Order Statistics from a Pareto Sample

Why Pareto Order Statistics Need Care

Pareto samples model positive quantities with a heavy upper tail, such as some loss sizes or large measurements. In such samples, the largest observations can have very different moment properties from the individual observations: taking a high rank does not automatically make a mean finite. The key is to identify how many observations must lie in the upper tail for a particular order statistic to be large.

Let $X_1,\ldots,X_n$ be independent and identically distributed with a Pareto Type I distribution, scale $x_m>0$, and shape $\alpha>0$. Write their increasing order statistics as $X_{(1)}\leq\cdots\leq X_{(n)}$. Because the distribution is continuous, ties occur with probability zero. The $j$th largest observation is therefore $X_{(n-j+1)}$, with $j=1$ denoting the maximum.

Definition: A Pareto Type I random variable with scale $x_m>0$ and shape $\alpha>0$ has distribution function and density $$ F(x)= \begin{cases} 0, & x<x_m,\\ 1-(x_m/x)^\alpha, & x\geq x_m, \end{cases} \qquad f(x)=\alpha x_m^\alpha x^{-\alpha-1},\quad x\geq x_m. $$ Its survival function is $1-F(x)=(x_m/x)^\alpha$ for $x\geq x_m$.

Finite-Sample Laws

For any continuous distribution with CDF $F$ and density $f$, the density of $X_{(k)}$ is $\frac{n!}{(k-1)!(n-k)!}F(x)^{k-1}[1-F(x)]^{n-k}f(x)$. The factors describe the $k-1$ observations below $x$, the $n-k$ above $x$, and the observation at $x$. Substituting the Pareto functions gives the Pareto order-statistic law.

Theorem: For $1\leq k\leq n$, the CDF and density of the $k$th smallest Pareto observation are $$ P(X_{(k)}\leq x)= \begin{cases} 0, & x<x_m,\\ \displaystyle\sum_{r=k}^{n}{n\choose r}F(x)^r[1-F(x)]^{n-r}, & x\geq x_m, \end{cases} $$ and, for $x\geq x_m$, $$ f_{X_{(k)}}(x)= \frac{n!}{(k-1)!(n-k)!} \left[1-(x_m/x)^\alpha\right]^{k-1} (x_m/x)^{\alpha(n-k)} \alpha x_m^\alpha x^{-\alpha-1}. $$

For the CDF, the event $X_{(k)}\leq x$ is exactly the event that at least $k$ of the $n$ observations are at most $x$. The number at most $x$ has a binomial distribution with parameters $n$ and $F(x)$, which gives the sum. For the density, among the $n$ observations one is in a small interval around $x$, exactly $k-1$ are below it, and $n-k$ are above it. The number of assignments to those roles is $n!/((k-1)!(n-k)!)$. Multiplying the corresponding probabilities per unit interval and taking the interval width to zero gives the displayed density.

Worked Example: The Maximum of Three Pareto Observations

Let $X_1,X_2,X_3$ be independent Pareto variables with $x_m=1$ and $\alpha=2$. Find the CDF and density of the maximum, and calculate the probability it is at most 2.

For the maximum, $k=n=3$, so its CDF is $F(x)^3$. Here $F(x)=1-x^{-2}$ for $x\geq1$, giving

P(X_{(3)}\leq x)= \begin{cases} 0, & x<1,\\ (1-x^{-2})^3, & x\geq1. \end{cases}

Differentiating the CDF for $x>1$ gives the density:

f_{X_{(3)}}(x)=3(1-x^{-2})^2(2x^{-3}) =6x^{-3}(1-x^{-2})^2,\quad x\geq1.

At $x=2$, $F(2)=1-2^{-2}=3/4$. Thus $P(X_{(3)}\leq2)=(3/4)^3=27/64$. Equivalently, all three observations must be at most 2, so the probability is $(3/4)^3$, as obtained from independence.

The CDF also has a direct expression for the $j$th largest observation. The event $X_{(n-j+1)}\leq x$ means that at most $j-1$ observations exceed $x$. Its upper-tail probability is therefore a binomial tail with success probability $(x_m/x)^\alpha$. In particular, the maximum has CDF $F(x)^n$ and density $nF(x)^{n-1}f(x)$.

Moments and the Strict Existence Condition

The moment calculation is especially transparent through the probability integral transform. For a continuous distribution, $U_i=F(X_i)$ are independent Uniform$(0,1)$ variables. The $k$th order statistic $U_{(k)}$ has beta density proportional to $u^{k-1}(1-u)^{n-k}$. Since a Pareto quantile is $F^{-1}(u)=x_m(1-u)^{-1/\alpha}$, the Pareto order statistic is a transformation of this beta variable.

Theorem: For real $s$, the $s$th moment of the $k$th smallest observation exists as a finite expectation if and only if $s<\alpha(n-k+1)$. Under this condition, $$ E[X_{(k)}^s] =x_m^s\frac{B(n-k+1-s/\alpha,k)}{B(n-k+1,k)}, $$ where $B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}\,dt$ for $a,b>0$.

To derive the formula, set $T=1-U_{(k)}$. The beta density of $U_{(k)}$ is $\frac{u^{k-1}(1-u)^{n-k}}{B(k,n-k+1)}$ on $0<u<1$. Changing from $u$ to $t=1-u$ shows that $T$ has beta density $\frac{t^{n-k}(1-t)^{k-1}}{B(n-k+1,k)}$. The quantile relation gives $X_{(k)}=x_mT^{-1/\alpha}$, so

E[X_{(k)}^s] =\frac{x_m^s}{B(n-k+1,k)} \int_0^1 t^{n-k-s/\alpha}(1-t)^{k-1}\,dt.

Near $t=1$, the factor $(1-t)^{k-1}$ is integrable because $k\geq1$. Near $t=0$, the integral is finite exactly when the exponent $n-k-s/\alpha$ is greater than $-1$, or $s<\alpha(n-k+1)$. If this condition holds, the integral is $B(n-k+1-s/\alpha,k)$, proving the formula. If the condition fails, the integral diverges at zero, so the moment is not finite. $\square$

For the $j$th largest observation, put $k=n-j+1$. The condition becomes $s<\alpha j$, and the moment formula becomes

E[X_{(n-j+1)}^s] =x_m^s\frac{B(j-s/\alpha,n-j+1)}{B(j,n-j+1)}, \qquad s<\alpha j.

For the maximum, $j=1$, so its $s$th moment is finite precisely when $s<\alpha$. This is the same threshold as for one Pareto observation: a maximum has a heavier chance of being large than a typical observation, but it does not gain a higher moment threshold as the sample size increases.

Worked Example: Mean of the Second Largest Observation

Suppose $n=5$, $x_m=2$, and $\alpha=3/2$. Find the mean of the second largest observation, $X_{(4)}$.

Here $j=2$. The mean condition is $1<\alpha j=(3/2)2=3$, so the mean is finite. The moment formula gives

E[X_{(4)}] =2\frac{B(2-1/(3/2),4)}{B(2,4)} =2\frac{B(4/3,4)}{B(2,4)}.

For positive $a$, $B(a,4)=6/[a(a+1)(a+2)(a+3)]$. Therefore $B(4/3,4)=6/[(4/3)(7/3)(10/3)(13/3)]=243/1820$, while $B(2,4)=1/20$. Hence

E[X_{(4)}] =2\frac{243/1820}{1/20} =\frac{486}{91}\approx5.34.

The result exceeds the scale $x_m=2$, as it must, since every observation is at least 2.

Worked Example: The Mean of the Maximum

Let $n=4$, $x_m=2$, and $\alpha=3$. Compute the mean of the maximum.

For $j=1$, the condition is $1<\alpha=3$, so the mean exists. Using the moment formula, $E[X_{(4)}]=2\cdot4B(2/3,4)$, since $B(1,4)=1/4$. Expanding the beta function,

B(2/3,4) =\frac{6}{(2/3)(5/3)(8/3)(11/3)} =\frac{243}{440}.

Thus $E[X_{(4)}]=8(243/440)=243/55\approx4.42$. This is larger than the scale and is also larger than the mean of one observation, $x_m\alpha/(\alpha-1)=3$.

Worked Example: A Boundary Case With Infinite Mean

Take $n=5$, $x_m=1$, and $\alpha=1/2$. Determine whether the maximum and the second largest observation have finite means.

The maximum corresponds to $j=1$, and its mean condition is $1<\alpha j=1/2$, which fails. Thus the maximum has infinite mean.

The second largest corresponds to $j=2$, and its condition is $1<\alpha j=(1/2)2=1$. This is equality, not a strict inequality, so the condition still fails. The second largest also has infinite mean. In the beta-integral proof, the integrand for its mean behaves like $t^{-1}$ as $t$ approaches zero, whose integral diverges.

Using the Rank to Read the Tail

The rank $j$ quantifies how many upper-tail observations are needed for an observation to reach the $j$th largest position. The moment threshold $\alpha j$ follows directly: the more extreme the rank requirement, the greater the number of observations that must simultaneously be large, and the larger the range of finite moments. For example, a moment of order $s$ for the $j$th largest exists exactly when the tail shape multiplied by $j$ exceeds $s$.

This distinction matters in finite-sample calculations. The maximum's mean does not become finite merely by taking a larger sample; its condition remains $1<\alpha$. By contrast, for a fixed rank $j>1$, the threshold is $\alpha j$. When considering a sequence of increasingly large samples, the rank itself must be stated carefully: the second largest always has $j=2$, while an order statistic whose rank changes with $n$ is a different sequence of random variables.

Common Mistake

A frequent error is to use a non-strict moment condition, writing $s\leq\alpha j$. That incorrectly includes the boundary. At equality, the beta integral has a $t^{-1}$ singularity at zero, and $\int_0^\varepsilon t^{-1}\,dt$ diverges. Thus finiteness requires $s<\alpha j$, not $s\leq\alpha j$. In particular, when $\alpha=1/2$, the second largest has an infinite mean because $1=(1/2)2$ is equality.

Key takeaway: For a Pareto sample, the $j$th largest observation has a finite $s$th moment exactly when $s<\alpha j$. Check the strict inequality before evaluating the beta-function formula; at the boundary, the moment diverges.

Check Your Understanding

Use the rank from the largest observation when applying the moment condition.

For a sample of size $n$, write the CDF of the maximum in terms of the Pareto CDF $F$.
For $\alpha=2/3$, which of the maximum and the second largest has a finite mean? Explain using the strict condition.
For $n=6$, what is the moment-existence condition for the third largest observation to have a finite $s$th moment?
Why does the beta integral diverge when $s=\alpha(n-k+1)$?
For a Pareto sample with $\alpha=2$, does the maximum have a finite second moment? State the boundary comparison.