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Math Help - Relative efficiency

  1. #1
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    Relative efficiency

    Suppose that Y_1, \ Y_2, \ \dotso,\ Y_n is a random sample from a normal distribution with mean \mu and variance \sigma^2. Given two unbiased estimators, find their relative efficiency.

    \sigma^2_1 = S^2 = \frac{1}{n-1} \sum^n_{i=1} (Y_i-\overline{Y})^2 and \sigma^2_2 =\frac{1}{2}(Y_1-Y_2)^2

    by definition the relative efficiency is \frac{V[Y_2]}{V[Y_1]} so in this case I should have:

    \frac{\sigma^2_2}{\sigma^2_1} = \frac{\frac{1}{2}(Y_1-Y_2)^2}{\frac{1}{n-1} \sum^n_{i=1} (Y_i-\overline{Y})^2}

    I was thinking of expanding it out but all I got was \frac{\sigma^2_2}{\sigma^2_1}.
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  2. #2
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    Quote Originally Posted by lllll View Post
    Suppose that Y_1, \ Y_2, \ \dotso,\ Y_n is a random sample from a normal distribution with mean \mu and variance \sigma^2. Given two unbiased estimators, find their relative efficiency.

    \sigma^2_1 = S^2 = \frac{1}{n-1} \sum^n_{i=1} (Y_i-\overline{Y})^2 and \sigma^2_2 =\frac{1}{2}(Y_1-Y_2)^2

    by definition the relative efficiency is \frac{V[Y_2]}{V[Y_1]} so in this case I should have:

    \frac{\sigma^2_2}{\sigma^2_1} = \frac{\frac{1}{2}(Y_1-Y_2)^2}{\frac{1}{n-1} \sum^n_{i=1} (Y_i-\overline{Y})^2}

    I was thinking of expanding it out but all I got was \frac{\sigma^2_2}{\sigma^2_1}.
    Var(\hat{\sigma_1}^2):

    It's well known that \frac{(n-1)S^2}{\sigma^2} has a \chi^2 distribution with n - 1 degrees of freedom. It's also well known that the variance of a \chi^2 distribution with \nu degrees of freedom is 2\nu. Therefore:

    Var\left(\frac{(n-1)S^2}{\sigma^2}\right) = 2(n-1)

    \Rightarrow \frac{(n-1)^2}{\sigma^4} Var(S^2) = 2(n-1)

    \Rightarrow Var(S^2) = \frac{2 \sigma^4}{n-1}.

    An approach similar to the one used below can also be used but it's more tedious.

    ----------------------------------------------------------------------------------------------------------------

    Var(\hat{\sigma_2}^2):

    Var\left( \frac{1}{2} (Y_1 - Y_2)^2 \right) = \frac{1}{4} Var((Y_1 - Y_2)^2) = \frac{1}{4} \left( E((Y_1 - Y_2)^4) - [E((Y_1 - Y_2)^2)]^2\right).


    E((Y_1 - Y_2)^2) = E(Y_1^2 - 2Y_1 Y_2 + Y_2^2) = E(Y_1^2) - 2 E(Y_1) \cdot E(Y_2) + E(Y_2^2)  = (\sigma^2 + \mu^2) - 2 \mu^2 + (\sigma^2 + \mu^2) = 2 \sigma^2.


    E((Y_1 - Y_2)^4) = E(Y_1^4 - 4Y_1^3 Y_2 + 6 Y_1^2 Y_2^2 - 4 Y_1 Y_2^3 + Y_2^4) = E(Y_1^4) - 4E(Y_1^3) E(Y_2) + 6 E(Y_1^2) E(Y_2^2) - 4 E(Y_1) E(Y_2^3) + E(Y_2^4)

    I'll cheat and get the moments E(Y^4) and E(Y^3) from here: http://en.wikipedia.org/wiki/Normal_distribution (the other moments were shown in my reply to your previous unbiased estimator question)

    = (\mu^4 + 6\mu^2 \sigma^2 + 3\sigma^4) - 4(\mu^3 + 3\mu \sigma^2) \mu + 6 (\mu^2 + \sigma^2)^2  - 4 \mu (\mu^3 + 3\mu \sigma^2) + (\mu^4 + 6\mu^2 \sigma^2 + 3\sigma^4) = 12 \sigma^4.


    Therefore Var\left( \frac{1}{2} (Y_1 - Y_2)^2 \right) = \frac{1}{4} \left(12 \sigma^4 - 4 \sigma^4 \right) = 2 \sigma^4.

    --------------------------------------------------------------------------------------------------------------

    Therefore \frac{Var(\hat{\sigma_2}^2)}{Var(\hat{\sigma_1}^2)  } = n - 1.
    Last edited by mr fantastic; December 7th 2008 at 12:37 AM.
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