My understanding is that the central limit theorem applies as long as the variance of the random variable is less than infinity. Is this equivalent to saying that all moments are finite? If not, what is an example of a random variable where its variance is less than infinity, so the central limit theorem applies, but its other moments are infinite?

  • $\begingroup$ Look at the family of Student t distributions for common examples where the third and higher moments are infinite or undefined. (Use the absolute values of these variables to remove the "or undefined" clause.) And please note that a finite variance guarantees a finite mean. $\endgroup$
    – whuber
    Dec 13, 2018 at 22:56
  • $\begingroup$ @whuber If the nth moment is finite, does that mean that all moments < n are also finite? Any suggestions where I can read about this please? $\endgroup$ Dec 13, 2018 at 23:24
  • $\begingroup$ Yes it does, and the Wikipedia page on moments can get you started: en.wikipedia.org/wiki/Moment_(mathematics). $\endgroup$
    – jbowman
    Dec 14, 2018 at 2:57
  • $\begingroup$ You can read about it at stats.stackexchange.com/questions/244202/…, which provides several proofs of a stronger statement about moments. $\endgroup$
    – whuber
    Dec 14, 2018 at 14:45
  • $\begingroup$ @jbowman all moments >n surely? it doesn't imply non existence of lower order moments. $\endgroup$
    – Glen_b
    Dec 17, 2018 at 2:30

1 Answer 1


Finite variance does not imply finite higher moments: A simple illustration can be done with the Pareto distribution, which has tails that obey the power law. Let $X \sim \text{Pareto}(1, \alpha)$ be a unit-scale Pareto random variable, which has density function:

$$f_X(x) = \alpha x^{-\alpha-1} \quad \quad \quad \text{for all }x \geqslant 1.$$

The $k$th raw moment of this distribution is:

$$\begin{equation} \begin{aligned} \mathbb{E}(X^k) = \int \limits_\mathscr{X} x^k f_X(x) dx &= \int \limits_1^\infty x^k \alpha x^{-\alpha-1} dx \\[6pt] &= \alpha \int \limits_1^\infty x^{k-\alpha-1} dx \\[6pt] &= \begin{cases} \infty & & \text{if } \alpha \leqslant k, \\[6pt] \frac{1}{\alpha-k} & & \text{if } \alpha > k. \\[6pt] \end{cases} \end{aligned} \end{equation}$$

We can see here that the raw moments of this distribution exist (i.e., are finite) for all $k < \alpha$. Hence, taking $2 < \alpha \leqslant k$ for some integer $k \geqslant 3$, the variance exists but the $k$th central moment does not exist.


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