If I try to predict the volatility with: predict(garch.fitted) I get
the
"conditional standard deviation predictions" (an interval) for every
period
return in r.
Q2) Is this a 1 period prediction (i.e. the volatility prediction
for the
next day)? How do I get a prediction for let's say 2 periods in the
future?
No these are not predictions for the next day. The term "predict" is
maybe somewhat misleading. "Fitted Values" is more appropriate, and in
fact both functions give the same results, e.g.
all.equal(predict(garch.fitted)[,1],fitted.values(garch.fitted)[,1])
[1] TRUE
For a two-step-ahead forecast I would use the predict function on the
fGARCH Object and would set the n.ahead argument to 2:
predict(fgarch.fitted, n.ahead = 2)
meanForecast meanError standardDeviation
1 0.0005044188 0.008436248 0.008436248
2 0.0005044188 0.008482511 0.008482511
Q3) Why is the meanForecast for all 10 future periods the same?
Because you have just modeled the variance equation and the mean
equation has only a constant. Garch will help you forecasting the
variance of the returns but not the returns themselves. If you want
return forecasts you should model the mean equation, e.g.
fgarch.fitted <- fGarch::garchFit(~ arma(2,1)+garch(1,1), data =
as.zoo(na.omit(r)), trace = FALSE)
predict(fgarch.fitted, n.ahead = 5)
meanForecast meanError standardDeviation
1 -0.0009371963 0.008495794 0.008495794
2 -0.0003940789 0.008556624 0.008539586
3 0.0004374707 0.008608828 0.008582793
4 0.0005284133 0.008651758 0.008625429
5 0.0005093462 0.008693970 0.008667503
Hope this helps
Zeno
-----Original Message-----
From: r-sig-finance-bounces at stat.math.ethz.ch
[mailto:r-sig-finance-bounces at stat.math.ethz.ch] On Behalf Of Mark
Breman
Sent: Dienstag, 9. November 2010 12:51
To: Patrick Burns
Cc: r-sig-finance at stat.math.ethz.ch
Subject: Re: [R-SIG-Finance] Questions on fitted garch(1,1)
Hi everyone,
As a followup on my three fitted garch questions;
Yes indeed, if I give the estimated coefficients of fgarch.fitted as
initial
values to garch() the results are much more inline with each other:
coef(garch(na.omit(r), order = c(1,1),
control=garch.control(start=c(1.3705e-06,8.1789e-02,9.0995e-01))))
***** ESTIMATION WITH ANALYTICAL GRADIENT *****
I INITIAL X(I) D(I)
1 1.370500e-06 1.000e+00
2 8.178900e-02 1.000e+00
3 9.099500e-01 1.000e+00
IT NF F RELDF PRELDF RELDX STPPAR D*STEP
NPRELDF
0 1 -1.093e+04
1 13 -1.093e+04 4.54e-09 1.67e-08 1.1e-09 4.3e+13 2.1e-09
3.63e+05
2 24 -1.093e+04 -1.30e-14 2.64e-14 8.1e-15 1.6e+00 1.5e-14
-1.22e-03
***** FALSE CONVERGENCE *****
FUNCTION -1.093479e+04 RELDX 8.076e-15
FUNC. EVALS 24 GRAD. EVALS 2
PRELDF 2.641e-14 NPRELDF -1.222e-03
I FINAL X(I) D(I) G(I)
1 1.372553e-06 1.000e+00 -1.965e+04
2 8.178900e-02 1.000e+00 2.046e+00
3 9.099500e-01 1.000e+00 -1.478e+01
a0 a1 b1
1.372553e-06 8.178900e-02 9.099500e-01
Very nice, thank you!
Now only question 2 and 3 remain a mystery to me.
Kind regards,
-Mark-
2010/11/4 Patrick Burns <patrick at burns-stat.com>
Taking Alexios' advice will probably
help, but I wouldn't be surprised if
they are still not the same. The
garch likelihood is extremely tricky
to optimize -- specialized optimizers
would be the ideal.
Assuming you can give starting values
to the optimizer in both cases, you can
give the other's answer to each and see
what happens.
On 04/11/2010 21:02, alexios wrote:
The default for fGarch::garchFit is to estimate the mean, assuming a
non-zero mean dataset is passed, as can be seen from the returned
"mu"
parameter ( a call to 'args(garchFit)' should highlight the
'include.mean' argument ). The default for tseries garch is to assume
zero mean data.
Demean the data before passing to tseries garch by the estimated 'mu'
value from 'fgarch.fitted' and you should obtain closer results.
-Alexios
On 11/4/2010 8:44 PM, Mark Breman wrote:
Hi everyone,
I'm trying to predict volatility using a garch(1,1) model and I'm
running
into some interesting issues I cannot find an answer for.
Suppose I have the following code:
library(quantmod)
library(tseries)
library(fGarch)
# get some returns
r = dailyReturn(Ad(getSymbols("SPY", from="2000-01-01",
to="2010-11-04",
auto.assign=F)))
# fit models
garch.fitted = garch(na.omit(r), order = c(1,1))
fgarch.fitted = fGarch::garchFit(~ garch(1,1), data =
as.zoo(na.omit(r)),
trace = FALSE)
Now if I look at the values of the coefficients for both fitted
models
they
are not the same:
coef(garch.fitted)
a0 a1 b1
1.339912e-06 8.165689e-02 9.101754e-01
coef(fgarch.fitted)
mu omega alpha1 beta1
4.990616e-04 1.370880e-06 8.307080e-02 9.086447e-01
Q1) Why are they different? What do I have to change to make the
coefficient
values of fgarch.fitted the same as garch.fitted?
If I try to predict the volatility with: predict(garch.fitted) I get
the
"conditional standard deviation predictions" (an interval) for every
period
return in r.
Q2) Is this a 1 period prediction (i.e. the volatility prediction
for the
next day)? How do I get a prediction for let's say 2 periods in the
future?
If I use predict(fgarch.fitted) I get:
predict(fgarch.fitted)
meanForecast meanError standardDeviation
1 0.0004990616 0.006677350 0.006677350
2 0.0004990616 0.006751925 0.006751925
3 0.0004990616 0.006825079 0.006825079
4 0.0004990616 0.006896859 0.006896859
5 0.0004990616 0.006967315 0.006967315
6 0.0004990616 0.007036491 0.007036491
7 0.0004990616 0.007104428 0.007104428
8 0.0004990616 0.007171167 0.007171167
9 0.0004990616 0.007236745 0.007236745
10 0.0004990616 0.007301198 0.007301198
Q3) Why is the meanForecast for all 10 future periods the same?
Thanks,
-Mark-
predict(garch)
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