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Math Help - Unbiased estimators

  1. #1
    Senior Member chella182's Avatar
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    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?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by chella182 View Post
    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
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by chella182 View Post
    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
    Last edited by CaptainBlack; August 12th 2009 at 03:32 PM.
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  4. #4
    Senior Member chella182's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    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}?
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by chella182 View Post
    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
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  6. #6
    Senior Member chella182's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    <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
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  7. #7
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    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 )
    Last edited by kobylkinks; August 13th 2009 at 11:14 AM.
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by chella182 View Post
    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
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  9. #9
    Senior Member chella182's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    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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