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Math Help - How to simplify this sample mean expression?

  1. #1
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    How to simplify this sample mean expression?

     {1 \over n}\sum_{i=1}^{n} {X_i}^2 - ({1 \over n} \sum_{i=1}^{n} {X_i})^2
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  2. #2
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    There is not much "simplification" to be done, but you may want to write one of the following:

    {1 \over n}\sum_{i=1}^{n} {X_i}^2 - ({1 \over n} \sum_{i=1}^{n} {X_i})^2<br />
= {1\over n}\sum_{i=1}^n \left(X_i - \frac{1}{n}\sum_{j=1}^n X_j\right)^2<br />

    (which you get by expanding to square in the right-hand side into X_i^2 - 2X_i{1\over n}\sum_{j=1}^n X_j + \left({1\over n}\sum_{j=1}^n X_j\right)^2). It is the other expression for the empirical variance.

    or:

    {1 \over n}\sum_{i=1}^{n} {X_i}^2 - ({1 \over n} \sum_{i=1}^{n} {X_i})^2<br />
= {1\over n}\sum_{i=1}^n X_i^2 - {1\over n^2}\sum_{1\leq i, j\leq n}X_iX_j ={1\over n}\sum_{i=1}^n X_i^2 - {1\over n^2}\sum_{i=1}^n X_i^2 - {1\over n^2}\sum_{1\leq i\neq j\leq n} X_iX_j = \left(\frac{1}{n}-\frac{1}{n^2}\right)\sum_{i=1}^n X_i^2 - \frac{2}{n^2} \sum_{1\leq i<j \leq n} X_i X_j
    (which is the expansion of the square of the sum. In the last expression, every product X_iX_j appears exactly once)

    I hope it answers you question.
    Laurent.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by chopet View Post
     {1 \over n}\sum_{i=1}^{n} {X_i}^2 - ({1 \over n} \sum_{i=1}^{n} {X_i})^2
    This is sample mean square minus the square sample mean, which is the sample variance.

    RonL
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