1. ## Estimators question

Dear All, I am having some trouble with this question. Note that the random variable X has a population mean of
μ.

THe markscheme gives α (alpha) to be 1/2 but why?

2. ## Re: Estimators question

1) should be easy enough. What is the definition of an unbiased estimator? Therefore what must $E[U]$ equal? Given the expression for $U$ in terms of $\alpha$ what must $\alpha$ equal?

we'll work through this 1 bit at a time.

3. ## Re: Estimators question

After giving it more time, I see how to do this question now! OK the only part I don't understand is the last part of the question, (v), where it asks me for a more efficient estimator of μ. The markscheme isn't particularly helpful. I have included it.

4. ## Re: Estimators question

Originally Posted by bbear123
After giving it more time, I see how to do this question now! OK the only part I don't understand is the last part of the question, (v), where it asks me for a more efficient estimator of μ. The markscheme isn't particularly helpful. I have included it.
markscheme seems clear enough.

3 iid rvs, you should almost intuitively know the best unbiased estimator of the mean is the sample average, i.e.

$\bar{\mu} = \dfrac 1 3 (X_1+X_2+X_3)$

the variance of $\bar{\mu}$ is

$V[\bar{\mu}] = \dfrac 1 9 \left(\sigma^2 + \sigma^2 + \sigma^2\right) = \dfrac {\sigma^2}{3}$

$\dfrac 1 3 < \dfrac 3 8$

What isn't clear?

5. ## Re: Estimators question

Ah OK, I see how it is done now. Thank you!

6. ## Re: Estimators question

Originally Posted by bbear123
Ah OK, I see how it is done now. Thank you!
as an aside you might try reworking the problem with

$U = \alpha X_1 + \beta X_1 + (1- \alpha - \beta) X_3$

I believe in this case you'll derive that $\alpha = \beta = \dfrac 1 3$

7. ## Re: Estimators question

Hey bbear123.

Hint - You know E[Xn] = mu [from your first post] and if U is unbiased estimator for mu then E[U] = mu as well.

This should help summarize what you need to do to find the unknowns.

Hint - you will need two equations in two unknowns to get alpha and beta and I would recommend looking at the Variance operator to do so.