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I am working on a problem where I have continuous probability distributions $p$ over a bounded domain $D$, i.e., $\forall x$ $p(x) \geq 0$ and $\|p\|_1 = \int_D p(x) dx = 1$. However, I also want $p$ to have bounded function 2-norm, i.e., $\|p\|_2 = \sqrt{\int_D p(x)^2 dx} \leq c$ for some finite $c$. Is there a name for such a class of distributions? Edit: Assuming $p \in L^2$, can one bound $\|p\|_2$ with some other quantities involving $p$? I provide some examples in the following paragraph.

The trivial bound is $\|p\|_2 \leq \|p\|_\infty {\rm Vol}(D)$. I am looking for something better, if possible. It seems that exponential distributions with large-enough spread have low 2-norm. Also, for uniform distributions one can easily get $\|p\|_2$ as a function of the volume of the domain $D$. I am looking for any resources / references along this direction. Please help.

Edit: As pointed out in the comments, I looked at the Plancherel Identity, and realize that $p \in L^2$ iff the characteristic function of the random variable $X$ having density $p$ is in $L^2$. However, I am looking for a simpler bound on $\|p\|_2$, or a condition on $p$ that is simpler than assuming the characteristic function is square integrable.

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    $\begingroup$ The distribution function $p$ is in $L^2$ if and only if the associated characteristic function is in $L^2$, this follows from Plancherel identity. See here, here or here for details. $\endgroup$ Commented Sep 8, 2022 at 5:16
  • $\begingroup$ Thanks, I looked at Plancherel @StratosFair. I am looking for a simpler bound on $\|p\|_2$, or a simpler condition on $p$ than assuming the characteristic function is in $L^2$. Edited question to reflect the same. $\endgroup$
    – Ambar
    Commented Sep 8, 2022 at 16:49

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