DBrower, idiot at large

fscyclist

Wow, CF, you're smarter than I thought. You're exactly right, false positives are dependent on the prevalance ("real incidence" in australian) of the condition in your population and the sensitivity and specificity of your test. That's why a B sample is drawn which will significantly lower your rate of false positives. The other technique is to use a different confirmatory test such as the IRMS for testosterone. In Flandis' case, he had a more sensitive test conducted twice (ratio test) followed by a very specific test (IRMS). What you hope for in this type of diagnostic testing is a high positive predicitive value - that a positive test accurately reflects the condition - in this case doping.

The problem with dope testing is that they usually have a very low negative predictive value - in other words a negative test is not indicative of someone being clean.

TheDarkLord

I don't agree with your analysis here, and it has to do with how you interpret "test accuracy" with conditional probabilities. Your analysis assumes that the definition of "accuracy of test" = Probability(+ve test/doping) where I'm using the standard conditional probability notation, i.e. probability that test gives a positive given that athlete is doping. If that were 99%, no wonder your false positive rate is so high because the thresholds for triggering off positives has to be very low to catch all the dopers. I would say that the real definition of test accuracy is P(doping/+ve test), i.e. probability that the athlete is doping given positive test result. In other words, if you say that a test is 99% accurate, then the probability of athlete being clean given a positive test is 1%; or false postive rate is 1%.

A completely separate parameter is P(+ve test/doping), which measures how efficient a test is in catching the cheaters. Now, from Bayes theorem, P(+ve test/doping) * P(doping) = P(doping/+ve test) * P(+ve test). The last quantity, P(+ve test) measures the incidence of positive tests. P(+ve test/doping) is dependent on the test itself, as is P(doping/+ve test). It is INDEPENDENT of the fraction of the peloton that cheats.

Any dope tester would be seriously concerned if your numbers were really true.

jimmypop

Brower is basking in the glow of "celebrity". He's also too proud to admit that he's wrong. This is not news, and it's the same song other apologists sing.

He's a cynic like I'm a Landis supporter. Much like CampyBob is a "realist".

Crankyfeet

Au contraire.

P(doping/+ve test) is dependent on P(doper).

That was my point in the example I gave.

fscyclist

Cranky is right. Here is a brief explanation of positive predictive value. http://en.wikipedia.org/wiki/Positive_predictive_value

TheDarkLord

I don't disagree with the concept. But his numbers cannot be right. To give an analogy - let us have a testosterone test which defines "+ve" when the T/E ratio is say 1.5 or something low like that. It will probably catch 95% of the cheaters in addition to giving a very large false positive rate. By Cranky's definition, the test is 95% accurate, while I would have to disagree.

TheDarkLord

No, it isn't. It is only dependent on the test. Just in case the use of the term "doper" is unclear in my text, what I was referring to as P(doper/+ve test) = probability that the rider who has triggered the positive test is really cheating. It is 1 minus the false positive rate, which is the probability that a positive test is triggered by a rider who is not doping. P(doper) as in fraction of peloton doping only comes in when you calculate the actual incidence of doping positives.

If the criteria for positives in the test is made more stringent, then it means that any positives determined by the test are more likely to be from real cheaters, which is equivalent to P(cheating/+ve test) going closer to unity. This is the best I can explain this thing; I hope I have cleared some ambiguity.

RAGT

Go to the extremes - it's easy to see that the probabilities must be dependent upon p(doper). Let's take the probability of a false positive or a false negative with the Bruyneel-test (look into the eyes):

a: All are doped:

Then a false positive is simply not possible (p=0)

b: None are doped:

Then a false negative is simply not possible (p=0)

Wayne666

Assuming the test is the least bit valid, can we all agree that when you have a low percentage of true positives it is unlikely that you will have very many false positives? Add to this the fact that there has been almost certainly a large amount of false negatives, it further decrease the probability that a positive woud be false rather than true.

It also means the "I never tested positive" offers little assurance of a rider being clean.

So the only way the doping appologists can hide behind a generalized false positive argument would be to assert that in reality there was very little doping going on and hence few cases of false negatives, which at this point, seems simply silly, no?

TheDarkLord

I don't think we are contradicting each other. I was referring to the properties of the test itself. Any test has (a) a chance to produce false positives (due to natural variation for example) and (b) the chance to miss true positives. The two are linked together to a large extent - make it more stringent, and you reduce the chance of false positives, but also reduce ability to catch all cheaters. These two quantities (or related quantities) are given in my posts as P(cheating given +ve test) or what I wrote as P(doping/+ve test), and P(+ve test given cheating) or what I wrote as P(+ve test/doping). These two quantities are inherent to the test.

Once you have these conditional probabilities, then, the probability of a false positive when applied to the rider sample is dependent on the fraction of doping in the sample.

TheDarkLord

Let me give one more shot at this: What we want is P(rider is clean/test is +ve). This is 1 - P(rider is dirty/test is +ve).

By Bayes theorem, P(rider is dirty/test is +ve) = P(test is +ve/rider is dirty) * P(rider is dirty) / P(test is +ve).

P(test is +ve) = P(test is +ve/rider is dirty) * P(rider is dirty) + P(test is +ve/rider is clean) * P(rider is clean).

I think so far Cranky and I are in agreement. The point where we disagree is that P(test is +ve/rider is clean) is not [1 - P(test is +ve/rider is dirty)]. IMO, these two variables are only weakly correlated, and are quantities dependent on the test. For instance, let us consider a testosterone test that triggers a positive when T/E ratio is 100. Then, whether a rider is clean or dirty, the test will not give a positive, and both P(test is +ve/rider is clean) and P(test is +ve/rider is dirty) are essentially zero.

italiano

I understand little but Cranky is right.....false positive CAN be conditined like that......funny.........his argument sound as Dbrauer, idiot at large according to bro

__________________
For inches and centimetres, let fools contend."
-- Damian Grammaticus

TheDarkLord

Troll.

italiano

Sciense star….gazer..check TBV yourself....easy

__________________
For inches and centimetres, let fools contend."
-- Damian Grammaticus