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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: $\hat{\sigma}_X^2 = \sum_{i=1}^n \frac{(X_i-\mu)^2} n$ for variance.

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

I begin ...

$\operatorname{E}\left[\hat{\sigma}_X^2\right]=\dfrac 1 n \operatorname{E} \left[\sum_{i=1}^n (X_i-\mu)^2\right]$

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

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

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    I edit my question $\mu$ is a theorical mean2012-09-02

2 Answers 2

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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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    I don't see that this answer contributes anything. If the question is how to find the expected value of the given random variable, the straightforward way is this: $\operatorname{E} \left( \frac 1 n \sum_{i=1}^n (X_i-\mu)^2 \right) = \frac 1 n \operatorname{E} \sum_{i=1}^n (X_i - \mu)^2 = \frac 1 n \sum_{i=1}^n \operatorname{E}((X_i-\mu)^2) = \frac 1 n \sum_{i=1}^n \sigma^2 = \frac 1 n n\sigma^2 = \sigma^2.$2016-12-16
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\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}

Note:

  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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    1. see https://en.wikipedia.org/wiki/Variance which explains variance calculation in terms of expectation. 2. see https://onlinecourses.science.psu.edu/stat414/node/167 3. see https://en.wikipedia.org/wiki/Variance in "Sum of uncorrelated variables" section.2016-12-17