survreg with measurement uncertainties

It seems a line through the origin doesn't fit the data very well.  That may be throwing off the fitting routine.

data = c(144.53, 1687.68, 5397.91)
err = c(8.32, 471.22, 796.67)
model = c(71.60, 859.23, 1699.19)
id = c(1, 2, 3)

# display
plot(data ~ model)
library("Hmisc")
errbar(model, add=TRUE, y = data, yplus= data+err, yminus=data-err, errbar.col=2)
abline(lm(data ~ model - 1))
abline(lm(data ~ model - 1, weights = 1/err^2), col=2)

Making err[1] larger, allows convergence:

err[1] <- 80.32

survreg(Surv(time = data)~model-1+cluster(id), weights=1/(err^2), dist='gaussian', init=c(2.1))

Call:
survreg(formula = Surv(time = data) ~ model - 1 + cluster(id),
    weights = 1/(err^2), dist = "gaussian", init = c(2.1))

Coefficients:
 model
2.6055

Scale= 139.299

Loglik(model)= 0   Loglik(intercept only)= 0
n= 3


And this matches the standard linear model call:

lm(data ~ model - 1, weights = 1/err^2)

Call:
lm(formula = data ~ model - 1, weights = 1/err^2)

Coefficients:
model
2.606


-----Original Message-----
From: Kyle Penner [mailto:kpenner at as.arizona.edu]
Sent: Wednesday, June 12, 2013 3:49 PM
To: Terry Therneau
Cc: r-help at r-project.org
Subject: Re: [R] survreg with measurement uncertainties

Hi Terry,

Thanks for your quick reply.  I am talking about uncertainty in the response.  I have 2 follow up questions:

1) my understanding from the documentation is that 'id' in cluster(id) should be the same when the predictors are not independent.  Is this correct?  (To be more concrete: my data are brightnesses at different wavelengths.  Each brightness is an independent measurement, so the elements of id should all be different?)

2) I tested survreg with uncertainties on an example where I already know the answer (and where I am not using limits), and it does not converge.  Below is the code I used, does anything jump out as incorrect?

data = c(144.53, 1687.68, 5397.91)
err = c(8.32, 471.22, 796.67)
model = c(71.60, 859.23, 1699.19)
id = c(1, 2, 3)

This works (2.9 is the answer from simple chi_sq fitting):

survreg(Surv(time = data, event = c(1,1,1))~model-1, dist='gaussian',
init=c(2.9))

This does not converge (2.1 is the answer from chi_sq fitting):

survreg(Surv(time = data, event = c(1,1,1))~model-1+cluster(id), weights=1/(err^2), dist='gaussian', init=c(2.1))

And this does, but the answer it returns is wonky:

data[2] = 3*err[2] # data[2] is very close to 3*err[2] already survreg(Surv(time = data, event = c(1,2,1))~model-1+cluster(id), weights=1/(err^2), dist='gaussian', init=c(2.1))

Thanks,

Kyle

On Wed, Jun 12, 2013 at 6:51 AM, Terry Therneau <therneau at mayo.edu> wrote:

I will assume that you are talking about uncertainty in the response.
Then one simple way to fit the model is to use case weights that are
proprional to 1/variance, along with +cluster(id) in the model
statement to get a correct variance for this case.  In linear models
this would be called the "White" or "Horvitz-Thompsen" or "GEE working
independence" variance estimate, depending on which literature you
happen to be reading (economics, survey sampling, or biostat).

Now if you are talking about errors in the predictor variables, that
is a much harder problem.

Terry Therneau



On 06/12/2013 05:00 AM, Kyle Penner wrote:

Hello,

I have some measurements that I am trying to fit a model to.  I also
have uncertainties for these measurements.  Some of the measurements
are not well detected, so I'd like to use a limit instead of the
actual measurement.  (I am always dealing with upper limits, i.e.
left censored data.)

I have successfully run survreg using the combination of well
detected measurements and limits, but I would like to include the
measurement uncertainty (for the well detected measurements) in the
fitting.  As far as I can tell, survreg doesn't support this.  Does
anyone have a suggestion for how to accomplish this?

Thanks,

Kyle

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