Does the existence of the mean imply the existence of the variance?

Question

Can a probability distribution have a (finite) mean, but an infinit variance?

Answer 1

Here's some intuition, for the case of continuous real variables with densities.

In that case, you're asking whether there's a density $f$ such that

$$\mathbb{E}[X]=\int_{-\infty}^\infty x\cdot f(x)\,dx$$

exists while

$$\mathbb{E}[X^2]=\int_{-\infty}^\infty x^2\cdot f(x)\,dx$$

fails to exist.

The answer is sure: $f$, which must decay quickly enough so that $\int_{-\infty}^\infty f(x)\,dx=1$, might decay quickly enough so that $x\cdot f(x)$ is integrable, even though $x^2\cdot f(x)$ is not. For example, taking $f$ to be proportional to $x^{-3}$ for $x>1$ will ensure $f(x)$ and $x\cdot f(x)$ are integrable, but $x^2\cdot f(x)$ isn't.

Kavi uses similar intuition in his discrete counterexample.

More generally, for any positive integer $k$, we can easily use this idea to construct simple examples where only the first $k$ moments exist.

Answer 2

If $X$ takes values $1,2,...$ with probabilities $\frac c {1^{3}},\frac c {2^{3}},...$ (where $c$ is chosen such that these numbers add up to $1$) then the mean is finite but variance is $\infty$.