# What does it mean for a probability distribution to have a finite fourth moment?

For example, suppose that some probability distribution X has a finite fourth moment. What distinguishes this distribution from another one, Y, which does not have a finite fourth moment?

I am to understand that this gives us greater control over the "tails" of the distribution, but I don't understand how so.

• Try to write mathematically what it really mean. I think that it will be clearer to you. Oct 6, 2015 at 20:28
• See my answer to this: stats.stackexchange.com/questions/94402/… Oct 7, 2015 at 7:10

Let $X$ and $Y$ be two random variables on a probability space $(\Omega,\mathcal{A},\mathbb{P})$. If $X$ has a finite fourth moment, i.e. $\mathbb{E}(|X|^4)<\infty$, then the tail of the distribution of $X$

$$\mathbb{P}(|X| \geq r)$$

satisfies the inequality

$$\mathbb{P}(|X| \geq r) \leq \frac{C}{r^4} \tag{1}$$

for all $r>0$ where $C>0$ is some fixed constant. This shows that the tail $\mathbb{P}(|X| \geq r)$ decays at least as $\frac{1}{r^4}$ for large $r$.

Just think of the case that the distribution of $Y$ has a density (with respect to Lebesgue measure) given by $$f(y) = C \frac{1}{y^p} 1_{(1,\infty)}(y).$$ Then $$\mathbb{E}(|Y|^4) = C \int_1^{\infty} y^{4-p} \, dy$$ is finite if, and only if, $p>5$. Note also that

$$\mathbb{P}(|Y| \geq r) = C \int_r^{\infty} \frac{1}{y^p} \,dy = C \frac{1}{1-p} r^{1-p}$$

for $r \geq 1$, i.e. $(1)$ holds for $p \geq 5$.

Remark: In $(1)$, we have seen that a necessary condition for the existence of the fourth moment is that the tail decays at least as $\frac{1}{r^4}$ for $r \gg 1$. In fact, one can show that the identity

$$\mathbb{E}(|X|^4) = 4 \int_{(0,\infty)} \mathbb{P}(|X| \geq r) r^{3} \, dr$$

holds for any random variable $X$. In particular, the fourth moment is finite if, and only if, $(1,\infty) \ni r \mapsto \mathbb{P}(|X| \geq r) r^3$ is integrable (with respect to Lebesgue measure).