91Ó°ÊÓ

A distribution has a finite mean and an infinite variance.

Consider the probability density function, \(f\left( x \right) = \left\{ {\begin{aligned}{*{20}{l}}{{x^{ - \frac{5}{2}}};{\rm{ for }}x \in \left( {1,\infty } \right)}\\{0;{\rm{ otherwise}}}\end{aligned}} \right.\).

Now, we should calculate the mean and variance of the random variable X with the PDF as shown below.

\(\begin{aligned}{}E\left( X \right) &= \int\limits_1^\infty {xf\left( x \right)dx} \\ &= \int\limits_1^\infty {x{x^{ - \frac{5}{2}}}dx} \\ &= \int\limits_1^\infty {{x^{ - \frac{3}{2}}}dx} \\ &= - 2\left( {{x^{ - \frac{1}{2}}}} \right)_1^\infty \\ &= 2\end{aligned}\)

Therefore, X has the finite mean 2.

\(\begin{aligned}{}E\left( {{X^2}} \right) &= \int\limits_1^\infty {{x^2}f\left( x \right)dx} \\ &= \int\limits_1^\infty {{x^2}{x^{ - \frac{5}{2}}}dx} \\ &= \int\limits_1^\infty {{x^{ - \frac{1}{2}}}dx} \\ &= 2\left( {{x^{\frac{1}{2}}}} \right)_1^\infty \\ &= {\rm{ undefined}}\end{aligned}\)

\(\begin{aligned}{}Var\left( X \right) &= E\left( {{X^2}} \right) - {\left( {E\left( X \right)} \right)^2}\\ &= {\rm{undefined}}\end{aligned}\)

This means that X has an infinite variance.

Thus, the probability density function \(f\left( x \right)\) has a finite mean and an infinite variance.