Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Hint: think of Pareto random variables. – Dilip Sarwate Nov 13 '12 at 4:17
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.