How to find distribution of kth highest of n iid's?

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For example, suppose there are five people whose heights are independent and exponentially distributed with mean 5.5 feet.

I want to be able to solve problems like 1) Find the expected height of the second tallest guy. 2) Find the probability that the third tallest guy is more than twice as tall as the shortest guy.

I think I should be able to find the solutions to problems like these if I know how to find the height distributions of each of the people.

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Hint: condition on height of the shortest, and use the "lack of memory" property of the exponential distribution.

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Let's note $\mathbf{X}_{î:n}$ the random variable of the $i^{th}$ smallest measure out of a sample having a total of $n$ measures. As an example, the smallest person is noted $\mathbf{X}_{1:n}$ and the tallest is noted $\mathbf{X}_{n:n}$.

We need the distribution of the measures (you said exponential in the problem, but the following can be for any distribution). Let's note $F_{\mathbf{X}}(x)$ the probability that a measure is x or smaller ($P(\mathbf{X} \le x$). Then what is needed is $F_{\mathbf{X}_{i:n}}(x)$.

The problem of the density of $i^{th}$ smallest can be conceived as the distribution of a sample in which $i-1$ observations are smaller than $x$, $n-1$ observations are larger than $x$, and finally, one observation is precisely $x$. Thus, the density if based on $F_{\mathbf{X}}$ raised to the power of $i-1$, based on $1-F_{\mathbf{X}}$ raised to the power of $n-i$, and finally, based on $f_{\mathbf{X}}$. The only difficulty is that we don't know which observations in the sample will be smaller than x, which will be larger, and which one will be equal. So we need to consider all the possible arrangements of the dataset, which is achieved by weighting the result with $$\frac{n!}{(i-1)!(n-k)!}.$$ Hence, the distribution is

$$ F_{\mathbf{X}_{i:n}}(x) = \frac{n!}{(i-1)!(n-k)!} \left(F_{\mathbf{X}}(x) \right)^{i-1} \times \left(1-F_{\mathbf{X}}(x) \right)^{n-i} \times f_{\mathbf{X}}(x). $$

Once you have the distribution, you can integrate this to get the density, and find expected value as usual with

$$ \int_A x f_{\mathbf{X}_{i:n}}(x) \, \mathbf{d}x $$ where $A$ is the domain of the random variable $\mathbf{X}$.