Re: [R] Comparison between GARCH and ARMA
Have you tried to Monte Carlo ARMA and GARCH? If you plot the resulting series in various ways, I suspect the differences will be apparent to you. If you'd like more from this list, I suggest you illustrate your question with commented, minimal, self-contained, reproducible code, as suggested in the posting guide www.R-project.org/posting-guide.html. Hope this helps. Spencer Graves [EMAIL PROTECTED] wrote: Hi i was just going by this thread, i thought of igniting my mind and got something wierd so i thought of making it wired. i think whether you take ARMA or GARCH. In computer science these are feedback systems or put it simply new values are function of past values. In ARMA case it is the return series and the error series. In case of GARCH it is the errors and stdev or returns and shock with propotionality of coeficient. In any case we are trying to find the returns only. What if i put stdev in ARMA and returns in GARCH ? I want to ask what it would end up showing me. For me both are having similar structure having two parts : 1. regression depending on past values 2. trying to reduce errors by averaging them i hope i am correct. please correct me where i am wrong. thanks and regards Email(Office) :- [EMAIL PROTECTED] , Email(Personal) :- [EMAIL PROTECTED] Wensui Liu [EMAIL PROTECTED] Sent by: [EMAIL PROTECTED] 08-11-06 12:24 AM To Leeds, Mark (IED) [EMAIL PROTECTED] cc r-help@stat.math.ethz.ch, Megh Dal [EMAIL PROTECTED] Subject Re: [R] Comparison between GARCH and ARMA 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) [EMAIL PROTECTED] 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: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] 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
Re: [R] Comparison between GARCH and ARMA
Hi i was just going by this thread, i thought of igniting my mind and got something wierd so i thought of making it wired. i think whether you take ARMA or GARCH. In computer science these are feedback systems or put it simply new values are function of past values. In ARMA case it is the return series and the error series. In case of GARCH it is the errors and stdev or returns and shock with propotionality of coeficient. In any case we are trying to find the returns only. What if i put stdev in ARMA and returns in GARCH ? I want to ask what it would end up showing me. For me both are having similar structure having two parts : 1. regression depending on past values 2. trying to reduce errors by averaging them i hope i am correct. please correct me where i am wrong. thanks and regards Email(Office) :- [EMAIL PROTECTED] , Email(Personal) :- [EMAIL PROTECTED] Wensui Liu [EMAIL PROTECTED] Sent by: [EMAIL PROTECTED] 08-11-06 12:24 AM To Leeds, Mark (IED) [EMAIL PROTECTED] cc r-help@stat.math.ethz.ch, Megh Dal [EMAIL PROTECTED] Subject Re: [R] Comparison between GARCH and ARMA 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) [EMAIL PROTECTED] 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: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] 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
Re: [R] Comparison between GARCH and ARMA
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: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] 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
Sponsored Link
Degrees online in as fast as 1 Yr - MBA, Bachelor's, Master's, Associate
Click now to apply http://yahoo.degrees.info
__
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.
This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}
__
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.
Re: [R] Comparison between GARCH and ARMA
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) [EMAIL PROTECTED] 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: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] 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 Sponsored Link Degrees online in as fast as 1 Yr - MBA, Bachelor's, Master's, Associate Click now to apply http://yahoo.degrees.info __ 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. This is
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 [EMAIL PROTECTED] 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) [EMAIL PROTECTED] 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: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] 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 Sponsored Link Degrees online in as fast as 1 Yr - MBA, Bachelor's, Master's,
Re: [R] Comparison between GARCH and ARMA
but not the return squared squared which is what was written previously. . From: Hannu Kahra [mailto:[EMAIL PROTECTED] 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 [EMAIL PROTECTED] 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) [EMAIL PROTECTED] 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: [EMAIL PROTECTED] [mailto: [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] ] 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
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) [EMAIL PROTECTED] wrote: but not the return squared squared which is what was written previously. . -- *From:* Hannu Kahra [mailto:[EMAIL PROTECTED] *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 [EMAIL PROTECTED] 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) [EMAIL PROTECTED] 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: [EMAIL PROTECTED] [mailto: [EMAIL PROTECTED] 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
