Joint Distribution of Order Statistics

The insurer wants $P(\text{max loss} > 800{,}000)$ among 5 independent claims, that is the distribution of $X_{(5)}$, the fifth order statistic. Order statistics have their own distributions that differ dramatically from the original.

Given iid random variables $X_1, \ldots, X_n$, the order statistics sort them:

$$X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)}$$

• $X_{(1)} = \min(X_1, \ldots, X_n)$
• $X_{(n)} = \max(X_1, \ldots, X_n)$

For iid continuous variables with CDF $F(x)$ and PDF $f(x)$:

$$F_{X_{(k)}}(x) = \sum_{j=k}^{n} \binom{n}{j} [F(x)]^j [1-F(x)]^{n-j}$$

HIGH-FREQUENCY: Two special cases dominate the exam:

Maximum ($k = n$):

$$F_{X_{(n)}}(x) = [F(x)]^n$$

All $n$ observations must be at most $x$.

Minimum ($k = 1$):

$$F_{X_{(1)}}(x) = 1 - [1-F(x)]^n = 1 - [S(x)]^n$$

Equivalently, $P(X_{(1)} > x) = [S(x)]^n$, all observations must exceed $x$.

$$f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!} [F(x)]^{k-1}[1-F(x)]^{n-k} f(x)$$

Common mistakes

• Confusing max CDF with min CDF. Maximum: $[F(x)]^n$. Minimum survival: $[S(x)]^n$. Swapping them gives the wrong answer. The max CDF is the product of CDFs; the min survival is the product of survivals.
• Forgetting the multinomial coefficient. The PDF of $X_{(k)}$ has $n!/[(k-1)!(n-k)!]$. For $n=5, k=3$, this is 30, not 1. Omitting it makes the PDF fail to integrate to 1.
• Using wrong Beta parameters. The k-th uniform order statistic is Beta$(k, n-k+1)$, not Beta$(k, n-k)$. For $n=5, k=2$: Beta(2, 4), not Beta(2, 3). Mean is $2/6 = 1/3$, not $2/5$. Trap: using $k/n$ instead of $k/(n+1)$.

Bottom line

• Order statistics sort the sample: $X_{(1)} \leq \cdots \leq X_{(n)}$, so $X_{(n)}$ is the maximum and $X_{(1)}$ the minimum.
• Maximum CDF: $[F(x)]^n$. Minimum survival: $[1-F(x)]^n$.
• PDF of $X_{(k)}$: coefficient $n!/[(k-1)!(n-k)!]$ times $[F(x)]^{k-1}[1-F(x)]^{n-k}f(x)$.
• Uniform order statistics: $X_{(k)} \sim \text{Beta}(k, n-k+1)$ with mean $k/(n+1)$.

Exam shortcut

First identify which order statistic: min, max, or general k-th. For min or max, the CDF approach is faster than the PDF. For Uniform(0,1), immediately write Beta$(k, n-k+1)$ with mean $k/(n+1)$. "Max = all below, Min = none below." $P(\max \leq x) = [F(x)]^n$. $P(\min > x) = [S(x)]^n$. "Exponential minimums: rates add." Min of $n$ iid Exp(rate $\lambda$) is Exp(rate $n\lambda$). Mean divides by $n$.

The full lesson (about 2,861 words, 19 min read) adds 3 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• 3f

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