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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.2012-11-13

2 Answers 2

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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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    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.2012-11-13
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An example is a random variable $X$ having a student-t distribution with $\nu = 2$ degrees of freedom

Its mean is $E[X] = 0$ for $\nu > 1$, but its second moment $E[X^2] = Var[X] = \infty$ for $1 < \nu \le 2$

Edit: Finite positive? $X+1$ I guess