Is Autocorrelated

Since vt = ut — ut—1, it is easy to show that E(vt) = E(ut — ut—1) = E(ut) — E(ut—1) = 0, since E(u) = 0, for each t. Now, var(vt) = var(ut — ut—1) = var (ut) + var(ut—1) = 2a2, since the variance of each ut is a2 and the us are independently distributed. Hence, vt is homoscedastic. But cov (vt, vt— 1) = E(vtvt—1) = E[(u — ut—1)(ut—1 — ut—2)] = —a 2

which is obviously nonzero. Therefore, although the u's are not autocorre-lated, the v's are.

12A.2 PROOF OF EQUATIONS (12.2.3), (12.2.4), AND (12.2.5)

Therefore,

because the us and e's are uncorrelated.

CHAPTER TWELVE: AUTOCORRELATION 505

Since var (ut) = var (ut—1) = a2 and var(et) = of, we get

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