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Given an ordered i.i.d sample $X_{(1)}, \dots, X_{(n)}$ from a continuous distribution $F(x)$. How can it be shown that:

(1) $\text{Pr}(X_{(k)} \leq x) = \text{P}r(N(x) \geq k)$

where $N(x)$ is the number of sample values less than $x$; furthermore, $N(x) \sim \text{Bin}(n, F(x))$

and

(2) $p(x) = \frac{n!}{(k-1)!(n-k)!} (F(x))^{k-1}(1-F(x))^{n-k}f(x)$

where $f(x)$ is the density of $F(x)$.

Beginning with the right-hand side of the equality given in (1), given $N(x) \sim \text{Bin}(n, F(x))$, it would seem that

$\text{Pr}(N(x) \geq k) = 1 - F_{\text{Bin}(n, F(x))}(k-1)$

where $F_{\text{Bin}(n, F(x))}(k)$ is the CDF for the binomial distribution evaluated at $k$ with parameters $n$ and $F(x)$.

From here, because (1) is an equality the left-hand side and right-hand side should be identical. Given that the term on the left-hand side in (1) is the CDF of $X_{(k)}$, the PDF ($p(x)$) could be found by solving:

$\frac{d}{dx}(1 - F_{\text{Bin}(n, F(x))}(x))$

Hence, $p(x)$ given in (2) should simply be $f_{\text{Bin}(n, F(x))}$(x).

Obviously, my attempt above isn't congruent with the solution. I was hoping somebody could help with this problem.

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    $\begingroup$ The statements in (1) are trivialities following immediately from the definitions. Deriving (2) rigorously does take some work and you're going about it in a good way, right up to the result following "Hence," which is not correct. Use the Chain Rule in the differentiation. $\endgroup$
    – whuber ♦
    Commented Mar 5, 2013 at 15:48
  • $\begingroup$ @whuber: I started writing down the derivative of the Binomial w.r.t $x$ applying the product and the chain rule but one does not directly arrive at the formula (if I am not wrong). All the teaching notes that can be found use the $\epsilon$ approach as in the link that I provide. It is in deed harder than it looks ... those order statistics ;)! $\endgroup$
    – Richi W
    Commented Mar 5, 2013 at 16:22
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    $\begingroup$ @Richard It helps to know that $\sum_{i=k}^n \binom{n}{i} (1-x)^{n-i-1} x^{i-1} (i - n x)$ = $k x^{k-1} \binom{n}{k} (1-x)^{n-k}$. $\endgroup$
    – whuber ♦
    Commented Mar 5, 2013 at 16:32
  • $\begingroup$ Yes .. this helps ;) Then you don't need the $\epsilon$ thing. @user9171 listen to what whuber says, he is right. Just do standard calculus and apply his hint here. $\endgroup$
    – Richi W
    Commented Mar 5, 2013 at 16:37
  • $\begingroup$ I've used the product and chain rules on this function, but I've what I've obtained is not identical to the desired result. I'm pretty sure the calculus is accurate, but perhaps there is an algebraic oversight. With the exception of $F(x)^{1}$ (which I have defined $\frac{d}{dx}F(x)^{1} = f(x)$), I have treated occurrences of $F(x)^{n}$ and $(1-x)^{n}$ as if they were $x^{n}$ and $(1-x)^{n}$ respectively and differentiated w.r.t. $x$. I have arrived at the following derivative: $$ k{n \choose k}[(k-1)F(x)^{k-2}(1-F(x))^{n-k}-(n-k)F(x)^{k-1}(1-F(x))^{n-k-1}f(x)] $$ $\endgroup$
    – user9171
    Commented Mar 6, 2013 at 1:05

1 Answer 1

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If you write $$ Pr(X_{(k)}\le x) = Pr(\{ k \text{ of the } X_i \text{ are } \le x\} \cup \{ k+1 \text{ of the } X_i \text{ are } \le x \} \cup \cdots \cup \{ \text{all of the } X_i \text{ are } \le x\}) = Pr(\#\{i|X_i \le x\} \ge k) $$ then (1) is clear. Recall the definition of ordering.

Then we consider $P[N(x)=k]$ (and not $P[N(x)\ge k])$. As this is an iid sample we get $Pr(X_i\le x) = F(x)$ by definition and by independence and counting the number permutations of $i$ such that $X_i \le x$ you get the Binomial distribution with $p = F(x)$.

I tried to derive the density directly but it didn't wokr out. Googling "order statistics" leads you to papers like this where the density is derived. It is too long for here, I guess.

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