1
$\begingroup$

If larger values have larger weights then show that for $n$ observations $x_1,x_2,...,x_n$ with respective weights $w_1,w_2,...,w_n$,
$\overline{x_w}>\overline{x}$, where $\overline{x_w}=\frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i}$ and $\overline{x}=\frac{\sum_{i=1}^nx_i}{n}$

The result is somewhat clear intuitively but I am not being able to show it analytically. I was thinking of using Chebyshev's sum inequality to solve this.

Let $x_1<x_2<...<x_n$, then by the problem it follows that $w_1<w_2<...<w_n$.

Then does it follow from Chebyshev's inequality that $\frac{1}{n}\sum_{i=1}^nw_ix_i>(\frac{1}{n}\sum_{i=1}^nx_i)(\frac{1}{n}\sum_{i=1}^nw_i)$

$\Rightarrow \frac{1}{n}\sum_{i=1}^nw_ix_i>\frac{1}{n}\sum_{i=1}^nx_i$ ?

$\endgroup$
1
$\begingroup$

You are almost there; the weighted mean is $$\frac{\sum w_ix_i}{\sum w_i}$$. Just divide both sides by $\sum w_i$ and you have the result you want.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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