What is the probability that... (order statistics question continued)...?

Question

What is the probability that the larger of two random observations drawn from any continuous pdf (?) will exceed the 60th percentile?

Theorem: Let Y be a continuous random variable with probability density function f_Y(y). If a random sample of size n is drawn from f_Y(y), the marginal pdf for the ith order statistic is given by:

f_Y'_i(y) = n!/[(i-1)!(n-i)!] * [F_y(y)]^(i-1) * [1-F_Y(y)]^(n-i) * f_Y(y)

for 1<=i <= n.

The answer to the question is 0.64. I don't understand how to reach it? How do I know what f_Y(y) equals? Does the solution involve some delicious integration?

Correction:

f_Y'_i(y) = n!/[(i-1)!(n-i)!] * [F_Y(y)]^(i-1) * [1-F_Y(y)]^(n-i) * f_Y(y)

for 1<=i <= n.

Answer 1

Dear metalgoomba,

I think the solution is simpler than you are trying to make it. No integration is needed. Let o1 and o2 be observation 1 and observation 2, respectively. Since percentiles refer to the portion of possible outcomes that are less than or equal to them, p(o1 <= 0.6) = p(o2 <= 0.6) = 0.6 . Similarly, the probabilities of exceeding the 60th percentile are given as p(o1 > 0.6) = p(o2 > 0.6) = 0.4 .

There are only three possibilities: (i) both observations fall within the 60th percentile, (ii) one observation falls within the 60th percentile and one exceeds it, or (iii) both observations exceed the 60th percentile. It's easy to see that in the second and third cases, the larger of the two observations is exceeding the 60th percentile.

So how do we get the probability of these two cases? The easy way is to compute the probability of the first case (the case in which the larger observation does not exceed the 60th percentile), then subtract it from 1. Thus, the probability of both observations falling within the 60th percentile is simply 0.6 x 0.6 = 0.36, so the probability of the larger observation exceeding the 60th percentile is 1 - 0.36 = 0.64 .

I don't know why you would want to apply the theorem you cite, given the simplicity of the solution. The place where the theorem is written should also define the symbols; I could only guess at them. My thoughts are that f_Y may be a probability density function, and F_Y a (cumulative) probability distribution function.

Answer 2

I would like to know also. But your clue is defining what 60th percentile is and assuming y with i = n1 and i =n2 for the larger of the two.