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Math Help - Expanded variance formula

  1. #1
    Member courteous's Avatar
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    Angry Expanded variance formula

    Textbook says that
    we can convert \sigma^2=\frac{1}{N}\sum_{i=1}^{N} (x_i-\mu)^2 into less cumbersome \sigma^2=\frac{1}{N}\sum_{i=1}^{N}x_i^2 -\mu^2. ( \mu is arithmetic mean.)
    I've tried about 5 (or more) times already.

    \sigma^2=\frac{1}{N}\sum_{i=1}^{N} (x_i-\mu)^2=\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2+\sum_{i=1}^{N} (-2x_i\mu)+\sum_{i=1}^{N}\mu^2\Big]=\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2-2\mu\sum_{i=1}^{N} x_i+\mu^2\Big] =\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2-2\mu(\mu*N)+\mu^2\Big]=\frac{1}{N}\sum_{i=1}^{N}x_i^2-2\mu^2+\frac{\mu^2}{N}

    How come I get \frac{\mu^2}{N}?
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  2. #2
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    Quote Originally Posted by courteous View Post
    Textbook says that I've tried about 5 (or more) times already.

    \sigma^2=\frac{1}{N}\sum_{i=1}^{N} (x_i-\mu)^2=\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2+\sum_{i=1}^{N} (-2x_i\mu)+\sum_{i=1}^{N}\mu^2\Big]=\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2-2\mu\sum_{i=1}^{N} x_i+\mu^2\Big] =\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2-2\mu(\mu*N)+\mu^2\Big]=\frac{1}{N}\sum_{i=1}^{N}x_i^2-2\mu^2+\frac{\mu^2}{N}

    How come I get \frac{\mu^2}{N}?
    You missed one factor.
    \sigma^2=\frac{1}{N}\sum_{i=1}^{N} (x_i-\mu)^2=\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2+\sum_{i=1}^{N} (-2x_i\mu)+\sum_{i=1}^{N}\mu^2\Big]=\frac{1}{N} \Big[\sum_{i=1}^{N}x_i^2-2\mu(\mu*N)+\color{blue}N\mu^2\Big]
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  3. #3
    Newbie Bill Zimmerly's Avatar
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    Wink

    Quote Originally Posted by courteous View Post
    Textbook says that I've tried about 5 (or more) times already.

    \sigma^2=\frac{1}{N}\sum_{i=1}^{N} (x_i-\mu)^2=\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2+\sum_{i=1}^{N} (-2x_i\mu)+\sum_{i=1}^{N}\mu^2\Big]=\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2-2\mu\sum_{i=1}^{N} x_i+\mu^2\Big] =\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2-2\mu(\mu*N)+\mu^2\Big]=\frac{1}{N}\sum_{i=1}^{N}x_i^2-2\mu^2+\frac{\mu^2}{N}

    How come I get \frac{\mu^2}{N}?
    What Plato said!
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