Uniform Order Statistics: Mean and Variance
You draw $n$ i.i.d. samples from a $\text{Uniform}(0,1)$ distribution and sort them in increasing order: $U_{(1)} \le U_{(2)} \le \cdots \le U_{(n)}$.
- Compute $E[U_{(k)}]$ and $\text{Var}(U_{(k)})$ in closed form for \le k \le n$.
- For fixed $k$, find $\lim_{n \to \infty} E[U_{(k)}]$ and give an intuitive explanation of the result.
Hints
- The $k$-th order statistic of $n$ uniform draws follows a well-known named distribution. Think about what family has support on $(0,1)$ and is parameterized by two shape parameters.
- If $U_{(k)} \sim \text{Beta}(k, \, n-k+1)$, the mean and variance are just the standard Beta moment formulas.
- For the limit, note that $E[U_{(k)}] = k/(n+1)$. With $k$ fixed and $n \to \infty$, this ratio vanishes -- think about why the $k$-th smallest of many points must cluster near the boundary.
Worked Solution
How to Think About It: The $k$-th order statistic out of $n$ uniform draws is the value that sits in the $k$-th position from the bottom. Before doing any math, ask yourself: where should this point land on average? If you have $n$ draws on $[0,1]$, they divide the interval into roughly $n+1$ equal gaps. The $k$-th point should sit near $k/(n+1)$. That is the answer for the mean, and you can say this in seconds. For the variance, the order statistic is squeezed by its neighbors, so it should have less variance than a single uniform draw (which has variance
/12$), especially for middle order statistics.\cdot 3 / (25 \cdot 6) = 6/150 = 0.04$. Compare to a single uniform's variance ofQuick Estimate: Take $n = 4$, $k = 2$. We expect $E[U_{(2)}] \approx 2/5 = 0.4$. The variance should be
/12 \approx 0.083$ -- roughly half, which makes sense since the second order statistic is constrained by the first and third. For the limit with $k$ fixed, as $n \to \infty$, $E[U_{(k)}] = k/(n+1) \to 0$. This is intuitive: among a huge number of draws, the $k$-th smallest is pinned near zero.Approach: The key fact is that the $k$-th order statistic of $n$ i.i.d. $\text{Uniform}(0,1)$ random variables follows a $\text{Beta}(k, n-k+1)$ distribution. Once you know this, the moments fall out immediately from the Beta distribution formulas.
Formal Solution:
*Step 1: Density of $U_{(k)}$*
The PDF of the $k$-th order statistic out of $n$ i.i.d. $\text{Uniform}(0,1)$ draws is
$f_{U_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!} \, x^{k-1}(1-x)^{n-k}, \quad 0 < x < 1.$
This is exactly the $\text{Beta}(k, \, n - k + 1)$ density. To see why: for $U_{(k)}$ to be near $x$, we need exactly $k-1$ draws below $x$ (probability $x^{k-1}$), one draw at $x$, and $n-k$ draws above $x$ (probability $(1-x)^{n-k}$), times the multinomial count.
*Step 2: Expected value*
For $X \sim \text{Beta}(\alpha, \beta)$, $E[X] = \alpha / (\alpha + \beta)$. With $\alpha = k$ and $\beta = n - k + 1$:
$E[U_{(k)}] = \frac{k}{k + (n - k + 1)} = \frac{k}{n+1}.$
This confirms the "equal spacing" intuition: the $n$ order statistics have means at
/(n+1), \, 2/(n+1), \ldots, n/(n+1)$, evenly dividing $[0,1]$.*Step 3: Variance*
For $X \sim \text{Beta}(\alpha, \beta)$, $\text{Var}(X) = \alpha \beta / ((\alpha + \beta)^2(\alpha + \beta + 1))$. Substituting:
$\text{Var}(U_{(k)}) = \frac{k(n - k + 1)}{(n+1)^2(n+2)}.$
Note that for $k = 1$ or $k = n$ (the min or max), the variance is $n / ((n+1)^2(n+2))$, and the median order statistic ($k \approx n/2$) has the largest variance among all order statistics, roughly $n/(4(n+1)^2)$ for large $n$.
*Step 4: Limiting behavior*
For fixed $k$ as $n \to \infty$:
$\lim_{n \to \infty} E[U_{(k)}] = \lim_{n \to \infty} \frac{k}{n+1} = 0.$
Also, $\text{Var}(U_{(k)}) \sim k \cdot n / (n^2 \cdot n) = k/n^2 \to 0$, so $U_{(k)}$ concentrates at $0$.
Interpretation: When $k$ is fixed and $n$ grows, $U_{(k)}$ is the $k$-th smallest out of more and more draws. As you throw in more points on $[0,1]$, the smallest few cluster near zero. By the law of large numbers, the minimum converges to $0$, and the $k$-th smallest is only slightly above the minimum -- it too converges to $0$. In contrast, if $k$ grows proportionally to $n$ (say $k = \lfloor n \cdot q \rfloor$ for fixed $q \in (0,1)$), then $E[U_{(k)}] \to q$, recovering the $q$-th quantile of the uniform distribution.
Answer:
$E[U_{(k)}] = \frac{k}{n+1}, \qquad \text{Var}(U_{(k)}) = \frac{k(n-k+1)}{(n+1)^2(n+2)}.$
For fixed $k$, $\lim_{n \to \infty} E[U_{(k)}] = 0$: the $k$-th smallest draw out of $n$ is pushed to zero as $n$ grows.
Intuition
Order statistics of uniforms are one of the most fundamental objects in probability, and the fact that they follow Beta distributions is worth internalizing permanently. The intuition is beautifully simple: $n$ uniform draws partition $[0,1]$ into $n+1$ gaps, and by symmetry each gap has the same expected length
/(n+1)$. The $k$-th order statistic sits after $k$ gaps, so its mean is $k/(n+1)$. The variance formula tells you that middle order statistics wiggle more than extreme ones (the $k(n-k+1)$ numerator is maximized near $k = n/2$), which makes sense -- the min and max are pinned near the boundaries.This shows up constantly in quant work. Quantile estimation, VaR calculations, and extreme value theory all rely on order-statistic behavior. The limiting result -- that for fixed $k$, the $k$-th order statistic goes to zero -- is the foundation for extreme value theory. In practice, if you are looking at the worst $k$ days out of $n$ trading days, those losses cluster near the tail boundary, and the clustering gets tighter as $n$ grows. Confusing fixed-rank order statistics ($k$ fixed) with quantile-type order statistics ($k/n$ fixed) is a common mistake that leads to wrong intuitions about tail risk.