Answer to Question #87367 in Statistics and Probability for Shivam Nishad

Question

Let X be a random variable with density function
f(X) = {2/x^3, if x≥1
{0, otherwise
Show that E(X) exists and E(X) = 2 but Var(X) does not exist.

Answer 1

$$f(x)= \begin{matrix} 2/x^3 & x\geqslant1, \\ 0, & otherwise \end{matrix}$$

$$E(X)=\displaystyle\int_{1}^\infin x{2 \over x^3}dx=[{-2 \over x}]\begin{matrix} \infin \\ 1 \end{matrix}=2$$

$$Var(X)=E(X^2)-(E(X))^2$$

$$\displaystyle\int_{1}^\infin x^2{2 \over x^3}dx=2[ln|x|]\begin{matrix} \infin \\ 1 \end{matrix}=\infin$$

Therefore, Var(X) does not exist.