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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}{i1}\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  
Newbie Joined: Oct 2019 From: USA Posts: 3 Thanks: 0  Quote:
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: $s_{i+1}^2=\frac{1}{i}\sum_{k=1}^{i+1}(x_ka_{i+1})^2$, however $(x_ka_{i+1})^2=(x_ka_i+a_ia_{i+1})^2=(x_ka_i)^2+2(x_ka_i)(a_ia_{i+1})+(a_ia_{i+1})^2$ 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 $i$), $\frac{i1}{i}s_i^2$, $\frac{2(a_ia_{i+1})}{i}\sum_{k=1}^i(x_ka_i)=0$, and $(a_ia_{i+1})^2$. 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  
Newbie Joined: Oct 2019 From: USA Posts: 3 Thanks: 0  Quote:
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 
Global Moderator Joined: May 2007 Posts: 6,835 Thanks: 733  

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