The fourth moment of a centered random variable is at least equal to the square of its variance

Question

Let X be a random variable with mean µ and variance $\sigma^2$. Show that

$E(X-\mu)^4\geq \sigma^4$

and use this to show that the kurtosis of $X$ is at least $-2$.

This looks like a form of Chebyshev's equation:

$P(|X-\mu|\geq a) \leq \frac{\sigma^2}{a^2}$

But does not relate the inequality in the probability.

I tried expanding $E(X-\mu)^4$ out, but that seemed like a dead end.

Can someone guide me on how to solve this? If it does deal with Chebyshev, can someone explain in detail how to get the inequality out of the absolute value probability?

Answer 1

Setting $Y=X-\mu$ it must actually be shown that $\mathbb EY^4\geq(\mathbb EY^2)^2$.

This is a direct consequence of: $$\mathbb EY^4-(\mathbb EY^2)^2=\mathsf{Var}Y^2\geq0$$

Answer 2

You don't need Chebyshev. Note that $E(X-\mu)^4=\operatorname{Var}(X-\mu)^2+E^2(X-\mu)^2$. The first term $\ge 0$; the second is $\sigma^4$.