1. ## 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.)
$\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]$$\displaystyle =\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2-2\mu(\mu*N)+\mu^2\Big]=\frac{1}{N}\sum_{i=1}^{N}x_i^2-2\mu^2+\frac{\mu^2}{N} How come I get \displaystyle \frac{\mu^2}{N}? 2. Originally Posted by courteous Textbook says that I've tried about 5 (or more) times already. \displaystyle \sigma^2=\frac{1}{N}\sum_{i=1}^{N} (x_i-\mu)^2=\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2+\sum_{i=1}^{N} (-2x_i\mu)+\sum_{i=1}^{N}\mu^2\Big]=\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2-2\mu\sum_{i=1}^{N} x_i+\mu^2\Big]$$\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 $\displaystyle \frac{\mu^2}{N}$?
$\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}$$\displaystyle \Big[\sum_{i=1}^{N}x_i^2-2\mu(\mu*N)+\color{blue}N\mu^2\Big] 3. Originally Posted by courteous Textbook says that I've tried about 5 (or more) times already. \displaystyle \sigma^2=\frac{1}{N}\sum_{i=1}^{N} (x_i-\mu)^2=\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2+\sum_{i=1}^{N} (-2x_i\mu)+\sum_{i=1}^{N}\mu^2\Big]=\frac{1}{N}\Big[\sum_{i=1}^{N}x_i^2-2\mu\sum_{i=1}^{N} x_i+\mu^2\Big]$$\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 $\displaystyle \frac{\mu^2}{N}$?