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  4. The variance of random variable X i.e., σx2 or operatornamevar(X) is equal to

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Q. The variance of random variable $X$ i.e., $\sigma_x^2$ or $\operatorname{var}(X)$ is equal to

Probability - Part 2

Solution:

We know that, $\text{Var}(X)-\sum_{i=1}^n\left(x_i-\mu\right)^2 p\left(x_i\right)$
$=\displaystyle\sum_{i=1}^n\left(x_i^3+\mu^2-2 \mu x_i\right) \mu\left(x_i\right)$
$ =\displaystyle\sum_{i=1}^n x_i^2 p\left(x_i\right)+\displaystyle\sum_{i=1}^n \mu^2 p\left(x_i\right)-\displaystyle\sum_{i=1}^n 2 \mu x_i p\left(x_i\right) $
$ =\displaystyle\sum_{i=1}^n x_i^2 p\left(x_i\right)+\mu^2 \displaystyle\sum_{i=1}^n p\left(x_i\right)-2 \mu \displaystyle\sum_{i=1}^n x_j p\left(x_i\right) $
$ =\displaystyle\sum_{i=1}^n x_i^2 p\left(x_i\right)+\mu^2-2 \mu^2\left[\because \displaystyle\sum_{i=1}^n p\left(x_i\right)=1 \text { and } \mu=\displaystyle\sum_{i=1}^n x_j p\left(x_j\right)\right]$
$=\displaystyle\sum_{i=1}^n x_i^2 p\left(x_i\right)-\mu^2$
or $ \text{Var}(X)=\displaystyle\sum_{i=1}^n x_i^2 p\left(x_j\right)-\left(\displaystyle\sum_{i=1}^n x_i p\left(x_i\right)\right)^2 $
or $\text{Var}(X)=E\left(X^2\right)-[E(X)]^2, $
where, $ E\left(X^2\right)=\displaystyle\sum_{i=1}^n x_i^2 p\left(x_i\right)$