Is the fourth moment always greater than the square of the variance?

Question

Let X be a random variable with 0 mean and symmetric distribution. Then $Var(X^2)=E[X^4]-E[X^2]^2$ where $E[X^2]=\sigma^2$ is the variance. If both second and forth moment are finite, am I allowed to deduce that the fourth moment is always greater than the square of the variance? Or we can't say anything about this?

Answer 1

By Cauchy-Schwarz, the following holds (whenever $E[Y^2]<\infty$). \begin{align} E[Y] = E[Y \cdot 1] &\le \sqrt{E[Y^2] E[1^2]} \\ E[Y]^2 &\le E[Y^2] \end{align}

Taking $Y=X^2$ answers your question.