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L1592_Frame_C41 Page 367 Tuesday, December 18, 2001 3:24 PM

Why Autocorrelation Distorts the Variance Estimates
Suppose that the system generating the data has the true underlying relation η = β 0 + β 1 x, where x could
be any independent variable, including time as in a times series of data. We observe n values: y 1 = η +
e 1 ,…, y i−2 = η + e i−2 , y i−1 = η + e i−1 , y i = η + e i ,…, y n = η + e n . The usual assumption is that the residuals
(e i ) are independent, meaning that the value of e i is not related to e i−1 , e i−2 , etc. Let us examine what
happens when this is not true.
Suppose that the residuals (e i ), instead of being random and independent, are correlated in a simple
way that is described by e i = ρ e i−1 + a i , in which the errors (a i ) are independent and normally distributed
2
with constant variance σ . The strength of the autocorrelation is indicated by the autocorrelation
coefficient (ρ), which ranges from −1 to +1. If ρ = 0, the e i are independent. If ρ is positive, successive
values of e i are similar to each other and:

e i = ρe i−1 + a i
e i−1 = ρe i−2 + a i−1
e i−2 = ρe i−3 + a i−2

and so on. By recursive substitution we can show that:
(
e i = ρρe i−2 + a i−1 ) + a i = ρ e i−2 + ρa i−1 + a i
2
and

3
2
e i = ρ e i−3 + ρ a i−2 + ρa i−1 + a i
This shows that the process is “remembering” past conditions to some extent, and the strength of this
memory is reflected in the value of ρ.
Reversing the order of the terms and continuing the recursive substitution gives:

e i = a i + ρa i−1 + ρ a i−2 + ρ a i−3 + ⋅⋅⋅ρ a i−n
3
n
2
The expected values of a i , a i−1 ,… are zero and so is the expected value of e i . The variance of e i and the
variance of a i , however, are not the same. The variance of e i is the sum of the variances of each term:
(
(
σ e = Var a i + ρ Var a i−1 ) + ρ Var a i−2 ) + ⋅⋅⋅ + ρ Var a i−n ) + ⋅⋅⋅
()
(
2
4
2n
2
2
By definition, the a’s are independent so σ a = Var(a i ) = Var(a i−1 ) = … = Var(a i−n ). Therefore, the variance
of e i is:
2
4
σ e = σ a 1 +( ρ + ρ + … + ρ + … )
2
2n
2
For positive correlation (ρ > 0), the power series converges and:
1
2n
4
( 1 + ρ + ρ + … + ρ + … ) = --------------
2
1 ρ– 2
Thus:
1
σ e = σ a --------------
2
2
1 ρ– 2
© 2002 By CRC Press LLC