Consider a random sample $X$ with pdf $ f(x)=\frac{k(p)}{x^{p}} $ for $x>1$, $p>0$ and $k(p) $ is a positive constant. Find set of possible values of $p$ for which variance is finite and fourth moment doesn't exist.
On direct solving, using the fact that integral over the entire range should be equal to 1, I found that if $0<p<1$ then the value of integrand becomes infinity. Does this mean that $p>1$?. How do I go about this problem further?
Note that $\Bbb E X^n=\int_1^\infty k(p)x^{n-p}dx=\frac{k(p)}{k(p-n)}$ provided $p>n+1$ for convergence, otherwise this expectation is infinite. So when is the mean $\mu:=\Bbb EX$ finite? When does the variance $\sigma^2:=\Bbb E X^2-\mu^2$ exist? What about the kurtosis $\sigma^{-4}\Bbb E(X-\mu)^4$?