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Thread: New estimator

  1. #1
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    New estimator

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

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

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

    Since I managed to find one equation which is: $\displaystyle \alpha + \beta =1$. What is another equations which allow me to minimise the variance in order to find suitable $\displaystyle \alpha$ and $\displaystyle \beta$?
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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, $\displaystyle \mu$, 2 surveys were conducted independently and the statistics were noted. ($\displaystyle \overline{X}_1,\overline{X}_2,\sigma _{\overline{X}_1}, \sigma _{\overline{X}_2} $ are obtained). Assume that $\displaystyle \overline{X_1}$ and $\displaystyle \overline{X_2}$ are unbaised. For some $\displaystyle \alpha$ and $\displaystyle \beta$, the two estimates can be combined to give a better estimator:

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

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

    Since I managed to find one equation which is: $\displaystyle \alpha + \beta =1$. What is another equations which allow me to minimise the variance in order to find suitable $\displaystyle \alpha$ and $\displaystyle \beta$?
    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 $\displaystyle \sigma$'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 $\displaystyle \alpha$ and $\displaystyle \beta$ that make the combined estimate unbiased. I get $\displaystyle \alpha + \beta =1$

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

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

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

    $\displaystyle \sigma _{\overline{X}}$ is the standard errors. So $\displaystyle \sigma^2 _{\overline{X}}=Var(\overline{X})$
    Last edited by noob mathematician; Sep 11th 2009 at 04:35 AM.
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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 $\displaystyle \alpha$ and $\displaystyle \beta$ that make the combined estimate unbiased. I get $\displaystyle \alpha + \beta =1$

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

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

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

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

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

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

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

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

    Subst the answer into the 2nd equation:

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

    However I'm not sure why the new variance estimator becomes $\displaystyle \frac{xy}{x+y}$?
    Last edited by noob mathematician; Sep 11th 2009 at 04:38 AM.
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