# Unbiased estimators

• April 29th 2009, 01:28 PM
chella182
Unbiased estimators
I think that's what this question's gearing towards anyway... I'm not 100% sure, hence how I haven't a clue what to do, even after skimming through my course notes. The question goes like this...

Suppose that $X_{1}$, $X_{2}$,..., $X_{n}$ are a random sample from a population with mean $\mu$ and variance $\sigma^{2}$.

i) Verify that $X_{1}+X_{2}-2X_{3}+X_{4}$ is unbiased and has variance $7\sigma^{2}$;

ii) Verify that $\frac{2X_{1}+X_{2}+X_{3}+...+X_{n}}{n+1}$ is unbiased and has variance $\frac{(n+3)\sigma^{2}}{(n+1)^{2}}$
• April 29th 2009, 06:23 PM
matheagle
Quote:

Originally Posted by chella182
I think that's what this question's gearing towards anyway... I'm not 100% sure, hence how I haven't a clue what to do, even after skimming through my course notes. The question goes like this...

Suppose that $X_{1}$, $X_{2}$,..., $X_{n}$ are a random sample from a population with mean $\mu$ and variance $\sigma^{2}$.

i) Verify that $X_{1}+X_{2}-2X_{3}+X_{4}$ is unbiased and has variance $7\sigma^{2}$;

ii) Verify that $\frac{2X_{1}+X_{2}+X_{3}+...+X_{n}}{n+1}$ is unbiased and has variance $\frac{(n+3)\sigma^{2}}{(n+1)^{2}}$

The expected value of each x_i is mu hence
$E\biggl(X_{1}+X_{2}-2X_{3}+X_{4}\biggr)=EX_1+EX_2-2EX_3+EX_4=\mu+\mu-2\mu+\mu=\mu$
The variance of each is sigma^2 and by independence..

$V\biggl(X_{1}+X_{2}-2X_{3}+X_{4}\biggr)=V(X_1)+V(X_2)+4V(X_3)+V(X_4)=7 \sigma^2$.

I'll leave 2 for you.
• April 30th 2009, 04:09 AM
chella182
Ah right, cheers :) sounds simple, I feel daft for posting it now :p

& I've done part (ii) successfully too. Thankyouu!