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Math Help - New estimator

  1. #1
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    New estimator

    In order to estimate a population mean, \mu, 2 surveys were conducted independently and the statistics were noted. ( \overline{X}_1,\overline{X}_2,\sigma _{\overline{X}_1}, \sigma _{\overline{X}_2} are obtained). Assume that \overline{X_1} and \overline{X_2} are unbaised. For some \alpha and \beta, the two estimates can be combined to give a better estimator:

    X=\alpha \overline{X}_1 + \beta \overline{X}_2

    What choice of \alpha and \beta will minimize the variances, given that \alpha + \beta =1?

    Since I managed to find one equation which is: \alpha + \beta =1. What is another equations which allow me to minimise the variance in order to find suitable \alpha and \beta?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by noob mathematician View Post
    In order to estimate a population mean, \mu, 2 surveys were conducted independently and the statistics were noted. ( \overline{X}_1,\overline{X}_2,\sigma _{\overline{X}_1}, \sigma _{\overline{X}_2} are obtained). Assume that \overline{X_1} and \overline{X_2} are unbaised. For some \alpha and \beta, the two estimates can be combined to give a better estimator:

    X=\alpha \overline{X}_1 + \beta \overline{X}_2

    What choice of \alpha and \beta will minimize the variances, given that \alpha + \beta =1?

    Since I managed to find one equation which is: \alpha + \beta =1. What is another equations which allow me to minimise the variance in order to find suitable \alpha and \beta?
    Do these surveys have the same sample size (or do we know the sample sizes), and are they otherwise identical?

    CB
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  3. #3
    MHF Contributor matheagle's Avatar
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    You need a relationship between the variances and are these \sigma's populations st deviations or sample st deviations?
    This is not complete.
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  4. #4
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    Hi,

    There are actually 2 questions on this (However there aren't much more clues to that)

    Qn1: Find the conditions on \alpha and \beta that make the combined estimate unbiased. I get \alpha + \beta =1

    Qn2: What choice of \alpha and \beta minimizes the variances, subject to the condition of unbiasedness above.
    Answer I suppose: (it's true that \alpha + \beta is still 1)

    \alpha= \frac{\sigma^2 _{\overline{X}_2}}{\sigma^2 _{\overline{X}_2}+\sigma^2 _{\overline{X}_1}}, \beta= \frac{\sigma^2 _{\overline{X}_1}}{\sigma^2 _{\overline{X}_2}+\sigma^2 _{\overline{X}_1}}

    I'm not sure whether is the population needed since we already know their statistics.

    \sigma _{\overline{X}} is the standard errors. So \sigma^2 _{\overline{X}}=Var(\overline{X})
    Last edited by noob mathematician; September 11th 2009 at 05:35 AM.
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  5. #5
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    Quote Originally Posted by noob mathematician View Post
    Hi,

    There are actually 2 questions on this (However there aren't much more clues to that)

    Qn1: Find the conditions on \alpha and \beta that make the combined estimate unbiased. I get \alpha + \beta =1

    Qn2: What choice of \alpha and \beta minimizes the variances, subject to the condition of unbiasedness above.
    Ans I suppose: (it's true that \alpha + \beta is still 1)

    \alpha= \frac{\sigma^2 _{\overline{X}_2}}{\sigma^2 _{\overline{X}_2}+\sigma^2 _{\overline{X}_1}}, \beta= \frac{\sigma^2 _{\overline{X}_1}}{\sigma^2 _{\overline{X}_2}+\sigma^2 _{\overline{X}_1}}

    I'm not sure whether is the population needed since we already know their statistics.

    \sigma _{\overline{X}} is the standard errors. So \sigma^2 _{\overline{X}}=Var(\overline{X})
    Hi,
    With the given answer: \alpha= \frac{\sigma^2 _{\overline{X}_2}}{\sigma^2 _{\overline{X}_2}+\sigma^2 _{\overline{X}_1}}, \beta= \frac{\sigma^2 _{\overline{X}_1}}{\sigma^2 _{\overline{X}_2}+\sigma^2 _{\overline{X}_1}}
    I've tried to work backward in order to attain the 2nd equation:

    X=\alpha\overline{X}_1+\beta\overline{X}_2
    Var(X)=\alpha^2 Var(\overline{X}_1)+\beta^2Var(\overline{X}_2)=\al  pha^2\sigma^2 _{\overline{X}_1}+\beta^2\sigma^2 _{\overline{X}_2} (Equation 2)

    Let x=\sigma^2 _{\overline{X}_1}, y=\sigma^2 _{\overline{X}_2}

    Subst the answer into the 1st equation: (\alpha +\beta=1)

    \frac{y}{x+y}+\frac{x}{x+y}=1 so it's valid.

    Subst the answer into the 2nd equation:

    Var(X)=(\frac{y}{x+y})^2x+(\frac{x}{x+y})^2y
    =\frac{xy^2}{(x+y)^2}+\frac{x^2y}{(x+y)^2}
    =\frac{xy}{x+y}

    However I'm not sure why the new variance estimator becomes \frac{xy}{x+y}?
    Last edited by noob mathematician; September 11th 2009 at 05:38 AM.
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