R help - Oct 2011 - Question on estimating standard errors with noisy signals using the quantreg package

Dear all,

My question might be more of a statistics question than a question on R,
although it's on how to apply the 'quantreg' package. Please accept
my
apologies if you believe I am strongly misusing this list. 

To be very brief, the problem is that I have data on only a random draw, not
all of doctors' patients. I am interested in the, say, median number of
patients of doctors. Does it suffice to use the "nid" option in
summary.rq?

More specifically, if the model generating the number of patients, say, r_i,
of doctor i is
   r_i = const + u_i,
then I think I would obtain the median of the number of doctors' patients
using rq(r~1, ...) and plugging this into summary.rq() using the option
se="iid".

Unfortunately, I don't observe r_i in the data but, instead, in the data I
only have a fraction p of these r_i patients. In fact, with (known)
probability p a patient is included in the data. Thus, for each doctor i the
number of patients IN THE DATA follows a binomial distribution with
parameters r_i and p. For each i I now have s_i patients in the data where
s_i is a draw from this binomial distribution. That is, the problem with the
data is that I don't observe r_i but s_i.

Simple montecarlo experiments confirm my intuition that standard errors
should be larger when using the "noisy" information s_i/p instead of
(the
unobserved) r_i.

My guess is that I can consistently estimate any quantile of the number of
doctors' patients AND THEIR STANDARD ERRORS using the quantreg's rq
command:
rq(I(s/p)~1, ...) and the summary.rq() command with option se="nid". 

Am I correct? I am greatful for any help on this issue.

Best regards,
Thorsten Vogel

On Oct 31, 2011, at 7:30 AM, Thorsten Vogel wrote:
> Dear all,
> 
> My question might be more of a statistics question than a question on R,
> although it's on how to apply the 'quantreg' package. Please
accept my
> apologies if you believe I am strongly misusing this list. 
> 
> To be very brief, the problem is that I have data on only a random draw,
not
> all of doctors' patients. I am interested in the, say, median number of
> patients of doctors. Does it suffice to use the "nid" option in
summary.rq?
> 
> More specifically, if the model generating the number of patients, say,
r_i,
> of doctor i is
>   r_i = const + u_i,
> then I think I would obtain the median of the number of doctors'
patients
> using rq(r~1, ...) and plugging this into summary.rq() using the option
> se="iid".
How big are the r_i?  I presume that they are big enough so that you don't
want to worry about the integer "features" of the data?  Are there
really no
covariates?  If so then you are fine with the iid option, but if not, probably
better to use "nid".   If the r_i can be small, it is worth
considering the dithering
approach of Machado and Santos-Silva (JASA, 2005).
> 
> Unfortunately, I don't observe r_i in the data but, instead, in the
data I
> only have a fraction p of these r_i patients. In fact, with (known)
> probability p a patient is included in the data. Thus, for each doctor i
the
> number of patients IN THE DATA follows a binomial distribution with
> parameters r_i and p. For each i I now have s_i patients in the data where
> s_i is a draw from this binomial distribution. That is, the problem with
the
> data is that I don't observe r_i but s_i.
Is it reasonable to assume that the p is the same across doctors?  This seems
to be some sort of compound Poisson problem to me, but I may misunderstand
your description.
> 
> Simple montecarlo experiments confirm my intuition that standard errors
> should be larger when using the "noisy" information s_i/p instead
of (the
> unobserved) r_i.
> 
> My guess is that I can consistently estimate any quantile of the number of
> doctors' patients AND THEIR STANDARD ERRORS using the quantreg's rq
command:
> rq(I(s/p)~1, ...) and the summary.rq() command with option
se="nid".
> 
> Am I correct? I am greatful for any help on this issue.
> 
> Best regards,
> Thorsten Vogel
> 
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