if the return is assumed zero, the epsilon t sequared is the variance of
the return but the epsilon_t is not the return. i think we should
have a conversation off line because i don't want to bother the whole
list with this. what the original email said is still incorrect,
( see below ) but, if all you are saying that the variance,
sigma_squared,  is expectation of epslion_t squared term is when the
mean is zero, i agree with you.
 
the return  itself, r_t, is still a  normal random variable times this
"innovation" as you put it so you can't build an arma or an arima or
whatever for the return. if you want to build one for the return squared
( dont' subtract a mean because it's already zero ) , 
that's fine but then you are really building one for the variance.  so,
the original should have had epsilon_t squared in the transformed
equation,
NOT r_t ^2. r_t is still equal to epsilon_t*z_t and e(r_t)^2=
sigma_squared but r_t^2 is not the same as the epsilon_t^2.   if this
isn't clear, 
i don't know what else to say. 
 
 
 
 
 
 
 
 
 

________________________________

From: Hannu Kahra [mailto:hkahra@gmail.com] 
Sent: Tuesday, November 07, 2006 4:15 PM
To: Leeds, Mark (IED)
Cc: Wensui Liu; r-help@stat.math.ethz.ch; Megh Dal
Subject: Re: [R] Comparison between GARCH and ARMA


Except in the case of zero means, that is point (7) in your previous
mail.


On 11/7/06, Leeds, Mark (IED) < Mark.Leeds@morganstanley.com
<mailto:Mark.Leeds@morganstanley.com> > wrote: 

	but not the return squared squared which is what was written
previously.
	.
	 

________________________________

	From: Hannu Kahra [mailto:hkahra@gmail.com] 
	Sent: Tuesday, November 07, 2006 3:54 PM
	To: Wensui Liu
	Cc: Leeds, Mark (IED); r-help@stat.math.ethz.ch; Megh Dal
	Subject: Re: [R] Comparison between GARCH and ARMA
	
	
	
	A GARCH model can be regarded as an application of the ARMA idea
to the squared innovation series. See, e.g. Ruey S. Tsay, Analysis of
Financial Time Series, Wiley, 2nd edition, page 114.
	
	Hannu
	
	
	On 11/7/06, Wensui Liu <liuwensui@gmail.com> wrote: 

		Mark,
		
		I totally agree that it doesn't make sense to compare
arma with garch.
		
		but to some extent, garch can be considered arma for
conditional
		variance. similarly, arch can be considered ma for
conditional 
		variance.
		
		the above is just my understanding, which might not be
correct.
		
		thanks.
		
		On 11/7/06, Leeds, Mark (IED)
<Mark.Leeds@morganstanley.com> wrote: 
		> 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@stat.math.ethz.ch
		> [mailto: r-help-bounces@stat.math.ethz.ch
<mailto:r-help-bounces@stat.math.ethz.ch> ] On Behalf Of Megh Dal
		> Sent: Tuesday, November 07, 2006 2:24 AM
		> To: r-help@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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		> https://stat.ethz.ch/mailman/listinfo/r-help
		> PLEASE do read the posting guide 
		> http://www.R-project.org/posting-guide.html
		> and provide commented, minimal, self-contained,
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		>
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		>
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		>
		> ______________________________________________
		> R-help@stat.math.ethz.ch mailing list
		> https://stat.ethz.ch/mailman/listinfo/r-help
		> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
		> and provide commented, minimal, self-contained,
reproducible code.
		>
		
		
		--
		WenSui Liu
		(http://spaces.msn.com/statcompute/blog )
		Senior Decision Support Analyst
		Cincinnati Children Hospital Medical Center
		
		______________________________________________
		R-help@stat.math.ethz.ch mailing list 
		https://stat.ethz.ch/mailman/listinfo/r-help
		PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html 
		and provide commented, minimal, self-contained,
reproducible code.
		


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