III. HOME WORK PROBLEMS
4. SEEING IS BELIEVING
Suppose A and B are i.i.d. random variables with mean
a and b. Prove that
Prob(a > b | A > B) >
Prob(a < b | A > B)
i.e. what is observed to be greater is more likely to be actually greater,
or the observed order is the actual order.
SOLUTION:
We model the uncertainty of a and b as random variables, i.e.,
A = a + v and B = b + w where
v and w are i.i.d. random variables. Since we have no prior
knowledge of the relative magnitude of a vs. b. We let
Prob(a > b)= Prob(a < b) = 0.5. Then,
Prob(a > b | A > B) = Prob(a >
b, A > B) / Prob(A > B) =
Prob(A > B | a > b) Prob(a >
b) / Prob(A > B)
and
Prob(a < b | A > B) = Prob(A >
B | a < b) Prob(a < b) /
Prob(A > B).
This means we must show that
Prob(A > B | a > b) > Prob(A >
B | a < b)
which is obvious.