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Assume $X$ is a random variable from a population with normal distribution. Using the likelihood function I get the expression below: $\hat{\sigma_X}^2 = \sum_{i=1}^{n}{\dfrac{(X_i-\mu)^2}{n}}$ for variance.

I want prove that, $E[\hat{\sigma_X}^2]=\dfrac{n-1}{n}\sigma_X^2$.

I begin ...

$E[\hat{\sigma_X}^2]=\dfrac{1}{n}E[\sum_{i=1}^{n}{(X_i-\mu)^2}]$

$\dfrac{1}{n}(E[(X_1-\mu)^2]+E[(X_2-\mu)^2]\cdots E[(X_n-\mu)^2])$

pdta: $\mu$ is a theorical mean (not estimator of mean)

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I don't know what you mean by pdta, but $\frac{1}{n} \sum_{i=1}^n X_i$ is an estimator $\hat{\mu}$ for the mean, not the actual mean $\mu_X$. –  Robert Israel Sep 2 '12 at 6:10
    
I think he knows that Robert. It is just that he is using non-standard notation. –  Michael Chernick Sep 2 '12 at 13:07
    
I edit my question $\mu$ is a theorical mean –  juaninf Sep 2 '12 at 14:47

1 Answer 1

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Hint: expand out $(X_i - \mu)^2 = X_i^2 - 2 X_i \mu + \mu^2$. Now compute $E[X_i^2]$, $E[X_i \mu]$ and $E[\mu^2]$.

EDIT: If $\mu$ is the actual mean rather than an estimator, then the statement is wrong: $E[(X_i - \mu)^2] = \sigma^2$ and $E[\hat{\sigma}_X^2] = \sigma^2$.

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yes ... $E[X_i^2]$-$2\mu E[X_i]$ + $\mu^2 E[1]$ ... other step please ... –  juaninf Sep 2 '12 at 14:41
    
I edit my question $\mu$ is a theorical mean –  juaninf Sep 2 '12 at 14:54

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