# A random sample X

• Dec 29th 2009, 08:32 PM
zorro
A random sample X
Question : A random sample $\displaystyle X_1,X_2,X_3 ..... X_n$ is taken. Show that the sample mean $\displaystyle \bar X$ is an unbiased estimate of the population mean $\displaystyle \mu$
• Dec 29th 2009, 11:33 PM
mr fantastic
Quote:

Originally Posted by zorro
Question : A random sample $\displaystyle X_1,X_2,X_3 ..... X_n$ is taken. Show that the sample mean $\displaystyle \bar X$ is an unbiased estimate of the population mean $\displaystyle \mu$

You will find the proof in almost any Stats textbook and undoubtedly Google will turn up proofs in abundance. Also, see here:http://www.mathhelpforum.com/math-he...stimators.html (and no doubt using the Search tool will turn up more threads).
• Dec 31st 2009, 02:17 AM
zorro
I was able to find one but dont know if its correct or no
Quote:

Originally Posted by mr fantastic
You will find the proof in almost any Stats textbook and undoubtedly Google will turn up proofs in abundance. Also, see here:http://www.mathhelpforum.com/math-he...stimators.html (and no doubt using the Search tool will turn up more threads).

Since $\displaystyle f(x) = \frac{1}{ \sigma \sqrt{2 \pi}} . e^{- \frac{(x- \mu)^2}{2 \sigma}}$

it follows that

$\displaystyle ln f(x) = - ln \ \sigma \sqrt{2 \pi} - \frac{1}{2} \left( \frac{x- \mu}{\sigma} \right)^2$

so that

..
..
....
$\displaystyle = \frac{\sigma^2}{n}$..............Is the theorem u were talking about???
• Dec 31st 2009, 09:29 PM
mr fantastic
Quote:

Originally Posted by zorro
Since $\displaystyle f(x) = \frac{1}{ \sigma \sqrt{2 \pi}} . e^{- \frac{(x- \mu)^2}{2 \sigma}}$

it follows that

$\displaystyle ln f(x) = - ln \ \sigma \sqrt{2 \pi} - \frac{1}{2} \left( \frac{x- \mu}{\sigma} \right)^2$

so that

..
..
....
$\displaystyle = \frac{\sigma^2}{n}$..............Is the theorem u were talking about???

I have no idea how you would think I was talking about anything like that. Did you click on the link in my first post? Have you done a Google search? Key words: mean unbiased estimator
• Jan 1st 2010, 01:24 AM
zorro
thanks Mr fantastic ......
i have got it now .......Thank u