Hi : I'm a R novice but I consider myself reasonably versed in time
series related material and
I have never heard of an equivalence between Garch(1,1) for volatility
and an ARMA(1,1) in the squared returns
and I'm almost sure there isn't.
There are various problems with what you wrote.
1) r(t) = h(t)*z(t) not h(i) but that's not a big deal.
2) you can't write the equation in terms of r(t) because r(t) h(t)*z(t) and
h(t) is UPDATED FIRST
And then the relation r(t) = h(t)*z(t) is true ( in the sense of the
model ). So, r(t) is
a function of z(t) , a random variable, so trying to use r(t) on the
left hand side of the volatility
equation doesn't make sense at all.
3) even if your equation was valid, what you wrote is not an ARMA(1,1).
The AR term is there but the MA term
( the beta term ) Has an r_t-1 terms in it when r_t-1 is on the left
side. An MA term in an ARMA framework
multiples lagged noise terms not the lag of what's on the left side.
That's what the AR term does.
4) even if your equation was correct in terms of it being a true
ARMA(1,1) , you
Have common coefficients on The AR term and MA term ( beta ) so you
would need contraints to tell the
Model that this was the same term in both places.
5) basically, you can't do what you
Are trying to do so you shouldn't expect to any consistency in estimates
Of the intercept for the reasons stated above.
why are you trying to transform in such a way anyway ?
Now back to your original garch(1,1) model
6) a garch(1,1) has a stationarity condition that alpha + beta is less
than 1
So this has to be satisfied when you estimate a garch(1,1).
It looks like this condition is satisfied so you should be okay there.
7) also, if you are really assuming/believe that the returns have mean
zero to begin with, without subtraction,
Then you shouldn't be subtracting the mean before you estimate
Because eseentially you will be subtracting noise and throwing out
useful
Information that could used in estimating the garch(1,1) parameters.
Maybe you aren't assuming that the mean is zero and you are making the
mean zero which is fine.
I hope this helps you. I don't mean to be rude but I am just trying to
get across that what you
Are doing is not valid. If you saw the equivalence somewhere in the
literature,
Let me know because I would be interested in looking at it.
mark
-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Megh Dal
Sent: Tuesday, November 07, 2006 2:24 AM
To: r-help at stat.math.ethz.ch
Subject: [R] Comparison between GARCH and ARMA
Dear all R user,
Please forgive me if my problem is too simple.
Actually my problem is basically Statistical rather directly R related.
Suppose I have return series ret
with mean zero. And I want to fit a Garch(1,1)
on this.
my is r[t] = h[i]*z[t]
h[t] = w + alpha*r[t-1]^2 + beta*h[t-1]
I want to estimate the three parameters here;
the R syntax is as follows:
# download data:
data(EuStockMarkets)
r <- diff(log(EuStockMarkets))[,"DAX"]
r = r - mean(r)
# fit a garch(1,1) on this:
library(tseries)
garch(r)
The estimated parameters are given below:
***** ESTIMATION WITH ANALYTICAL GRADIENT *****
Call:
garch(x = r)
Coefficient(s):
a0 a1 b1
4.746e-06 6.837e-02 8.877e-01
Now it is straightforward to transform Garch(1,1)
to a ARMA like this:
r[t]^2 = w + (alpha+beta)*r[t-1]^2 + beta*(h[t-1] -
r[t-1]^2) - (h[t] - r[t]^2)
= w + (alpha+beta)*r[t-1]^2 + beta*theta[t-1] + theta[t]
So if I fit a ARMA(1,1) on r[t]^2 I am getting following result;
arma(r^2, order=c(1,1))
Call:
arma(x = r^2, order = c(1, 1))
Coefficient(s):
ar1 ma1 intercept
9.157e-01 -8.398e-01 9.033e-06
Therefore if the above derivation is correct then I should get a same
intercept term for both Garch and ARMA case. But here I am not getting
it. Can anyone explain why?
Any input will be highly appreciated.
Thanks and regards,
Megh
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