Displaying 18 results from an estimated 18 matches for "b_j".
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2011 Jul 03
3
Hint improve my code
...code below. I am worried that the parameters I want to
be estimated are "not being found" when I ran my code. Is there a way I can
code them so that R recognize that they should be estimated.
This is the error I am getting.
> out1=optim(llik,par=start.par)
Error in pnorm(au_j, mean = b_j * R_m, sd = sigma_j) :
object 'au_j' not found
#Yet al_j,au_j,sigma_j and b_j are just estimates that balance the
likelihood function?
llik=function(R_j,R_m)
if(R_j< 0)
{
sum[log(1/(2*pi*(sigma_j^2)))-(1/(2*(sigma_j^2))*(R_j+al_j-b_j*R_m))^2]
}else if(R_j>0)
{
sum[log(1/(2*pi*(si...
2011 Jul 23
1
Extend my code to run several data at once.
...ay you help me adjust it to
accomodate several rows of R_j and print the 200 results.
***Please do not get intimidated by the maths in the code.***
my code
######
afull=read.table("D:/hope.txt",header=T)
library(optimx)
llik = function(x)
{
al_j=x[1]; au_j=x[2]; sigma_j=x[3]; b_j=x[4]
sum(na.rm=T,
ifelse(a$R_j< 0, log(1 / ( sqrt(2*pi) * sigma_j) )-
(1/( 2*sigma_j^2 ) ) * (
(a$R_j+al_j-b_j*a$R_m)^2 ) ,
ifelse(a$R_j>0 , log(1 / ( sqrt(2*pi) * sigma_j) )-
(1/( 2*sigma_j^2 ) ) * (
(a$R_j+al_j-b...
2011 Jul 04
3
loop in optim
Hi
May you help me correct my loop function.
I want optim to estimates al_j; au_j; sigma_j; b_j by looking at 0 to 20,
21 to 40, 41 to 60 data points.
The final result should have 4 columns of each of the estimates AND 4 rows
of each of 0 to 20, 21 to 40, 41 to 60.
###MY code is
n=20
runs=4
out=matrix(0,nrow=runs)
llik = function(x)
{
al_j=x[1]; au_j=x[2]; sigma_j=x[3]; b_j=x[4]...
2011 Jul 01
2
Help fix last line of my optimization code
...args list of 'min' being evaluated was:
(interval)####
My data and my code looks like below.
R_j R_m
0.002 0.026567296
0.01 0.003194435
. .
. .
. .
. .
0.0006 0.010281122
a=read.table("D:/ff.txt",header=T)
attach(a)
llik=function(R_j,R_m)
#The parameters al_j, au_j, b_j ,
and sigma_j need to be estimated and there are no initial estimates to
them.
if(R_j< 0)
{
LF=sum[log(1/(2*pi*(sigma_j^2)))-(1/(2*(sigma_j^2))*(R_j+al_j-b_j*R_m))^2]
}else if(R_j>0)
{
LF=sum[log(1/(2*pi*(sigma_j^2)))-(1/(2*(sigma_j^2))*(R_j+au_j-b_j*R_m))^2]
}else
{
LF=sum[(log(pnorm((au_...
2011 Jul 06
1
Group Data indexed by n Variables
...%m")
MonthDayCombs = paste(Months, Days)
AvgDemand = data.matrix(by(Data$RescaledDemand, DayMonthCombs, mean))
On 4 July 2011 10:34, EdBo <n.bowora@gmail.com> wrote:
> Hi
>
> May you help me correct my loop function.
>
> I want optim to estimates al_j; au_j; sigma_j; b_j by looking at 0 to 20,
> 21 to 40, 41 to 60 data points.
>
> The final result should have 4 columns of each of the estimates AND 4 rows
> of each of 0 to 20, 21 to 40, 41 to 60.
>
> ###MY code is
>
> n=20
> runs=4
> out=matrix(0,nrow=runs)
>
> llik = function(x)...
2010 Oct 18
0
specifying lme function with a priori hypothesis concerning between-group variation in slopes
...case of a single group. The model is: Y_i= a +bX_i +
error where I indexes the different values of X and Y in this group . The
a priori hypothesis of the slope is: b=K. This is easily tested with a
t-test (b-K=0).
Now imagine that there are j groups. For each group j the model is: Y_ij=
a_j + b_jX_ij + error. Both the intercepts (a) and the slopes (b) are
allowed to vary between groups. The a priori (null) hypothesis of interest
involved the between-group values of the slopes and is: b_j=Kj where Kj is
specified a priori for each group j based on theoretical considerations but
whose valu...
2012 Aug 27
2
randomLCA
Can anybody, please, explain me how many parameter are estimated using
randomLCA?
For examples, model "dentistry.lca2random" estimate 1 scale (or
variance, b_j) parameter and 2 position parameters (a_cj)? Doesn't
it?
Do I need at least 4 diagnostic tests for such a model?
What happens if I specify options blocksize and byclass? How many
diagnostic tests (or rater) I need?
Extract from see "randomLCA examples", by Ken Beath.
> dentis...
1999 Dec 10
1
orthogonal and nested model
I'm working with a orthogonal and nested model (mixed).
I have four factors, A,B,C,D;
A and B are fixed and orthogonal
C is nested in AB interaction
and finally, D is nested in C.
I would like to model the following
Y_ijklm=Mu+A_i+B_j+AB_ij+C_k(ij)+D_l(k(ij))+Error_m(...)
I used the next command
>summary(aov(abund~A*B + C % in % A:B + D % in % C % in % A:B ,datos))
Is it the correct formula syntax?
Thaks for all.
Marcelo.
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
r-help mailing list...
2007 Aug 05
0
null hypothesis for two-way anova
Dear R community,
Confused by some of my lab results I ask for the definition of the null
hypothesis of a two-way analysis of variance in R (anova() and aov()).
Starting with the following model
y = a_i + b_j , i in A and j in B
is the tested null hypothesis
H_0: a_i = 0 for all i in A
or
H_0: a_m = a_n for any m and n in A?
Consequently the same questions for interaction effects. Starting with
the model
y = a_i + b_j + f_ij , i in A and j in B
is the tested null hypothesis
H_0: f_ij = 0 for al...
2012 Feb 02
0
glmer question
I would like to fit the following model:
logit(p_{ij}) = \mu + a_i + b_j
where a_i ~ N(0, \sigma_a^2) , b_j ~ N(0, \sigma_b^2) and \sigma_a
= \sigma_b.
Is it possible to fit a model with such a constraint on the variance
components in glmer?
--
View this message in context: http://r.789695.n4.nabble.com/glmer-question-tp4351829p4351829.html
Sent from the R h...
2012 Oct 29
2
Two-way Random Effects with unbalanced data
Hi there,
I am looking to fit a two-way random effects model to an *unblalanced*
layout,
y_ijk = mu + a_i + b_j + eps_ijk,
i=1,...,R, j=1,...,C, k=1,...,K_ij.
I am interested first of all in estimates for the variance components,
sigsq_a, sigsq_b and sigsq_error.
In the balanced case, there are simple (MM, MLE) estimates for these; In the
unbalanced setup, this is much more complicated because orthogon...
2006 Oct 24
1
Variance Component/ICC Confidence Intervals via Bootstrap or Jackknife
I'm using the lme function in nmle to estimate the variance components
of a fully nested two-level model:
Y_ijk = mu + a_i + b_j(i) + e_k(j(i))
lme computes estimates of the variances for a, b, and e, call them v_a,
v_b, and v_e, and I can use the intervals function to get confidence
intervals. My understanding is that these intervals are probably not
that robust plus I need intervals on the intraclass correlation
coefficie...
2000 Jan 01
0
Re: Tests in linear regression
...verything, and merry christmas.
FrSa> Best wishes, Francisco Sabido.
Dear Francisco,
I think everything you need is already part of R:
summary(lm.sml <- lm(y ~ x1))
summary(lm.big <- lm(y ~ x1 + x2 + x3 + x4))
anova(lm.sml, lm.big)
now will do a test of b2 = b3 = b4 = 0
if b_j is the coefficient of x_j.
Look at the online help ?lm and ?anova
and at
demo(lm.glm)
where models "l1" and "l0" are compared that way.
---------- ---------- ----------
Happy new year to everyone!
---------- ---------- ----------
Martin Maechler <maechler at stat...
2005 Dec 12
0
marginal effects in glm's
Hi,
I wonder if there is a function in (some package of) R which computes
marginal effects of the variables in a glm, say, for concretness, a
probit model. By marginal effects of the covariate x_j I mean
d P(y=1 | x),
which is approx
g(xB)B_j dx_j
where g is the pdf of the normal distribution, x is the vector of
covariates (at some points, say, the mean values) and B is the estimated
vector of coefficients. Of course, it isn't difficult to write such a
function, but that's exactly why I find it strange the fact that I
didn&...
2004 Nov 09
0
Is nesting {} inside \eqn OK?
...earlier, here
is a fuller excerpt from the input file:
--------------------------------------------
With \eqn{J} possible outcomes and \eqn{p_j}{p(j)} the probability of
the \eqn{j}'th outcome,
the formula is \deqn{\newcommand{\B}{{\bf \beta}}\newcommand{\X}{{\bf
X}}
p_j = \frac{e^{\X\B_j}}{\displaystyle\sum_{k=0}^J e^{\X\B_k}}.}{
p(j) = exp[X*b(j)]/sum{exp[X*b(k)], k=0 to J}.}
\eqn{{\bf\beta}_j}{b(j)} is the vector of coefficients for outcome
\eqn{j} and
\eqn{{\bf X}}{X} are the covariates.
--------------------------------------------
By the way, the \newcommand is not gl...
2010 Dec 26
0
GLS with corAR(1) correlation structure residual/standard error calculation
...se terms are
the e's. I cannot replicate the reported residuals using this approach. I
also do not know how Z_0 should be calculated, i.e. what does the first step
of this recursive procedure look like?
>From the residuals, I also cannot replicate the reported standard errors. I
am using se(b_j) = sqrt(sigma^2/sum(x_i-x_mean)^2) where sigma =sqrt(SSR/df)
Any help on this or explanation of how GLS works would be much appreciated.
Any clarification would be much appreciated.
[[alternative HTML version deleted]]
2007 Apr 14
6
[LLVMdev] Regalloc Refactoring
On Thu, 12 Apr 2007, Fernando Magno Quintao Pereira wrote:
>> I'm definitely interested in improving coalescing and it sounds like
>> this would fall under that work. Do you have references to papers
>> that talk about the various algorithms?
>
> Some suggestions:
>
> @InProceedings{Budimlic02,
> AUTHOR = {Zoran Budimlic and Keith D. Cooper and Timothy
2020 Feb 27
2
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