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I think there is something wrong with the following reasoning, but I don't know where.

Assumptions:

  • Working in $\mathbb{R}_+$

  • $n \in \mathbb{N}_{\geq 3}$

  • $C \in \mathbb{N}_{\geq 2}$

  • $\varphi$ is a concave increasing function with bounded derivative ($\varphi'(x) \leq C$ for every $x \in \mathbb{R}_+$)

  • $X$ is a positive bounded random variable, with bounded mean and bounded variance.

First of all, as $\varphi$ is a concave increasing function, it satisfies the following property for every $x, y \in \mathbb{R}_+$ $$\varphi(y) \le \varphi(x) + \varphi'(x)(y-x).$$ Moreover, by Jensen's inequality, as $\varphi$ is concave, we have $$\mathbb{E}[\varphi(X)] \le \varphi(\mathbb{E}[X]) \implies \varphi(\mathbb{E}[X]) - \mathbb{E}[\varphi(X)] \geq 0. \tag{1}$$ But we also have $$ \begin{align*} \varphi(\mathbb{E}[X]) - \mathbb{E}[\varphi(X)] &= \mathbb{E}[\varphi(\mathbb{E}[X]) - \varphi(X)]\\ &\leq \mathbb{E}[\varphi'(X)(\mathbb{E}[X]-X)]\\ &\leq C \mathbb{E}[\mathbb{E}[X]-X]\\ &= C(\mathbb{E}[X]-\mathbb{E}[X])\\ &= 0\\ \implies\varphi(\mathbb{E}[X]) - \mathbb{E}[\varphi(X)] &\leq 0. \tag{2} \end{align*} $$ Equations $(1)$ and $(2)$ together yield $$\varphi(\mathbb{E}[X]) - \mathbb{E}[\varphi(X)] = 0. \tag{3}$$

So what am I doing wrong?

2 Answers 2

3

You don't (necessarily) have

$$\varphi'(X)\bigl(\mathbb{E}[X] - X\bigr) \leqslant C\bigl(\mathbb{E}[X] - X\bigr),$$

since $\mathbb{E}[X] - X$ can be negative (and except in trivial cases, it is negative on a set of positive measure). Thus the step

$$\mathbb{E}[\varphi'(X)(\mathbb{E}[X] - X)] \leqslant C\,\mathbb{E}[\mathbb{E}[X] - X]$$

is invalid.

2

$E[\phi'(X)(E[X]-X)] \leq CE[E[X]-X]$ does not follow because $E[X]-X$ may be negative.