero, and (3) both period and group effects are zero. Use an F test in each case. The ordinary least squares regression results are

R2 = .92803, |
e'e = 146.761, |
40 observations |

Variable |
Coefficient |
Standard Error |

X1 |
.446845 |
.07887 |

X2 |
1.83915 |
.1534 |

Constant |
3.60568 |
2.555 |

Period 1 |
-3.57906 |
1.723 |

Period 2 |
-1.49784 |
1.716 |

Period 3 |
2.00677 |
1.760 |

Period 4 |
-3.03206 |
1.731 |

Period 5 |
-5.58937 |
1.768 |

Period 6 |
-1.49474 |
1.714 |

Period 7 |
1.52021 |
1.714 |

Period 8 |
-2.25414 |
1.737 |

Period 9 |
-3.29360 |
1.722 |

Group 1 |
-.339998 |
1.135 |

Group 2 |
4.39271 |
1.183 |

Group 3 |
5.00207 |
1.125 |

Estimated covariance matrix for the slopes:

Pi P2

p1 .0062209 P2 .00030947 .023523 For testing the hypotheses that the sets of dummy variable coefficients are zero, we will require the sums of squared residuals from the restrictions. These are

Regression Sum of squares

All variables included 146.761

Period variables omitted 318.503

Group variables omitted 369.356

Period and group variables omitted 585.622 The F statistics are therefore,

(1) F[9,25] = [(318.503 - 146.761)/9]/[146.761/25] = 3.251

(2) F[3,25] = [(369.356 - 146.761)/3]/[146.761/25] = 12.639

(3) F[12,25] = [(585.622 - 146.761)/12]/[146.761/25] = 6.23

The critical values for the three distributions are 2.283, 2.992, and 2.165, respectively. All sample statistics are larger than the table value, so all of the hypotheses are rejected. □

8. Two way random effects model: We modify the random effects model by the addition of a time specific disturbance. Thus, yit = a + P'x,, + &it + u + vt, where E[e J = E[u,] = E[vJ = 0, E^ut] = E[ev] = E[uv] = 0, for all ij,t,s, Var[eit] = CTe2Cov[6it,e;S] = 0 for all tj,s, Var[ui] = au2Cov[ui,u;] = 0 for all ij Var[vt] = ctv2, Cov[vt,vs] = 0 for all t,s. Write out the full covariance matrix for a data set with n=2 and T=2. The covariance matrix would be i = 1, t = 1 i = 1, t = 2 i = 2, t = 1 i = 2, t = 2

y1 |
x1 |
p + |
S1 | |

= | ||||

y2 |
x2 |
e 2 |

satisfies the groupwise heteroscedastic regression model of Section

11.7.2. All variables have zero means. The following sample second moment matrix is obtained from a sample of 20 observations:

>1 |
> 2 |
X1 |
X2 | |

>"1 |
20 |
6 |
4 |
3 |

>"2 |
6 |
10 |
3 |
6 |

4 |
3 |
5 |
2 | |

X2 |
3 |
6 |
2 |
10 |

(a) Compute the two separate OLS estimates of p, their sampling variances, the estimates of ct12 and ct22, and the R2s in the two regressions.

(b) Carry out the Lagrange Multiplier test of the hypothesis that ct12 = ct22

(c) Compute the two step FGLS estimate of p and an estimate of its sampling variance. Test the hypothesis that p equals one.

(d) Carry out the Wald test of equal disturbance variances.

(e) Compute the maximum likelihood estimates of p, ct12, and ct22 by iterating the FGLS estimates to convergence.

(f) Carry out a likelihood ratio test of equal disturbance variances.

(g) Compute the two step FGLS estimate of p assuming that the model in (14-7) applies. [That is, allow for cross sectional correlation.] Compare your results to those of part (c).

The two separate regressions are as follows:

Sample 1 Sample 2

e'e = y' y - bx'y 20 - 4(4/5) = 84/5 10 - 6(6/10) = 64/10

R2 = 1 - e' e/y' y 1 - (84/5)/20 = .16 1 - (64/10)/10 = .36

s2 = e' e/(n-1) (84/5)/19 = .88421 (64/10)/19 = .33684

Est.Var[b] = s2/x' x .88421/5 = .17684 .33684/10 = .033684

To carry out a Lagrange multiplier test of the hypothesis of equal variances, we require the separate and common variance estimators based on the restricted slope estimator. This, in turn, is the pooled least squares estimator. For the combined sample, we obtain b = [x1 y + x2 y2]/[xj 'x1 + x2 x2] = (4 + 6) / (5 + 10) = 2/3. Then, the variance estimators are based on this estimate. For the hypothesized common variance, e'e = (y1 y + y2 y2) - b(xx 'y: + x2 y2) = (20 + 10) - (2/3)(4 + 6) = 70/3, so the estimate of the common variance is e' e/40 = (70/3)/40 = .58333. Note that the divisor is 40, not 39, because we are comptuting maximum likelihood estimators. The individual estimators are e1 ei/20 = (y1 y1 - 2b(x1 y0 + b2(x1 x0)/20 = (20 - 2(2/3)4 + (2/3)25)/20 = .84444 and e2' e2/20 = (y2' y2 - 2b(x2' y2) + b2(x2' x2))/20 = (10 - 2(2/3)6 + (2/3)210)/20 = .32222. The LM statistic is given in Example 16.3,

LM = (772)[(s12/s2 - 1)2 + (s22/s2 - 1)2] = 10[(.84444/.58333 - 1)2 + (.32222/.58333 - 1)2] = 4.007. This has one degree of freedom for the single restriction. The critical value from the chi-squared table is 3.84, so we would reject the hypothesis.

In order to compute a two step GLS estimate, we can use either the original variance estimates based on the separate least squares estimates or those obtained above in doing the LM test. Since both pairs are consistent, both FGLS estimators will have all of the desirable asymptotic properties. For our estimator, we

used ct 12 = ej e/T from the original regressions. Thus, ct 1 = .84 and ct 2 = .32. The GLS estimator is p = [(1/ct 12 )x1 'y1 + (1/ ct 22)x2'y2]/[ (1/ct 12 )x1'x1 + (1/ CT 22)x2'x2] = [4/.84 + 6/.32]/[5/.84 + 10/.32] = .632.

The estimated sampling variance is 1/[ (1/ ct 1 )x1' x1 + (1/ ct 2 )x2'x2] = .02688. This implies an asymptotic standard error of (.02688)2 = .16395. To test the hypothesis that p = 1, we would refer z = (.632 - 1) / .16395 = -2.245 to a standard normal table. This is reasonably large, and at the usual significance levels, would lead to rejection of the hypothesis.

The Wald test is based on the unrestricted variance estimates. Using b = .632, the variance

estimators are ct ^ = [y/yj - 2b(x1'y1) + b2(x1'x1)]/20 = .847056 and ct 22= [y2y2 - 2b(x2'y2) + b2(x/x2)]/20 = .320512

while the pooled estimator would be ct 2= [y' y - 2b(x' y) + b2(x' x)]/40 = .583784. The statistic is given at the end of Example 16.3, W = (772)[( CT / CT 12 - 1)2 + (CT / CT 22 - 1)2]

= 10[(.583784/.847056 - 1)2 + (.583784/.320512 - 1)2] = 7.713. We reach the same conclusion as before.

To compute the maximum likelihood estimators, we begin our iterations from the two separate

ordinary least squares estimates of b which produce estimates ct i2 = .84 and ct 22= .32. The iterations are

Iteration 0 1

CT 1

.840000 .847056 .847071 .847071

CT 2

.320000 .320512 .320506 .320506

.632000 .631819 .631818 converged

Now, to compute the likelihood ratio statistic for a likelihood ratio test of the hypothesis of equal variances, we refer x2 = 40ln.58333 - 20ln.847071 - 20ln.320506 to the chi-squared table. (Under the null hypothesis, the pooled least squares estimator is maximum likelihood.) Thus, x2 = 4.5164, which is roughly equal to the LM statistic and leads once again to rejection of the null hypothesis.

Finally, we allow for cross sectional correlation of the disturbances. Our initial estimate of b is the pooled least squares estimator, 2/3. The estimates of the two variances are .84444 and .32222 as before while the cross sectional covariance estimate is ei' e2/20 = [y1 y2 - b(x1 y2 + x2 yi) + b2(x1 x2)]/20 = .14444. Before proceeding, we note, the estimated squared correlation of the two disturbances is r = .14444 / [(.84444)(.32222)]1/2 = .277, which is not particularly large. The LM test statistic given in (16-14) is 1.533, which is well under the critical value of 3.84. Thus, we would not reject the hypothesis of zero cross section correlation. Nonetheless, we proceed. The estimator is shown in (16-6). The two step FGLS and iterated maximum likelihood estimates appear below.

Iteration |
CT 12 |
CT 22 |
A CT 12 |
P |

0 |
.84444 |
.32222 |
.14444 |
.5791338 |

1 |
.8521955 |
.3202177 |
.1597994 |
.5731058 |

2 |
.8528702 |
.3203616 |
.1609133 |
.5727069 |

3 |
.8529155 |
.3203725 |
.1609873 |
.5726805 |

4 |
.8529185 |
.3203732 |
.1609921 |
.5726788 |

5 |
.8529187 |
.3203732 |
.1609925 |
converged |

Because the correlation is relatively low, the effect on the previous estimate is relatively minor. □

Because the correlation is relatively low, the effect on the previous estimate is relatively minor. □

10. Suppose that in the model of Section 15.2.1, X,- is the same for all i. What is the generalized least squares estimator of p? How would you compute the estimator if it were necessary to estimate ct,2? If all of the regressor matrices are the same, the estimator in (15-6) reduces to p = (X' X)-1 S h {(1/ct2)/[S n=1 (1/ct,2)] }X'y, = S ,'=1 „„

a weighted average of the ordinary least squares estimators, b,- = (X' X)-1X' y,- with weights

Wi = (1/ct2)/[E n=1 (1/ct,2)]. If it were necessary to estimate the weights, a simple two step estimator could be based on individual variance estimators. Either of s2 = e'eJT based on separate least squares regressions (with different estimators of p) or based on residuals computed from a common pooled ordinary least squares slope estimator could be used.

11. Repeat Exercise for the model of Section 13.9.1.

The estimator is shown in (15-11). If all of the X matrices are the same, the estimator reduces to a weighted average of the OLS estimators again. Using (15-11) directly with a common X,

P = [S jX'XJ^S^X'y,] = [1/S j][S j[X'X]-1X'y, = [1/S,(SIaj)][S,(SIa%] The disturbance variances and covariances can be estimated as suggested in the previous exercise. ~

12. The following table presents a hypothetical panel of data:

12. The following table presents a hypothetical panel of data:

y |
x |
y |
x |
y |
x | |||||||

t= 1 |
30. |
27 |
24 |
.31 |
38. |
71 |
28 |
.35 |
37. |
03 |
21 |
.16 |

t= 2 |
35. |
59 |
28 |
.47 |
29. |
74 |
27 |
.38 |
43. |
82 |
26 |
.76 |

t= 3 |
17. |
90 |
23 |
.74 |
11. |
29 |
12 |
.74 |
37. |
12 |
22 |
.21 |

t= 4 |
44. |
90 |
25 |
.44 |
26. |
17 |
21 |
.08 |
24. |
34 |
19 |
.02 |

t= 5 |
37. |
58 |
20 |
.80 |
5. |
85 |
14 |
.02 |
26. |
15 |
18 |
.64 |

t= 6 |
23. |
15 |
10 |
.55 |
29. |
01 |
20 |
.43 |
26. |
01 |
18 |
.97 |

t= 7 |
30. |
53 |
18 |
.40 |
30. |
38 |
28 |
.13 |
29. |
64 |
21 |
.35 |

t= 8 |
39. |
90 |
25 |
.40 |
36. |
03 |
21 |
.78 |
30. |
25 |
21 |
.34 |

t= 9 |
20. |
44 |
13 |
.57 |
37. |
90 |
25 |
.65 |
25. |
41 |
15 |
.86 |

t=10 |
36. |
85 |
25 |
.60 |
33. |
90 |
11 |
.66 |
26. |
04 |
13 |
.28 |

(a) Estimate the groupwise heteroscedastic model of Section 11.7.2. Include an estimate of the asymptotic variance of the slope estimator. Use a two step procedure, basing the FGLS estimator at the second step on residuals from the pooled least squares regression.

(b) Carry out the Wald, Lagrange multiplier, and likelihood ratio tests of the hypothesis that the variances are all equal. For the likelihood ratio test, use the FGLS estimates.

(c) Carry out a Lagrange multiplier test of the hypothesis that the disturbances are uncorrelated across individuals.

The various least squares estimators of the parameters are

Sample 1 |
Sample 2 |
Sample 3 |
Pooled | |

a |
11.6644 |
5.42213 |
1.41116 |
8.06392 |

(9.658) |
(10.46) |
(7.328) | ||

b |
.926881 |
1.06410 |
1.46885 |
1.05413 |

(.4328) |
(.4756) |
(.3590) | ||

e'e |
452.206 |
673.409 |
125.281 | |

(464.288) |
(732.560) |
(171.240) |
(1368.088) |

(Values of e'e in parentheses above are based on the pooled slope estimator.) The FGLS estimator and its estimated asymptotic covariance matrix are

(Values of e'e in parentheses above are based on the pooled slope estimator.) The FGLS estimator and its estimated asymptotic covariance matrix are

22.8049 -1.0629 -1.0629 0.05197

Note that the FGLS estimator of the slope is closer to the 1.46885 of sample 3 (the highest of the three OLS estimates). This is to be expected since the third group has the smallest residual variance. The LM test statistic is based on the pooled regression,

LM = (10/2){[(464.288/10)/(1368.088/30) - 1]2 + ...} = 3.7901 To compute the Wald statistic, we require the unrestricted regression. The parameter estimates are given above. The sums of squares are 465.708, 785.399, and 145.055 for i = 1, 2, and 3, respectively. For the common estimate of ct2, we use the total sum of squared GLS residuals, 1396.162. Then,

W = (10/2){[(1396.162/30)/(465.708/10) - 1]2 + ...} = 25.21. The Wald statistic is far larger than the LM statistic. Since there are two restrictions, at significance levels of 95% or 99% with critical values of 5.99 or 9.21, the two tests lead to different conclusions. The likelihood ratio statistic based on the FGLS estimates is x2 = 30ln(1396.162/30) - 10ln(465.708/10) ... = 6.42 which is between the previous two and between the 95% and 99% critical values.

The correlation matrix for the residuals from the pooled OLS regression is

so the LM statistic is LM = 10[(-.0704)2 + (-.7619)2 + (-.0825)2] = 5.9225. The 95% critical value from the chi-squared distribution with 3 degrees of freedom is 7.82, so we would not reject the hypothesis of uncorrelated disturbances.

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