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Let $N\subset\mathbb{R}$ be finite and define $$ A(N) = \sum_{i \in\mathbb{Z} }\min\{ 2 ^i, |N\cap[2^i,2^{i+1})| \}, $$ where $\mathbb{Z}=\{0,\pm1,\pm2,\ldots\}$ and $|\cdot|$ denotes set cardinality.

If $X\sim\mathrm{uniform}(N)$ is the corresponding random variable, then $||X||_A:=A(N)$ acts as a quasi-norm of sorts. It appears to be homogeneous and nearly subadditive, in the sense that $ ||X+Y||_A \le c(||X||_A+||Y||_A)$ for some universal $c>0$.

Question: Can $||X||_A$ be related to any classic norm of $X$? Probably not an $L_p$ norm, but perhaps an Orlicz norm? I'm mostly interested in upper-bounding $||X||_A$ by a classic random variable norm.

Update: What I said about homogeneity is incorrect: it appears to be "nearly" homogeneous in the sense that $ ||\alpha X||_A \le c\alpha|| X||_A$ for some universal $c>0$.

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    $\begingroup$ Are you sure that what you wrote for negative $i$ in the definition of $A(N)$ is what you really meant? $\endgroup$
    – fedja
    Commented Nov 24, 2023 at 14:29
  • $\begingroup$ Hi @fedja yes I think so... What do you think a more logical definition for $A(N)$ would be? I actually don't mind just summing over the non-negative $i$ but then you lose homogeneity. $\endgroup$
    – Aryeh Kontorovich
    Commented Nov 25, 2023 at 16:37
  • $\begingroup$ So the negative part of $N$ is just ignored in the definition of $A(N)$ but not ignored in the definition of $X$? Or is $N$ contained in $(0,+\infty)$? $\endgroup$
    – fedja
    Commented Nov 25, 2023 at 19:51

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