1. ## 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.

2. Originally Posted by chengbin
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.
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