How can I show that the fractional moments of the (unit variance) Laplacian distribution are higher than of the standard normal distribution, for moments higher than 2?

Formally, if $l \sim Laplace(0,1/\sqrt{2})$ and $g \sim Normal(0,1)$, show that, for $p\in \Re^+$,

$E[|l|^p] > E[|g|^p] \iff p > 2$.

  • 2
    $\begingroup$ Write down the integrals and make a variable substitution $|t|=x^2$ for the integral corresponding to Laplacian moments. $\endgroup$ – mpiktas Aug 17 '16 at 12:23
  • $\begingroup$ Their ratio can be computed in closed form: it is $1/\Gamma(p/2+1)$. $\endgroup$ – whuber Aug 17 '16 at 13:14

Thanks for the help in the comments, here is a solution:

$E[|l|^p] = \int_{-\infty}^{\infty} |t|^p \frac{\sqrt{2}}{2} e^{-\sqrt{2}|t|} dt$

Do a variable substitution $|t| = \frac{x^2}{2 \sqrt{2}}$:

$E[|l|^p] = \frac{\sqrt{\pi}}{2^{3p+1}} \int_{-\infty}^{\infty} \frac{|x|^{2p+1}}{\sqrt{2 \pi}} e^{-x^2/2} dx = \frac{\sqrt{\pi}}{2^{3p+1}} E[|g|^{2p+1}$]

We use the expression for the absolute moment for the normal, $E[|g|^p] = 2^{p/2} \frac{\Gamma (\frac{p+1}{2})}{\sqrt{\pi}}$, to calculate the ratio between the Laplace and normal moments:

$\frac{E[|l|^p]}{E[|g|^p]} = \frac{\sqrt{\pi}}{2^{3p+1}} \frac{E[|g|^{2p+1}]}{E[|g|^p]} = \frac{\sqrt{\pi}}{2^{p}} \frac{\Gamma(p+1)}{\Gamma(\frac{p+1}{2})} = \Gamma(\frac{p}{2} + 1)$.

Since $\Gamma(x) \leq 1$ for $1 \leq x \leq 2$ and $\Gamma(x) > 1$ for $x > 2$, we have, for $p \in \Re^+$:

$\frac{E[|l|^p]}{E[|g|^p]} > 1 \iff p > 2$.

  • $\begingroup$ Your double arrow is incorrect. When $p$ exceeds $2$, the ratio exceeds $1$: but the ratio also exceeds $1$ for other $p$ as well. $\endgroup$ – whuber Aug 17 '16 at 17:37
  • $\begingroup$ That's right, the ratio would also exceed 1 for negative p, but it in this problem I was only considering $p \ge 0$. $\endgroup$ – Carlos Stein Aug 17 '16 at 17:50

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