RE: [R] Label using equivalent of \mathbb{R}
> -Original Message-
> From: Simon Cullen
> Sent: Thursday, August 26, 2004 1:26 PM
> To: [EMAIL PROTECTED]
> Subject: [R] Label using equivalent of \mathbb{R}
>
> Hi,
>
> I'm trying to label the horizontal axis of a plot with a symbol that is
> the equivalent of \mathbb{R} in LaTeX. I've had a look through the help
> pages for plotmath and for Hershey and haven't found the symbol. Could
> someone give me a pointer, please?
>
[Dietrich Trenkler]
> plot(rnorm(10),xlab=expression(bold(x)),ylab=expression(bold(y)))
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RE: [R] Two factor ANOVA with lm()
> -Original Message-
> From: Prof Brian Ripley
> Sent: Monday, August 23, 2004 1:15 PM
> To: Trenkler, Dietrich
> Subject: Re: [R] Two factor ANOVA with lm()
>
> On Mon, 23 Aug 2004, Trenkler, Dietrich wrote:
>
> [...]
>
> > outset? Or put another way: Why is it that lm() uses the corner point
> > constraints by default? Where can I find a documentation for this
> > behavior?
>
> In almost any piece of documentation on linear models in R, including the
> FAQ and `An Introduction to R', which says
>
> The main reason for mentioning this is that R and S have different
> defaults for unordered factors, S using Helmert contrasts. So if you
> need to compare your results to those of a textbook or paper which used
> S-PLUS, you will need to set
>
> options(contrasts = c("contr.helmert", "contr.poly"))
>
> This is a deliberate difference, as treatment contrasts (R's default)
> are thought easier for newcomers to interpret.
>
> Now, what does the posting guide say about doing your homework?
[Dietrich Trenkler] I swear I didn't need it for a homework.
I just overlooked the self-evident... (blush)
Thank you.
D. Trenkler
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[R] Two factor ANOVA with lm()
The following is a data frame
> "jjd" <- structure(list(Observations = c(6.8, 6.6, 5.3, 6.1,
7.5, 7.4, 7.2, 6.5, 7.8, 9.1, 8.8, 9.1), LevelA = structure(c(1,
1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3), .Label = c("A1", "A2",
"A3"), class = "factor"), LevelB = structure(c(1, 1, 2, 2,
1, 1, 2, 2, 1, 1, 2, 2), .Label = c("B1", "B2"), class = "factor")),
.Names = c("Observations", "LevelA", "LevelB"), row.names = c("1",
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12"),
class = "data.frame")
representing data from
@BOOK{Dobson02,
author = {Annette J. Dobson},
year = 2002,
title = {An Introduction to Generalized Linear Models},
edition = {2.},
publisher = {Chapman \& Hall/CRC},
address = {Boca Raton, Florida, 33431}
}
page 101. To reproduce the estimates c(6.7,0.75,1.75,-1.0,0.4,1.5)
given on page 103 in a two factor ANOVA entering
> jja1 <- lm(Observations~LevelA*LevelB,data=jjd)
> summary(jja1)
I get
Call:
lm(formula = Observations ~ LevelA * LevelB, data = jjd)
Residuals:
Min 1Q Median 3QMax
-6.500e-01 -2.000e-01 -3.469e-17 2.000e-01 6.500e-01
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 6.7000 0.3512 19.078 1.34e-06 ***
LevelAA20.7500 0.4967 1.510 0.1818
LevelAA31.7500 0.4967 3.524 0.0125 *
LevelBB2 -1. 0.4967 -2.013 0.0907 .
LevelAA2:LevelBB2 0.4000 0.7024 0.569 0.5897
LevelAA3:LevelBB2 1.5000 0.7024 2.136 0.0766 .
---
Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
Residual standard error: 0.4967 on 6 degrees of freedom
Multiple R-Squared: 0.9065, Adjusted R-squared: 0.8286
F-statistic: 11.64 on 5 and 6 DF, p-value: 0.00481
This is fine. But why do I get these estimates?
Entering
> model.matrix(jja1)
delivers
(Intercept) LevelAA2 LevelAA3 LevelBB2 LevelAA2:LevelBB2
LevelAA3:LevelBB2
11000 0
0
21000 0
0
31001 0
0
41001 0
0
51100 0
0
61100 0
0
71101 1
0
81101 1
0
91010 0
0
10 1010 0
0
11 1011 0
1
12 1011 0
1
attr(,"assign")
[1] 0 1 1 2 3 3
attr(,"contrasts")
attr(,"contrasts")$LevelA
[1] "contr.treatment"
attr(,"contrasts")$LevelB
[1] "contr.treatment"
which shows that internally lm() seems to use corner point constraints
of the form
\[\alpha_1=\beta_1=(\alpha\beta)_{11}=
(\alpha\beta)_{12}=(\alpha\beta)_{12}=(\alpha\beta)_{31}=0\]
in the model $E[Y_{jkl}]=\mu+\alpha_j+\beta_k+(\alpha\beta)_{jk}$
$j=1,2,3$, $k=1,2$, $l=1,2$, Dobson, page 102.
My question is: how can I incorporate restrictions like
$\alpha_1+\alpha_2+\alpha_3=0$, $\beta_1+\beta_2=0$,
$(\alpha\beta)_{21}+\alpha\beta)_{22}=0$,
$(\alpha\beta)_{31}+(\alpha\beta)_{32}=0$ and
$(\alpha\beta)_{11}+(\alpha\beta)_{21}+(\alpha\beta)_{31}=0$ from the
outset? Or put another way: Why is it that lm() uses the corner point
constraints by default? Where can I find a documentation for this
behavior?
I know that I can use something like lm(y~X) where y <- c(6.8, 6.6,
5.3, 6.1, 7.5, 7.4, 7.2, 6.5, 7.8, 9.1, 8.8, 9.1) and X is an
appropriate design matrix. But I wonder if there is a more direct way.
Many thanks in advance.
D. Trenkler
--
Dietrich Trenkler Universität Osnabrück
FB Wirtschaftswissenschaften
Rolandstr.8 D-49069 Osnabrück
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RE: [R] Erlang distribution
> -Original Message-
> From: Nils Aschenbruck
> Sent: Thursday, August 19, 2004 11:51 AM
> To: [EMAIL PROTECTED]
> Subject: [R] Erlang distribution
>
>
> Hello,
>
> is there a packet that supports the Erlang distribution?
>
> I want to use this distribution for tests against empirical data. Thus, is
> there a packet that also supports "fitdistr" (Maximum-likelihood fitting)
> for this distribution?
>
[Dietrich Trenkler] Hi Nils,
you do not need a special package because the Erlang is a special
gamma distribution.
For instance
"derlang" <- function(x, k, l = 1) {
f <- dgamma(x, k, l)
f
}
delivers the density of the Erlang(k,1) distribution.
HTH
D. Trenkler
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RE: [R] Bug in qnorm or pnorm?
> -Original Message- > From: Deepayan Sarkar > Sent: Friday, August 06, 2004 3:31 PM > To: [EMAIL PROTECTED] > Subject: Re: [R] Bug in qnorm or pnorm? > > On Friday 06 August 2004 08:13, Trenkler, Dietrich wrote: > > > Given that pnorm(8.30) delivers 1 shouldn't we get Inf > > for x<-8.30;x-qnorm(pnorm(x)) ? > > Why? > > > pnorm(8.30) > [1] 1 > > qnorm(pnorm(8.30)) ## same as qnorm(1) > [1] Inf > > 8.30 - qnorm(pnorm(8.30)) ## same as 8.30 - Inf > [1] -Inf > > This seems perfectly acceptable to me for all reasonable definitions of > Inf. > > Deepayan > > [Dietrich Trenkler] Yes of course, you're right. I meant qnorm(pnorm(8.3)) should deliver Inf -- as it does. Thank you. Dietrich. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Bug in qnorm or pnorm?
> -Original Message- > From: Trenkler, Dietrich > Sent: Friday, August 06, 2004 3:13 PM > To: 'r-help' > Subject: [R] Bug in qnorm or pnorm? > > I found the following strange behavior using qnorm() and pnorm(): > > > x<-8.21;x-qnorm(pnorm(x)) > [1] 0.0004638484 > > x<-8.22;x-qnorm(pnorm(x)) > [1] 0.01046385 > > x<-8.23;x-qnorm(pnorm(x)) > [1] 0.02046385 > > x<-8.24;x-qnorm(pnorm(x)) > [1] 0.03046385 > > x<-8.25;x-qnorm(pnorm(x)) > [1] 0.04046385 > > x<-8.26;x-qnorm(pnorm(x)) > [1] 0.05046385 > > x<-8.27;x-qnorm(pnorm(x)) > [1] 0.06046385 > > x<-8.28;x-qnorm(pnorm(x)) > [1] 0.07046385 > > x<-8.29;x-qnorm(pnorm(x)) > [1] 0.08046385 > > x<-8.30;x-qnorm(pnorm(x)) > [1] -Inf > > > Given that pnorm(8.30) delivers 1 shouldn't we get Inf > for x<-8.30;x-qnorm(pnorm(x)) ? > > Thanks in advance. > [Dietrich Trenkler] Oops, forgot to mention: > unlist(R.Version()) platform archos system "i386-pc-mingw32""i386" "mingw32" "i386, mingw32" status major minor year "" "1" "9.1" "2004" month day language "06" "21" "R" D. Trenkler __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Bug in qnorm or pnorm?
I found the following strange behavior using qnorm() and pnorm(): > x<-8.21;x-qnorm(pnorm(x)) [1] 0.0004638484 > x<-8.22;x-qnorm(pnorm(x)) [1] 0.01046385 > x<-8.23;x-qnorm(pnorm(x)) [1] 0.02046385 > x<-8.24;x-qnorm(pnorm(x)) [1] 0.03046385 > x<-8.25;x-qnorm(pnorm(x)) [1] 0.04046385 > x<-8.26;x-qnorm(pnorm(x)) [1] 0.05046385 > x<-8.27;x-qnorm(pnorm(x)) [1] 0.06046385 > x<-8.28;x-qnorm(pnorm(x)) [1] 0.07046385 > x<-8.29;x-qnorm(pnorm(x)) [1] 0.08046385 > x<-8.30;x-qnorm(pnorm(x)) [1] -Inf Given that pnorm(8.30) delivers 1 shouldn't we get Inf for x<-8.30;x-qnorm(pnorm(x)) ? Thanks in advance. Dietrich Trenkler -- Dietrich Trenkler Universität Osnabrück FB Wirtschaftswissenschaften Rolandstr.8 D-49069 Osnabrück [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] GHK simulator
Dear R-community,
not to re-invent the wheel I wonder if someone of you
has ever written a function to compute the GHK smooth recursive
simulator to estimate multivariate normal probabilities. See for instance
page 194 of
@BOOK{Greene97,
author = {William H. Greene},
year = 1997,
title = {Econometric Analysis},
edition = {3rd},
publisher = {Prentice-Hall},
address = {New Jersey 07458}
}
Thank you.
Dietrich Trenkler
--
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FB Wirtschaftswissenschaften
Rolandstr.8 D-49069 Osnabrück
[EMAIL PROTECTED]
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RE: [R] plotting a table together with graphs
> -Original Message- > From: Federico Calboli > Sent: Tuesday, July 13, 2004 6:06 PM > To: r-help > Subject: [R] plotting a table together with graphs > > Dear All, > > I would like to ask how to add a table to a "matrix" of graphs. > > I have three non linear regression graphs plotted together after: > > par(mfrow=c(2,2)) > > which leaves an empty bottom right corner. I would like to use the space > to add a table (at the moment that's problem number one, adding a "nice" > table will come later). I know it is possible to print tables through > LaTeX and the Design/Hmisc libraries, although I would not have a clue > about printing it together with graphs, but I'd like something "quicker" > if at all possible. > [Dietrich Trenkler] I understand you are using LaTeX. So have a look at the psfrag package. HTH D. Trenkler __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] help on ks.test
> -Original Message-
> From: Peter Dalgaard
> Sent: Thursday, May 06, 2004 4:32 PM
> To: Janete Borges
> Cc: [EMAIL PROTECTED]
> Subject: Re: [R] help on ks.test
>
> "Janete Borges" <[EMAIL PROTECTED]> writes:
>
> > Dear All
> >
> > I need to test the goodness-of-fit of a (Negative) Exponential
> Distribution
> > to a dataset. The parameter of the distribution is unknown. What is the
> > appropriate test to do? I've tried the ks.test, although I think this
> > isn't the appropriate one, as I don't know the population parameter.
> > Can anybody help me?
> >
> > Thanks in advance,
> > Janete
>
> The bias of the K-S test with estimated parameters is well known to be
> substantial, but I haven't heard about correction terms except (I
> think) for the normal distribution.
[Dietrich Trenkler] There is a Lilliefors-version of the KS-test
for the exponential distribution. See e.g.
@ARTICLE{Lilliefors69a,
author = {H. W. Lilliefors},
year = 1969,
title = {On the {K}olmogorov-{S}mirnov Test for Exponential
Distribution with Mean Unknown Variance Unknown},
journal = {Journal of the American Statistical Association},
volume = 64,
pages = {387--389},
keywords = {Lilliefors Test for Exponentiality; Goodness-of-Fit;
Kolmogorov's Test}
}
or
@ARTICLE{Mason86,
author = {Andrew L. Mason and C.B. Bell},
year = 1986,
title = {New {L}illiefors and {S}rinivasan Tables with
Applications},
journal = {Communications in Statistics, Part B--Simulation and
Computation},
volume = 15,
pages = {451--477},
comment = {BIB 2},
keywords = {Lilliefors Test; Goodness-of-Fit; Simulation}
}
HTH
Let me stress that the KS-test may not be very powerful.
Dietrich
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RE: [R] matrix exponential: M^0
> -Original Message- > From: Federico Calboli > Sent: Tuesday, January 20, 2004 5:40 PM > To: r-help > Subject: [R] matrix exponential: M^0 > > I would like to ask why the zeroeth power of a matrix gives me a matrix > of ones rather than the identity matrix: > > > D<-rbind(c(0,0,0),c(0,0,0),c(0,0,0)) > > D<-as.matrix(D) > > D > [,1] [,2] [,3] > [1,]000 > [2,]000 > [3,]000 > > > D^0 > [,1] [,2] [,3] > [1,]111 > [2,]111 > [3,]111 > > I would have expected the identity matrix here. > > I find the same result with every other square matrix I used. [Dietrich Trenkler] M^0 means appying ^0 to each element of M. Matrix multiplication can be achieved by A%*%B. In this way A%*%A is not the same as A^2. Dietrich __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] weibull test
> -Original Message-
> From: Meriema BELAIDOUNI
> Sent: Wednesday, December 18, 2002 11:19 AM
> To: [EMAIL PROTECTED]
> Subject: [R] weibull test
>
> Hello
> What is the appropriate method to test if a given distribution is a
> weibull
> thank you
> meriema
[Dietrich Trenkler] The following articles may be of interest for
you:
@ARTICLE{Chandra81,
author = {M. Chandra and N.D. Singpurwalla and M.A. Stephens},
year = 1981,
title = {Kolmogorov Statistics for Tests of Fit for the
Extreme-Value and
Weibull Distributions},
journal = {Journal of the American Statistical Association},
volume = 76,
pages = {729--731},
keywords = {Extreme-Value Distribution; Goodness of Fit;
Kolmogorov-Smirnov
Tests; Kuiper Statistic; Weibull Distribution}
}
@ARTICLE{Lockhart94,
author = {Richard A. Lockhart and Michael A. Stephens},
year = 1994,
title = {Estimation and Tests of Fit for the Three-Parameter
Weibull
Distribution},
journal = {Journal of the Royal Statistical Society B},
volume = 56,
pages = {491--500},
keywords = {Empirical Distribution Function; Empirical
Distribution Function
Tests; Goodness of Fit; Reliability; Survival Analysis}
}
Hope this helps.
D. Trenkler
--
Dietrich Trenkler Universität Osnabrück
FB Wirtschaftswissenschaften
Rolandstr.8 D-49069 Osnabrück
[EMAIL PROTECTED]
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