Regression Diagnostics and Specification Tests
8.1 Since H = PX is idempotent, it is positive semidefinite with b0H b > 0 for any arbitrary vector b. Specifically, for b0 = (1,0,.., 0/ we get hn > 0. Also, H2 = H. Hence,
n
hii =J2 hb2 > h2i > 0.
j=i
From this inequality, we deduce that hjy — h11 < 0 or that h11(h11 — 1/ < 0. But h11 > 0, hence 0 < h11 < 1. There is nothing particular about our choice of h11. The same proof holds for h22 or h33 or in general hii. Hence, 0 < hii < 1 for i = 1,2,.., n.
8.2 A Simple Regression With No Intercept. Consider yi = xi" + ui for i = 1,2,.., n
a. H = Px = x(x0x)_1x0 = xx0/x0x since x0x is a scalar. Therefore, hii =
n
x2/ x2 for i = 1,2,.., n. Note that the xi’s are not in deviation form as
i=1
in the case of a simple regression with an intercept. In this case tr(H/ =
n
tr(Px/ = tr(xx0//x0x = tr(x0x//x0x = x0x/x0x = 1. Hence, hii = 1.
n
xj 2 — xi2
j=1
n
From(8.18), (n2)s(i/2 = (n—1/s2 — = (n—1/s2 — e2 I ^ x’2/xj2 I.
/V /V x^e’ / n 1/2
From (8.19), DFBETASi=("—"«//sp/v’C^x)3^^ • E xi2 /sa).
2^xj i=1 /
c. From (8.21), DFFITi = y — yw = x0[" — "(i/] =
B. H. Baltagi, Solutions Manual for Econometrics, Springer Texts in Business and Economics, DOI 10.1007/9783642545481_8, © SpringerVerlag Berlin Heidelberg 2015
det [X0X] = x0x = X x2 and (1 – hii) = 1 ( x2 E x2^ = E x2 / E x2.
i—1 V i—1 J j^i 1 i—1
The last term is the ratio of the two determinants. Rearranging terms, one
can easily verify (8.27). From (8.26),
8.3 From(8.17)s;?i) = E ^yt — xt0"(i^ substituting (8.13), one gets
where hit = xt,(X, X) 1xi. Adding and subtracting the ith term of this summation, yields
8.4 Obtaining ei* from an Augmented Regression
a. In order to get "* from the augmented regression given in (8.5), one can premultiply by Pdi as described in (8.14) and perform OLS. The Frisch – WaughLovell Theorem guarantees that the resulting estimate of "* is the same as that from (8.5). The effect of Pdi is to wipe out the ith observation from the regression and hence " * = "(i) as required.
b. In order to get ® from (8.5), one can premultiply by Px and perform OLS on the transformed equation Pxy = Pxdi® + Pxu. The FrischWaugh – Lovell Theorem guarantees that the resulting estimate of ® is the same as that from (8.5). OLS yields ® = (d/Pxd^ 1 d/Pxy. Using the fact that e = Pxy, d/e = ei and di0Pxdi = di0Hdi = hii one gets ф = ei/(1 — hii) as required.
c. The FrischWaughLovell Theorem also states that the residuals from (8.5) are the same as those obtained from (8.14). This means that the ith observation residual is zero. This also gives us the fact that ф = yi — xi0"(i) = the forecasted residual for the ith observation, see below (8.14). Hence the RSS from (8.5) = P (yt — x/"рЛ since the ith observation contributes
a zero residual. As in (8.17) and (8.18) and problem 8.3 one can substitute (8.13) to get
®ф*
using the fact that PxX = 0. Hence " * = (X0X) 1X0y = "ols and
b. ф* = (DpPxD^) _ DpPxy = (DpPxD^ _ Dpe = (bp^Dp) _ ep since e = Pxj and Dpe = ep
c. The residuals are given by
y — X"ols — PxDp^DpPxD^ 1 ep = e — PxDp (DpPxDp) ^ so that the residuals sum of squares is
e0e + ep(DpPxDp)1DpPxD^DpPxD^)_ ep — 2e0PxDp (DpPxDp)_.
since Pxe = e and e0Dp = ep. From the FrischWaughLovell Theorem, (8.6) has the same residuals sum of squares as (8.6) premultiplied by Px, i. e., Pxy = PxDp®* + Pxu. This has also the same residuals sum of squares
as the augmented regression in problem 8.5, since this also yields the above regression when premultiplied by Px.
d. From (8.7), the denominator residual sum of squares is given in part (c). The numerator residual sum of squares is e0e—the RSS obtained in part (c). This yields ep (DpPxDp) ep as required.
e. For problem 8.4, consider the augmented regression y = X"* + Pxdi® + u. This has the same residual sum of squares by the FrischWaughLovell Theorem as the following regression:
Pxy = Pxdi® + Pxu
This last regression also have the same residual sum of squares as y = X"* + di® + u
Hence, using the first regression and the results in part (c) we get Residual Sum of Squares = (Residual Sum of Squares with di deleted)
■e, di (d0Pxdi) 1 d0e
when Dp is replaced by di. But d0e = ei and d0Pxdi = 1 — hii, hence this last term is e?/(1 — hii) which is exactly what we proved in problem 8.4, part (c). The Fstatistic for ® = 0, would be exactly the square of the tstatistic in problem 8.4, part (d).
8.6 Let A = X0X and a = b = x/. Using the updated formula
Note that X0 = [x1,.., xn] where Xi is kxl, so that
n
where hii = xi,(X, X) 1xi. This verifies (8.12). Note thatX0y = xjYj and
X(i/y(i) = X) xjyj = X0y – xiyi.
j^i
Therefore, postmultiplying ^X(i/X(i^ by X^y^) we get
Л, i – , , ,_i (X0X)1 xix0"
" (i) = (X'(i)X(i^ X(i/y(i) = " (X0X) xiyi
(X0X)1 xix0 (X0X)1 xiyi
1 hii
and
– – (X0X)1xix0" /1 – hii C hiA rv/v,1 x y
""(i) = H 1 — hii J(XX xiyi
(X’X)1 xi (y – xi") (X0X)1 xiei
which is identical.
8.8 det X0(i)X(i)l = det [X0X – xix0] = det [{Ik – xix0(X0X)1} X0X]. Let a = xi and b0 = x0(X0X)1. Using det[Ik – ab0] = 1 – b0a one gets
det [Ik – xix0(X0X)1] = 1 – x0(X0X)1xi = 1 – hii. Using det(AB) = det(A)
8.9 The cigarette data example given in Table 3.2.
a. Tables 8.1 and 8.2 were generated by SAS with PROC REG asking for all the diagnostic options allowed by this procedure.
b. For the New Hampshire (NH) observation number 27 in Table 8.2, the leverage h^H = 0.13081 which is larger than 2h = 2k/n = 0.13043. The internally studentized residual eNH is computed from (8.1) as follows: eNH 0.15991
sV 1 – hNH 0.16343V1 – 0.13081
The externally studentized residual eNH is computed from (8.3) as follows: eNH 0.15991
s<NH)VT=hNS 0.16323^/1 – 0.13081
where s(2NH) is obtained from (8.2) as follows:
(n – k)s2 – eNn/(1 – hNH)
42
both Єкн and e*NH) are less than 2 in absolute value. From (8.13), the change in the regression coefficients due to the omission of the NH observation is given by " – "(nh) = (X, X)_1xNHeNH/(1 – hNH) where (X’X)1 is given in the empirical example and xJNH = (1,0.15852, 5.00319) with eNH = 0.15991 and hNH = 0.13081. This gives (" – "(NH))/ = (0.3174, 0.0834, 0.0709). In order to assess whether this change is large or small, we compute DFBETAS given in (8.19). For the NH observation, these are given by
" 1 – ° 1 ,(nh) _ 0.3174
s(nh)T(XX)^ 0.163235 л/3°9298І69
Similarly, DFBETAS^ = 0.2573 and DFBETASNH,3 = 0.3608. These are not larger than 2 in absolute value. However, DFBETASNH,1 and DFBETASnh,3 are both larger than 2/pn = 2/V46 = 0.2949 in absolute value.
The change in the fit due to omission of the NH observation is given by (8.21). In fact,
DFFITnh = y NH — y(NH) = xNH[" — " (NH)]
/0.3174
or simply
hNHeNH (0.13081/(0.15991)
1 — hNH (1 — 0.13081)
Scaling it by the variance of y(NH) we get from (8.22)
This is not larger than the size adjusted cutoff of 2/Уk/n = 0.511. Cook’s distance measure is given by (8.25) and for NH can be computed as follows:
COVRATIO omitting the NH observation can be computed from (8.28) as
= 1.1422
which means that COVRATIONH — 1 = 0.1422 is less than 3k/n = 0.1956. Finally, FVARATIO omitting the NH observation can be computed from (8.29) as
c. Similarly, the same calculations can be obtained for the observations of the states of AR, CT, NJ and UT.
d. Also, the states of NV, ME, NM and ND.
8.10 The Consumption—Income data given in Table 5.3.
The following Stata output runs the ols regression of C on Y and generates the influence diagnostics one by one. In addition, it highlights the observations with the diagnostic higher than the cut off value recommended in the text of the chapter.
regcy 

Source 
SS 
df 
MS 
Number of obs F(1,47) Prob > F Rsquared Adj Rsquared Root MSE 
= 49 = 7389.28 = 0.0000 = 0.9937 = 0.9935 = 437.63 
Model Residual 
1.4152e+09 9001347.76 
1 47 
1.4152e+09 191518.037 

Total 
1.4242e+09 
48 
29670457.8 
c 
Coef. 
Std. Err. 
t 
P> t 
[95% Conf. Interval] 
y _cons 
.979228 1343.314 
.0113915 219.5614 
85.96 6.12 
0.000 0.000 
.9563111 1.002145 1785.014 901.6131 
. predict h, hat
. predict e if e(sample), residual. predict est1 ife(sample), rstandard. predict est2 if e(sample), rstudent. predict dfits, dfits. predict covr, covratio. predict cook if e(sample), cooksd. predict dfbet, dfbeta(y)
. list e est1 est2 cook h dfits covr, divider
e 
est1 
est2 
cook 
h 
dfits 
covr 

1. 
635.4909 
1.508036 
1.529366 
.0892458 
.0727745 
.4284583 
1.019567 
2. 
647.5295 
1.536112 
1.55933 
.0917852 
.0721806 
.4349271 
1.015024 
3. 
520.9777 
1.234602 
1.241698 
.0575692 
.0702329 
.3412708 
1.051163 
4. 
498.7493 
1.179571 
1.184622 
.0495726 
.0665165 
.3162213 
1.053104 
5. 
517.4854 
1.222239 
1.228853 
.0510749 
.064003 
.3213385 
1.045562 
6. 
351.0732 
.8263996 
.8235661 
.0208956 
.0576647 
.2037279 
1.075873 
7. 
321.1447 
.7538875 
.7503751 
.0157461 
.0525012 
.1766337 
1.075311 
8. 
321.9283 
.7540409 
.7505296 
.0144151 
.0482588 
.1690038 
1.070507 
9. 
139.071 
.3251773 
.3220618 
.0024888 
.0449572 
.0698759 
1.08818 
10. 
229.1068 
.5347434 
.5306408 
.0061962 
.0415371 
.1104667 
1.07598 
11. 
273.736 
.6382437 
.6341716 
.0083841 
.0395361 
.1286659 
1.068164 
12. 
17.04481 
.0396845 
.0392607 
.0000301 
.0367645 
.0076702 
1.083723 
13. 
110.857 
.25773 
.2551538 
.0011682 
.0339782 
.047853 
1.077618 
14. 
.7574701 
.0017584 
.0017396 
4.95e08 
.0310562 
.0003114 
1.077411 
15. 
311.9394 
.7226375 
.7189135 
.0072596 
.0270514 
.1198744 
1.049266 
16. 
289.1532 
.6702327 
.6662558 
.006508 
.0281591 
.1134104 
1.053764 
17. 
310.0611 
.7183715 
.7146223 
.0072371 
.0272825 
.119681 
1.049793 
18. 
148.776 
.3443774 
.3411247 
.0015509 
.0254884 
.0551685 
1.065856 
19. 
85.67493 
.1981783 
.1961407 
.0004859 
.0241443 
.0308519 
1.067993 
20. 
176.7102 
.4084195 
.4047702 
.0019229 
.0225362 
.0614609 
1.060452 
21. 
205.995 
.4759794 
.4720276 
.0025513 
.022026 
.0708388 
1.057196 
22. 
428.808 
.9908097 
.9906127 
.0110453 
.0220072 
.1485998 
1.023316 
23. 
637.0542 
1.471591 
1.490597 
.0237717 
.0214825 
.2208608 
.9708202 
24. 
768.879 
1.77582 
1.818907 
.034097 
.0211669 
.267476 
.928207 
25. 
458.3625 
1.058388 
1.059773 
.0118348 
.0206929 
.1540507 
1.015801 
26. 
929.7892 
2.146842 
2.23636 
.0484752 
.020602 
.3243518 
.8671094 
27. 
688.1302 
1.589254 
1.616285 
.0272016 
.0210855 
.2372122 
.9548985 
28. 
579.1982 
1.338157 
1.349808 
.019947 
.0217933 
.2014737 
.9874383 
29. 
317.75 
.734245 
.7305942 
.0061013 
.0221336 
.1099165 
1.043229 
30. 
411.8743 
.9525952 
.9516382 
.0111001 
.0238806 
.1488479 
1.028592 
31. 
439.981 
1.01826 
1.018669 
.0133716 
.0251442 
.1635992 
1.02415 
32. 
428.6367 
.9923085 
.9921432 
.0130068 
.0257385 
.1612607 
1.027102 
33. 
479.2094 
1.109026 
1.111808 
.0158363 
.0251048 
.1784142 
1.015522 
34. 
549.0905 
1.271857 
1.280483 
.0222745 
.0268017 
.2124977 
1.000132 
35. 
110.233 
.2553027 
.2527474 
.0008897 
.0265748 
.041761 
1.069479 
36. 
66.3933 
.1538773 
.1522699 
.0003404 
.0279479 
.0258193 
1.072884 
37. 
32.47656 
.0753326 
.0745313 
.0000865 
.0295681 
.0130097 
1.075499 
38. 
100.64 
.2336927 
.2313277 
.0008919 
.031631 
.0418084 
1.075547 
39. 
122.4207 
.2847169 
.2819149 
.001456 
.0346758 
.0534312 
1.077724 
40. 
103.4364 
.2414919 
.2390574 
.0012808 
.0420747 
.0501011 
1.087101 
year 
c 
est2 
dfbet 
h 
dfits 

1. 
1959 
8776 
1.529366 
.3634503 
.0727745 
.4284583 
2. 
1960 
8837 
1.55933 
.3683456 
.0721806 
.4349271 
47. 
2005 
26290 
1.955164 
.4722213 
.0744023 
.5543262 
48. 
2006 
26835 
1.611164 
.4214261 
.0831371 
.4851599 
49. 
2007 
27319 
1.562965 
.4323753 
.0900456 
.4916666 
. list 
year c est2 dfbet h covr if abs(covr1) 
>(3*(2/49)) 

year 
c 
est2 
dfbet 
h 
covr 

26. 
1984 
16343 
2.23636 
.0314578 
.020602 
.8671094 
The 
Gasoline data used 
in Chap. 10. 
The following 
SAS output gives the 
diagnostics for two countries: Austria and Belgium. 
a. AUSTRIA:
Dep Var 
Predict 
Std Err 
Lower95% 
Upper95% 
Lower95% 
Upper95% 

Obs 
Y 
Value 
Predict 
Mean 
Mean 
Predict 
Predict 
1 
4.1732 
4.1443 
0.026 
4.0891 
4.1996 
4.0442 
4.2445 
2 
4.1010 
4.1121 
0.021 
4.0680 
4.1563 
4.0176 
4.2066 
3 
4.0732 
4.0700 
0.016 
4.0359 
4.1041 
3.9798 
4.1602 
4 
4.0595 
4.0661 
0.014 
4.0362 
4.0960 
3.9774 
4.1548 
5 
4.0377 
4.0728 
0.012 
4.0469 
4.0987 
3.9854 
4.1603 
6 
4.0340 
4.0756 
0.013 
4.0480 
4.1032 
3.9877 
4.1636 
7 
4.0475 
4.0296 
0.014 
3.9991 
4.0601 
3.9407 
4.1185 
8 
4.0529 
4.0298 
0.016 
3.9947 
4.0650 
3.9392 
4.1205 
9 
4.0455 
4.0191 
0.019 
3.9794 
4.0587 
3.9266 
4.1115 
10 
4.0464 
4.0583 
0.015 
4.0267 
4.0898 
3.9690 
4.1475 
11 
4.0809 
4.1108 
0.016 
4.0767 
4.1450 
4.0206 
4.2011 
12 
4.1067 
4.1378 
0.022 
4.0909 
4.1847 
4.0420 
4.2336 
13 
4.1280 
4.0885 
0.015 
4.0561 
4.1210 
3.9989 
4.1782 
14 
4.1994 
4.1258 
0.022 
4.0796 
4.1721 
4.0303 
4.2214 
15 
4.0185 
4.0270 
0.017 
3.9907 
4.0632 
3.9359 
4.1180 
16 
4.0290 
3.9799 
0.016 
3.9449 
4.0149 
3.8893 
4.0704 
17 
3.9854 
4.0287 
0.017 
3.9931 
4.0642 
3.9379 
4.1195 
18 
3.9317 
3.9125 
0.024 
3.8606 
3.9643 
3.8141 
4.0108 
19 
3.9227 
3.9845 
0.019 
3.9439 
4.0250 
3.8916 
4.0773 
Rstudent
0.9821
0.3246
0.0853
0.1746
0.9387
1.1370
0.4787
0.6360
0.7554
0.3178
0.8287
0.9541
1.1008
2.6772
0.2318
1.4281
1.2415
0.6118
1.9663
X3
Dfbetas
0.5587
0.1009
0.0008
0.0118
0.0387
0.1219
0.1286
0.2176
0.3220
0.0607
0.1027
0.2111
0.1143
1.2772
0.0407
0.1521
0.1923
0.2859
0.0226
b. BELGIUM:
Dep Var 
Predict 
Std Err 
Lower95% 
Upper95% 
Lower95% 
Upper95% 

Obs 
Y 
Value 
Predict 
Mean 
Mean 
Predict 
Predict 
1 
4.1640 
4.1311 
0.019 
4.0907 
4.1715 
4.0477 
4.2144 
2 
4.1244 
4.0947 
0.017 
4.0593 
4.1301 
4.0137 
4.1757 
3 
4.0760 
4.0794 
0.015 
4.0471 
4.1117 
3.9997 
4.1591 
4 
4.0013 
4.0412 
0.013 
4.0136 
4.0688 
3.9632 
4.1192 
5 
3.9944 
4.0172 
0.012 
3.9924 
4.0420 
3.9402 
4.0942 
6 
3.9515 
3.9485 
0.015 
3.9156 
3.9814 
3.8685 
4.0285 
7 
3.8205 
3.8823 
0.017 
3.8458 
3.9189 
3.8008 
3.9639 
8 
3.9069 
3.9045 
0.012 
3.8782 
3.9309 
3.8270 
3.9820 
9 
3.8287 
3.8242 
0.019 
3.7842 
3.8643 
3.7410 
3.9074 
10 
3.8546 
3.8457 
0.014 
3.8166 
3.8748 
3.7672 
3.9242 
11 
3.8704 
3.8516 
0.011 
3.8273 
3.8759 
3.7747 
3.9284 
12 
3.8722 
3.8537 
0.011 
3.8297 
3.8776 
3.7769 
3.9304 
13 
3.9054 
3.8614 
0.014 
3.8305 
3.8922 
3.7822 
3.9405 
14 
3.8960 
3.8874 
0.013 
3.8606 
3.9142 
3.8097 
3.9651 
15 
3.8182 
3.8941 
0.016 
3.8592 
3.9289 
3.8133 
3.9749 
16 
3.8778 
3.8472 
0.013 
3.8185 
3.8760 
3.7689 
3.9256 
17 
3.8641 
3.8649 
0.016 
3.8310 
3.8988 
3.7845 
3.9453 
18 
3.8543 
3.8452 
0.014 
3.8163 
3.8742 
3.7668 
3.9237 
19 
3.8427 
3.8492 
0.028 
3.7897 
3.9086 
3.7551 
3.9433 
Std Err 
Student 
Cook’s 

Obs 
Residual 
Residual 
Residual 
2101 2 
D 
Rstudent 

1 
0.0329 
0.028 
1.157 
** 
0.148 
1.1715 

2 
0.0297 
0.030 
0.992 
* 
0.076 
0.9911 

3 
0.00343 
0.031 
0.112 
0.001 
0.1080 

4 
0.0399 
0.032 
1.261 
** 
0.067 
1.2890 

5 
0.0228 
0.032 
0.710 
* 
0.016 
0.6973 

6 
0.00302 
0.031 
0.099 
0.001 
0.0956 

7 
0.0618 
0.030 
2.088 
**** 
0.366 
2.3952 

8 
0.00236 
0.032 
0.074 
0.000 
0.0715 

9 
0.00445 
0.029 
0.156 
0.003 
0.1506 

10 
0.00890 
0.031 
0.284 
0.004 
0.2748 

11 
0.0188 
0.032 
0.583 
* 
0.011 
0.5700 

12 
0.0186 
0.032 
0.575 
* 
0.010 
0.5620 

13 
0.0440 
0.031 
1.421 
** 
0.110 
1.4762 

14 
0.00862 
0.032 
0.271 
0.003 
0.2624 

15 
0.0759 
0.030 
2.525 
***** 
0.472 
3.2161 

16 
0.0305 
0.031 
0.972 
* 
0.043 
0.9698 

17 
0.00074 
0.030 
0.025 
0.000 
0.0237 

18 
0.00907 
0.031 
0.289 
0.004 
0.2798 

19 
0.00645 
0.020 
0.326 
0.053 
0.3160 
Hat Diag 
Cov 
INTERCEP 
X1 
X2 
X3 

Obs 
H 
Ratio 
Dffits 
Dfbetas 
Dfbetas 
Dfbetas 
Dfbetas 
1 
0.3070 
1.3082 
0.7797 
0.0068 
0.3518 
0.0665 
0.4465 
2 
0.2352 
1.3138 
0.5497 
0.0273 
0.2039 
0.1169 
0.2469 
3 
0.1959 
1.6334 
0.0533 
0.0091 
0.0199 
0.0058 
0.0296 
4 
0.1433 
0.9822 
0.5272 
0.1909 
0.0889 
0.1344 
0.2118 
5 
0.1158 
1.3002 
0.2524 
0.0948 
0.0430 
0.0826 
0.1029 
6 
0.2039 
1.6509 
0.0484 
0.0411 
0.0219 
0.0356 
0.0061 
7 
0.2516 
0.4458 
1.3888 
0.0166 
0.9139 
0.6530 
1.0734 
8 
0.1307 
1.5138 
0.0277 
0.0028 
0.0169 
0.0093 
0.0186 
9 
0.3019 
1.8754 
0.0990 
0.0335 
0.0873 
0.0070 
0.0884 
10 
0.1596 
1.5347 
0.1198 
0.0551 
0.0959 
0.0243 
0.0886 
11 
0.1110 
1.3524 
0.2014 
0.0807 
0.1254 
0.0633 
0.1126 
12 
0.1080 
1.3513 
0.1956 
0.0788 
0.0905 
0.0902 
0.0721 
13 
0.1792 
0.9001 
0.6897 
0.5169 
0.0787 
0.5245 
0.1559 
14 
0.1353 
1.4943 
0.1038 
0.0707 
0.0596 
0.0313 
0.0366 
15 
0.2284 
0.1868 
1.7498 
1.4346 
1.1919 
0.7987 
0.7277 
16 
0.1552 
1.2027 
0.4157 
0.3104 
0.1147 
0.2246 
0.0158 
17 
0.2158 
1.6802 
0.0124 
0.0101 
0.0071 
0.0057 
0.0035 
18 
0.1573 
1.5292 
0.1209 
0.0564 
0.0499 
0.0005 
0.0308 
19 
0.6649 
3.8223 
0.4451 
0.1947 
0.0442 
0.3807 
0.1410 
Sum of Residuals 0
Sum of Squared Residuals 0.0175
Predicted Resid SS (Press) 0.0289
SAS PROGRAM
Data GASOLINE;
Input COUNTRY $ YEAR Y X1 X2 X3; CARDS;
DATA AUSTRIA; SET GASOLINE; IF COUNTRY=‘AUSTRIA’;
Proc reg data=AUSTRIA;
Model Y=X1 X2 X3 / influence p r cli clm;
RUN;
DATA BELGIUM; SET GASOLINE; IF COUNTRY=‘BELGIUM’;%
Proc reg data=BELGIUM;
Model Y=X1 X2 X3 / influence p r cli clm;
RUN;
8.13 Independence of Recursive Residual. This is based on Johnston (1984, p. 386).
a. Using the updating formula given in (8.11) with A = (X0Xt) and a = —b = x(+1,weget
(X0X. C xt+ixj+i)"1 = (X0X.):1
— (XXt)_1 xt+i (l C xt+i (XXt)_1 x.+i)^ x0+i (^Xt)_1
1 C x0+i (Х. Х.Г1 xt+i
which is exactly (8.31).
b. "t+1 = (Xt+iXt+i):1 X0+i Yt+1
Replacing X0,1 by (X0,xt+i) andYt+1 by ( :  yields
yt+1
"t+i = (X0+iXt+i) (X0Yt C xt+iyt+i)
Replacing (X0+1Xt+^ 1 by its expression in (8.31) we get
(X0Xt):1 xt+ix0+i" t
ft+i
(X0X.) 1 xt+ix0+1 (Х. Х.) 1 xt+i yt+i
ft+i
(X0Xt) xt+ixt+i"t ft+i — ft+i C 1
ft+i ft+i
= " t— (X0Xt):1 xt+i (у.+1 — x0+i" t) /ft+i where we used the fact that x.+1 (X0Xt) 1 xt+1 = ft+i — 1.
c. Using (8.30), wt+i = (yt+i – xt+1"t) /fi where ft+i = 1 + x0+1 (X(Xt) 1 xt+i. Defining vt+i = vft+1 wt+i we get vt+i = yt+i –
x0+1"t = xt+1(" — "t) + ut+1 for t = k, ..,T — 1. Since ut ~ IIN(0, o2), then wt+i has zero mean and var(wt+1) = o2. Furthermore, wt+1 are linear in the y’s. Therefore, they are themselves normally distributed. Given normality of the wt+i’s it is sufficient to show that cov(wt+1, ws+1) = 0 for t ф s; t = s = k,.., T — 1. But ft+i is fixed. Therefore, it suffices to show that cov(vt+1, vs+1) = 0 for t ф s.
Using the fact that "t = (X0Xt) 1 X0Yt = " + (X0Xt) 1 Xt0ut where ut = (u1,.., ut), we get vt+i = ut+i — x0+1 (X(Xt) 1 X{ut. Therefore,
E (vt+ivs+i) = ^[u,+i – x0+1(X0Xt)1X0u^[us+1 – x£+i (Х^Х^)1 X^]} = E (ut+iu+i) – x0+i (X0Xt)1 X0E (utus+i)
– E (ut+ius) Xs (Х^Х8)1 xs+i + x0+i (Х0Х0)1 X0E (utus) Xs (xsXs)1 xs+i.
Assuming, without loss of generality, that t < s, we get E(ut+1us+1) = 0, E(ut+1us+1) = 0 since t < s < s + 1, E (ut+1us) = (0,.., o2, ..0) where the o2 is in the (t + 1)th position, and 0u11
ut
Substituting these covariances in E(vt+1vs+1) yields immediately zero for the first two terms and what remains is
E (vt+ivs+i) = 02xt+i (XsXs)1 xs+1 + O2x0+i (Х0Х,)1 X0 [It, 0] Xs (xsXs)1 xs+1
But [It, 0]Xs = Xt for t < s. Hence
E(vt+ivs+i) = – o2xt+i(XsXs)1xs+i + o2xt+i(XsXs)1 xs+1 = 0 fort Ф s.
A simpler proof using the C matrix defined in (8.34) is given in problem 8.14.
8.14 Recursive Residuals are Linear Unbiased With Scalar Covariance Matrix (LUS).
a. The first element of w = Cy where C is defined in (8.34) is
Wk+i = —xk+1(XkXt) 1XkYk^/ft+1 + yk+1/VfC7
where Yk = (yi, ..,yk). Hence, using the fact that "k = (XkXk) 1 XkYk, we get wk+1 = (yk+1 — xk+1"^ /fn which is (8.30) for t = k. Similarly, the tth element of w = Cy from (8.34) is wt = —x( (Xt_1Xt_^ 1 X{_ 1Yt_1 Д/ft + yt/vf where Y(_ 1 = (y1,.., yt_1). Hence, using the fact that"t_1 = (X(_1Xt_1)_1 X(_1Yt_1,wegetwt = (yt — x("t_^/pft which is (8.30) for t = t — 1. Also, the last term can be expressed in the same way.
wt = —xT(XT_1Xt_1)_1XT_1Yt_1/pfT+Ут/PfT = (yT — xT"t_i )/pfT which is (8.30) for t = T — 1.
b. The first row of CX is obtained by multiplying the first row of C by
Xk
X
x
This yields —xk+1 (XkXk) 1 XkXk^fk+7 + xk+1/Vfk+7 = °.
Similarly, the tth row of CX is obtained by multiplying the tth row of C by
This yields — X (X(_1Xt_^ 1 X(_1Xt_1 Д/f) + x( / Vf = 0.
Also, the last row of CX is obtained by multiplying the Tth row of C by
X = [Xt,_1 xT
This yields —xT (XT_1Xr_1) 1 XT_1Xr_1/pfT C xT/pfT = 0.
This proves that CX = 0. This means that w = Cy = CX" + Cu = Cu since CX = 0.
Hence, E(w) = CE(u) = 0. Note that
0 0 .. so that the first diagonal element of CC0 is
xk+1(XkXk)1XkXk(XkXk)1xk+1 1 1 + xk+1(XkXk)1 xk+1
fk+1 fk+1 fk+1
similarly, the tth diagonal element of CC0 is x0(X01 Xt1)1 X01Xt1(X01Xt1)1xt 1 ft
ft
and the Tth diagonal element of CC0 is
xT(XT1Xt1)1XT1Xt1(XT1Xt1)1xt 1 ^ fp
fr
Using the fact that
Xk
Xt1 =
t1.
one gets the (1,t) element of CC0 by multiplying the first row of C by the tth column of C0. This yields
xk+1(XkXk)1XkXk(X(_1Xt1)1 xt _ xk+1(X(_1Xt1)1xt = 0 pfk+1 pft pfk+1 pft
Similarly, the (1,T) element of CC0 is obtained by multiplying the first row of C by the Tth column of C0. This yields xk+1 (XkXk) 1 XkXk (XT1 Xt – 1) 1 xr xk+1(XT1XT1)1xr
/fk+w0f^
Similarly, one can show that the (t,1)th element, the (t, T)th element, the (T,1)th element and the (T, t)th element of CC0 are all zero. This proves that CC0 = IT_k. Hence, w is linear in y, unbiased with mean zero, and var(w) = var(Cu) = C E(uu0)C0 = o2CC0 = o2IT_k, i. e., the variancecovariance matrix is a scalar times an identity matrix. Using Theil’s (1971) terminology, these recursive residuals are (LUS) linear unbiased with a scalar covariance matrix. A further result holds for all LUS
residuals. In fact, C0C = Px see Theil (1971, p. 208). Hence, TT
w2 = w0w = y0C0Cy = y0Pxy
t=k+i
Therefore, the sum of squares of the (T — k) recursive residuals is equal to the sum of squares of the T least squares residuals.
c. If u ~ N(0, o2IT) then w = Cu is also Normal since it is a linear function of normal random variables. From part (b) we proved that w has mean zero and variance o2IT_k. Hence, w ~ N(0, o2IT_k) as required.
d. Let us express the (t + 1) residuals as follows:
Yt+i — Xt+i" t+i = Yt+i — Xt+i" t — Xt+i(" t+i — " t) so that
RSSt+i = (Yt+i — Xt+i"t+i)0(Yt+i — Xt+i" t+i)
= (Yt+i — Xt+i" t)0(Yt+i — Xt+i" t) + (" t+i — " t )0Xt+iXt+i (" t+i — " t) — 2(" t+i — " t)0 X0+i(Yt+i — Xt+i" t)
where ft+i = 1 C x(+1 (X0Xt) 1 xt+i is defined in (8.32) and wt+1 is the recursive residual defined in (8.30). Next, we make the substitution for ("t+1 — "t) from (8.32) into the second term of RSSt+1 to get (" t+1 — " t),X0+1Xt+1(" t+1 — " t) = (yt+1 — x+1" t),x;+1(X( Xt)1(X;+1 Xt+1)(X0Xt)1xt+1(yt+1 — x0+1" t)
Postmultiplying (8.31) by xt+1 one gets (X0+1Xt+1)1xt+1 = (X0Xt) 1 xt+1 — (XtX*) xt+1c
1 C c
where c = x(+1 (X(xt) 1 xt+1. Collecting terms one gets (X0+1Xt+1) xt+1 = (X0Xt) xt+1/ft+1
where ft+1 = 1 C c.
Substituting this above, the second term of RSSt+1 reduces to
x0+1 (X0Xt)1 xt+1w2+1.
For the third term of RSSt+1, we observe that X0+^Yt+1 — X0+1"^ = [X0, xt+1] yt — X;"t * = X0(Yt — Xt"t) C xt+1(yt+1 — x]+1"t).
yt+1 xt+1"t
The first term is zero since Xt and its least squares residuals are orthogonal.
Hence, Xt+1(Yt+1 — Xt+1"t) = xt+1wt+1 ft+1 and the third term of RSSt+1 becomes — 2("t+1 — "t),xt+1wt+^/f’t+7 using (8.32), this reduces to — 2×0+1 (X£Xt) 1 xt+1w2+1. Adding all three terms yields
RSSt+1 = RSSt C ft+1w2+1 C xt+1 (X0Xt)1 xt+1w2+1
— 2×0+1 (X0XO 1 xt+1w0+1 = RSSt C ft+1w0!+1 — (ft+1 — 1)w0!+1 = RSSt C w0!+1 as required.
8.15 The Harvey and Collier (1977) Misspecification ttest as a Variable Additions Test. This is based on Wu (1993).
a. The Chow Ftest for Ho; у = 0 in (8.44) is given by
F RRSS — URSS _ y0Pxy — y, P[x, z]y
URSS/(T — k — 1) y0P[X, z]y/(T — k — 1).
Using the fact that z = C0iTk where C is defined in (8.34) and iTk is a vector of ones of dimension (T — k). From (8.35) we know that CX = 0, CC0 = ITk and C0C = Px. Hence, z0X = iTkCX = 0 and
Therefore, URSS = y0P[X, z]y = y0y — y0Pxy — y0Pzy and RRSS = y0PXy = y0y — y0Pxy and the Fstatistic reduces to
f = ___________ y0PY__________
y0(Px — Pz)y/(T — k — 1)
which is distributed as F(1, T — k — 1) under the null hypothesis Ho; у = 0. b. The numerator of the Fstatistic is
y0z(z0z) 1z0y = y0C0 iTk (iTkCC0it^ 1 iTkCy
ButCC0 = ITkand iTkCC0iTk = iTkiTk = (T—k). Therefore y0Pzy = y0C0iTkiT_kCy/(T—k). The recursive residuals are constructed as w = Cy, see below (8.34). Hence
T
T _
where w = ^2 wt/(T — k). Using Px = C0C, we can write
t=k+1
T
y0Pxy = y0C0Cy = w0w = ^ w2
t=k+1
Hence, the denomentator of the Fstatistic is y0(Px — Pz)y/(T — k — 1) =
T
J2 w2 — (T — k)w2
t=k+1
T
wheresww = ^2 (wt — w)2/(T — k — 1) was givenbelow(8.43).Therefore,
t=k+1
8.16 The Gasoline data model given in Chap. 10.
a. Using Eviews, and the data for AUSTRIA one can generate the regression
of y on X1, X2 and X3 and the corresponding recursive residuals. LS // Dependent Variable is Y Sample: 1960 1978 Included observations: 19
Variable 
Coefficient 
Std. Error 
tStatistic 
Prob. 
C 
3.726605 
0.373018 
9.990422 
0.0000 
X1 
0.760721 
0.211471 
3.597289 
0.0026 
X2 
0.793199 
0.150086 
5.284956 
0.0001 
X3 
0.519871 
0.113130 
4.595336 
0.0004 
0.001047
0.010798
0.042785
0.035901
0.024071
0.009944
0.007583
0.018521
0.006881
0.003415
0.092971
0.013540
0.069668
0.000706
0.070614
b. EViews also gives as an option a plot of the recursive residuals plus or minus twice their standard error. Next, the CUSUM plot is given along with 5% upper and lower lines.
Test Equation:
LS // Dependent Variable is Y Sample: 1960 1977 Included observations: 18
Variable Coefficient! 
Std. Error 
tStatistic 
Prob. 

C 
4.000172 
0.369022 
10.83991 
0.0000 
X1 
0.803752 
0.194999 
4.121830 
0.0010 
X2 
■0.738174 
0.140340 
5.259896 
0.0001 
X3 
0.522216 
0.103666 
5.037483 
0.0002 
Rsquared 
0.732757 
Mean dependent var 
4.063917 

Adjusted Rsquared 0.675490 
S. D. dependent var 
0.063042 

S. E. of regression 
0.035913 
Akaike info criterion 
6.460203 

Sum squared resid 
0.018056 
Schwarz criterion 
6.262343 

Log likelihood 
36.60094 
Fstatistic 
12.79557 

DurbinWatson stat 
2.028488 
Prob(Fstatistic) 
0.000268 
8.17 The Differencing Test in a Regression with Equicorrelated Disturbances. This is based on Baltagi (1990).
a. For the equicorrelated case £2 can be written as £2 = o2(1 — p)[ET + 9JT] where ET = IT — JT, JT = JT/T and 9 = [1 + (T — 1)p]/(1 — p).
£_1 = [Et C (1/9)Jt]/o2(1 — p) i0£_1 = i0/9o2(1 — p)
and L = Et/o2(1 — p). Hence " = (X, ETX)_1X, ETY, which is the OLS estimator of ", since ET is a matrix that transforms each variable into deviations from its mean. That GLS is equivalent to OLS for the equicorrelated case, is a standard result in the literature.
Also, D£ = o2(1 — p)D since Di = 0 and D£D0 = o2(1 — p)DD0. Therefore, M = PD//o2(1 — p) where PD = D,(DD,)_1D. In order to show that M = L, it remains to show that PD/ = ET or equivalently that PD/ C JT = IT from the definition of ET. Note that both PD/ and JT are symmetric idempotent matrices which are orthogonal to each other. (DJt = 0 since Di = 0/. Using Graybill’s (1961) Theorem 1.69, these two properties of Pd and JT imply a third: Their sum is idempotent with rank equal to the sum of the ranks. But rank of PD/ is (T1) and rank of JT is 1, hence their sum is idempotent of rank T, which could only be IT. This proves the Maeshiro and Wichers (1989) result, i. e., " = ", which happens to be the OLS estimator from (1) in this particular case.
b. The Plosser, Schwert and White differencing test is based on the difference between the OLS estimator from (1) and the OLS estimator from (2). But OLS on (1) is equivalent to GLS on (1). Also, part (a) proved that GLS on (1) is in fact GLS from (2). Hence, the differencing test can be based upon the difference between the OLS and GLS estimators from the differenced equation. An alternative solution is given by Koning (1992).
8.18 a. Stata performs Ramsey’s RESET test by issuing the command estat ovtest after running the regression as follows:
. reg Iwage wks south smsa ms exp exp2 occ ind union fem blk ed
(We suppress the regression output since it is the same as Table 4.1 in the text and solution 4.13).
. estat ovtest
Ramsey RESET test using powers of the fitted values of lwage Ho: model has no omitted variables F(3, 579) = 0.79 Prob>F = 0.5006
This does not reject the null of no regression misspecification.
b. Stata also performs White’s (1982) Information matrix test by issuing the command estat imtest after running the regression as follows:
. estat imtest
Cameron & Trivedi’s decomposition of IMtest
Source I 
chi2 
df 
p 
Heteroskedasticity  
103.28 
81 
0.082 
Skewness  
5.37 
12 
0.9444 
Kurtosis I ………………………. — 
1.76 
1 
0.1844 
Total  
110.41 
94 
0.1187 
This does not reject the null, even though heteroskedasticity seems to be a problem.
8.20 a. Stata performs Ramsey’s RESET test by issuing the command estat ovtest after running the regression as follows:
. reg lnc lnrp lnrdi
(We suppress the regression output since it is the same as that in solution 5.13).
. estat ovtest
Ramsey RESET test using powers of the fitted values of lnc Ho: model has no omitted variables F(3, 40) = 3.11 Prob>F = 0.0369
This rejects the null of no regression misspecification at the 5% level. b. Stata also performs White’s (1982) Information matrix test by issuing the command estat imtest after running the regression as follows:
. estat imtest
Cameron & Trivedi’s decomposition of IMtest
Source I 
chi2 
df 
p 
Heteroskedasticity  
15.66 
5 
0.0079 
Skewness  
3.59 
2 
0.1664 
Kurtosis  
0.06 
1 
0.8028 
………………………. Total  
19.31 
8 
0.0133 
This rejects the null, and seems to indicate that heteroskedasticity seems to be the problem. This was confirmed using this data set in problem 5.13.
References
Johnston, J. (1984), Econometric Methods (McGrawHill: New York).
Koning, R. H. (1992), “The Differencing Test in a Regression with Equicorrelated Disturbances,” Econometric Theory, Solution 90.4.5, 8: 155156.
Theil, H. (1971), Principles of Econometrics (Wiley: New York).
White, H. (1982), “Maximum Likelihood Estimation of Misspecified Models,” Econometrica, 50: 125.
Wu, P. (1993), “Variable Addition Test,” Econometric Theory, Problem 93.1.2, 9: 145146.
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