Peter draws n = 100 independent realizations of a continuous rv and ranks them i

Question

Peter draws n = 100 independent realizations of a continuous rv and ranks
them in increasing order from 1 to 100. Subsequently, Paula draws a single
value from the same population and inserts this value into the rank order
created earlier by Peter. For example, if her value is such that 50 of Peter’s
draws are smaller and 50 are larger, then the rank associated with her draw
would be 51 — that is, overall, her value would be the 51st in increasing order.
Or, if her value is smaller than all 100 of Peter’s, then the rank 1 would be
associated with it.
a. Is it more likely that Paula’s value will occupy rank 51 than rank 1?
b. Derive for general n the probability that Paula’s value will occupy rank
k, where 1 k n + 1.

Explanation / Answer

a. Consider first the scenario when Peter draws just two values; let us call
these values a and b. Let us call x the value obtained by Paula. Clearly, because
the draws are independent and are obtained from the same population, all six
orders are equally likely: abx, axb, bax, bxa, xab, xba, where, e.g., bxa stands
for the order b < x < a. Two of them are such that Paula’s draw (x) is first,
second, or third. Thus, all three ranks for Paula’s draw are equally likely.
b. In general, there are (n + 1)! equally likely orders (i.e., permutations)
of the n values drawn by Peter and the single value drawn by Paula. How
many of them are such that Paula’s value ends up at rank k? If rank k is
reserved for Paula’s value then there remain n! orders in which the n values
drawn by Peter can be arranged around this rank. For example, if n = 3, and
Paula’s value (call it x) is fixed at rank 2, Peter’s three values (say, a, b, c) can
be arranged as follows: axbc, axcb, bxac, bxca, cxab, cxba. These are those six
outcomes in which Paula’s value x ends up at rank no. 2. Obviously, exactly
the same generative procedure could be used to put Paula’s value x at any
rank from 1 to 4. In general, then, of the (n + 1)! permutations there are n!
such that Paula’s value appears at rank no. k. Thus, the probability for each
possible rank k of Paula’s draw is equal to n!/(n+1)! = 1/(n+1) . Note that this
value is independent of the actual rank, k.
Specifically, with n = 100 each rank for Paula’s draw from 1 to 101 is
equally likely. Intuitively, most people think that it is much more likely that
Paula’s draw will eventually occupy some middle, intermediate rank, such as
k = 51, rather than an extreme rank such as k = 1 or k = 101. But that is
not correct: any rank is equally likely.

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