Can anything beat a GARCH(1,1)? 2. Spurious regressions–non-sta

50 Pages·2010·0.17 MB·English

Today’s Agenda
1. Will (cid:2)nish volatility modelling later
(cid:150) Can anything beat a GARCH(1,1)?
2. Spurious regressions(cid:150)non-stationary processes
3. Aside: Wold’s Decomposition-Stationary Processes
4. GMM(cid:150)Simple Introduction
(cid:150) ARCH estimation
(cid:150) Other volatility estimations
(cid:150) CCAPM estimation
5. MLE(cid:150)Simple Introduction
(cid:150) GARCH estimation
6. Announcements
1 Spurious Regressions and Non-Stationary Time
Series
Thus far, we have been looking at stationary time
(cid:15)
series. We have focused on r ; (cid:27)2; assuming
t t
that they are stationary. For some series, this
assumption is more tenable than for others.
But suppose you want to work with non-stationary
(cid:15)
time-series, i.e. prices, volume, number of investors
in a particular fund, number of funds, etc. Those
processes are inherently non-stationary.
Let p be the log-price. We know that
t
(cid:15)
p = p + "
t t 1 t
(cid:0)
or p is an AR(1) process with an unit-root.
t
This process is non-stationary. We cannot apply
(cid:15)
the CLT.
But we are still interested in testing the null (cid:30) = 1
(cid:15)
versus (cid:30) < 1:
Problem. Under the null, the process is non-
(cid:15)
stationary.
Under the alternative, the process is stationary.
(cid:15)
It turns out that (Funtional CLT)
(cid:15)
t
1 1
p = "
t s
pT pT
s=1
X
[rT]
1
= " W (r)
s
pT )
s=1
X
where W (r) is a Brownian motion on [0; 1] : I.e.,
W(r) N(0; r); 0 < r < 1:
(cid:24)
Note that this result is asymptotic. We don’t have
(cid:15)
to assume that the " s are normal (hence the
0
Functional CLT, or FCLT).
Q: Can’t we standardize the non-stationary
(cid:15)
processes by a power of T in order for them to
converge.
A: Yes.
(cid:15)
Let’s get a (cid:147)(cid:3)avor(cid:148) of how things work:
(cid:15)
Recall that
(cid:15)
p p p ((cid:30)p + " )
^ t t 1 t 1 t 1 t
(cid:30) = =
(cid:0) (cid:0) (cid:0)
p2 p2
P t 1 P t 1
(cid:0) (cid:0)
p "
t 1 t
= (cid:30)P+ P
(cid:0)
p2
P t 1
(cid:0)
If (cid:30) < 1; we had
P
(cid:15)
1
" p
^ T t t 1 p
(cid:30) = (cid:30) + (cid:30)
(cid:0)
1 p2 !
TP t 1
(cid:0)
pT (cid:30)^ (cid:30) N 0; (cid:27)2
P
^
(cid:0) (cid:24) (cid:30)
(cid:16) (cid:17) (cid:16) (cid:17)
But if (cid:30) = 1; the results do not hold. But
(cid:15)
" p
^ t t 1
(cid:30) = (cid:30) +
(cid:0)
p2
P t 1
(cid:0)
^
T (cid:30) (cid:30) O (1)
p P
(cid:0) )
(cid:16) (cid:17)
^
In other words, (cid:30) is super-consistent.
(cid:15)
^
Q: But since we don’t know the distribution of (cid:30);
(cid:15)
can we use this result for testing?
A: Yes, if we simulate the distribution.
(cid:15)
Dickey-Fuller (DF) Test:
(cid:15)
H : (cid:30) = 1
o
H : (cid:30) < 1
a
^
(cid:30) 1
The test is: t =
(cid:0)
(^)
(cid:15) se (cid:30)
Suppose that " follows an AR(p) process. The
t
(cid:15)
distribution of the DF test is in(cid:3)uenced by those
parameters. Not good.
To get rid of those parameters, we run the following
(cid:15)
regression:
p = (cid:30)p + (cid:16) (cid:1)p + (cid:16) (cid:1)p + ::: + (cid:16) (cid:1)p + (cid:29)
t t 1 1 t 1 2 t 2 k t=k t
(cid:0) (cid:0) (cid:0)
^
(cid:30) 1
Then, t = :
(cid:0)
(^)
(cid:15) se (cid:30)
This is called the Augmented DF,or ADF test.
(cid:15)
Summary of ADF test(cid:150)Testing for a unit root:
(cid:15)
(cid:150) Regress p on p ; (cid:1)p ; (cid:1)p ; :::; (cid:1)p
t t 1 t 1 t 2 t=k
(cid:0) (cid:0) (cid:0)
^
(cid:30) 1
(cid:150) Form: t =
(cid:0)
(^)
se (cid:30)
(cid:150) Get the critical value from simulations.
So, is working with non-stationary variables that
(cid:15)
easy?
NO:
(cid:15)
Suppose p1 and p2 are the prices of the same asset
t t
(cid:15)
traded on two markets. Then, it must be the case
that
1 2
p = p
t t
Empirically, this is almost true. We (cid:2)nd
(cid:15)
1 2
p p = "
t t t
(cid:0)
where " is almost iid, and E (" ) = 0:
t t
How do we take advantage of this?
(cid:15)
Regress:
(cid:15)
1 2
p = (cid:13)p + "
t t t
If
(cid:15)
" > 0
t
1 2
p > (cid:13)p
t t
The asset is (cid:147)too expensive(cid:148) in market 1.
(cid:15)
Easy, right?
(cid:15)
Wrong!
(cid:15)
Suppose we have two assets, with prices p and
t
(cid:15)
q : One might be tempted to look for arbitrage
t
strategies as in
p = (cid:13)q + "
t t t
If (cid:13) > 0; there is a relationship, and we can trade.
(cid:15)
No.
(cid:15)
Suppose
(cid:15)
p = p + u
t t 1 t
(cid:0)
q = q + v
t t 1 t
(cid:0)
cov (u v ) = 0
t t
Note: The two log-prices represent two independent
(cid:15)
discrete Brownian motions.
But:
(cid:15) T T t t
p q v u
t=1 t t t=1 s=1 s s=1 s
(cid:13)^ = =
T q2 T t v 2
P t=1 t P (cid:0)Pt=1 s(cid:1)=(cid:0)1Ps (cid:1)
1 T t t
v u
TP2 t=1 s=1 s P s=(cid:0)P1 s (cid:1)
= = O (1)
p
2
1 T t
v
PT2 (cid:0)Pt=1 s(cid:1)=(cid:0)1Ps (cid:1)
Similarly:
P (cid:0)P (cid:1)
(cid:15)
(cid:13)^
t =
se ((cid:13)^)
is not consistent.
The R2 does not converge to 0; as T :
(cid:15) ! 1
Illustration: spurious.m
(cid:15)
Co-integration:
Suppose that p and q are unit root (integrated)
t t
(cid:15)
processes, but there is a linear combination of p
t
and q that is stationary. That is, there exists a
t
vector (cid:13) = 1 (cid:13) such that
1
(cid:0)
p (cid:13)q = "
t t t
(cid:2) (cid:3) (cid:0)
and " is a stationary process. Then, p and q are
t t t
said to be cointegrated.
There are formal tests for cointegration, but they
(cid:15)
have low power against the alternative. Why?
(Think spurious correlation).
Cointegration is only between contemporaneous
(cid:15)
variables. I.e. p (cid:13)q is a cointegrating vector.
t t
(cid:0)
But (cid:1)p is not. However, the latter is also a way of
t
stationarizing the process.
Cointegration occurs (cid:147)naturally(cid:148) in economics. It is
(cid:15)
dictated by theory.