Assume $X = \mbox{random variable} X>0$, $\mathbb{E}X<\infty \implies \exists \phi:\mathbb R_+\to\mathbb R_+$, superlinar, convex with $\mathbb{E}\phi(X)<\infty$
superlinear means $\lim\frac{\phi(x)}{x}=\infty, x\to\infty$
I have a proof for this statement but I would like to know if there exists also a geometric/heuristic argument for this statement.