The motivation stems from this question on CV.

A website receives \(1\, \text{order}\,/3 \, \text{min}\). The time until the next order follows an exponential distribution with a \(\lambda\) rate parameter of \(3 \, \text{min}\). Knowing that there were \(10\) clients placing orders online from that site last night we want to calculate the expected waiting time for the first and last orders placed after 5pm (night-time period = time \(0\)). And we want to find out the time of the \(k\)-th order.

Every order placed cames at a time. This time is the random variable. Since there are 10 orders, we have \(X_1, X_2, X_3,...X_{10}\). To make the derivation more generalized, we can consider \(X_n\) as the last entry. They are independent of each other, and they are governed by the *pdf* \(\lambda\,e^{-\lambda\,x}\), and *cdf* \(1 \,-\, e^-{-\lambda\,x}\), where \(x\) stands for time. These will be referred as the *common* density and probability functions, because they are shared by all the variables.

The point now is to describe a different variable: the *order* of the observations. So if we imagine that the callers carry a bib number from \(1\) to \(10\) that links to the times they place their orders, client \(X_1\) may end up being the third one to place the order, when ordering the times of the \(10\) clients, in which case the new variable will be denoted as \(X_{(3)}\). Notice the small parentheses.

The cdf of this new variable for the last order, \(n=10\) will be:

\(F_{x_{(n)}}(x) = \Pr(X_{n}\leq x) = \Pr\left(\text{all}\,X_i \leq x\right) = \displaystyle \prod_{k=1}^n \Pr(X_n \leq x) = F(x)^n\), since they are independent and identically ditributed.

In the case of the exponential distribution: \(\left (1 - e^{-\lambda\,x \,}\right)^n.\)

On the other hand the pdf is:

\(\frac{d}{dx}\,F_{x_{(n)}}(x) = n\,(F(x))^{n-1}\, f(x) = n\,\left (1 - e^{-\lambda\,x \,}\right)^{n-1}\,\lambda\,e^{-\lambda x}.\)

The pdf and cdf for the first order \((n=1)\) entered on the website are:

\[\begin{align} F_{x_{(1)}}(x) &= P(\text{min}\, X_i \leq x)\\[2ex] &= 1 - P(X_{(1)} \geq x)\\[2ex] &= 1 - (1 - F(x))^n\\[2ex] &= 1 - (1 - (1 - e^{-\lambda x}))^n\\[2ex] &= 1 - e^{\,-\lambda x n}. \end{align}\]

The pdf for the minimum will be:

\[\begin{align} f_{X_{(1)}}(x)&=\large \frac{d}{dx}\,F_{x_{(1)}}(x)\\[2ex] &= - n \left(1 - F(x)\right)^{n-1}\,(-f(x))\\[2ex] &= n \left(1 - F(x)\right)^{n-1}\,(f(x))\\[2ex] &= n(1 - (1 - e^{-\lambda x}))^{n-1}\, \lambda\,e^{-\lambda x}\\[2ex] &= n(e^{-\lambda x (n-1)})\, \lambda\,e^{-\lambda x}\\[2ex] &= n\, \lambda \, e^{-\lambda x n}. \end{align}\]

What is the joint pdf for all the order statistics:

For the actual observations evaluated at the the observation points:

\[\begin{align} \Pr(X_1 \in [x_1, x_1 + dx_1], X_2 \in [x_2, x_2 + dx_2], \cdots, X_n \in [x_n, x_n + dx_n]) &= \displaystyle \prod_{k=1}^n \, P(X_k \in [X_k, x_k + dx_k])\\[2ex] &= \displaystyle \prod_{k=1}^n \, f(x_k)\,dx_k\\[2ex] &=f(x_1)\,f(x_2)\,\cdots,f(x_n)\,dx_1\,dx_2,\cdots,\,dx_n. \end{align}\]

The first part is the joint pdf for \(\large f_{x_1,x_2,\cdots,x_n}\) evaluated at \(\large x_1, x_2,\cdots, x_n\), or

\(\large f_{x_1,x_2,\cdots,x_n\,(x_1, x_2,\cdots, x_n)}\large \,dx_1\,dx_2\,dx_3,\cdots\,dx_n\) given the independence of the marginals.

Now, for the *order statistics* we need to see the indifference to the actual relationship between subject and order:

\[\begin{align} f_{x_{(1)},x_{(2)},x_{(3)}\cdots,x_{(n)}\,(x_1, x_2,\cdots, x_n)}\,dx_1\,dx_2\,dx_3,\cdots\,dx_n &=\Pr\left(X_{(1)} \in [x_1, x_1 + dx_1], X_{(2)} \in [x_2, x_2 + dx_2], \cdots, X_{(n)} \in [x_n, x_n + dx_n]\right)\\[2ex] &= n! f(x_1)\,f(x_2)\,\cdots,f(x_n)\large \,dx_1\,dx_2\,dx_3,\cdots\,dx_n \end{align}\]

And the pdf is \(\large n!\, f(x_1)\,f(x_2)\,\cdots,f(x_n).\)

For the marginal pdf of the \(\large K\)-th order statistic:

We look at the probability of the observed value to be in the \(k\) small interval. There are \(n\) choices for the one value in the small interval (\(X_{(k)}\)) with probability \(f(x)\,dx\). There are \(k - 1\) to the left; and \(n - k\) to the right. So we choose the ones to the left with the combination below, and we look at the probability of being to the left of \(x\): \(\large F(x)^{k-1}\,(1-F(x))^{n-k})\) in a binomial:

\[ f_{x{(k)}}(x)\,dx = n\,\binom{n-1}{k-1} (f(x)\,dx) \,F(x)^{k-1}\,\left(1-F(x)\right)^{n-k}\]

Simplifying,

\[f_{x{(k)}}(x) = n\,\binom{n-1}{ k-1} \, \,F(x)^{k-1}\,\left(1-F(x)\right)^{n-k}\, f(x)\]

In the case of the exponential distribution:

\[ \begin{align} f_{x{(k)}} &= n\,\binom{n-1}{ k-1} \, \,\left(1 - e^{\,-\lambda x}\right)^{k-1}\,\left(1-(1 - e^{\,-\lambda x})\right)^{n-k}\, f(x)\\[2ex] &=n\,\binom{n-1}{ k-1} \, \,(1 - e^{\,-\lambda x})^{k-1}\,( e^{\,-\lambda x})^{n-k}\, f(x)\\[2ex] &=n\,\binom{n-1}{k-1} \, \,(1 - e^{\,-\lambda x})^{k-1}\,( e^{\,-\lambda x})^{n-k}\, \lambda e^{\,-\lambda x}\\[2ex] &=\lambda \, n\,\binom{n-1}{k-1} \, \,(1 - e^{\,-\lambda x})^{k-1}\,( e^{\,-\lambda x})^{n-k}\, e^{\,-\lambda x}\\[2ex] &=\lambda \, n\,\binom{n-1}{ k-1} \, \,(1 - e^{\,-\lambda x})^{k-1}\,e^{\,-\lambda x(2 +n - k)} \end{align} \]

Integrating:

And \(_2F_1\)(a,b; c; x) is the hypergeometric function.