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Thread: Sampling without replacement

  1. #1
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    Sampling without replacement

    Hi,

    I'm trying to show that the sample variance obtained from a random sampling without replacement is

    $\displaystyle \frac{\sigma^2}{n}\left(1 - \frac{n - 1}{N - 1}\right) \quad
    \mbox{where} \quad \sigma^2 = \frac{1}{N} \sum_i^N \left(x_i - \mu\right)^2 $

    and \mu is the sample mean. A lot of books cite the factor $\displaystyle \left(1 - \frac{n - 1}{N - 1}\right) $ as a correction for small samples, but I couldn't derive it. Could someone help me here, please?
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  2. #2
    Senior Member Sambit's Avatar
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    Have you taken into account the Covariance term while calculating the variance? The Covariance term becomes 0 when sampling is done With Replacement; but that will not vanish in case of Without Replacement.
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  3. #3
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    Hi Sambit, thank you for your response.

    Yes, I considered the covariance, but I'm probably doing something wrong. I got the following result

    $\displaystyle
    Cov(x_i,x_j) = E(x_i x_j) - E(x_i)E(x_j) = x_i x_j \frac{n}{N}\frac{n-1}{N-1} - x_ix_j\left(\frac{n}{N}\right)^2.
    $

    The problem is that it does not depend on $\displaystyle \sigma $. I'm missing something... or doing something completely wrong.
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  4. #4
    Senior Member Sambit's Avatar
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    The right expression for Covariance is $\displaystyle -\frac{\sigma^2}{N-1}$. Consult any book, you will find it there.
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  5. #5
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    Thanks again, Sambit.

    I saw this expression, but I don't know how to derive it. Most books only mention it but does not show how to obtain this formula. I was reading Kish's Survey Sampling book, but he just mention it as a correction for finite sample with replacement. I found some links from the University of Texas, but they also just mention the covariance without showing all calculations behind.

    I'm stuck. I have no clue how to calculate this covariance. Could you give some hint, please?
    This problem is driving me crazy!
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  6. #6
    Senior Member Sambit's Avatar
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    $\displaystyle Cov(y_i,y_j)$$\displaystyle
    =E[\{y_i-\bar{Y}\}\{y_j-\bar{Y}\}]$$\displaystyle
    =\sum_{\alpha=1}^N\sum_{\beta=1}^N(Y_{\alpha}-\bar{Y})(Y_{\beta}-\bar{Y}).P(y_i=Y_{\alpha},y_j=Y_{\beta})$$\displaystyle
    =\sum_{\alpha=1}^N\sum_{\beta=1}^N(Y_{\alpha}-\bar{Y})(Y_{\beta}-\bar{Y})\frac{1}{N(N-1)} $ where$\displaystyle \alpha\neq\beta$
    $\displaystyle
    =\frac{1}{N(N-1)}\sum_{\alpha=1}^N\sum_{\beta=1}^N(Y_{\alpha}-\bar{Y})(Y_{\beta}-\bar{Y})$ where$\displaystyle \alpha\neq\beta$
    $\displaystyle
    =\frac{1}{N(N-1)}[\sum_{\alpha=1}^N\sum_{\beta=1}^N(Y_{\alpha}-\bar{Y})(Y_{\beta}-\bar{Y})-\sum_{\alpha=1}^N(Y_{\alpha}-\bar{Y})^2]$$\displaystyle
    =\frac{1}{N(N-1)}\sum_{\alpha=1}^N(Y_{\alpha}-\bar{Y})\sum_{\beta=1}^N(Y_{\beta}-\bar{Y})-N\sigma^2$$\displaystyle
    =\frac{1}{N(N-1)}[0-N\sigma^2]$$\displaystyle
    =-\frac{\sigma^2}{N-1}$

    Here $\displaystyle i,j$ denote sample members and $\displaystyle \alpha,\beta$ denote population members.
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  7. #7
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    Thank you very much for your help, Sambit. Now I can sleep in peace!
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