1. ## Unbiased estimators

Suppose that $X_{1}, X_{2},...,X_{n}$ are independently and identically distributed variables in a random sample drawn from a population with mean $\mu=20$ and variance $\sigma^{2}=9$.

a) Determine the mean and variance of $Y=\frac{3X_{1}+2X_{2}+X_{3}+...+X_{n}}{n+3}$.

b) Construct your own unbiased estimator for the population mean, and calculate its variance.

I'm not entirely sure what I'm doing for a) and I don't know what to do for b) at all. Can anyone help?

2. Originally Posted by chella182
Suppose that $X_{1}, X_{2},...,X_{n}$ are independently and identically distributed variables in a random sample drawn from a population with mean $\mu=20$ and variance $\sigma^{2}=9$.

a) Determine the mean and variance of $Y=\frac{3X_{1}+2X_{2}+X_{3}+...+X_{n}}{n+3}$.
Use:

$E\left(\sum_{i=1}^n \lambda_i X_i\right)=\sum_{i=1}^n \lambda_i E(X_i)$

That will give you the mean straight off. Then as the $X_i$'s are iid:

$Var\left(\sum_{i=1}^n \lambda_i X_i\right)=\sum_{i=1}^n \lambda_i^2 Var(X_i)$

CB

3. Originally Posted by chella182
Suppose that $X_{1}, X_{2},...,X_{n}$ are independently and identically distributed variables in a random sample drawn from a population with mean $\mu=20$ and variance $\sigma^{2}=9$.

a) Determine the mean and variance of $Y=\frac{3X_{1}+2X_{2}+X_{3}+...+X_{n}}{n+3}$.

b) Construct your own unbiased estimator for the population mean, and calculate its variance.

I'm not entirely sure what I'm doing for a) and I don't know what to do for b) at all. Can anyone help?
For b) why not make like simple for yourself at select an estimator:

$m=X_1$

for $\mu$ ?

(The question does not require that the estimator be particularly good in whatever sense you might like to choose, only that it be unbiased)

CB

4. Originally Posted by CaptainBlack
Use:

$E\left(\sum_{i=1}^n \lambda_i X_i\right)=\sum_{i=1}^n \lambda_i E(X_i)$

That will give you the mean straight off. Then as the $X_i$'s are iid:

$Var\left(\sum_{i=1}^n \lambda_i X_i\right)=\sum_{i=1}^n \lambda_i^2 Var(X_i)$

CB
Okay, I know all of that, but would that make the mean $\frac{80+40n}{n+3}$?

5. Originally Posted by chella182
Okay, I know all of that, but would that make the mean $\frac{80+40n}{n+3}$?
$
Y=\frac{3X_{1}+2X_{2}+X_{3}+...+X_{n}}{n+3}
$

so:

$E(Y)=\frac{1}{n+3} \left(3E(X_1)+2E(X_2)+ E(X_3) + .. + E(X_n)\right)$ $=\frac{1}{n+3} \left(3(20)+2(20)+ (20) + .. + (20)\right)=20$

CB

6. Originally Posted by CaptainBlack
$
Y=\frac{3X_{1}+2X_{2}+X_{3}+...+X_{n}}{n+3}
$

so:

$E(Y)=\frac{1}{n+3} \left(3E(X_1)+2E(X_2)+ E(X_3) + .. + E(X_n)\right)$ $=\frac{1}{n+3} \left(3(20)+2(20)+ (20) + .. + (20)\right)=20$

CB
Sorry, I don't understand how you've gotten 20

7. ## Hidden sense

I think I track the hidden sense of your problem. They expect you to compare different population mean estimators including sample mean that looks like (X_1+...X_n)/n because it has the smallest variance among unbiased pop mean estimators (Gauss-Markov) provided X_i have identical variances. You can choose any combination g_1*X_1+...+g_n*X_n with g_1+...+g_n=1 as unbiased pop mean estimate. If they have the same means but different variances the optimal weights g_i are proportional to
1/(sigma_i)^2. Let's say e.g. g_i=1/(sigma_i)^2 / ( 1/(sigma_1)^2+...+1/(sigma_n)^2 )

8. Originally Posted by chella182
Sorry, I don't understand how you've gotten 20
$E(X_1)=E(X_2)= ... =E(X_n)=20$

Now count up how many expectations you have, there are $n$ terms but the first two add an extra $3 \times 20$ on so you have:

$E(Y)=\frac{(n+3)20}{n+3}=20$

CB

9. Originally Posted by CaptainBlack
$E(X_1)=E(X_2)= ... =E(X_n)=20$

Now count up how many expectations you have, there are $n$ terms but the first two add an extra $3 \times 20$ on so you have:

$E(Y)=\frac{(n+3)20}{n+3}=20$

CB
Ah right, yeah, thankyou.