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 $X_1,\dots ,X_n$, the order statistics sort them:

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

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

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

$$F_{X_{(k)}}(x)=\sum _{j=k}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)[F(x){]}^j[1-F(x){]}^{n-j}$$

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: $X_{(1)}\le \cdots \le X_{(n)}$, 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." $P(\max \le x)=[F(x){]}^n$. $P(\min >x)=[S(x){]}^n$. "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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