Interval Forecasts for MA and AR Models
Assume that yt is a covariance stationary
time series.
Let yˆ T  h ,T denote the h-step ahead forecast of
y formed at time T (=E(yT+h│yT,yT-1,…)).
The h-step ahead forecast error is:
eT  h ,T  yT  h  yˆT  h ,T
Note that E(eT+h,T) = 0, that is, the h-step
ahead forecasts we have constructed are
unbiased.
Proof:
E (eT  h,T )  E ( yT  h  yˆT  h ,T )
= E(yT+h)- E[E(yT+h│yT,yT-1,…)]
= E(yT+h)- E(yT+h)
=0
Let  h denote the variance of the h-step
forecast error, i.e.,
2
 h2  E(eT2h,T )
Assume that the ε’s (and, therefore, the
y’s) are normally distributed:
εt ~NWN(0,σ2)
Then
yˆT  h ,T  2 h
is an approximate 95% forecast interval for
yT+h . More generally,
yˆT  h,T  Z1  h
is (1-2α)*100 percent forecast interval for
yT+h, where Z1-α is the (1-α)*100 percentile
of the N(0,1) distribution.
I. yt ~ WN
If
yt = εt, εt ~ NWN(0,σ2)
then
yˆ T  h ,T  0 for all h > 0
eT+h,T = εT+h
 h2  E(eT2h,T )  E( T2 H )   2
So, the 95% FI for yT+h is simply
[-2σ,2σ]
What is σ? Since yt = εt,
σ2 = E(εt2)=E(yt2). So, a natural estimator
of σ2 is the sample mean of the yt2’s, i.e.,
1 T 2
̂ 
 yt
T 1
II. yt ~ MA(1)
Suppose
yt = εt + θεt-1
εt ~ NWN(0,σ2)
h = 1:
yT+1 = εT+1 + θεT
yˆT 1,T   T
eT+1,T = εT+1
12  E(eT21,T )  E( T21 )   2
So, the 95% FI for yT+1 is
θεT + 2σ
In practice, we would replace θ and σ with
estimates:
1) Get an estimate of θ from the estimated
MA(1) model, ˆ
2) Recall that for the MA(1) model
Var(yt) = (1+ θ2)σ2
So,
σ2 = Var(yt)/ (1+ θ2)
and, we can estimate σ2 by
1 T 2
ˆ  [  y t ] /(1  ˆ 2 )
T 1
2
Note: In interpreting the resulting interval as
a “95-percent” interval, we are ignoring the
effects of using estimates of θ and σ.
h = 2:
yT+2 = εT+2 + θεT+1
yˆT  2,T  0
eT+2,T = εT+2 + θεT+1
 22  E(eT22,T )  E( T22   2 T21   T 1 T 2 )  (1   2 ) 2
So, the 95% FI for yT+2 is
0  2 (1   2 ) 2
i.e.,
[2 (1   2 ) 2 ,2 (1   2 ) 2 ]
(To make this operational, we can replace θ
and σ with the estimates we used for the
one-step ahead interval. As before, we are
not accounting for “parameter uncertainty”
when we make these substitutions.)
For h > 2:
yT+h = εT+h + θεT+h-1
yˆT  h ,T  0
eT+h,T = εT+h + θεT+h-1
 h2  E(eT2h,T )  E( T2h   2 T2h1   T h T h1 )  (1   2 ) 2
So, the 95% FI for yT+h is
0  2 (1   2 ) 2
i.e.,
[2 (1   2 ) 2 ,2 (1   2 ) 2 ]
(To make this operational, we can replace θ
and σ with the estimates we used for the
one-step ahead interval. As before, we are
not accounting for “parameter uncertainty”
when we make these substitutions.)
III. yt ~ MA(q)
Suppose
yt = εt + θ1εt-1 +…+ θqεt-q
εt ~ NWN(0,σ2)
for h < q:
yˆ T  h ,T   h T   h1 T 1  ...   q  T q  h
eT+h,T = εT+h + θ1εT+h-1 + …+ θh-1εT+1
 h2  (1  12  ...   h21 ) σ2
So, the 95% FI for yT+h:
yˆT h  2 (1  12  ...   h21 ) 2
for h > q:
yˆT  h ,T  0
eT+h,T = εT+h + θ1εT+h-1 + …+ θqεT+h-q
 h2  (1  12  ...   q2 ) σ2
So, the 95% FI for yT+h:
yˆT  h  2 (1  12  ...   q2 ) 2
To make these operational, we replace the
θ’s with the estimates obtained from fitting
the MA(q). We estimate σ2 by using
 2  var( y) /(1  12  ...  q2 )