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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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Hint: think of Pareto random variables. –  Dilip Sarwate Nov 13 '12 at 4:17
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1 Answer 1

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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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Are there any limitations on $\alpha$ or will this work for any positive value of $\alpha$, say $\alpha = 5$? –  Dilip Sarwate Nov 13 '12 at 4:30
    
Distribution is defined for $\alpha > 0$ but the question is satisfied for $\alpha \in (1,2]$ as then expectation is finite and variance infinite.., for $\alpha \leq 1$ expectation is infinite, variance doesn't exist.. but –  TheJoker Nov 13 '12 at 6:41
    
So maybe you could include the restriction on $\alpha$ in your answer so that anyone who just looks at the answer has the complete correct answer to the question of a random variable with finite mean and divergent $E[X^2]$. For $\alpha > 2$, both $E[X]$ and $E[X^2]$ are finite. –  Dilip Sarwate Nov 13 '12 at 12:22
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