Springer Texts in Business and Economics

Regressions with Non-zero Mean Disturbances

a. For the gamma distribution, E(ui) = 0 and var(ui) = 0. Hence, the distur­bances of the simple regression have non-zero mean but constant variance. Adding and subtracting 0 on the right hand side of the regression we get Yi = (a + 0) + "Xi + (ui - 0) = a* + "Xi + ui* where a* = a + 0

and ui* = ui — 0 with E (ui*) = 0 and var (ui*) = 0. OLS yields the BLU estimators of a* and " since the ui* disturbances satisfy all the require­ments of the Gauss-Markov Theorem, including the zero mean assumption. Hence, E(aols) = a* = a + 0 which is biased for a by the mean of the disturbances 0. ButE(s2) = var(u*) = var (ui) = 0. Therefore,

E Cols - s2) = E (aols) - E(s2) = a + 0 - 0 = a.

b. Similarly, for the x2 distribution, we have E(ui) = v and var(u^ = 2v. In this case, the same analysis applies except that E(aols) = a + v and E(s2) = 2v. Hence, E((Ools - s2/2) = a.

c. Finally, for the exponential distribution, we have E(ui) = 0 and var(ui) = 02. In this case, plim aols = a + 0 and plim s2 = 02. Hence, plim (aols - s) = plim aols - plim s = a.

5.10 The Heteroskedastic Consequences of an Arbitrary Variance for the Initial Dis­turbance of an AR(1) Model. Parts (a), (b), and (c) are based on the solution given by Kim (1991).

a. Continual substitution into the process ut = put_i + et yields

t-i

ut = p‘uo + ^ p£t-j. j=0

Exploiting the independence conditions, E(uo-j) = 0 for all j, and E(ei£j) = 0 for all i ф j, the variance of ut is

t-i t-i

ot2 = var(ut) = p2t var(uo) + ^ p2j var (-н) = p2toF2/x + a©2 ^ p2j

j=0 j=0

= p2to2/x + o2[(1 - p2t)/(1 - p2)]

= [{1/(1 - p2) + [1/x - 1/(1 - p2)]p2t} oe2 .

This expression for o2 depends on t. Hence, for an arbitrary value of x, the disturbances are rendered heteroskedastic. If x = 1 - p2 then o2 does not depend on t and o2 becomes homoskedastic.

b. Using the above equation for ot2, define

a = CTt2 - СТ-1 = [(1/x) - 1/(1 - p2)] p2t(1 - p_2)CTe2 = bc

whereb = (1/x) —1/(1 — p2) andc = p2t(1-p-2)o2. Note that c < 0. Ifx >

1 - p2, then b < 0 implying that a >0 and the variance is increasing. While

if x < 1 - p2, then b >0 implying that a <0 and the variance is decreasing.

If x = 1 - p2 then a = 0 and the process becomes homoskedastic.

Эст2 2oF2p2tlog p 2

Alternatively, one can show that —- = — --------- — [(1 - p2) - x]

@t x(1 - p2)

where @abx/@x = babx log a, has been used. Since log p < 0, the right hand side of the above equation is positive (ot2 is increasing) if x > 1 - p2, and is negative otherwise.

t-1

p*uo C J2 Pj ©t-j, and noting that E(ut) = 0 for all t, finds, for

j=0

t-1 t-s-1

cov(ut, ut-s) = E(utut-s) = p‘uo C ^2 Pi£‘-i I P* suo + ^2 p©t-s-j

i=0 j=0

t-s-1

= p2t-svar(uo) + ps p2i var(et-s-i)

i=0

d.

The Bias of the Standard Errors of OLS Process with an Arbitrary Variance on the Initial Observation. This is based on the solution by Koning (1992). Consider the time-series model

Also, note that in the stationary case (x = 1 — p2), we have:

since each term in the double summation is non-negative. Hence, the true variance of "ols (the left hand side of the last equation) is greater than the

estimated variance of "ols (the right hand side of the last equation). There­fore, the true t-ratio corresponding to Ho; " = 0 versus Hi; " ф 0 is lower than the estimated t-ratio. In other words, OLS rejects too often.

From the equation for ot2 it is easily seen that 9ot2/Эх < 0, and hence, that the true variance of "ols is larger than the estimated var("ols) if х < 1 — p2. Hence, the true t-ratio decreases and compared to the stationary case, OLS rejects more often. The opposite holds for х > 1 — p2.