3
$\begingroup$

I would appreciate your help as I climb the stats learning curve!

I want to prove the following:

"Let $x_1, x_2, ... , x_n$ be any numbers and let $\overline x = (x_1 + x_2 + ... + x_n)/n$

Then show:

$(n-1)s^2 =\sum(x_i-\overline x)^2 = \sum(x_i^2 - n\overline x^2)$, where $s^2$ is the observed value of the sample variance $S^2$.

So far I have:

\begin{align} s^2 &= \frac{\sum(x_i - \overline x)^2}{n-1} \\ (n-1)s^2 &= \sum(x_i - \overline x)^2 \\ &= \sum(x_i^2 - 2x_i\overline x + \overline x^2) \\ &= \sum(x_i^2) - 2\frac{\sum(x_i)\sum(x_i)}{n} +\sum(\overline x^2) \\ &= \sum(x_i^2) - 2\frac{\sum(x_i)\sum(x_i)}{n} + \sum(\frac{\sum(x_i)}{n}) \end{align}

As you can probably tell, I'm new to stats and would greatly appreciate any help. This is where I am stuck. I can't seem to figure out how to get from this point to the desired result, esp. given all the sum notations. Could someone point me in the right direction?

Also, I realize that what you are given in the problem is super important, so if it seems like I left out information, please tell me! I look to this community for help in my journey up this learning curve.

$\endgroup$
2
  • 2
    $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ May 23, 2016 at 19:40
  • 1
    $\begingroup$ Thanks gung. Just updated and read the wiki. Very excited! :) $\endgroup$ May 23, 2016 at 19:50

1 Answer 1

3
$\begingroup$

I think there's a parenthesis confounding you

\begin{align} s^2 &= \frac{\sum(x_i - \overline x)^2}{n-1} \\ (n-1)\cdot s^2 &= \sum(x_i - \overline x)^2 \\ &= \sum(x_i^2 - 2x_i\overline x + \overline x^2) \\ &= \sum(x_i^2)-2\overline x \cdot\sum x_i +n\cdot\overline x^{2} \\ &= \sum(x_i^2)-2n\cdot\overline x^{2} +n\cdot\overline x^{2}\\ &= \sum(x_i^2)-n\cdot\overline x^{2} \end{align}

$\endgroup$
5
  • $\begingroup$ Hi @Firebug, thank you so much for your solution. Just a quick question. It seems you are treating $\overline x$ as a constant here when you pull it out of the summation in the third step. Is that acceptable? I rewrote $\overline x$ as $\frac{\sum(x_i)}{n}$ because I thought it couldn't be treated as a constant. In addition, I believe you have $\sum(x_i) = n\overline x$. Is that justified as well? Thank you kindly! $\endgroup$ May 23, 2016 at 20:26
  • 1
    $\begingroup$ Yes, it is acceptable, because $\sum_j \frac{\sum_i x_i}{n} = \sum_j \overline x$ doesn't depend at all on the index $j$. $\endgroup$
    – Firebug
    May 23, 2016 at 20:28
  • 1
    $\begingroup$ And $\overline x = \frac{\sum_i x_i}{n} \therefore \sum x_i = n \cdot \overline x$. $\endgroup$
    – Firebug
    May 23, 2016 at 20:31
  • $\begingroup$ Thank you so much for your solutions Firebug. So appreciated. $\endgroup$ May 23, 2016 at 20:36
  • $\begingroup$ sorry didn't mean to leave a comment :X $\endgroup$ May 26, 2016 at 23:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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