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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 ...


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

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
up vote 1 down vote accepted

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

\begin{align} E[\frac{1}{n}\sum_{i=1}^n (X_i-\bar X)^2]&= \frac1n \sum_{i=1}^n E[X_i -\bar X]^2\\ &=\frac1n \sum_{i=1}^n E[ (X_i -\bar X) (X_i -\bar X)]\\ &= \frac1n \sum_{i=1}^n E[X_i (X_i -\bar X)]- \bar X\sum_{i=1}^n \space E[(X_i -\bar X)]\\ &= \frac1n \sum_{i=1}^n E[X_i (X_i- \bar X)]\\ &= \frac1n \sum_{i=1}^n E[{X_i}^2]- E[\bar X \space X_i] \\ &=\frac1n\left( \sum_{i=1}^n E[{X_i}^2]- \ E\left[\ \bar X \space\sum_{i=1}^nX_i\right]\right)\\ &=\frac1n\left( \sum_{i=1}^n E[{X_i}^2]- n E\left[ \bar X \space\sum_{i=1}^n \frac{X_i} {n}\right]\right) \\ &=\frac1n\left( \sum_{i=1}^n E[{X_i}^2]- n E\left[ \bar X^2\right]\right) \\ &=\frac1n\left(n \sigma^2_x + n {\mu}^2- n \left[ \frac{\sigma^2_x}{n}+\mu^2\right] \right)\\ &=\frac{n-1}{n} \sigma^2_x\\ \end{align}


  1. $E[X^2]=\sigma^2 + \mu^2$

  2. $E[\bar X]=\mu$

  3. $\sigma^2_\bar x= \frac{\sigma^2_x}{n}$

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