Unbiased estimators

**chella182** · August 12th 2009, 01:33 PM

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?

**CaptainBlack** · August 12th 2009, 02:11 PM

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

**CaptainBlack** · August 12th 2009, 02:14 PM

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

**chella182** · August 12th 2009, 02:50 PM

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 $\frac{80+40n}{n+3}$ ?

**CaptainBlack** · August 12th 2009, 11:42 PM

Originally Posted by chella182

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

$<br /> Y=\frac{3X_{1}+2X_{2}+X_{3}+...+X_{n}}{n+3}<br />$

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

**chella182** · August 13th 2009, 07:38 AM

Originally Posted by CaptainBlack

$<br /> Y=\frac{3X_{1}+2X_{2}+X_{3}+...+X_{n}}{n+3}<br />$

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

**kobylkinks** · August 13th 2009, 10:02 AM

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 )

**CaptainBlack** · August 13th 2009, 11:05 AM

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

**chella182** · August 13th 2009, 12:33 PM

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.

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