Discussion in 'Scientific Statistics Math' started by Art Kendall, Oct 26, 2010.

1. ### Art KendallGuest

Art Kendall, Oct 26, 2010

2. ### Bruce WeaverGuest

Hi Art. What I see here is the same thing that Altman and Bland
discuss in this BMJ Statistics Note on predictive values of diagnostic
tests. Notice how the predictive values are related to prevalence of
disease. The formulae are much more readable in the free PDF.

http://www.bmj.com/content/309/6947/102.1.full?maxtoshow=?eaf

Cheers,
Bruce

Bruce Weaver, Oct 26, 2010

3. ### PaulGuest

I'm not a Bayesian statistician (in fact, I'm not a statistician, just
a user),
but as I read the post I failed to see any estimates of a cost
function. The
author also seemed to be treating the NSA monitoring in isolation --
either it
identifies me as a terrorist or it doesn't -- whereas I suspect that
the results
of the monitoring are combined with other sources of information to
prioritize
suspects. So while I neither defend nor condemn phone monitoring, I
find the
author's arguments less than persuasive.

For that matter, his calculations and conclusions all eventually trace
back to
the needle-in-a-haystack problem -- 1,000 (or fewer) terrorists in a
population
of 300 million. Does that mean that we should never look for needles
in haystacks?
Or at least not unless we have tests with zero false positive/false
negative rates?
Does that mean an end to organ donation?

/Paul

Paul, Oct 26, 2010
4. ### Barry W BrownGuest

The argument assumes that there is random screening for terrorists --
clearly a futile process if there are only 1000 terrorists in 300m
people. One assumes more intelligent strategies such as infiltration
of dissident groups, etc.

Barry W Brown, Oct 26, 2010
5. ### Rich UlrichGuest

In a slightly paranoid vein -- I could imagine that some law officers
might use the "1000 terrorists" as an excuse to run the system,
while happily developing investigations of 100 times as many
money-launderers, a few million illegal immigrants, and so on.

By the way, I don't think that Bayesian statisticians have much
a simple understanding of how base-rates affect probability of
observing an event. Epidemiologists are one large set of people
with a lot of practice at that.

Rich Ulrich, Oct 28, 2010
6. ### illywhackerGuest

The same point applies to the medical case. If A is the proposition
'has abnormality' and T the proposition 'test positive', then it is
elementary that:

P(A | T) = P(T | A) P(A) / P(T) ,

where

P(T) = P(T | A) P(A) + P(T | ~A) P(~A) ,

where ~ is negation. Let P(T | A) = p, and P(T | ~A) = r, then

P(A | T) = p P(A) / (p P(A) + r P(~A)) .

If P(A) is interpreted as frequency of occurrence in the population,
then even if p = 1 and r is small, the probability that the patient
has the abnormality given a positive test may be small if the
abnormality is rare.

However, as always (and this is one of the major strengths of the
Bayesian approach), we need to indicate *all* our prior knowldge in
the notation or we will make errors. We have some prior knowledge -
all of medicine for a start. Denote all this knowledge by K. K will
include the fact that the patient came to see the doctor; that they
have other symptoms; and so on. BAyes theorem now becomes:

P(A | T, K) = P(T | A, K) P(A | K) / [P(T | A, K)P(A | K) + P(T | ~A,
K)P(~A | K)] .

It may be reasonable that P(T | A, K) and P(T | ~A, K) are independent
of K, but P(A | K) is certianly dependent on it. It is certainly not
just the frequency in the population.

A similar flaw infects the analysis at the beginning of the paper
cited by Bruce, although their hearts are in the right place. The idea
here is to estimate the accuracy of the test (the numbers p and r
above). These are not really probabilities, but (very) succinct
summaries of biological and medical circumstances. They must be
estimated, or marginalized out, if we wish to apply the results of the
study to compute P(A | T). As well as raising the issue of our
uncertainty about p and r given the results of the study, which
naturally affect our certainty about the results of the test, this
also brings in other questions about our prior knowledge. In
particular, it requires us to think about

P({A_{i}})

where {A_{i}} is the set of propositions about abnormality pertaining
to the individuals in the population. It is normally assumed without

P({A_{i}}) = \prod_{i} P(A_{i}) ,

where P(A_{i}) does not depend on i. This may be reasonable, but it
may well not be.

ilywhacker;

illywhacker, Nov 3, 2010
7. ### Herman RubinGuest

The problem is that you are getting into it in the middle.

The general decision problem under uncertainty requires rhat onr
look at all the aspects of the problem in all states of nature.
It can be shown that this corresponds to minimizing the loss-prior
combination by the use of Bayes Theorem, so this is the foundation
for rational Bayesian behavior. The mathematics is the easy part.

.........................

Herman Rubin, Dec 8, 2010