
\documentclass[12pt,epsf]{article}


\usepackage{verbatim}

\newcommand{\comments}[1]{}

\renewcommand\floatpagefraction{.9}
\renewcommand\topfraction{.9}
\renewcommand\bottomfraction{.9}
\renewcommand\textfraction{.1}
\setcounter{totalnumber}{50}
\setcounter{topnumber}{50}
\setcounter{bottomnumber}{50}

\usepackage{hyperref} \hypersetup{breaklinks=true, pagecolor=white,
colorlinks=
true, linkcolor=black, hyperfootnotes= false, urlcolor=blue } \urlstyle{
rm}

\usepackage{graphicx}

\usepackage{amsmath}

 \usepackage{amssymb}

\usepackage{hyperref}

  \reversemarginpar
  \topmargin  -.3in
  \oddsidemargin -.1in
  \textheight  8.5in
  \textwidth 7in
  \baselineskip 16pt



         \begin{document}


\begin{raggedright}


 \parindent 24pt

 \parskip 10pt

\noindent
8 October 2007\\
 Eric Rasmusen, visitor, Nuffield College,  erasmuse@indiana.edu\\

\bigskip

\bigskip

\noindent
{\bf Notes on Standard Errors in the Case Control Method and the Meaning
of Repeated Sampling}

\bigskip

   A standard counterintuitive result in statistics is that if the
true model is logit, then it is okay to use a sample selected on the
Y's, which is what the "case-control method" amounts to. You may select
1000 observations with Y=1 and 1000 observations with Y=0 and do
estimation of the effects of every variable but the constant in the
usual way, without any sort of weighting. This was shown in Prentice \&
Pyke (1979). They also purport to show that the standard errors may be
computed in the usual way--- that is, using the curvature (2nd
derivative) of the likelihood function. This, I was skeptical of. If the
constant is misestimated, how can you deduce the variance of the
disturbance term, and if you can't deduce that, how can you deduce the
standard error of any of the coefficients?   Nowhere have I seen a clear
demonstration or an intuition for the result, so I thought there might
be a crucial unnoticed mistake in the math somewhere, as  is not unknown
in famous papers (e. g.  Hotelling on location, Tullock on
overdissipation,  Viner on average cost curves, and Rothschild-Stiglitz
on risk).

 Since I did not follow all the steps of the Prentice-Pyke proof and so
did not know of any error in what they did, I tried doing a Monte Carlo
study which seemed to confirm my intuition.

Since then, however, I have seen where my Monte Carlo study went wrong,
and now I believe Prentice and Pyke.  I am writing these notes up
because they will help me remember where I went wrong and perhaps will
be helpful to  other people too.  I'll distribute these to people with
whom I've discussed the idea too, so they can see why I've abandoned it.

1.Here is some  intuition-- a bit shaky, I fear.    Suppose that a
coefficient is  estimated
correctly by some estimator. We want to estimate the estimator's
standard error, to know how variable the estimate would be if we
repeated the estimation with different  disturbances.  For this, we need
to know how noisy the data is. We do not need to know how noisy the data
in the  whole population is, however, just how noisy in the kind of
sample we draw. If our procedure is to draw a biased sample, then we
need to know what will happen in other biased samples, not in the
population. It  is okay to use the sample for this purpose. In using a
standard error, we  are not generalizing anything to the population (not
estimating goodness of fit, for example), we are just generalizing to
repeated samples.

2.   Here is how to  think about repeated sampling  and how to do a
Monte Carlow
study.  What I did was to construct a population of 60,000 data points,
drawing X from a uniform distribution on [0,1] and a disturbance epsilon
from a logit density with an  alpha "constant" coefficient of -4 and a
beta X coefficient of 0.  If   alpha; + epsilon $< 0$ then Y=0; if
alpha + epsilon  $\geq  0$ then Y = 1.  That yields   1,039 points
with
Y= 1, about 1.7\% of them.

Our estimation procedure is to combine  two   random samples of 1,000
observations with Y=0 and  1,000 observations with Y=1 and do a logit
estimate of alpha and beta.  We would expect the estimate  of alpha   to
be wrong-- not close to 0.017-- and the estimate of beta to be right--
close to 0.000-- since we have a large enough sample that consistent
estimates ought to be close to the true parameters.

  The maximum likelihood estimate would give us standard errors based on
the second derivative of the likelihood function or  on bootstrapping.
In repeated sampling, we would expect the standard deviation of the
alpha estimates not to  be close to the average of the estimates of its
standard error.  The question to be investigated is whether the  the
standard deviation of the  beta estimates is  close to the average of
the estimates of its standard error.

   So far, so good.  Where I made my mistake, I think, is in the
definition of "repeated sampling".  Ordinarily in frequentist thinking,
in repeated sampling we keep the X values the same in each sample, and
we draw new disturbances, which combine with the fixed X's to give new
Y's. That also amounts to conditioning on the X's, though we wouldn't
have had to condition the X's, since our estimator should work fine even
if we changed the X's in each sample too. (If we did change the X's,
though, that change the information content in each sample--- a sample
in which X only varied between .3 and .4 would have less information and
yield worse estimates than one with X varying widely between .02 and
.94. So in small samples, especially, we'd have to make some allowance
for that.)

Here, though, we can't keep the X's fixed.  If we did, then although our
first sample would have 1,000 observations with Y=1, our succeeding
samples would have about 34. We wouldn't be using the case-control
method.

So what we have to do is to think about repeated samples with 1,000 Y=0
observations and 1,000 Y=1's.  Turning our usual thinking upside down,
we need to keep the Y's fixed, draw new disturbances, and let the X's
vary.  This is especially hard to think about here, because knowing Y
and epsilon does not tell us X-- remember, Y is coarse and contains less
information than alpha + beta*X + epsilon,and beta is zero here too,
making things even worse.

 The best way to proceed is to think about repeating the entire
scientific  procedure, including the sampling as well as the estimation.
The way I did this was to take 100 n=2000 samples from the 60,000-point
population, each time  combining equal-sized subsamples with Y=0 and
with Y=1.

Recall, however, that there are only 1,037 Y=1 values in the entire
population. Thus, my repeated sampling had to be with replacement, and
was using the same  Y=1 observations over and over. It is OK to use the
same X values repeatedly, but these observations also had the same
epsilon values each time, so the samples are not independent in the way
needed for the law of large numbers to work. The standard errors
computed by maximum  likelihood came out  wrong--- not equal to the
standard deviation of the estimates, but that is to be expected when the
draws are not independent.

   Realizing this, I also tried doing the procedure with 100 n=200
samples instead of 100 n=2000 samples. I still used sampling with
replacement, but now there was less overlap between replacements,  less
dependence between samples. And now the estimated standard errors were
close to the standard deviations.

   This, I expect is what would happen if I did the kind of repeated
sampling that is our thought experiment for the kind of real studies
that use the case-control method. That thought experiment is to take
repeated draws of 60,000-point populations, with the same X's each time
but with different epsilons and hence Y's. Each of the 100 Monte Carlo
samples would be from a different population draw.


 Prentice,  R. L. \&  R. Pyke (1979) ``Logistic Disease Incidence Models
and Case-Control Studies,'' {\it   Biometrika},  66(3):  403-411  (
December 1979).

\end{raggedright}
\end{document}
