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Let $x_1,x_2,..x_{21}$ be a random sample from a distribution having variance 5. Let $$\bar X = \frac1{21}\sum_{i=1}^{21}X_i\quad\text{and}\quad S=\frac1{21}\sum_{i=1}^{21}(X_i-\bar X)^2$$ What is the value of $E(S)$ ?

I am unable to get why $\bar X$ is given and how to take expectation of $S$. After squaring and taking expectation of each term?

amoeba
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Tesla
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  • you are asked to compute standard deviance. You may find some hints here: http://stats.stackexchange.com/questions/3931/intuitive-explanation-for-dividing-by-n-1-when-calculating-standard-deviation – Yurii Jan 01 '16 at 06:29
  • @Yurii I have read complete post refered by you , hints are not useful to me for this question . – Tesla Jan 01 '16 at 07:11
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    $\bar X$ is not given but simply defined. Have you tried the decomposition$$S=\frac1{21}\sum_{i=1}^{21}X_i^2-\bar X^2$$before computing the expectation? – Xi'an Jan 01 '16 at 08:29
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    It is [well known](http://stats.stackexchange.com/questions/3931) that $(21/20) S$ is an unbiased estimator of $5$. Because expectation is linear, $S$ must be an unbiased estimator of $5\times 20/21$. – whuber Jan 01 '16 at 14:39

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