My Math Forum Unbiased estimator proof

February 7th, 2010, 01:44 PM   #1
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Unbiased estimator proof

First, here's the question:
Quote:
 Given a random sample of size $n$ from a population that has known mean $\mu$ and the finite variance ${\sigma}^2$, show that ${\frac{1}{n}}\sum_{i=1}^{n}{(X_i - \mu)^2}$ is an unbiased estimator of $\sigma^2$.
So I have to show that the expected value of this is equal to $\sigma^2$, for it to be unbiased estimator. Are you allowed to do the following:
$E({\frac{1}{n}}\sum_{i=1}^{n}{(X_i - \mu)^2}) = {\frac{1}{n}}E(\sum_{i=1}^{n}{(X_i - \mu)^2}) = {\frac{1}{n}}\sum_{i=1}^{n}E({(X_i - \mu)^2}) = {\frac{1}{n}}\sum_{i=1}^{n}\sigma^2 = {\frac{1}{n}}n\sigma^2= \sigma^2$.

It seems "too easy" this way, which is what makes me think this isnt correct. Even though this is an odd numbered question in my book, there is no answer for it. Maybe the authors thought it was "trivial".

Thanks for any help

 February 7th, 2010, 02:29 PM #2 Global Moderator   Joined: May 2007 Posts: 6,641 Thanks: 625 Re: Unbiased estimator proof So it's easy! What's wrong with that?? It happens to be correct. A more interesting problem is to show that if you use the sample mean rather than the true mean, you have to change the divisor from n to n-1.
 February 7th, 2010, 02:45 PM #3 Member   Joined: Sep 2008 Posts: 41 Thanks: 0 Re: Unbiased estimator proof Hi mathman, Actually, the reasoning for dividing by n-1 instead of n in the definition of sample variance S^2 is a few pages before these exercises! It was shown that this is done to make S^2 an unbiased estimator of population variance. My main concern about my proof above was that I knew (from a theorem) that $E(\sum_{i=1}^{n}{X_i)} = \sum_{i=1}^{n}E({X_i)}$, as well as other linear combinations like $E(\sum_{i=1}^{n}{a_iX_i)} = \sum_{i=1}^{n}a_iE({X_i)}$. But I wasnt sure if this was true for more "complicated" things like $E(\sum_{i=1}^{n}{(X_i-\mu)^2})$ being equivalent to $\sum_{i=1}^{n}{E((X_i-\mu)^2})$. As I said, there is a theorem for the first of those equations, but I cant find one for the last equation. It seems logical, but I just wanted to confirm that this sort of property always holds for similar expressions. Thanks!
 February 8th, 2010, 02:40 PM #4 Global Moderator   Joined: May 2007 Posts: 6,641 Thanks: 625 Re: Unbiased estimator proof The average of a finite sum is always the sum of the averages, no matter what the terms look like.
 February 8th, 2010, 02:48 PM #5 Member   Joined: Sep 2008 Posts: 41 Thanks: 0 Re: Unbiased estimator proof Yeah, I started thinking about that after I first replied. There isnt anything special about whether its simple X or a function of X, its still just the sum of them. Thanks!
 November 10th, 2012, 02:50 AM #6 Newbie   Joined: Nov 2012 Posts: 2 Thanks: 0 Re: Unbiased estimator proof The proof as a whole can be found under http://economictheoryblog.wordpress.com ... exlatexs2/ Regards
November 10th, 2012, 07:51 PM   #7
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Re: Unbiased estimator proof

Quote:
 Originally Posted by IsidoreBeautrelet The proof as a whole can be found under http://economictheoryblog.wordpress.com ... exlatexs2/
C'est quoi ton point, Isidore?
Ce post date de: Sun Feb 07, 2010 4:44 pm

September 12th, 2014, 01:59 AM   #8
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Quote:
 Originally Posted by Denis C'est quoi ton point, Isidore? Ce post date de: Sun Feb 07, 2010 4:44 pm
I was looking for the answer myself and thought the answer provided in the forum was not sufficient. Hope this answers your question

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