# Mean and variance of the order statistics of a discrete uniform sample without replacement

In answering calculate the mean and variance of the highest number drawn on lottery based on the lowest number drawn, I couldn't find the mean and variance of the order statistics of a discrete uniform sample without replacement anywhere, so I figured I'd derive them here for future reference.

Draw $k$ distinct numbers uniformly from $1$ to $n$ and order them as $X_{(1)}\lt\cdots\lt X_{(k)}$. What are the mean and variance of the $X_{(i)}$?

## 1 Answer

Uniformly randomly permute $n-k$ white balls and $k$ red balls and take the positions of the red balls as the numbers drawn. Consider the indicator variables $Y_{im}$ that are $1$ if the $m$-th white ball is to the left of the $i$-th red ball and $0$ otherwise. Then $X_{(i)}=\sum_mY_{im}+i$.

To find the mean, we need $\def\xp#1{\mathbb E\left[#1\right]}\xp{Y_{im}}$, the probability that a given white ball is to the left of the $i$-th red ball. This happens if among the $k$ red balls and the $1$ given white ball the rank of the white ball is at most $i$, with probability $\frac i{k+1}$. Thus we have

\begin{align} \xp{X_{(i)}} &= \xp{\sum_mY_{im}+i} \\ &= \sum_m\xp{Y_{im}}+i \\ &= i\left(\frac{n-k}{k+1}+1\right) \\ &=i\cdot\frac{n+1}{k+1}\;. \end{align}

To find the variance, we need $\xp{Y_{il}Y_{im}}$, the probability that two particular white balls are both to the left of the $i$-th red ball. This happens if among the $k$ red balls and the $2$ particular white balls the ranks of the white balls are at most $i+1$, with probability $\frac{\binom{i+1}2}{\binom{k+2}2}=\frac{i(i+1)}{(k+1)(k+2)}$. Thus we have

\begin{align} \operatorname{Var}\left[X_{(i)}\right] &= \operatorname{Var}\left[X_{(i)}-i\right] \\ &= \xp{\left(X_{(i)}-i\right)^2}+\xp{X_{(i)}-i}^2 \\ &= \xp{\left(\sum_mY_{im}\right)^2}-\xp{\sum_mY_{im}}^2 \\ &= \xp{\sum_mY_{im}}+\xp{\sum_{l\ne m}Y_{il}Y_{im}}-\xp{\sum_mY_{im}}^2 \\ &= i\cdot\frac{n-k}{k+1}+(n-k)(n-k-1)\frac{i(i+1)}{(k+1)(k+2)}-\left(i\cdot\frac{n-k}{k+1}\right)^2 \\ &= i\cdot\frac{(n-k)}{k+1}\left(1+\frac{(n-k-1)(i+1)}{k+2}-\frac{i(n-k)}{k+1}\right) \\ &= i(k+1-i)\cdot\frac{(n-k)(n+1)}{(k+1)^2(k+2)} \;. \end{align}

As expected, we have symmetry under the transformation $i\to k+1-i$.