Given a probability triple $(\Omega, \mathcal{F}, \mu)$ of Lebesgue measure $[0,1]$, find a random variable $X : \Omega \to \mathbb{R}$ such that the expected value $E(X)$ converges to a finite, positive value, but $E(X^2)$ diverges.
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One answer is Pareto distribution with parameters $\alpha , x_0$ which are both positive. The distribution is given by: $$f_X(x)= \begin{cases} \alpha\,\frac{x_0^\alpha}{x^{\alpha+1}} & x \ge x_0, \\ 0 & x < x_0. \end{cases}$$ Note that $E[X] = \infty$ for $\alpha \leq 1$ and is finite elsewhere. The variance is not finite for $\alpha \in [1,2) $ Hence it satisfies your question for $(1,2)$ In general,$E[X^n]= \infty \ \ ;n\geq \alpha$. EDIT: Clarified the answer as suggested. |
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