Okay, so let’s take an example. Suppose there’s a disease with prevalence P(D) = 6%, and a test with true positive rate P(+|D) = 90% and false positive rate P(+|~D) = 10%.
We have seen a positive result on the test.
We take the information-theoretic version of Bayes theorem:
Now suppose the prevalence of the disease was 70%; then we find
inf(D|+) ~= 0.07 bits (= 95%)
Which makes sense, because the second test merely confirmed what was already likely; hence it is less informative (but even the first is not terribly, due to low prior).
Yeah, I can definitely see the appeal of this method. Great post, thanks!
1% of women at age forty who participate in routine screening have breast cancer. 80% of women with breast cancer will get positive mammographies. 9.6% of women without breast cancer will also get positive mammographies. A woman in this age group had a positive mammography in a routine screening. What is the probability that she actually has breast cancer?
This looks a wonderful post. I think I am going to get a notebook and go through it all step-by-step. May have some questions.
Okay, so let’s take an example. Suppose there’s a disease with prevalence P(D) = 6%, and a test with true positive rate P(+|D) = 90% and false positive rate P(+|~D) = 10%.
We have seen a positive result on the test.
We take the information-theoretic version of Bayes theorem:
inf(D|+) = inf(D) - iev(D,+)
inf(D|+) = log2[0.148] - log2[0.9*0.06] ~= 1.45 bits (= 36%)
Now suppose the prevalence of the disease was 70%; then we find
inf(D|+) ~= 0.07 bits (= 95%)
Which makes sense, because the second test merely confirmed what was already likely; hence it is less informative (but even the first is not terribly, due to low prior).
Yeah, I can definitely see the appeal of this method. Great post, thanks!
An even more convenient example:
It works! That was fun.