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 have quite a simple question but I can't for the life of me figure it out.

For a set of iid samples $\,\,X_1, X_2, .., X_n\,\,$ from distribution with mean $\,\mu$.

If you are given the sample variance as

$$ S^2 = \frac{1}{n-1}\sum\limits_{i=1}^{n}\left(X_i - \bar{X}\right)^2 $$

How can you write the following?

$$ S^2 = \frac{1}{n-1}\left[\sum\limits_{i=1}^{n}\left(X_i - \mu\right)^2 - n\left(\mu - \bar{X}\right)^2\right] $$

All texts that cover this just skip the details but I can't work it out myself. I get stuck after expanding like so

$$ S^2 = \frac{1}{n-1}\sum\limits_{i=1}^{n}\left(X_i^2 -2X_i\bar{X} + \bar{X}^2\right) $$

What am I missing?

Edit: A similarly equivalent expression is often given that I also can't derive but which may be more obvious is

$$ S^2 = \frac{1}{n-1}\left[\sum\limits_{i=1}^{n}X_i^2 - n\bar{X}^2\right] $$

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

$$\begin{align*} \frac{1}{n-1}\left[\sum\limits_{i=1}^{n}\left(X_i - \mu\right)^2 - n\left(\mu - \bar{X}\right)^2\right] &= \frac{1}{n-1}\sum\limits_{i=1}^{n}\left[\left(X_i - \mu\right)^2 - \left(\mu - \bar{X}\right)^2\right]\\ &= \frac{1}{n-1}\sum\limits_{i=1}^{n}\left(X_i - \bar{X}\right) \left(X_i + \bar{X} - 2\mu\right).\end{align*}$$ Now, $$\sum_{i=1}^n (X_i - \bar{X}) = \sum_{i=1}^n X_i - \sum_{i=1}^n \bar{X} = n\bar{X} - n\bar{X}= 0 $$ and so $$\begin{align*} \frac{1}{n-1}\sum\limits_{i=1}^{n}\left(X_i - \bar{X}\right) \left(X_i + \bar{X} - 2\mu\right) &= \frac{1}{n-1}\left[\sum\limits_{i=1}^{n}\left(X_i - \bar{X}\right)X_i + \sum\limits_{i=1}^{n}\left(X_i - \bar{X}\right)\left(\bar{X} - 2\mu\right)\right]\\ &= \frac{1}{n-1}\left[\sum\limits_{i=1}^{n}\left(X_i - \bar{X}\right)X_i + \left(\bar{X} - 2\mu\right)\sum\limits_{i=1}^{n}\left(X_i - \bar{X}\right) \right]\\ &= \frac{1}{n-1}\left[\sum\limits_{i=1}^{n}\left(X_i - \bar{X}\right)X_i\right]\\ &= \frac{1}{n-1}\left[\sum\limits_{i=1}^{n}\left(X_i - \bar{X}\right)X_i - \bar{X}\sum\limits_{i=1}^{n}\left(X_i - \bar{X}\right)\right]\\ &= \frac{1}{n-1}\sum\limits_{i=1}^{n}\left(X_i - \bar{X}\right)^2\\ &= S^2 \end{align*}$$

share|improve this answer
    
Crystal clear, thank you. (I'm a little bit surprised that this is often assumed obvious because it isn't to me) –  Chris Jun 2 '12 at 2:41
    
@Chris If you liked Dilip's answer (I know I did) then it'd be nice if you upvote it and, perhaps, accept it as best answer, though you can wait for this last some time in case other answers pop up –  DonAntonio Jun 2 '12 at 3:05
    
@Don: Note that the OP cannot yet upvote until someone else upvotes his question so that he has at least 15 rep. –  cardinal Jun 2 '12 at 18:05
add comment

$$S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X)^2$$ $$=\frac{1}{n-1}\sum_{i=1}^n(X_i-\mu +\mu-\bar X)^2$$

$$=\frac{1}{n-1}\sum_{i=1}^n[(X_i-\mu)^2 +2(X_i-\mu)(\mu-\bar X)+(\mu-\bar X)^2]$$ $$=\frac{1}{n-1}[\sum_{i=1}^n(X_i-\mu)^2 +2\sum_{i=1}^n(X_i-\mu)(\mu-\bar X)+\sum_{i=1}^n(\mu-\bar X)^2]$$ $$=\frac{1}{n-1}[\sum_{i=1}^n(X_i-\mu)^2 +2n\sum_{i=1}^n(\frac{X_i}{n}-\frac{\mu}{n})(\mu-\bar X)+\sum_{i=1}^n(\mu-\bar X)^2]$$ $$=\frac{1}{n-1}[\sum_{i=1}^n(X_i-\mu)^2 -2n(\mu-\bar X)(\mu-\bar X)+n(\mu-\bar X)^2]$$

Let me add that replace $\frac{\sum_{i=1}^n X_i}{n}$ by $\bar X$ in the last but one line. and flip it around to get the negative sign.

share|improve this answer
    
The middle term gives zero. Hence the result. Hmm... –  cardinal Jun 2 '12 at 2:01
    
I guess then, my trouble is understanding why the middle term is zero. Could you elaborate please? –  Chris Jun 2 '12 at 2:03
    
@Chris: It's not zero, hence, my comment. –  cardinal Jun 2 '12 at 2:03
1  
I have seen my mistake. I will fix it. –  smanoos Jun 2 '12 at 2:10
    
Thank you! I definitely need to get better at manipulating expressions inside summations. –  Chris Jun 2 '12 at 2:19
show 1 more comment

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.