ε
t
1
=
u
t
1
+ (
ρ

λ
)
u
t
2
+
ρ
(
ρ

λ
)
u
t
3
+
ρ
2
(
ρ

λ
)
u
t
4
+ .
..
Therefore, the middle term is zero and the third is simply
λσ
u
2
.
Thus,
Cov[
ε
t
,
ε
t
1
] =
σ
u
2
{[
ρ
(1 +
λ
2
 2
ρλ
)]/(1 
ρ
2
)

λ
]} =
σ
u
2
[(
ρ

λ
)(1 
λρ
)/(1 
ρ
2
)]
For lags greater than 1,
Cov[
ε
t
,
ε
tj
] =
ρ
Cov[
ε
t
1
,
ε
tj
] + Cov[
ε
tj
,
u
t
] 
λ
Cov[
ε
tj
,
u
t
1
].
Since
ε
tj
involves only
u
s up to its current period,
ε
tj
is uncorrelated with
u
t
and
u
t
1
if
j
is greater than 1.
Therefore, after the first lag, the autocovariances behave in the familiar fashion, Cov[
ε
t
,
ε
tj
]
=
ρ
Cov[
ε
t
,
ε
t

j
+1
]
The autocorrelation coefficient of the residuals estimates Cov[
ε
t
,
ε
t
1
]/Var[
ε
t
]
=
(
ρ

λ
)(1 
ρλ
)/(1 +
λ
2
 2
ρλ
).
3.
Since the regression contains a lagged dependent variable, we cannot use the DurbinWatson statistic
directly.
The
h
statistic in (1534) would be
h
= (1  1.21/2)[21 / (1  21(.18
2
)]
1/2
=
3.201.
The 95% critical
value from the standard normal distribution for this onetailed test would be 1.645.
Therefore, we would
reject the hypothesis of no autocorrelation.
4.
It is commonly asserted that the DurbinWatson statistic is only appropriate for testing for first order
autoregressive disturbances.
What combination of the coefficients of the model is estimated by the
DurbinWatson statistic in each of the following cases:
AR(1), AR(2), MA(1)?
In each case, assume that the
regression model does not contain a lagged dependent variable.
Comment on the impact on your results of
relaxing this assumption.
In each case,
plim
d
=
2  2
ρ
1
where
ρ
1
=
Corr[
ε
t
,
ε
t
1
].
The first order autocorrelations are as
follows: AR(1):
ρ
(see (159))
and AR(2):
θ
1
/(1 
θ
2
).
For the AR(2), a proof is as follows:
First,
ε
t
=
θ
1
ε
t
1
+
θ
2
ε
t
2
+
u
t
.
Denote Var[
ε
t
] as
c
0
and Cov[
ε
t
,
ε
t
1
] as
c
1
.
Then, it follows immediately that
c
1
=
θ
1
c
0
+
θ
2
c
1
since
u
t
is independent of
ε
t
1
.
Therefore
ρ
1
=
c
1
/
c
0
=
θ
1
/(1 
θ
2
).
For the MA(1):

λ
/ (1 +
λ
2
)
(See
(1543)).
To prove this, write
ε
t
=
u
t

λ
u
t
1
.
Then, since the
u
s are independent, the result follows just by
multiplying out
ρ
1
= Cov[
ε
t
,
ε
t
1
]/Var[
ε
t
]
=

λ
Var[
u
t
1
]/{Var[
u
t
] +
λ
2
Var[
u
t
1
]}
=

λ
/(1 +
λ
2
).
Applications
1.
Phillips Curve
> date;1950.1$
> peri;1950.12000.4$
> crea;dp=inflinfl[1]$
> crea;dy=loggdploggdp[1]$
> peri;1950.32000.4$
> regr;lhs=dp;rhs=one,unemp$;ar1;res=u$
++
 Ordinary
least squares regression
Weighting variable = none

 Dep. var. = DP
Mean=
.1926996283E01, S.D.=
2.818214558

 Model size: Observations =
202, Parameters =
2, Deg.Fr.=
200 
 Residuals:
Sum of squares= 1592.321197
, Std.Dev.=
2.82163 
 Fit:
Rsquared=
.002561, Adjusted Rsquared =
.00243 
 Model test: F[
1,
200] =