Show that
fY(y) = 1/(y^2), y >= 1
is a valid pdf but that Y does not have a finite expected value.
We know that valid $\displaystyle pdfs$ integrate to $\displaystyle 1.$
$\displaystyle \int_1^{\infty} \frac{1}{y^2}dy = \frac{-1}{y}|_{1}^{\infty} = 1$
$\displaystyle E[Y]= \int_1^{\infty} y\times \frac{1}{y^2}dy = \int_1^{\infty} \frac{1}{y}dy= \ln(y)|_{1}^{\infty} = \infty$
Let me know if you have questions.
Anonymous