Unbiased estimators?? Can't do it.

**Provoke** · March 10th 2010, 12:51 PM

Given X1,...,X7, a random sample from a population of mean mu and variance sigma^2. Prove that the following estimators are unbiased:

estimator 1 = (X1+...X7)/7 = mean = unbiased by definition,

my problem is with the second one:

estimator 2 = (2X1 - X6 + X4) / 2

Then they ask which estimator is the best, so that I kinda know, it will be the estimator with the lower variance but I can't solve for the variance because I have no numbers.

And then they ask for efficiency...needless to tell that I don't know that 1 either.

Much appreciate it guys!!

**Random Variable** · March 10th 2010, 01:08 PM

$E\Big(\frac{2X_{1}-X_{6}+X_{4}}{2}\Big) = E(X_{1}) - \frac{1}{2}E(X_{6}) + \frac{1}{2}E(X_{4})$

$= \mu - \frac{1}{2} \mu + \frac{1}{2} \mu = \mu$

$V\Big(\frac{2X_{1}-X_{6}+X_{4}}{2}\Big) = V(X_{1}) + \Big(-\frac{1}{2}\Big)^{2} V(X_{6})$ $+ \Big(\frac{1}{2}\Big)^{2} V(X_{4})$

$= \sigma^{2} + \frac{1}{4}\sigma^{2} + \frac{1}{4}\sigma^{2} = \frac{3}{2} \sigma^{2}$

**Provoke** · March 10th 2010, 01:53 PM

Thanks a lot. Show me the following if you don't mind :P

I wana calculate relative efficiency of the 2, they say it's:

MSE(estimator) = E( estimator - teta)^2

and relative efficiency is: MSE(estimator1)/MSE(estimator2)

I got trouble calculating the MSE of each estimator, those symbols mix me up badly lol :P

tyy

**Random Variable** · March 10th 2010, 02:24 PM

Let $\hat{\theta_{1}}$ be the first unbiased estimator. and let $\hat{\theta_{2}}$ be the second unbiased estimator.

then $\text{eff} \ (\hat{\theta_{1}}, \hat{\theta_{2}}) =\frac{V(\hat{\theta_{1}})}{V(\hat{\theta_{2}})}$

$= \frac{(1/7) \sigma^{2}}{(3/2) \sigma^{2}} = \frac{2}{21} \approx 0.0952$

**Provoke** · March 10th 2010, 02:29 PM

Originally Posted by Random Variable

Let $\hat{\theta_{1}}$ be the first unbiased estimator. and let $\hat{\theta_{2}}$ be the second unbiased estimator.

then $\text{eff} \ (\hat{\theta_{1}}, \hat{\theta_{2}}) =\frac{V(\hat{\theta_{1}})}{V(\hat{\theta_{2}})}$

$= \frac{(1/7) \sigma^{2}}{(3/2) \sigma^{2}} = \frac{2}{21} \approx 0.0952$

Oh so I don't have to use MSE I get it. But how did you get 1/7, when I calculated the V(mean) I did:

V( (X1+X2+X3+X4+X5+X6+X7)/7) = V(X1)/7 +V(X2)/7 + ...V(X7)/7 = sigma^2/7 * 7 = sigma^2

(

**Random Variable** · March 10th 2010, 07:13 PM

Originally Posted by Provoke

Oh so I don't have to use MSE I get it. But how did you get 1/7, when I calculated the V(mean) I did:

V( (X1+X2+X3+X4+X5+X6+X7)/7) = V(X1)/7 +V(X2)/7 + ...V(X7)/7 = sigma^2/7 * 7 = sigma^2

(

In general $MSE(\hat{\theta}) = Var(\hat{\theta}) + \big( E(\hat{\theta})-\theta)\big)^{2}$

There's probably a proof in your textbook.

In our case the estimators are unbiased, so $E(\hat{\theta})-\theta = \theta - \theta =0$

and $MSE(\hat{\theta}) = Var(\hat{\theta})$

**Random Variable** · March 10th 2010, 07:16 PM

$V(aX) = a^{2}V(X)$

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