Let us assume a hypothetical disease which afflicts people with 10% probability (i.e. there is a 10% chance that I have this disease right now).
Let us also assume a hypothetical test for this disease with 90% accuracy (i.e. in 9 out of 10 cases it correctly determines if somebody has the disease or not).
Finally, let us assume that I have just been tested and the result came back positive.
What is the probability that I actually have this disease?
added later: The answer is in the comments.
This kind of calculation should be interesting to doctors and their patients; but I was actually thinking about the stock market when I came up with it.
I assume investors would like to avoid market crashes and severe drawdowns, but they are quite rare: 1974, 1987, 2000, 2008 come to my mind. So they seem to happen (on average) more than ten years apart and usually the "crash phase" lasts for a few months only. In other words, the probability that a particular month will experience a "crash" is well below 10%. Therefore, if an investor wants to avoid those "disease months" consistently, she needs a predictor with accuracy much better than 90%.
added even later: Using Bayes' formula, one would calculate p(S|P) as probability to be Sick conditional on the test being Positive as
p(P|S) * p(S) / p(P)
In my experience (I have asked this kind of question several times in job interviews), if people have a problem with it, the problem is usually with the denominator; i.e. p(P) = p(P|S)*p(S) + p(P|N)*p(N), with N denoting Non-sick.
So let me help the Bayesians a little bit: It is usually more intuitive to calculate the ratio p(S|P) / p(N|P) because then p(P) falls out of the equation.
8 comments:
Bayes disease?
Perhaps, but there is a simple frequentist answer ...
Just guessing, I'd say you have a 50% chance of having the disease. I don't know what the probability of my guess being right is though until you tell me and then the probability of me knowing will go to 1.
Just to be a little more clear it seems to me that the four possibilities are 1%, 9%, 9%, and 81%. The two test positive possibilities are 9% and 9% and the test negative possibilities are excluded so you have a 50% probability of having the disease.
Correct.
Consider 100 ppl , on average 90 will be healthy and 10 will have the disease.
The test falsely finds 90*0.1 = 9 healthy ppl to have the disease and of the 10 sick people will identify correctly 10*0.9 = 9
Since I don't know to which of the two groups of 9 people each I belong, my chances are 50:50
I just meant that I was taught to use Bayes theorem to solve this problem.
>> Therefore, if an investor wants to avoid those "disease months" consistently, she needs a predictor with accuracy much better than 90%.
Makes one wonder whether those few who occasionally do get it right and are proclaimed geniuses might not also be very very lucky along with their genius.
Agreed, but I should mention that in my example the payout is asymmetric if one should bet on such a predictor; so even if the probability does not exceed 50% the expected profit could still be positive.
Of course, in reality there is (probably) not such good predictor with > 90% accuracy.
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