# Fourth moment bound for unit-variance distribution

Given that a random real variable $$X$$ has zero mean and variance equal to 1, can we bound its fourth moment $$\langle X^4\rangle$$ (assuming it exists)?

• Hint: consider a ternary variable taking on the values $\pm x$ each with probability $1/(2x^2)$ (for $x^2 \ge 1$) and otherwise taking the value $0.$ Compute its mean, variance, and fourth moment. BTW, if there exists any random variable for which the fourth moment doesn't exist, that moment must be infinite, demonstrating no bound is possible. – whuber Jan 30 at 18:40

No, you cant, at least not without further restrictions. One restrictions that could do is that the range of $$X$$ is a finite interval.
One simple example, see the coments for another example. Let $$X$$ be a discrete random variable such that $$\DeclareMathOperator{\P}{\mathbb{P}} \P(X=\pm a)=p/2, \qquad \P(X=0)=1-p.$$ Then $$\DeclareMathOperator{\E}{\mathbb{E}} \E X=0$$ and the variance $$\DeclareMathOperator{\V}{\mathbb{V}} \V X=p a^2$$ so the unit variance condition gives $$p=a^{-2}$$. But the the fourth moment $$\E X^4 = p a^4 = a^{-2} a^4 = a^2$$ and clearly this is unbounded when $$a$$ increases without bound.
• A $t$-distributed variate with three degrees of freedom does not have a fourth moment. You can make the fourth moment of a $t$ variate arbitrarily large by setting the degrees of freedom parameter equal to $4 + \epsilon, \, \epsilon \downarrow 0$. – jbowman Jan 30 at 18:31