Question

The variance of random variable $X$ i.e., ${\sigma }_x^2$ or $\mathrm{var}(X)$ is equal to

• A. $E\left(X^2\right)+{\left[E\left(X^2\right)\right]}^2$
• B. $E(X)-\left[E\left(X^2\right)\right]$
• C. $E\left(X^2\right)-[E(X){]}^2$
• D. None of the above

Answer 1

Answer: (C)

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