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Math Help - Statistics proof

  1. #1
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    Statistics proof

    Given that random variable X, its mean value \mu, its variance \sigma ^2, and the mean value X^2, which is \mu _{X^2}, prove that \sigma ^2=\mu _{X^2} - \mu _X^2.

    Letting X be the values x_1,x_2,...,x_n and P(X = x_i) =p_i, (i=1,2,...,n)

    \therefore \sigma ^2 = \sum_{i=1}^n (x_i-\mu _x)^2 \cdot p_i

     = \sum_{i=1}^n x_i^2 \cdot p_i - 2\mu _X \cdot \bigg (\sum_{i=1}^n x_i \cdot p_i \bigg ) + \mu _X^2 \bigg ( \sum_{i=1}^n p_i \bigg ) I'm pretty lost how to get this step and the rest.

    =\mu_{X^2}-2\mu _X \cdot \mu _X + \mu _X^2

    =\mu _{X^2} - \mu _X^2

    I should have taken a picture of the book. That was a ton of code to write, even though it doesn't look like much at all.
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  2. #2
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    Quote Originally Posted by chengbin View Post
    Given that random variable X, its mean value \mu, its variance \sigma ^2, and the mean value X^2, which is \mu _{X^2}, prove that \sigma ^2=\mu _{X^2} - \mu _X^2.

    Letting X be the values x_1,x_2,...,x_n and P(X = x_i) =p_i, (i=1,2,...,n)

    \therefore \sigma ^2 = \sum_{i=1}^n (x_i-\mu _x)^2 \cdot p_i

     = \sum_{i=1}^n x_i^2 \cdot p_i - 2\mu _X \cdot \bigg (\sum_{i=1}^n x_i \cdot p_i \bigg ) + \mu _X^2 \bigg ( \sum_{i=1}^n p_i \bigg ) I'm pretty lost how to get this step Mr F says: Just expand!

    and the rest. Mr F says: By definition: {\color{red}\sum_{i=1}^n p_i = 1}, {\color{red}\sum_{i=1}^n x_i p_i = \mu_X} and {\color{red}\sum_{i=1}^n x^2_i p_i = \mu_{X^2}}.

    =\mu_{X^2}-2\mu _X \cdot \mu _X + \mu _X^2

    =\mu _{X^2} - \mu _X^2

    I should have taken a picture of the book. That was a ton of code to write, even though it doesn't look like much at all.
    ..
    Last edited by mr fantastic; September 30th 2009 at 02:49 AM.
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