The usual sample variance $$ \mathrm{Var}(X) = \frac{1}{n-1} \sum_{i = 1}^n (X_i - \bar X)^2 \tag{1} $$ of an iid. sample $X_1, \dots, X_n$ is known to be a U-statistic with kernel $$ h(X_i, X_j) = \frac{1}{2} (X_i - X_j)^2. $$ This means that we can write $$ \mathrm{Var}(X) = \mathrm{Average}_{1 \le i < j \le n} h(X_i, X_j). \tag{2} $$ As a nice consequence of this fact, we immediately know $\mathrm{Var}(X)$ is a minimum variance unbiased estimator of the true variance of $X$.

The usual proof that (2) = (1) is relatively simple but lengthy. So I wanted to know if anyone knows a short, elegant proof of this identity?


1 Answer 1


Maybe this is the proof you meant. Others might find it useful as a reference (I came across the question looking for this) and besides it is only a few lines long. Taking the usual approach of averaging over all the ways the kernel may be formed and expanding the quadratic gives:

$\begin{align} U_n & = \binom{n}{2}^{-1} \sum_{i < j}^{n} \frac{1}{2} (X_i - X_j)^2 \\ & = \frac{1}{n(n-1)}\left \{ \frac{1}{2} \sum_{i \neq j}^{n} \left ( X_i^2 + X_j^2 - 2X_iX_j \right )\right \} \\ & = \frac{1}{n(n-1)} \left \{ n \sum_{i=1}^n X_i^2 - n^2 \overline{X}^2 \right \} \\ & = \frac{1}{n-1} \sum_{i=1}^{n} \left ( X_i - \overline{X} \right )^2 = S_n^2 \end{align}$

This is from section 3.2 in Mathematical Statistics by Shao.

  • $\begingroup$ Wow, thanks for reviving this old question! How do we get from the second to the third line? $\endgroup$
    – Michael M
    Aug 26, 2020 at 13:51
  • 1
    $\begingroup$ No worries, to get to third line just distribute the sum to all the terms in parentheses - we're always summing over $i,j$ but the first two terms only depend on $i$ or $j$ so you gain an $n$, for the final term multiply and divide by $n^2$ $\endgroup$ Aug 26, 2020 at 14:54
  • $\begingroup$ @ShakeelGavioli-Akilagun Thanks for your answers as well. But I got confused for your explanation of why we gain an n, not (n-1), as we move from the second to the third line above? We have the summation over i and j, where i is not equal to j. $\endgroup$
    – Rico
    Mar 31, 2021 at 7:26
  • $\begingroup$ @Rico: The $i=j$ terms are zero: $\sum_{i\neq j}\left(X_i^2+X_j^2-2X_iX_j\right) = \sum_{i, j}\left(X_i^2+X_j^2-2X_iX_j\right) - \underbrace{\sum_{i=j}\left(X_i^2+X_j^2-2X_iX_j\right)}_{=0}$. $\endgroup$
    – Flo
    Jan 26 at 9:42

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