Question

For a random sample Xi.X2,......, Xn, let

Answer 1

We have

$\sum_{i=1}^{n}\left ( X_{i}-\bar{X} \right )^{2}=\sum_{i=1}^{n}\left ( X^{2}_{i}+\bar{X}^{2}-2X_{i}\bar{X} \right )$

$=\left [\sum_{i=1}^{n} X^{2}_{i}+\sum_{i=1}^{n}\bar{X}^{2}-2\bar{X}\sum_{i=1}^{n}X_{i}\right ]=\left [\sum_{i=1}^{n} X^{2}_{i}+n\bar{X}^{2}-2\bar{X}\cdot n\bar{X} \right ] =\left [\sum_{i=1}^{n} X^{2}_{i}-n\bar{X}^{2} \right ]$

Here we will use:

$\sigma^{2}=E(X_{i}^{2})-[E(X_{i})]^{2}=E(X_{i}^{2})-\mu^{2}$
$E(X_{i}^{2})=\sigma^{2}+\mu^{2}$

And

$V(\bar{X})=E(\bar{X}^{2})-[E(\bar{X})]^{2}=E(\bar{X}^{2})-\mu^{2}$
$E(\bar{X}^{2})=\frac{\sigma^{2}}{n}+\mu^{2}$

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Given that

So,

$=\frac{1}{n} \left [ \sum_{i=1}^{n} (\sigma^{2}+\mu^{2}) -n\mu^{2} \right ]=\frac{1}{n} \left [ n\sigma^{2}+n\mu^{2} -n\left (\frac{\sigma^{2}}{n}+\mu^{2} \right )\right ]$

$=\frac{1}{n} \left [ n\sigma^{2} -\sigma^{2} \right ]=\frac{\sigma^{2}(n-1)}{n}$

hence,

$E(S'^{2})=rac{sigma^{2}(n-1)}{n}$