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)?
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2$\begingroup$ 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. $\endgroup$– whuber ♦Commented Jan 30, 2020 at 18:40
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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.
answered Jan 30, 2020 at 17:30
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2$\begingroup$ Thanks. But can you give a (counter-)example where the fourth moment diverges? $\endgroup$– a06eCommented Jan 30, 2020 at 17:42
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2$\begingroup$ 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 $4 + \epsilon, \, \epsilon \downarrow 0$. $\endgroup$– jbowmanCommented Jan 30, 2020 at 18:31
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$\begingroup$ Here is a nice counterexample :) math.stackexchange.com/a/2510884/472987 $\endgroup$ Commented Feb 18, 2020 at 2:10