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Associated with a convex function $\phi:[0,\infty)\mapsto[0,\infty)$ satisfying $\lim_{x\to 0} \frac{\phi(x)}{x} = 0, \lim_{x\to\infty}\frac{\phi(x)}{x} = \infty,$ the Orlicz norm of a random variable $X$ is defined as:

$$||X||_\phi:=\inf\left\{c>0: \mathbb{E}\left[\phi\left(\frac{|X|}{c}\right)\right]\leq 1\right\}~.$$

Let $Z,Y$ be independent random variables taking values in Polish spaces $\mathcal{Z}, \mathcal{Y}$ respectively, and let $f:\mathcal{Z}\times\mathcal{Y}\mapsto\mathbb{R}$ be any measurable function such that $||f(Z,Y)||_\phi<\infty$. Define

$$g(y):= ||f(Z,y)||_\phi,\ \forall\ y\in\mathcal{Y}~.$$

Conjecture: $$||g(Y)||_\phi\leq ||f(Z,Y)||_\phi~.$$

The above holds with equality when $\phi(x) = x^p$ for $p>1,$ in which case the Orlicz norm is simply the $L_p$ norm.

Is the conjecture true for any Orlicz norm? I tried out other choices of $\phi$ and $Z,Y,$ and found the inequality to hold in all cases.

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  • $\begingroup$ I think there are absolute values $|.|$ missing somewhere, and/or some condition that $X$, $Y$ and $f$ are nonnegative. (And you didn't define $\mathcal{X}$ and $\mathcal{Y}$ ). $\endgroup$
    – Jean Duchon
    Commented Aug 27, 2015 at 7:55
  • $\begingroup$ Maybe replace $X$ with $Z$ in the definition of the Orlicz norm, as $X$ is real there ? $\endgroup$
    – Jean Duchon
    Commented Aug 27, 2015 at 11:48
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    $\begingroup$ Thanks. Done. Actually, the conjecture is false, a counterexample was provided to me by Ramon van Handel (princeton.edu/~rvan) as follows: Let $Y,Z\sim\mathrm{Bernoulli}(1/2)$ taking values in $\{0,1\}.$ Let $\phi(x) = e^x-1,$ and let $f(z,y) = zy.$ In this case, $||g(Y)||_\phi = \frac{1}{(\log 3)^2}>||f(Z,Y)||_\phi = \frac{1}{\log 5}.$ $\endgroup$
    – Hedonist
    Commented Aug 27, 2015 at 14:53
  • $\begingroup$ But this $\phi$ doesn't satisfy $\lim_{x\to 0} \frac{\phi(x)}{x} = 0$, if it is of any relevance... $\endgroup$
    – Jean Duchon
    Commented Aug 27, 2015 at 15:33
  • $\begingroup$ Thanks for pointing that out Jean. I guess I don't quite need that condition (I am new to this field, and am not sure why that condition is used sometimes to define Orlicz norms, but not at other times). $\endgroup$
    – Hedonist
    Commented Aug 27, 2015 at 18:37

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