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Interesting thoughts. However, this argument is biased because it assumes that the performance of the applicants WHO WERE ACCEPTED is not biased by the selectio
by nartz 11y ago
Interesting thoughts. However, this argument is biased because it assumes that the performance of the applicants WHO WERE ACCEPTED is not biased by the selection process itself, and that the performance characteristics of the selected sample are representative of the performance characteristics of the total, which could be a weak assumption.
An attempt at translating to mathematics (feel free to correct me!):
X = event that person belongs to group x
Y = event that person belongs to group y
S = event that person is selected
W = event that person will perform like a 'winner'
for simplicity P(X) + P(Y) = 1
Naturally, 'unbiased' in this case is simply
P(S|X) = P(S1), and P(S|Y) = P(S2), i.e. that the selection process is independent of a certain variable X or Y
PG says we can measure the the performance of these selected applicant winners for each class, i.e. P(X|S,W).
I believe PG assumes that:
P(X|W) / P(Y|W) should equal P(X|S,W)/P(Y|S,W). We can see that these are different distributions, since the second is already conditioned on the selection process.
Simplified, PG assumes that P(X|S,W) = P(X|W) i.e. that conditioning on the selection process does not bias the winning results.
Its left for the reader exercise to determine the 'pathological' cases where this selection variable's distribution makes PG's assumption correct or incorrect.
However, this is simply theoretical - the actual distribution may or may not be 'pathological' and the assumptions made by PG could very well be good.