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1. The Hodrick-Prescott (HP) filter is defined as minimizer mˆ = (mˆ1, . . . , mˆT )′
of
Ψ (m; λ) =
T∑
t=1
(yt − mt)2
+ λ
T∑
t=3
(mt − 2mt−1 + mt−2)2
with respect to m = (m1, . . . , mT )′
. Rewrite this penalized least squares criterion in ma-
trix notation and show that the HP filter is a linear filter, i.e. that it may be represented
as
ˆm = A(λ)y
for some non-stochastic matrix A(λ) depending on the smoothing parameter λ > 0.
Solution When
we mark
y =



y1
...
yT


, m =



m1
...
mT


 and B =







1 −2 1 0 · · · 0
0 1 −2 1 0 · · · 0
... ... ...
...
...
...
...
... 0
0 · · · 1 −2 1







we can rewrite Hodric-Precsott filter as:
min
m
Ψ (m; λ) = (y − m)T
(y − m) + λ (Bm)T
(Bm) .
We can find minimum ˆm of equation when we derive that and set this derivative equal
to zero:
∂Ψ (m; λ)
∂m
= 2m − 2y + λ.2.BT
Bm= 0
When we prove that the second derivative is positive definite, we know that ˆm for
which
ˆm + λBT
Bm = y
is really minimimum for which we search.
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2. Because for any vector c ̸= 0 we have:
cT
(
I + λBT
B
)
c = cT
c
>0
+ λ (Bc)T
(Bc)
≥0
we know that the second derivative is positive definite i.e:
∂2
Ψ(m; λ)
∂m∂m
= 2
(
I + λBT
B
)
> 0.
Therefore we can say that our ˆm is
ˆm =
(
I + λBT
B
)−1
A(λ)
y = A(λ)y.
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3. Suppose that a time series {yt} can be decomposed into deterministic trend mt, a
seasonal component st and irregular (random) fluctuations ut such that yt = mt+st+ut
for t = 1, . . . , T. Show that the weighted moving average filter
y∗
t =
1
9
(−yt−2 + 4yt−1 + 3yt + 4yt+1 − yt+2) t = 3, . . . , T − 2
has the following properties:
a) A cubic time trend, mt =
∑3
j=0 βjtj
, passes the filter without distortion,
i.e. m∗
t = mt. (3 points)
b) A seasonal component st with period 3 (i.e. st = st+3 for all t) is eliminated by the
filter, that is s∗
t = const. (3 points)
Solution
We are only interested in what happens when we pass our series through filter consid-
ering separate somponents.
yt = mt + st + ut
weighted moving average filter
; ; ; ; ; y∗
t = m∗
t + s∗
t + u∗
t
So we will try to explain separate new components by using old ones.
a)
m∗
t = 1
9
(−mt−2 + 4mt−1 + 3mt + 4mt+1 − mt+2)
=
1
9
(
−
∑3
j=0 βj(t − 2)j
+ 4
∑3
j=0 βj(t − 1)j
+ 3
∑3
j=0 βjtj
+
+4
∑3
j=0 βj(t + 1)j
−
∑3
j=0 βj(t + 2)j
)
=
−β0 −β1(t − 2) −β2(t − 2)2
−β3(t − 2)3
+4β0 +4β1(t − 1) +4β2(t − 1)2
+4β3(t − 1)3
+3β0 +3β1(t) +3β2(t)2
+3β3(t)3
+4β0 +4β1(t + 1) +4β2(t + 1)2
+4β3(t + 1)3
−β0 −β1(t + 2) −β2(t + 2)2
−β3(t + 2)3
.
We sum all coeficients which belong to t1
, . . . , t3
respectively. (It’s easier to look
at that column by column.) And we get:
m∗
t =
1
9
(−9mt−2 + 36mt−1 + 27mt + 36mt+1 − 9mt+2) = mt.
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4. b)
s∗
t = 1
9
(−st−2 + 4st−1 + 3st + 4st+1 − st+2)
= 1
9
(−st+1 + 4st+2 + 3st + 4st+1 − st+2)
= 1
3
(st + st+1 + st+2) .
But also
s∗
t+1 = 1
3
(st+1 + st+2 + st+3)
= 1
3
(st+1 + st+2 + st )
and identical for next ones too. So it is obvious that s∗
t is the same for all
t ∈ {3, . . . T − 2}, thus constant.
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