# Problem understanding the theorem

• Jan 5th 2010, 01:47 PM
flintstone
Problem understanding the theorem
Theorem : Is S^2 is the variance of a random sample from a infinite population with the finite variance \sigma^2 , then E(S^2) = \sigma^2
Proof:

$E(S^2) = E \left[ \frac{1}{n-1} . \sum_{i=1}^{n} (X_i - \bar X)^2 \right]$..............Understood!!!!

$E(S^2) = \frac{1}{n-1} . E \left[\sum_{i=1}^{n} [(X_i - \mu) - (\bar X - \mu)]^2 \right]$.............Understood!!!!

$E(S^2) = \frac{1}{n-1} . \left[\sum_{i=1}^{n} E[(X_i - \mu)^2] - n E[(\bar X - \mu)^2] \right]$...................Understood!!!!

since $E[(X_i - \mu)^2] = \sigma^2$ and $E[(\bar X - \mu)^2] = \frac{\sigma^2}{n}$ it follows that
$

E(S^2) = \frac{1}{n-1} \left[\sum_{i=1}^{n} \sigma^2 - n \frac{\sigma^2}{n}\right]$
...............Understood!!!!

$= \sigma^2$....................Couldnt Understood!!!!

How did he get
$\sigma^2$
• Jan 5th 2010, 02:13 PM
matheagle
the sum of a constant n times is n times that constant...

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

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

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

$= \sigma^2$
• Jan 5th 2010, 07:30 PM
flintstone
thanks