Re: [R] mgcv: distribution of dev with Poisson data
Greg,
The deviance being chi^2 distributed on the residual degrees of freedom
works in some cases (mostly where the response itself can be reasonably
approximated as Gaussian), but rather poorly in others (noteably low
count data). This is what you are seeing in your simulations - in the
first case the Poisson mean is relatively high - so the normal
approximation to the Poisson is not too bad and the deviance is approx.
chi.sq on residual df. In the second case the Poisson mean is low, there
are lots of zeroes, and the approximation breaks down. So yes, the
Pearson estimator is probably a better bet in the latter case.
best,
Simom
ps. Note that the chi squared approximation for the *difference* in
deviance between two nested models does not suffer from this problem.
On 04/02/14 09:25, Greg Dropkin wrote:
mgcv: distribution of dev
hi
I can't tell if this is a simple error.
I'm puzzled by the distribution of dev when fitting a gam to Poisson
generated data.
I expected dev to be approximately chi-squared on residual d.f., i.e.
about 1000 in each case below.
In particular, the low values in the 3rd and 4th versions would suggest
scale 1, yet the data is Poisson generated.
The problem isn't caused by insufficient k values in the smoother.
Does this mean that with sparse data the distribution of dev is no longer
approx chi-sq on residual df?
Does it mean the scale estimate when fitting quasipoisson should be the
Pearson version?
thanks
greg
library(mgcv)
set.seed(1)
x1-runif(1000)
linp-2+exp(-2*x1)*sin(8*x1)
sum(exp(linp))
dev1-dev2-sums-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linp))
sums[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1[j]=m1$dev
dev2[j]=m2$dev
}
mean(sums)
sd(sums)
mean(dev1)
sd(dev1)
mean(dev2)
sd(dev2)
#dev slighly higher than expected
linpa--1+exp(-2*x1)*sin(8*x1)
sum(exp(linpa))
dev1a-dev2a-sumsa-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpa))
sumsa[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1a[j]=m1$dev
dev2a[j]=m2$dev
}
mean(sumsa)
sd(sumsa)
mean(dev1a)
sd(dev1a)
mean(dev2a)
sd(dev2a)
#dev a bit lower than expected
linpb--1.5+exp(-2*x1)*sin(8*x1)
sum(exp(linpb))
dev1b-dev2b-sumsb-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpb))
sumsb[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1b[j]=m1$dev
dev2b[j]=m2$dev
}
mean(sumsb)
sd(sumsb)
mean(dev1b)
sd(dev1b)
mean(dev2b)
sd(dev2b)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale closer, but also 1
linpc--2+exp(-2*x1)*sin(8*x1)
sum(exp(linpc))
dev1c-dev2c-sumsc-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpc))
sumsc[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1c[j]=m1$dev
dev2c[j]=m2$dev
}
mean(sumsc)
sd(sumsc)
mean(dev1c)
sd(dev1c)
mean(dev2c)
sd(dev2c)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$sig2
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale ok
--
Simon Wood, Mathematical Science, University of Bath BA2 7AY UK
+44 (0)1225 386603 http://people.bath.ac.uk/sw283
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] mgcv: distribution of dev with Poisson data
Hi Greg,
Yes, this sounds right - with quasipoisson gam uses `extended
quasi-likelihood' (see McCullagh and Nelder's GLM book) to allow
estimation of the scale parameter along with the smoothing parameters
via (RE)ML, and it could well be that this gives a biased scale estimate
with low counts (although the shape of the mean variance relationship is
unaffected by this).
It might well be more sensible to default to a Pearson estimate of the
scale parameter for 'gam' ('bam' and 'gamm' already do this), to avoid
this issue with quasipoisson and low counts. Your email reminded me that
someone had reported rather too high p-values for quasipoisson with
these sort of data, and looking back at my list of open issues I see
that `someone' was you as well! It's possible that moving to a Pearson
estimator of the scale will solve that problem too.
Thanks for this... very helpful.
best,
Simon
On 05/02/14 12:56, Greg Dropkin wrote:
thanks Simon
also, it appears at least with ML that the default scale estimate with
quasipoisson (i.e. using dev) is the scale which minimises the ML value of
the fitted model. So it is the best model but doesn't actually give the
correct mean-variance relation. Is that right?
thanks again
Greg
Greg,
The deviance being chi^2 distributed on the residual degrees of freedom
works in some cases (mostly where the response itself can be reasonably
approximated as Gaussian), but rather poorly in others (noteably low
count data). This is what you are seeing in your simulations - in the
first case the Poisson mean is relatively high - so the normal
approximation to the Poisson is not too bad and the deviance is approx.
chi.sq on residual df. In the second case the Poisson mean is low, there
are lots of zeroes, and the approximation breaks down. So yes, the
Pearson estimator is probably a better bet in the latter case.
best,
Simom
ps. Note that the chi squared approximation for the *difference* in
deviance between two nested models does not suffer from this problem.
On 04/02/14 09:25, Greg Dropkin wrote:
mgcv: distribution of dev
hi
I can't tell if this is a simple error.
I'm puzzled by the distribution of dev when fitting a gam to Poisson
generated data.
I expected dev to be approximately chi-squared on residual d.f., i.e.
about 1000 in each case below.
In particular, the low values in the 3rd and 4th versions would suggest
scale 1, yet the data is Poisson generated.
The problem isn't caused by insufficient k values in the smoother.
Does this mean that with sparse data the distribution of dev is no
longer
approx chi-sq on residual df?
Does it mean the scale estimate when fitting quasipoisson should be the
Pearson version?
thanks
greg
library(mgcv)
set.seed(1)
x1-runif(1000)
linp-2+exp(-2*x1)*sin(8*x1)
sum(exp(linp))
dev1-dev2-sums-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linp))
sums[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1[j]=m1$dev
dev2[j]=m2$dev
}
mean(sums)
sd(sums)
mean(dev1)
sd(dev1)
mean(dev2)
sd(dev2)
#dev slighly higher than expected
linpa--1+exp(-2*x1)*sin(8*x1)
sum(exp(linpa))
dev1a-dev2a-sumsa-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpa))
sumsa[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1a[j]=m1$dev
dev2a[j]=m2$dev
}
mean(sumsa)
sd(sumsa)
mean(dev1a)
sd(dev1a)
mean(dev2a)
sd(dev2a)
#dev a bit lower than expected
linpb--1.5+exp(-2*x1)*sin(8*x1)
sum(exp(linpb))
dev1b-dev2b-sumsb-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpb))
sumsb[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1b[j]=m1$dev
dev2b[j]=m2$dev
}
mean(sumsb)
sd(sumsb)
mean(dev1b)
sd(dev1b)
mean(dev2b)
sd(dev2b)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale closer, but also 1
linpc--2+exp(-2*x1)*sin(8*x1)
sum(exp(linpc))
dev1c-dev2c-sumsc-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpc))
sumsc[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1c[j]=m1$dev
dev2c[j]=m2$dev
}
mean(sumsc)
sd(sumsc)
mean(dev1c)
sd(dev1c)
mean(dev2c)
sd(dev2c)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$sig2
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale ok
--
Simon Wood, Mathematical Science, University of Bath BA2 7AY UK
+44 (0)1225 386603 http://people.bath.ac.uk/sw283
--
Simon Wood, Mathematical Science, University of Bath BA2 7AY UK
+44 (0)1225 386603 http://people.bath.ac.uk/sw283
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented,
Re: [R] mgcv: distribution of dev with Poisson data
thanks Simon
also, it appears at least with ML that the default scale estimate with
quasipoisson (i.e. using dev) is the scale which minimises the ML value of
the fitted model. So it is the best model but doesn't actually give the
correct mean-variance relation. Is that right?
thanks again
Greg
Greg,
The deviance being chi^2 distributed on the residual degrees of freedom
works in some cases (mostly where the response itself can be reasonably
approximated as Gaussian), but rather poorly in others (noteably low
count data). This is what you are seeing in your simulations - in the
first case the Poisson mean is relatively high - so the normal
approximation to the Poisson is not too bad and the deviance is approx.
chi.sq on residual df. In the second case the Poisson mean is low, there
are lots of zeroes, and the approximation breaks down. So yes, the
Pearson estimator is probably a better bet in the latter case.
best,
Simom
ps. Note that the chi squared approximation for the *difference* in
deviance between two nested models does not suffer from this problem.
On 04/02/14 09:25, Greg Dropkin wrote:
mgcv: distribution of dev
hi
I can't tell if this is a simple error.
I'm puzzled by the distribution of dev when fitting a gam to Poisson
generated data.
I expected dev to be approximately chi-squared on residual d.f., i.e.
about 1000 in each case below.
In particular, the low values in the 3rd and 4th versions would suggest
scale 1, yet the data is Poisson generated.
The problem isn't caused by insufficient k values in the smoother.
Does this mean that with sparse data the distribution of dev is no
longer
approx chi-sq on residual df?
Does it mean the scale estimate when fitting quasipoisson should be the
Pearson version?
thanks
greg
library(mgcv)
set.seed(1)
x1-runif(1000)
linp-2+exp(-2*x1)*sin(8*x1)
sum(exp(linp))
dev1-dev2-sums-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linp))
sums[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1[j]=m1$dev
dev2[j]=m2$dev
}
mean(sums)
sd(sums)
mean(dev1)
sd(dev1)
mean(dev2)
sd(dev2)
#dev slighly higher than expected
linpa--1+exp(-2*x1)*sin(8*x1)
sum(exp(linpa))
dev1a-dev2a-sumsa-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpa))
sumsa[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1a[j]=m1$dev
dev2a[j]=m2$dev
}
mean(sumsa)
sd(sumsa)
mean(dev1a)
sd(dev1a)
mean(dev2a)
sd(dev2a)
#dev a bit lower than expected
linpb--1.5+exp(-2*x1)*sin(8*x1)
sum(exp(linpb))
dev1b-dev2b-sumsb-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpb))
sumsb[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1b[j]=m1$dev
dev2b[j]=m2$dev
}
mean(sumsb)
sd(sumsb)
mean(dev1b)
sd(dev1b)
mean(dev2b)
sd(dev2b)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale closer, but also 1
linpc--2+exp(-2*x1)*sin(8*x1)
sum(exp(linpc))
dev1c-dev2c-sumsc-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpc))
sumsc[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1c[j]=m1$dev
dev2c[j]=m2$dev
}
mean(sumsc)
sd(sumsc)
mean(dev1c)
sd(dev1c)
mean(dev2c)
sd(dev2c)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$sig2
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale ok
--
Simon Wood, Mathematical Science, University of Bath BA2 7AY UK
+44 (0)1225 386603 http://people.bath.ac.uk/sw283
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] mgcv: distribution of dev with Poisson data
On 05/02/2014 12:56, Greg Dropkin wrote:
thanks Simon
also, it appears at least with ML that the default scale estimate with
quasipoisson (i.e. using dev) is the scale which minimises the ML value of
the fitted model. So it is the best model but doesn't actually give the
correct mean-variance relation. Is that right?
Well, estimates do not usually give the correct value
But ML is biased in many cases, sometimes badly so here, and most people
prefer the Pearson estimator of the dispersion. This is McCullagn
Nelder's book, even the first edition, but without much evidence:
Venables Ripley (2002, §7.5) has plots to back up their hunch.
thanks again
Greg
Greg,
The deviance being chi^2 distributed on the residual degrees of freedom
works in some cases (mostly where the response itself can be reasonably
approximated as Gaussian), but rather poorly in others (noteably low
count data). This is what you are seeing in your simulations - in the
first case the Poisson mean is relatively high - so the normal
approximation to the Poisson is not too bad and the deviance is approx.
chi.sq on residual df. In the second case the Poisson mean is low, there
are lots of zeroes, and the approximation breaks down. So yes, the
Pearson estimator is probably a better bet in the latter case.
best,
Simom
ps. Note that the chi squared approximation for the *difference* in
deviance between two nested models does not suffer from this problem.
On 04/02/14 09:25, Greg Dropkin wrote:
mgcv: distribution of dev
hi
I can't tell if this is a simple error.
I'm puzzled by the distribution of dev when fitting a gam to Poisson
generated data.
I expected dev to be approximately chi-squared on residual d.f., i.e.
about 1000 in each case below.
In particular, the low values in the 3rd and 4th versions would suggest
scale 1, yet the data is Poisson generated.
The problem isn't caused by insufficient k values in the smoother.
Does this mean that with sparse data the distribution of dev is no
longer
approx chi-sq on residual df?
Does it mean the scale estimate when fitting quasipoisson should be the
Pearson version?
thanks
greg
library(mgcv)
set.seed(1)
x1-runif(1000)
linp-2+exp(-2*x1)*sin(8*x1)
sum(exp(linp))
dev1-dev2-sums-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linp))
sums[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1[j]=m1$dev
dev2[j]=m2$dev
}
mean(sums)
sd(sums)
mean(dev1)
sd(dev1)
mean(dev2)
sd(dev2)
#dev slighly higher than expected
linpa--1+exp(-2*x1)*sin(8*x1)
sum(exp(linpa))
dev1a-dev2a-sumsa-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpa))
sumsa[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1a[j]=m1$dev
dev2a[j]=m2$dev
}
mean(sumsa)
sd(sumsa)
mean(dev1a)
sd(dev1a)
mean(dev2a)
sd(dev2a)
#dev a bit lower than expected
linpb--1.5+exp(-2*x1)*sin(8*x1)
sum(exp(linpb))
dev1b-dev2b-sumsb-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpb))
sumsb[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1b[j]=m1$dev
dev2b[j]=m2$dev
}
mean(sumsb)
sd(sumsb)
mean(dev1b)
sd(dev1b)
mean(dev2b)
sd(dev2b)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale closer, but also 1
linpc--2+exp(-2*x1)*sin(8*x1)
sum(exp(linpc))
dev1c-dev2c-sumsc-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpc))
sumsc[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1c[j]=m1$dev
dev2c[j]=m2$dev
}
mean(sumsc)
sd(sumsc)
mean(dev1c)
sd(dev1c)
mean(dev2c)
sd(dev2c)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$sig2
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale ok
--
Simon Wood, Mathematical Science, University of Bath BA2 7AY UK
+44 (0)1225 386603 http://people.bath.ac.uk/sw283
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
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, UKFax: +44 1865 272595
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] mgcv: distribution of dev with Poisson data
hi Simon
yes, I also got the right shape of the mean-variance relation but the
wrong estimate of the parameter.
thanks very much
Greg
Hi Greg,
Yes, this sounds right - with quasipoisson gam uses `extended
quasi-likelihood' (see McCullagh and Nelder's GLM book) to allow
estimation of the scale parameter along with the smoothing parameters
via (RE)ML, and it could well be that this gives a biased scale estimate
with low counts (although the shape of the mean variance relationship is
unaffected by this).
It might well be more sensible to default to a Pearson estimate of the
scale parameter for 'gam' ('bam' and 'gamm' already do this), to avoid
this issue with quasipoisson and low counts. Your email reminded me that
someone had reported rather too high p-values for quasipoisson with
these sort of data, and looking back at my list of open issues I see
that `someone' was you as well! It's possible that moving to a Pearson
estimator of the scale will solve that problem too.
Thanks for this... very helpful.
best,
Simon
On 05/02/14 12:56, Greg Dropkin wrote:
thanks Simon
also, it appears at least with ML that the default scale estimate with
quasipoisson (i.e. using dev) is the scale which minimises the ML value
of
the fitted model. So it is the best model but doesn't actually give
the
correct mean-variance relation. Is that right?
thanks again
Greg
Greg,
The deviance being chi^2 distributed on the residual degrees of freedom
works in some cases (mostly where the response itself can be reasonably
approximated as Gaussian), but rather poorly in others (noteably low
count data). This is what you are seeing in your simulations - in the
first case the Poisson mean is relatively high - so the normal
approximation to the Poisson is not too bad and the deviance is approx.
chi.sq on residual df. In the second case the Poisson mean is low,
there
are lots of zeroes, and the approximation breaks down. So yes, the
Pearson estimator is probably a better bet in the latter case.
best,
Simom
ps. Note that the chi squared approximation for the *difference* in
deviance between two nested models does not suffer from this problem.
On 04/02/14 09:25, Greg Dropkin wrote:
mgcv: distribution of dev
hi
I can't tell if this is a simple error.
I'm puzzled by the distribution of dev when fitting a gam to Poisson
generated data.
I expected dev to be approximately chi-squared on residual d.f., i.e.
about 1000 in each case below.
In particular, the low values in the 3rd and 4th versions would
suggest
scale 1, yet the data is Poisson generated.
The problem isn't caused by insufficient k values in the smoother.
Does this mean that with sparse data the distribution of dev is no
longer
approx chi-sq on residual df?
Does it mean the scale estimate when fitting quasipoisson should be
the
Pearson version?
thanks
greg
library(mgcv)
set.seed(1)
x1-runif(1000)
linp-2+exp(-2*x1)*sin(8*x1)
sum(exp(linp))
dev1-dev2-sums-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linp))
sums[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1[j]=m1$dev
dev2[j]=m2$dev
}
mean(sums)
sd(sums)
mean(dev1)
sd(dev1)
mean(dev2)
sd(dev2)
#dev slighly higher than expected
linpa--1+exp(-2*x1)*sin(8*x1)
sum(exp(linpa))
dev1a-dev2a-sumsa-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpa))
sumsa[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1a[j]=m1$dev
dev2a[j]=m2$dev
}
mean(sumsa)
sd(sumsa)
mean(dev1a)
sd(dev1a)
mean(dev2a)
sd(dev2a)
#dev a bit lower than expected
linpb--1.5+exp(-2*x1)*sin(8*x1)
sum(exp(linpb))
dev1b-dev2b-sumsb-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpb))
sumsb[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1b[j]=m1$dev
dev2b[j]=m2$dev
}
mean(sumsb)
sd(sumsb)
mean(dev1b)
sd(dev1b)
mean(dev2b)
sd(dev2b)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale closer, but also 1
linpc--2+exp(-2*x1)*sin(8*x1)
sum(exp(linpc))
dev1c-dev2c-sumsc-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpc))
sumsc[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1c[j]=m1$dev
dev2c[j]=m2$dev
}
mean(sumsc)
sd(sumsc)
mean(dev1c)
sd(dev1c)
mean(dev2c)
sd(dev2c)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$sig2
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale ok
--
Simon Wood, Mathematical Science, University of Bath BA2 7AY UK
+44 (0)1225 386603 http://people.bath.ac.uk/sw283
--
Simon Wood, Mathematical Science, University of Bath BA2 7AY UK
+44 (0)1225
Re: [R] mgcv: distribution of dev with Poisson data
Dear Prof Ripley
yes, but if the estimate is biased it's good to know what the bias is.
The problem illustrated in the simulations has nothing to do with ML,
though, as the default fitting method in mgcv when scale is unknown is
GCV and that is what was used, by default, here.
The point about ML was that I'd looked separately at what happens when
scale varies in fitting a quasipoisson model, and found that ML was
minimised (in an example) at the scale estimate (i.e. the default estimate
using dev). A very brief look at the code made that seem plausible as it
does involve the derivatives with respect to scale.
In my limited experience ML is actually better behaved than GCV or REML,
giving more conservative estimates and tighter CIs which behave better
under simulation.
re the Pearson estimates being preferable, that is also what Simon Wood's
book says on p172, another reason why I was puzzled that the default
estimate used dev instead.
thanks
Greg
On 05/02/2014 12:56, Greg Dropkin wrote:
thanks Simon
also, it appears at least with ML that the default scale estimate with
quasipoisson (i.e. using dev) is the scale which minimises the ML value
of
the fitted model. So it is the best model but doesn't actually give the
correct mean-variance relation. Is that right?
Well, estimates do not usually give the correct value
But ML is biased in many cases, sometimes badly so here, and most people
prefer the Pearson estimator of the dispersion. This is McCullagn
Nelder's book, even the first edition, but without much evidence:
Venables Ripley (2002, §7.5) has plots to back up their hunch.
thanks again
Greg
Greg,
The deviance being chi^2 distributed on the residual degrees of
freedom
works in some cases (mostly where the response itself can be
reasonably
approximated as Gaussian), but rather poorly in others (noteably low
count data). This is what you are seeing in your simulations - in the
first case the Poisson mean is relatively high - so the normal
approximation to the Poisson is not too bad and the deviance is
approx.
chi.sq on residual df. In the second case the Poisson mean is low, there
are lots of zeroes, and the approximation breaks down. So yes, the
Pearson estimator is probably a better bet in the latter case. best,
Simom
ps. Note that the chi squared approximation for the *difference* in
deviance between two nested models does not suffer from this problem.
On 04/02/14 09:25, Greg Dropkin wrote:
mgcv: distribution of dev
hi
I can't tell if this is a simple error.
I'm puzzled by the distribution of dev when fitting a gam to Poisson
generated data.
I expected dev to be approximately chi-squared on residual d.f., i.e.
about 1000 in each case below.
In particular, the low values in the 3rd and 4th versions would suggest
scale 1, yet the data is Poisson generated.
The problem isn't caused by insufficient k values in the smoother.
Does this mean that with sparse data the distribution of dev is no
longer
approx chi-sq on residual df?
Does it mean the scale estimate when fitting quasipoisson should be the
Pearson version?
thanks
greg
library(mgcv)
set.seed(1)
x1-runif(1000)
linp-2+exp(-2*x1)*sin(8*x1)
sum(exp(linp))
dev1-dev2-sums-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linp))
sums[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1[j]=m1$dev
dev2[j]=m2$dev
}
mean(sums)
sd(sums)
mean(dev1)
sd(dev1)
mean(dev2)
sd(dev2)
#dev slighly higher than expected
linpa--1+exp(-2*x1)*sin(8*x1)
sum(exp(linpa))
dev1a-dev2a-sumsa-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpa))
sumsa[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1a[j]=m1$dev
dev2a[j]=m2$dev
}
mean(sumsa)
sd(sumsa)
mean(dev1a)
sd(dev1a)
mean(dev2a)
sd(dev2a)
#dev a bit lower than expected
linpb--1.5+exp(-2*x1)*sin(8*x1)
sum(exp(linpb))
dev1b-dev2b-sumsb-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpb))
sumsb[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1b[j]=m1$dev
dev2b[j]=m2$dev
}
mean(sumsb)
sd(sumsb)
mean(dev1b)
sd(dev1b)
mean(dev2b)
sd(dev2b)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale closer, but also 1
linpc--2+exp(-2*x1)*sin(8*x1)
sum(exp(linpc))
dev1c-dev2c-sumsc-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpc))
sumsc[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1c[j]=m1$dev
dev2c[j]=m2$dev
}
mean(sumsc)
sd(sumsc)
mean(dev1c)
sd(dev1c)
mean(dev2c)
sd(dev2c)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$sig2
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale ok
--
Simon Wood,
[R] mgcv: distribution of dev with Poisson data
mgcv: distribution of dev
hi
I can't tell if this is a simple error.
I'm puzzled by the distribution of dev when fitting a gam to Poisson
generated data.
I expected dev to be approximately chi-squared on residual d.f., i.e.
about 1000 in each case below.
In particular, the low values in the 3rd and 4th versions would suggest
scale 1, yet the data is Poisson generated.
The problem isn't caused by insufficient k values in the smoother.
Does this mean that with sparse data the distribution of dev is no longer
approx chi-sq on residual df?
Does it mean the scale estimate when fitting quasipoisson should be the
Pearson version?
thanks
greg
library(mgcv)
set.seed(1)
x1-runif(1000)
linp-2+exp(-2*x1)*sin(8*x1)
sum(exp(linp))
dev1-dev2-sums-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linp))
sums[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1[j]=m1$dev
dev2[j]=m2$dev
}
mean(sums)
sd(sums)
mean(dev1)
sd(dev1)
mean(dev2)
sd(dev2)
#dev slighly higher than expected
linpa--1+exp(-2*x1)*sin(8*x1)
sum(exp(linpa))
dev1a-dev2a-sumsa-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpa))
sumsa[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1a[j]=m1$dev
dev2a[j]=m2$dev
}
mean(sumsa)
sd(sumsa)
mean(dev1a)
sd(dev1a)
mean(dev2a)
sd(dev2a)
#dev a bit lower than expected
linpb--1.5+exp(-2*x1)*sin(8*x1)
sum(exp(linpb))
dev1b-dev2b-sumsb-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpb))
sumsb[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1b[j]=m1$dev
dev2b[j]=m2$dev
}
mean(sumsb)
sd(sumsb)
mean(dev1b)
sd(dev1b)
mean(dev2b)
sd(dev2b)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale closer, but also 1
linpc--2+exp(-2*x1)*sin(8*x1)
sum(exp(linpc))
dev1c-dev2c-sumsc-rep(0,20)
for (j in 1:20)
{
y-rpois(1000,exp(linpc))
sumsc[j]-sum(y)
m1-gam(y~s(x1),family=poisson)
m2-gam(y~s(x1,k=20),family=poisson)
dev1c[j]=m1$dev
dev2c[j]=m2$dev
}
mean(sumsc)
sd(sumsc)
mean(dev1c)
sd(dev1c)
mean(dev2c)
sd(dev2c)
#dev much lower than expected
m1q-gam(y~s(x1),family=quasipoisson,scale=-1)
m1q$scale
m1q$sig2
m1q$dev/(1000-sum(m1q$edf))
#scale estimate 1
sum((y-fitted(m1q))^2/fitted(m1q))/(1000-sum(m1q$edf))
#Pearson estimate of scale ok
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