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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.

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1 Answer 1

We can assume without loss of generality that $\mathbb{E}[X]=1$. Define now an increasing and unbounded sequence $\{x_n\}_{n\in\mathbb{N}}$ through: $$ x_n = \inf\left\{x\in\mathbb{R}^+:\mathbb{P}[X>x]\leq\frac{1}{4^n}\right\}$$ (so $x_0=0$) and $$ i(x) = \max\left\{n\in\mathbb{N}:x_n\leq x\right\}.$$ You can take $$ \phi(x) = x\cdot\frac{2^{i(x)+1}(x-x_{i(x)})+2^{i(x)}(x_{i(x)+1}-x)}{x_{i(x)+1}-x_{i(x)}}$$ as a non-negative continous, increasing convex function and easily prove that with such choices $$ \mathbb{P}[\phi(X)>t]\leq 4\cdot\mathbb{P}[X>t],$$ so $\mathbb{E}[\phi(X)]$ is necessarily finite.

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