Displaying 20 results from an estimated 32 matches for "1.1e".
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1.1
2011 Jul 13
1
AR-GARCH with additional variable - estimation problem
Dear list members,
I am trying to estimate parameters of the AR(1)-GARCH(1,1) model. I have one
additional dummy variable for the AR(1) part.
First I wanted to do it using garchFit function (everything would be then
estimated in one step) however in the fGarch library I didn't find a way to
include an additional variable.
That would be the formula but, as said, I think it is impossible to add
2009 Feb 16
0
odd GARCH(1,1) results
Hi everybody,
I'm trying to fit a Garch(1,1) process to the DAX returns. My data
consists of about 2300 10day-logreturns in chronologically descending
order (see attachment). But if I use the garch function I get a very
high alpha_1 and a quite low beta, which doesn't make that much sense. I
think I am missing something, but have no idea what it might be. I'd
appreciate it a lot
2004 Oct 21
2
How to calculate a double integral ...?
Dear R-Friends,
How can I calculate a double integral like
\int_a^b \int_c^y g(x, y) dx dy
where a, b, c are constants, g(x, y), e.g.,
g(x, y) = tan(x + y).
I tried to nested integrate() and adapt(),
but none of them working, seemingly due to the
limits can not be specified constants.
Best regards,
C. Joseph Lu
Department of Statistics
National Cheng-Kung University
Voice:
2005 Jan 11
1
integrate() and complex values
[apologies for possible multiple post]
Hi
The manpage for integrate() does not mention imaginary numbers.
I found the following unexpected:
> integrate(function(x){0*x+1+1i},0,1)
1 with absolute error < 1.1e-14
>
(I would expect 1+1i here)
One can write a little wrapper, but it's not straightforward.
Would it be hard to accommodate such functions in integrate()?
Some other
2010 Dec 14
1
Forcing standard notation in Sweave tables
I'm hoping someone with some experience generating tables in Sweave will be able to solve this problem for me. I'm experiencing some inconsistency with the way floating point numbers are displayed -- for some tables, they will be in standard notation, e.g.
11±26
while in others, it uses exponential notation, e.g.
1.1e+01±2.6e+01
I am using code such as the following; note I have
2006 Jan 06
1
Problem with Integral of Indicator Function
Hi..
i was trying to integrate the indicator funtion but had problems when limits where negative or equal to the indicator condition
my function is
________________________
fun1<-function(x){
as.numeric(x>=2)
}
_________________________
which should be Ind(x>=2)*x
seems to work for the following two cases
2009 Feb 23
2
1.095e+09 for integers
I've had a very long file written out by R with write.table, with
fields of time values, converted from POSIXlt as.numeric. Among 2.5
million values, very few had 6 trailing zeroes, and those were output
in scientific notation as in the subject. Is this the default
behavior for long integers, and how can it be turned off (with all
digits for any integer field in write.table)? This
2004 Sep 13
1
an integration question
Dear all,
I'm stuck on a problem concerning integration..Results from the analytical
expression and numerical approximation (as returned by integrate()) do not
match.
It probably depends on some error of mine, so apologizes for this off-topic
question.
I'm interested in computing the integral of f where:
f<-function(x){exp(-3-.2*pmin(x-7.5,0))}
x<-seq(0,15,length=50)
plot(x,
2011 Mar 04
3
integrate a fuction
I'm having a function of the form
1> f<-function(x){
1+
1+ return(x^p)
1+
1+ }
,and I would like to integrate it with respect to x, where p should be any
constant.
One way would be to set a value for p globally and then call integrate
function:
p=2
integrate(f, lower = -1, upper = 1)
However, I would like to use 'integrate' inside a function, so I could call
it passing
2002 Sep 19
2
Rounding
Hi,
Suppose I have:
459
1789
23590
and I'd like to round them to:
400
1700
24000
On the other hand, say if I have:
232
1234
23120
that need to be rounded to:
300
1300
24000
I tried the round(), floor() or ceiling() and can't get what I want. Is
there any tricks I can use to achieve this goal?
Cheers,
Kevin
2011 Jun 23
2
new to R need urgent help!
hi all-
I am doing some research, have never used R before until today and need to
understand the following program for a project.
if some one could PLEASE help me understand this program ASAP i would
GREATLY appreciate it (any syntax/ statistic comments would be great)
PLEASE PLEASE HELP!! THANKYOU!!!
-on a side note, it seems to me that R doesnt include the pv, and it was
calculated
2010 Dec 10
1
survreg vs. aftreg (eha) - the relationship between fitted coefficients?
Dear R-users,
I need to use the aftreg function in package 'eha' to estimate failure times for left truncated survival data. Apparently, survreg still cannot fit such models. Both functions should be fitting the accelerated failure time (Weibull) model. However, as G?ran Brostr?m points out in the help file for aftreg, the parameterisation is different giving rise to different
2014 Sep 03
0
[PATCH 3/3] virtio_ring: unify direct/indirect code paths.
virtqueue_add() populates the virtqueue descriptor table from the sgs
given. If it uses an indirect descriptor table, then it puts a single
descriptor in the descriptor table pointing to the kmalloc'ed indirect
table where the sg is populated.
Previously vring_add_indirect() did the allocation and the simple
linear layout. We replace that with alloc_indirect() which allocates
the indirect
2001 Sep 18
1
case weights in coxph (survival)
Hi,
I am having trouble with the survival library, particualrily the coxph
function.
the following works
coxph(jtree9$cph.call,z,rep(1,dim(z)[1]))
Call:
coxph(formula = jtree9$cph.call, data = z, weights = rep(1, dim(z)[1]))
coef exp(coef) se(coef) z p
SM 0.2574 1.294 0.0786 3.274 1.1e-03
Sex -0.1283 0.880 0.1809 -0.709
2007 Feb 23
4
using "integrate" in a function definition
Dear list members,
I'm quite new to R, and though I tried to find the answer to my probably
very basic question through the available resources (website, mailing
list archives, docs, google), I've not found it.
If I try to use the "integrate" function from within my own functions,
my functions seem to misbehave in some contexts. The following example
is a bit silly, but
2014 Sep 03
8
[PATCH 0/3] virtio: simplify virtio_ring.
I resurrected these patches after prompting from Andy Lutomirski's
recent patches. I put them on the back-burner because vring_bench
had a 15% slowdown on my laptop: pktgen testing revealed a speedup,
if anything, so I've cleaned them up.
Rusty Russell (3):
virtio_net: pass well-formed sgs to virtqueue_add_*()
virtio_ring: assume sgs are always well-formed.
virtio_ring: unify
2014 Sep 03
8
[PATCH 0/3] virtio: simplify virtio_ring.
I resurrected these patches after prompting from Andy Lutomirski's
recent patches. I put them on the back-burner because vring_bench
had a 15% slowdown on my laptop: pktgen testing revealed a speedup,
if anything, so I've cleaned them up.
Rusty Russell (3):
virtio_net: pass well-formed sgs to virtqueue_add_*()
virtio_ring: assume sgs are always well-formed.
virtio_ring: unify
2005 Feb 01
1
Updates to CentOS-4Beta
1. There are updates to CentOS-4Beta for the i386 and x86_64 arches.
The following RPMS have been changed:
a. createrepo-0.4.2-1.noarch.rpm - This is an update from the upstream
maintainer.
b. yum-2.1.13-1.c4.noarch.rpm - This is an update from the upstream
maintainer.
c. firefox-1.0-6.centos4.3.i386.rpm - The original build did not strip
the library files of unnecessary symbols, causing
2011 Jun 24
0
understand GEE output for wald test
Hi
I'm having some difficulty understanding my output (below) from GEE. the
person who wrote the program included some comments about the '3-th term
gives diff between hyp/ox at time..', and after created an L vector to use
for a WALD test. I was wondering if someone could help me understand the
GEE output, the programmers comment, how L was determined, and its use in
the WALD
2007 Oct 19
1
X matrix deemed to be singular in counting process coxph
Dear all,
I have a question with respect to counting process formulation of the
coxph(survival) model.
I have two groups of observations for which I have partitioned each
observation into two distinct time intervals, namely, entry day till day 13,
and day 13 till death or censorship day (of course the latter only for the
observations that survived the first 13 day interval), and added a