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Math Help - unbiased estimator

  1. #1
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    unbiased estimator

    Let Y_1, \ Y_2, \ \dotso, \ Y_n be a random sample of size n from a normal population with mean \mu and variance \sigma^2. Assuming n=2k for some integer k, one possible estimator for \sigma^2 is given by:

    \hat{\sigma}^2=\frac{1}{2k}\sum^k_{i=1} (Y_{2i}-Y_{2i-1})^2

    show that \hat{\sigma}^2 is an unbiased estimator for \sigma^2

    Attempt:

    so when I expend it out it gives:

    \hat{\sigma}^2=\frac{1}{2k}\sum^k_{i=1} Y_{2i}^2-2(Y_{2i} \cdot Y_{2i-1})+ Y_{2i-1}^2 That term in the middle is throwing me off.

    I know that I eventually have to use the fact that

    E \bigg{[}\frac{1}{2k}\sum^k_{i=1} Y_{2i}^2-2(Y_{2i} \cdot Y_{2i-1})+ Y_{2i-1}^2 \bigg{]}

    and E[Y^2] = V[Y]+(E[Y])^2 \longrightarrow \sigma^2+\mu^2

    so taking that into consideration I would get:

    E \bigg{[}\frac{1}{2k}\sum^k_{i=1} Y_{2i}^2-2(Y_{2i} \cdot Y_{2i-1})+ Y_{2i-1}^2 \bigg{]} \longrightarrow E\bigg{[} \frac{1}{2k} \sum^k_{i=1}Y_{2i}^2\bigg{]} -E\bigg{[} \frac{1}{2k}\sum^k_{i=1}2(Y_{2i} \cdot Y_{2i-1})\bigg{]}  + E\bigg{[} \frac{1}{2k} \sum^k_{i=1}Y_{2i-1}^2\bigg{]}

    \frac{k}{2} E\bigg{[} \overline{Y}_{2i}^2\bigg{]} -E\bigg{[} (\mu_{2i} \cdot \mu_{2i-1})\bigg{]}  + \frac{k}{2} E\bigg{[} \overline{Y}_{2i-1}^2\bigg{]}

    I'm not sure if I should have the bottom indicator or if it should just be:

    \frac{k}{2} E\bigg{[} \overline{Y}^2\bigg{]} -E\bigg{[} \frac{1}{2} 2(\mu \cdot \mu)\bigg{]}  + \frac{k}{2} E\bigg{[} \overline{Y}^2\bigg{]} = {k} E\bigg{[} \overline{Y}^2\bigg{]} -E\bigg{[}(\mu^2)\bigg{]} although this doesn't seem correct.

    =k\bigg{(} \frac{\sigma^2}{k}+\mu^2\bigg{)} -\bigg{(} {\sigma^2}+\mu^2\bigg{)}= \mu^2(k-1).
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  2. #2
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    Quote Originally Posted by lllll View Post
    Let Y_1, \ Y_2, \ \dotso, \ Y_n be a random sample of size n from a normal population with mean \mu and variance \sigma^2. Assuming n=2k for some integer k, one possible estimator for \sigma^2 is given by:

    \hat{\sigma}^2=\frac{1}{2k}\sum^k_{i=1} (Y_{2i}-Y_{2i-1})^2

    show that \hat{\sigma}^2 is an unbiased estimator for \sigma^2

    Attempt:

    so when I expend it out it gives:

    \hat{\sigma}^2=\frac{1}{2k}\sum^k_{i=1} Y_{2i}^2-2(Y_{2i} \cdot Y_{2i-1})+ Y_{2i-1}^2 That term in the middle is throwing me off.

    I know that I eventually have to use the fact that

    E \bigg{[}\frac{1}{2k}\sum^k_{i=1} Y_{2i}^2-2(Y_{2i} \cdot Y_{2i-1})+ Y_{2i-1}^2 \bigg{]}

    and E[Y^2] = V[Y]+(E[Y])^2 \longrightarrow \sigma^2+\mu^2

    so taking that into consideration I would get:

    E \bigg{[}\frac{1}{2k}\sum^k_{i=1} Y_{2i}^2-2(Y_{2i} \cdot Y_{2i-1})+ Y_{2i-1}^2 \bigg{]} \longrightarrow E\bigg{[} \frac{1}{2k} \sum^k_{i=1}Y_{2i}^2\bigg{]} -E\bigg{[} \frac{1}{2k}\sum^k_{i=1}2(Y_{2i} \cdot Y_{2i-1})\bigg{]}  + E\bigg{[} \frac{1}{2k} \sum^k_{i=1}Y_{2i-1}^2\bigg{]}

    Mr F says: What follows is wrong. See below.

    \frac{k}{2} E\bigg{[} \overline{Y}_{2i}^2\bigg{]} -E\bigg{[} (\mu_{2i} \cdot \mu_{2i-1})\bigg{]}  + \frac{k}{2} E\bigg{[} \overline{Y}_{2i-1}^2\bigg{]}

    I'm not sure if I should have the bottom indicator or if it should just be:

    \frac{k}{2} E\bigg{[} \overline{Y}^2\bigg{]} -E\bigg{[} \frac{1}{2} 2(\mu \cdot \mu)\bigg{]}  + \frac{k}{2} E\bigg{[} \overline{Y}^2\bigg{]} = {k} E\bigg{[} \overline{Y}^2\bigg{]} -E\bigg{[}(\mu^2)\bigg{]} although this doesn't seem correct.

    =k\bigg{(} \frac{\sigma^2}{k}+\mu^2\bigg{)} -\bigg{(} {\sigma^2}+\mu^2\bigg{)}= \mu^2(k-1).
    E \left[ \frac{1}{2k} \sum^k_{i=1}Y_{2i}^2\right] -E\left[ \frac{1}{2k}\sum^k_{i=1}2(Y_{2i} \cdot Y_{2i-1})\right] + E\left[ \frac{1}{2k} \sum^k_{i=1}Y_{2i-1}^2\right]


     = \frac{1}{2k} \sum^k_{i=1} E \left[Y_{2i}^2\right] - \frac{1}{k}\sum^k_{i=1} E\left[ Y_{2i} \cdot Y_{2i-1} \right] + \frac{1}{2k} \sum^k_{i=1} E\left[ Y_{2i-1}^2\right]


     = \frac{1}{2k} \cdot k \left( \sigma^2 + \mu^2 \right) - \frac{1}{k}\sum^k_{i=1} E[ Y_{2i}] \cdot E[Y_{2i-1}] + \frac{1}{2k} \cdot k \left( \sigma^2 + \mu^2 \right)


    Note 1: Var(Y) = E(Y^2) - [E(Y)]^2 \Rightarrow \sigma^2 = E(Y^2) - \mu^2 \Rightarrow E(Y^2) = \sigma^2 + \mu^2.

    Note 2: The middle term reduces to this because the X_i's are independent.


     = \sigma^2 + \mu^2 - \frac{1}{k}\sum^k_{i=1} \mu \cdot \mu

     = \sigma^2 + \mu^2 - \frac{1}{k} \cdot k \mu^2 = \sigma^2.
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  3. #3
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    If I wanted to show that \hat{\sigma}^2 is a consistent estimator for \sigma^2, how would I go about it?
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  4. #4
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    Quote Originally Posted by lllll View Post
    If I wanted to show that \hat{\sigma}^2 is a consistent estimator for \sigma^2, how would I go about it?
    From my reply in this thread http://www.mathhelpforum.com/math-he...tml#post232762 it's known that Var((Y_1 - Y_2)^2) = 8 \sigma^4.

    Therefore Var(\hat{\sigma}^2) = \frac{1}{4k^2} \left( k [8 \sigma^4]\right) = \frac{2 \sigma^4}{k} = \frac{4 \sigma^4}{n}.

    Since \hat{\sigma^2} is an unbiased estimator it's sufficient to show that \lim_{n \rightarrow +\infty} Var(\hat{\sigma^2}) = 0 to prove consistency ....
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