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Suppose you have pdf $$f(x) = \begin{cases} \frac{8}{x^3} &, \text{ if $x\ge 2$} \\ 0 &, \text{ otherwise} \end{cases}$$

I have found that $\Bbb{E}(X)=4$ and am trying to find $\operatorname{Var}(X)$ using $\Bbb{E}(X^2)-(\Bbb{E}(X))^2$.

To find $\Bbb{E}(X^2)$, I've been using $$\int_{-\infty}^\infty u^2 f(u) du = \int_2^\infty \frac{8}{u} du = \lim_{t\to \infty}(8\ln t - 8\ln2)$$

However, $\lim_{t\to \infty}(\ln t)$ DNE, so does that mean that neither does $\operatorname{Var}(X)$?

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I think $Var(X)$ exists and $Var(X)=+\infty$

since $$E(X)=4$$ ,so if $$E(X^2)$$ exists , in hence $Var(X)$ exists.

$E(Y)$ exists if $E(Y^+)<\infty $ or $E(Y^-)<\infty$. on the other hands if both $E(Y^+)=\infty $ , $E(Y^-)=\infty$ so $E(Y)$ does not exist.

meaning-of-non-existence-of-expectation $$E((X^2)^{+})=E(X^2)=\infty$$ $$E((X^2)^{-})=E(\max(0,-X^2))=E(0)=0<\infty$$

so $E(X^2)$ exists and $E(X^2)=\infty$

This distribution is Pareto distribution that can see for $x_m=\alpha =2$ wikipedia say Variance $=\infty$ Pareto_distribution

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