Let \(Y_1<Y_2<\cdots, Y_n\) be the order statistics of n independent observations from a continuous distribution with cumulative distribution function \(F(x)\) and probability density function:

\( f(x)=F'(x) \)

where \(0<F(x)<1\) over the support \(a<x<b\). Then, the probability density function of the \(r^{th}\) order statistic is:

\(g_r(y)=\dfrac{n!}{(r-1)!(n-r)!}\left[F(y)\right]^{r-1}\left[1-F(y)\right]^{n-r}f(y)\)

over the support \(a<y<b\).

Proof

We'll again follow the strategy of first finding the cumulative distribution function \(G_r(y)\) of the \(r^{th}\) order statistic, and then differentiating it with respect to \(y\) to get the probability density function \(g_r(y)\). Now, if the event \(\{X_i\le y\},\;i=1, 2, \cdots, r\) is considered a "success," and we let \(Z\) = the number of such successes in \(n\) mutually independent trials, then \(Z\) is a binomial random variable with \(n\) trials and probability of success:

\(F(y)=P(X_i \le y)\)

Now, the \(r^{th}\) order statistic \(Y_r\) is less than or equal to \(y\) only if r or more of the \(n\) observations \(x_1, x_2, \cdots, x_n\) are less than or equal to \(y\), which implies:

\(G_r(y)=P(Y_r \le y)=P(Z=r)+P(Z=r+1)+ ... + P(Z=n)\)

which can be written using summation notation as:

\(G_r(y)=\sum_{k=r}^{n} P(Z=k)\)

Now, we can replace \(P(Z=k)\) with the probability mass function of a binomial random variable with parameters \(n\) and \(p=F(y)\). Doing so, we get:

\(G_r(y)=\sum_{k=r}^{n}\binom{n}{k}\left[F(y)\right]^{k}\left[1-F(y)\right]^{n-k}\)

Rewriting that slightly by pulling the \(n^{th}\) term out of the summation notation, we get:

\(G_r(y)=\sum_{k=r}^{n-1}\binom{n}{k}\left[F(y)\right]^{k}\left[1-F(y)\right]^{n-k}+\left[F(y)\right]^{n}\)

Now, it's just a matter of taking the derivative of \(G_r(y)\) with respect to \(y\). Using the product rule in conjunction with the chain rule on the terms in the summation, and the power rule in conjunction with the chain rule, we get:

\(g_r(y)=\sum_{k=r}^{n-1}{n\choose k}(k)[F(y)]^{k-1}f(y)[1-F(y)]^{n-k}\\ +\sum_{k=r}^{n-1}[F(y)]^k(n-k)[1-F(y)]^{n-k-1}(-f(y))\\+n[F(y)]^{n-1}f(y) \) (**)

Now, it's just a matter of recognizing that:

\(\left(\begin{array}{l}
n \\
k
\end{array}\right) k=\frac{n !}{\color{blue}\underbrace{\color{black}k !}_{\underset{\text{}}{{\color{blue}\color{red}\cancel {\color{blue}k}\color{blue}(k-1)!}}}\color{black}(n-k) !} \times \color{red}\cancel {\color{black}k}\color{black}=\frac{n !}{(k-1) !(n-k) !}\)

 and

\(\left(\begin{array}{l}
n \\
k
\end{array}\right)(n-k)=\frac{n !}{k !\color{blue}\underbrace{\color{black}(n-k) !}_{\underset{\text{}}{{\color{blue}\color{red}\cancel {\color{blue}(n-k)}\color{blue}(n- k -1)!}}}\color{black}} \times \color{red}\cancel {\color{black}(n-k)}\color{black}=\frac{n !}{k !(n-k-1) !}\)

Once we do that, we see that the p.d.f. of the \(r^{th}\) order statistic \(Y_r\) is just the first term in the summation in \(g_r(y)\). That is:

\(g_r(y)=\dfrac{n!}{(r-1)!(n-r)!}\left[F(y)\right]^{r-1}\left[1-F(y)\right]^{n-r}f(y)\)

for \(a<y<b\). As was to be proved! Simple enough! Well, okay, that's a little unfair to say it's simple, as it's not all that obvious, is it? For homework, you'll be asked to write out, for the case when \(n=6\) and r = 3, the terms in the starred equation (**). In doing so, you should see that for all but the first of the positive terms in the starred equation, there is a corresponding negative term, so that everything but the first term cancels out. After you get a chance to work through that exercise, then perhaps it would be fair to say simple enough!