On Sat, 11 Jun 2005, John Fox wrote:
Dear Marc,
I get the same results -- same coefficients, standard errors, and fitted
probabilities -- from multinom() and glm(). It's true that the deviances
differ, but they, I believe, are defined only up to an additive constant:
Yes. There are many variations on the definition of (residual) deviance,
but it compares -2 log likelihood with a `saturated' model. For grouped
data you have a choice: a separate term for each group or for each
observation. A binomial GLM uses the first but the second is more
normal in logistic regression (since it has a direct interpretation via
log-probability scoring).
multinom() is support software for a book (which the R posting guide does
ask you to consult): this is discussed with a worked example on pp 203-4.
dt
output factor n
1 m 1.2 10
2 f 1.2 12
3 m 1.3 14
4 f 1.3 14
5 m 1.4 15
6 f 1.4 12
dt.m <- multinom(output ~ factor, data=dt, weights=n)
# weights: 3 (2 variable)
initial value 53.372333
iter 10 value 53.115208
iter 10 value 53.115208
iter 10 value 53.115208
final value 53.115208
converged
dt2
m f factor
1 10 12 1.2
2 14 14 1.3
3 15 12 1.4
dt.b <- glm(cbind(m,f) ~ factor, data=dt2, family=binomial)
summary(dt.m)
Call:
multinom(formula = output ~ factor, data = dt, weights = n)
Coefficients:
Values Std. Err.
(Intercept) -2.632443 3.771265
factor 2.034873 2.881479
Residual Deviance: 106.2304
AIC: 110.2304
Correlation of Coefficients:
[1] -0.9981598
summary(dt.b)
Call:
glm(formula = cbind(m, f) ~ factor, family = binomial, data = dt2)
Deviance Residuals:
1 2 3
0.01932 -0.03411 0.01747
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -2.632 3.771 -0.698 0.485
factor 2.035 2.881 0.706 0.480
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 0.5031047 on 2 degrees of freedom
Residual deviance: 0.0018418 on 1 degrees of freedom
AIC: 15.115
Number of Fisher Scoring iterations: 2
predict(dt.b, type="response")
1 2 3
0.4524946 0.5032227 0.5538845
predict(dt.m, type="probs")
1 2 3 4 5 6
0.4524948 0.4524948 0.5032229 0.5032229 0.5538846 0.5538846
These fitted probabilities *are* correct: Since the members of each pair
(1,2), (3,4), and (5,6) have identical values of factor they are identical
fitted probabilities.
(Note, by the way, the large negative correlation between the coefficients,
produced by the configuration of factor values.)
I hope this helps,
John
--------------------------------
John Fox
Department of Sociology
McMaster University
Hamilton, Ontario
Canada L8S 4M4
905-525-9140x23604
http://socserv.mcmaster.ca/jfox
--------------------------------
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Marc Girondot
Sent: Saturday, June 11, 2005 3:06 AM
To: John Fox
Cc: [email protected]
Subject: [R] Problem with multinom ?
Thanks for your response.
OK, multinom() is a more logical in this context.
But similar problem occurs:
Let these data to be analyzed using classical glm with binomial error:
m f factor m theo f theo
-Ln L model -Ln L full interecept
f
10 12 1.2 0.452494473 0.547505527
1.778835688 1.778648963 2.632455675
-2.034882223
14 14 1.3 0.503222759 0.496777241 1.901401922 1.900820284
15 12 1.4 0.553884782 0.446115218 1.877062369 1.876909821
Sum -Ln L
5.557299979 5.556379068 Residual deviance
Deviance
11.11459996 11.11275814 0.001841823
If I try to use multinom() function to analyze these data,
the fitted parameters are correct but the residual deviance not.
dt<-read.table('/try.txt'. header=T)
dt
output factor n
1 m 1.2 10
2 f 1.2 12
3 m 1.3 14
4 f 1.3 14
5 m 1.4 15
6 f 1.4 12
dt.plr <- multinom(output ~ factor. data=dt. weights=n. maxit=1000)
# weights: 3 (2 variable)
initial value 53.372333
iter 10 value 53.115208
iter 10 value 53.115208
iter 10 value 53.115208
final value 53.115208
converged
dt.plr
Call:
multinom(formula = output ~ factor. data = dt. weights = n.
maxit = 1000)
Coefficients:
(Intercept) factor
-2.632443 2.034873
Residual Deviance: 106.2304
AIC: 110.2304
dt.pr1<-predict(dt.plr. . type="probs")
dt.pr1
1 2 3 4 5 6
0.4524948 0.4524948 0.5032229 0.5032229 0.5538846 0.5538846
Probability for 2. 4 and 6 are not correct and its explain
the non-correct residual deviance obtained in R.
Probably the problem I have is due to an incorrect data
format... could someone help me...
Thanks
(I know there is a simple way to analyze binomial data. but
in fine I want to use multinom() for 5 categories of outputs.
Thanks a lot
Marc
--
__________________________________________________________
Marc Girondot, Pr
Laboratoire Ecologie, Syst�matique et Evolution Equipe de
Conservation des Populations et des Communaut�s CNRS, ENGREF
et Universit� Paris-Sud 11 , UMR 8079 B�timent 362
91405 Orsay Cedex, France
Tel: 33 1 (0)1.69.15.72.30 Fax: 33 1 (0)1 69
15 56 96 e-mail: [EMAIL PROTECTED]
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Skype: girondot
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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