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Thread: Expanded variance formula

  1. #1
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    Angry Expanded variance formula

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

    $\displaystyle \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]$$\displaystyle =\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 $\displaystyle \frac{\mu^2}{N}$?
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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.

    $\displaystyle \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]$$\displaystyle =\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 $\displaystyle \frac{\mu^2}{N}$?
    You missed one factor.
    $\displaystyle \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}$$\displaystyle \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.

    $\displaystyle \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]$$\displaystyle =\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 $\displaystyle \frac{\mu^2}{N}$?
    What Plato said!
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