# Simple proof for sample variance as U-statistics

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?

\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}
• 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 $n^2$ Aug 26, 2020 at 14:54