# Prove $\sum_{i=1}^n$$w_i^2\geq\frac{1}{n} given \sum_{i=1}^n w_i=1 I was looking at my stats textbook and they claim that the sample variance of a weighted distribution involving i.i.d. x_is will be smallest when each of the weights is equal. I follow this argument up to the point where I reach \sum_{i=1}^n$$w_i^2\geq\frac{1}{n}$ given $\sum_{i=1}^n w_i=1$

(this result obtained due to the fact that $Var(\bar{x_w})= \sigma^2\sum_{i=1}^nw_i^2$ and $Var(\bar{x})=\frac{\sigma^2}{n}$, so setting them equal and cancelling the $\sigma^2$ on each side yields that inequality, where $\bar{x_w}$ is the weighted average). Trying out a few examples, it seems pretty obvious that the inequality holds, but the book offers no mathematical justification and I was hoping someone here could help put it more concretely - I'm not sure how to approach it myself.

Any thoughts?

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## 1 Answer

Use Cauchy-Schwarz: $$(\sum_{i=1}^n w_i \cdot 1)^2 \le (\sum_{i=1}^n w_i^2)(\sum_{i=1}^n 1^2).$$

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Sorry, I'm sure I'm missing the obvious here, but I don't see how this helps? –  user1257768 Jan 31 '13 at 22:38
The left sum is $\sum w_i = 1$ and the sum of $1$s is $n$. Divide both sides by $n$ to get the statement. –  Cocopuffs Jan 31 '13 at 22:40
Ooh, of course! Elegant and precise - thank you very much. –  user1257768 Jan 31 '13 at 22:42
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