Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was looking at my stats textbook and they claim that the sample variance of a weighted distribution involving i.i.d. $x_i$s 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?

share|improve this question

1 Answer 1

up vote 8 down vote accepted

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.