Re: [R] Fitting a Weibull/NaNs
On 19 Oct 2003 21:18:09 -0700, you wrote: The problem seems to be that some of the values of d$Age.Month are 0 and since the Weibull always has a value of 0 at 0, the log likelihood comes out insane. (I'm getting 0 values due to quantization error). OTOH when I remove the 0 values it works great, but that seems kind of ad hoc. Is there some standard fix for this? A standard fix is to replace zeros with some small positive number, but this isn't entirely satisfactory. It's definitely worthwhile doing a sensitivity test: e.g. do you get essentially the same answer using 0.01 and 0.0001 for the replacement value? If not, you might want to question the use of a model that predicts zero density in an area where you've got observations. Duncan Murdoch __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Fitting a Weibull/NaNs
I'm trying to fit a Weibull distribution to some data via maximum
likelihood estimation. I'm following the procedure described by Doug
Bates in his Using Open Source Software to Teach Mathematical
Statistics but I keep getting warnings about NaNs being converted to
maximum positive value:
llfunc - function (x) { -sum(dweibull(AM,shape=x[1],scale=x[2], log=TRUE))}
mle - nlm(llfunc,c(shape=1.5,scale=40), hessian=TRUE)
Warning messages:
1: NaNs produced in: dweibull(x, shape, scale, log)
2: NA/Inf replaced by maximum positive value
3: NaNs produced in: dweibull(x, shape, scale, log)
4: NA/Inf replaced by maximum positive value
Can someone offer some advice here?
Thanks,
-Ekr
--
[Eric Rescorla [EMAIL PROTECTED]
http://www.rtfm.com/
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Re: [R] Fitting a Weibull/NaNs
I have not used nlm, but that happens routinely with function
minimizers trying to test negative values for one or more component of
x. My standard approach to something like this is to parameterize
llfunc in terms of log(shape) and log(scale), as follows:
llfunc - function (x) { -sum(dweibull(AM,shape=exp(x[1]),scale=exp(x[2]), log=TRUE))}
Have you tried this? If no, I suspect the warnings will disappear
when you try this. If not, I suggest you rewrite llfunc to store
nlglk - (-sum(...)) and then print out x whenever nlglk is NA or Inf
or Nan.
hope this helps. spencer graves
Eric Rescorla wrote:
I'm trying to fit a Weibull distribution to some data via maximum
likelihood estimation. I'm following the procedure described by Doug
Bates in his Using Open Source Software to Teach Mathematical
Statistics but I keep getting warnings about NaNs being converted to
maximum positive value:
llfunc - function (x) { -sum(dweibull(AM,shape=x[1],scale=x[2], log=TRUE))}
mle - nlm(llfunc,c(shape=1.5,scale=40), hessian=TRUE)
Warning messages:
1: NaNs produced in: dweibull(x, shape, scale, log)
2: NA/Inf replaced by maximum positive value
3: NaNs produced in: dweibull(x, shape, scale, log)
4: NA/Inf replaced by maximum positive value
Can someone offer some advice here?
Thanks,
-Ekr
--
[Eric Rescorla [EMAIL PROTECTED]
http://www.rtfm.com/
__
[EMAIL PROTECTED] mailing list
https://www.stat.math.ethz.ch/mailman/listinfo/r-help
__
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Re: [R] Fitting a Weibull/NaNs
Spencer Graves [EMAIL PROTECTED] writes:
I have not used nlm, but that happens routinely with function
minimizers trying to test negative values for one or more
component of x. My standard approach to something like this is
to parameterize llfunc in terms of log(shape) and log(scale),
as follows: llfunc - function (x) {
-sum(dweibull(AM,shape=exp(x[1]),scale=exp(x[2]), log=TRUE))}
Have you tried this? If no, I suspect the warnings will
disappear when you try this.
This works. I've got some more questions, though:
(1) Does it introduce bias to work with the logs like this?
(2) My original data set had zero values. I added .5 experimentally,
which is how I got to this data set. This procedure doesn't work
on the original data set.
Instead I get (with the numbers below being the values
that caused problems):
[1] 0.41 3.70 1.00
[1] 0.41 3.70 1.00
[1] 0.410001 3.70 1.00
[1] 0.41 3.74 1.00
[1] 0.41 3.70 1.01
Warning messages:
1: NA/Inf replaced by maximum positive value
2: NA/Inf replaced by maximum positive value
3: NA/Inf replaced by maximum positive value
4: NA/Inf replaced by maximum positive value
Thanks,
-Ekr
--
[Eric Rescorla [EMAIL PROTECTED]
http://www.rtfm.com/
__
[EMAIL PROTECTED] mailing list
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Re: [R] Fitting a Weibull/NaNs
If the algorithm works properly, you should get exactly the same
answer using a linear or a log scale for the parameters.
The bigger question is not bias but the accuracy of a normal
approximation for confidence intervals and regions. I have evaluated
this by making contour plots of the log(likelihood). I use outer to
compute this over an appropriate grid of the parameters. Then I use
contour [or image with contour(..., add=TRUE)] to see the result.
After I get a picture, I may specify the levels, using, e.g.,
2*log(likelihood ratio) is approximately chi-square with 2 degrees of
freedom. The normality assumption says that the contours should be
close to elliptical. I've also fit log(likelihood) to a parabola in the
parameters, possibly after deleting points beyond the 0.001 level for
chi-square(2). If I get a good fit, I'm happy. If not, I try a
different parameterization.
When I've done this, I've found that I tend to get more nearly
normal contours by throwing the constraint to (-Inf) than leaving it at
0, i.e., by
Bates and Watts (1988) Nonlinear Regression Analysis and Its
Applications (Wiley) explain that parameter effects curvature seems to
be vastly greater than the intrinsic curvature of the nonlinear
manifold, onto which a response vector is projected by nonlinear least
square. This is different from maximum likelihood, but I believe that
this principle would still likely apply.
Does this make sense?
spencer graves
p.s. I don't understand what you are saying about 0.41 3.70
1.00 below. You are giving me a set of three numbers when you are
trying to estimate two parameters and getting NAs, Inf's and NaNs. I
don't understand. Are you printing out x when the log(likelihood) is
NA, NaN or Inf? If yes, is one component of x = 0?
Eric Rescorla wrote:
Spencer Graves [EMAIL PROTECTED] writes:
I have not used nlm, but that happens routinely with function
minimizers trying to test negative values for one or more
component of x. My standard approach to something like this is
to parameterize llfunc in terms of log(shape) and log(scale),
as follows: llfunc - function (x) {
-sum(dweibull(AM,shape=exp(x[1]),scale=exp(x[2]), log=TRUE))}
Have you tried this? If no, I suspect the warnings will
disappear when you try this.
This works. I've got some more questions, though:
(1) Does it introduce bias to work with the logs like this?
(2) My original data set had zero values. I added .5 experimentally,
which is how I got to this data set. This procedure doesn't work
on the original data set.
Instead I get (with the numbers below being the values
that caused problems):
[1] 0.41 3.70 1.00
[1] 0.41 3.70 1.00
[1] 0.410001 3.70 1.00
[1] 0.41 3.74 1.00
[1] 0.41 3.70 1.01
Warning messages:
1: NA/Inf replaced by maximum positive value
2: NA/Inf replaced by maximum positive value
3: NA/Inf replaced by maximum positive value
4: NA/Inf replaced by maximum positive value
Thanks,
-Ekr
__
[EMAIL PROTECTED] mailing list
https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Fitting a Weibull/NaNs
Spencer Graves [EMAIL PROTECTED] writes:
Bates and Watts (1988) Nonlinear Regression Analysis and Its
Applications (Wiley) explain that parameter effects curvature
seems to be vastly greater than the intrinsic curvature of the
nonlinear manifold, onto which a response vector is projected by
nonlinear least square. This is different from maximum
likelihood, but I believe that this principle would still likely
apply. Does this make sense? spencer graves
Some :)
p.s. I don't understand what you are saying about 0.41 3.70
1.00 below. You are giving me a set of three numbers when
you are trying to estimate two parameters and getting NAs,
Inf's and NaNs. I don't understand. Are you printing out
x when the log(likelihood) is NA, NaN or Inf? If yes, is
one component of x = 0? Eric Rescorla wrote:
Doh! Typographical error to R. I had the hessian=TRUE clause inside
the c(). Doesn't make any difference for the results, though.
I'm doing the following:
llfunc -
+ function (zzz) {
+ tmp - -sum(dweibull(d$Age.Month,shape=exp(zzz[1]),scale=exp(zzz[2]), log=TRUE))
+ if(is.infinite(tmp) | is.na(tmp)) { print(zzz);}
+ tmp
+
+ }
mle - nlm(llfunc,c(shape=.37,scale=4.0), hessian=TRUE)
[1] 0.37 4.00
[1] 0.37 4.00
[1] 0.370001 4.00
[1] 0.37 4.04
[1] 0.3701 4.
[1] 0.3700 4.0004
[1] 0.3702 4.
[1] 0.3701 4.0004
[1] 0.3700 4.0008
Warning messages:
1: NA/Inf replaced by maximum positive value
2: NA/Inf replaced by maximum positive value
3: NA/Inf replaced by maximum positive value
4: NA/Inf replaced by maximum positive value
5: NA/Inf replaced by maximum positive value
6: NA/Inf replaced by maximum positive value
7: NA/Inf replaced by maximum positive value
8: NA/Inf replaced by maximum positive value
I'm a little vague on how this is supposed to work, but when
I just compute
-sum(dweibull(d$Age.Month,shape=1.5,scale=40,log=TRUE))
I get Inf.
The problem seems to be that some of the values of d$Age.Month are 0
and since the Weibull always has a value of 0 at 0, the log likelihood
comes out insane. (I'm getting 0 values due to quantization error).
OTOH when I remove the 0 values it works great, but that seems
kind of ad hoc. Is there some standard fix for this?
Thanks much,
-Ekr
--
[Eric Rescorla [EMAIL PROTECTED]
http://www.rtfm.com/
__
[EMAIL PROTECTED] mailing list
https://www.stat.math.ethz.ch/mailman/listinfo/r-help
