My Math Forum Proof of sample variance formula

 October 3rd, 2019, 08:05 AM #1 Newbie   Joined: Oct 2019 From: USA Posts: 3 Thanks: 0 Proof of sample variance formula This formula should be checked for $\displaystyle i+1$: $\displaystyle s_{i}^2= \frac{1}{i-1}\sum\limits_{k=1}^{i} (x_k-\bar{x}_i)^2$ The result should be this: $\displaystyle s_{i+1}^2=\left(1-\frac{1}{i}\right)s_{i}^2+(i+1)(\bar{x}_{i+1}-\bar{x}_{i})^2$ Assumed as known, this is: $\displaystyle \bar{x}_{i+1}=\bar{x}_i+\frac{x_{i+1}-\bar{x}_i}{i+1}$ $\displaystyle \bar{x}_i=\frac{1}{i}\sum\limits_{k=1}^{i} x_k$ Does anyone have any idea how best to show this? I first started using $\displaystyle i + 1$ in $\displaystyle s_{i}^2$ so that this becomes $\displaystyle s_{i+1}^2$ and then I would try to resolve the sum on the right hand side. Does anyone have any ideas here?
 October 3rd, 2019, 01:12 PM #2 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 I have not tried it, but I believe brute force should work. The original formula is derived from showing $E(s_i^2)=\sigma_i^2$. Last edited by mathman; October 3rd, 2019 at 01:16 PM.
October 4th, 2019, 04:39 AM   #3
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 Originally Posted by mathman I have not tried it, but I believe brute force should work. The original formula is derived from showing $E(s_i^2)=\sigma_i^2$.

If you say "brute force", can you specify what you mean with that? One idea of mine was to solve the $\displaystyle (x...)^2$ expression in the sum and reduce the sum(s) to maybe known formulas...?!

 October 4th, 2019, 04:07 PM #4 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 I haven't made any attempt, since the origin of the definition is as I had described, which gives a better understanding of its meaning.
 October 6th, 2019, 02:01 PM #5 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 Partial brute force: $s_{i+1}^2=\frac{1}{i}\sum_{k=1}^{i+1}(x_k-a_{i+1})^2$, however $(x_k-a_{i+1})^2=(x_k-a_i+a_i-a_{i+1})^2=(x_k-a_i)^2+2(x_k-a_i)(a_i-a_{i+1})+(a_i-a_{i+1})^2$ Now $\sum_{k=1}^{i+1}=\sum_{k=1}^i(..)+(x_{i+1}-a_{i+1})$ The sum term has three parts (after dividing by $i$), $\frac{i-1}{i}s_i^2$, $\frac{2(a_i-a_{i+1})}{i}\sum_{k=1}^i(x_k-a_i)=0$, and $(a_i-a_{i+1})^2$. Note: I use $a_i$ rather than $\bar{x}_i$ - more convenient. I'll let you finish. Last edited by mathman; October 6th, 2019 at 02:04 PM.
October 9th, 2019, 11:09 AM   #6
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 Originally Posted by mathman Partial brute force: $s_{i+1}^2=\frac{1}{i}\sum_{k=1}^{i+1}(x_k-a_{i+1})^2$, however $(x_k-a_{i+1})^2=(x_k-a_i+a_i-a_{i+1})^2=(x_k-a_i)^2+2(x_k-a_i)(a_i-a_{i+1})+(a_i-a_{i+1})^2$ Now $\sum_{k=1}^{i+1}=\sum_{k=1}^i(..)+(x_{i+1}-a_{i+1})$ The sum term has three parts (after dividing by $i$), $\frac{i-1}{i}s_i^2$, $\frac{2(a_i-a_{i+1})}{i}\sum_{k=1}^i(x_k-a_i)=0$, and $(a_i-a_{i+1})^2$. Note: I use $a_i$ rather than $\bar{x}_i$ - more convenient. I'll let you finish.

Du hast absolut recht. Man muss nur die Gleichung lösen und die unerwünschten Variablen ersetzen, damit nur der Balkenausdruck darin ist. Damit kommen Sie zu einer Lösung!

October 9th, 2019, 04:48 PM   #7
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 Originally Posted by naliM Du hast absolut recht. Man muss nur die Gleichung lösen und die unerwünschten Variablen ersetzen, damit nur der Balkenausdruck darin ist. Damit kommen Sie zu einer Lösung!
I have long since forgotten German. Translate please!

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