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I believe the following statement is true, but I can't find a proof anywhere. Does someone know if there is a reference or if it has a name?

Let $f : \mathbb{R} \to \mathbb{R}$ be convex. Let $X$ be a random variable so that $f(X)$ is defined. Let $\gamma \in [0,1]$. If $\mathbb{E}[X]=0$, then $\mathbb{E}[f(\gamma X)] \leq \mathbb{E}[f(X)]$.

Setting $\gamma=0$ is simply Jensen's inequality.

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  • $\begingroup$ @T.Bongers The case $\gamma=0$ is true and it is Jensen's Inequality: $$\mathbb{E}\big[f(0\cdot X)\big]=f(0)=f\big(\mathbb{E}[X]\big)\leq \mathbb{E}\big[f(X)\big]\,.$$ Perhaps, you didn't see the requirement that $\mathbb{E}[X]=0$ (I made that mistake too). $\endgroup$ – Batominovski Aug 27 '16 at 3:07
  • $\begingroup$ Yes, you're right. $\endgroup$ – user296602 Aug 27 '16 at 3:11
  • $\begingroup$ See Rudin RCA Chapter 3. Perhaps that would help. $\endgroup$ – Vim Aug 27 '16 at 4:54
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The result is a corollary of Jensen's inequality. If $\gamma\in[0,1]$ then by convexity $$f(\gamma X)=f(\gamma X + (1-\gamma)0)\le \gamma f(X)+(1-\gamma)f(0).$$ Taking expectations, $$ {\mathbb E}[f(\gamma X)]\le \gamma {\mathbb E}[f(X)]+(1-\gamma) f(0). $$ But $f(0)=f({\mathbb E}[X])$, and by Jensen this last is at most ${\mathbb E}[f(X)]$. Finally $1-\gamma\ge0$, so $$(1-\gamma)f(0)\le(1-\gamma){\mathbb E}[f(X)],$$ and the result is proved. (So the result is both a generalization and a corollary of Jensen...)

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  • $\begingroup$ In other words, the two are equivalent. $\endgroup$ – marty cohen Aug 27 '16 at 5:55
  • $\begingroup$ I was gonna say that. In other words, $\lambda\mapsto\Bbb E[\lambda X]$ is convex. $\endgroup$ – David C. Ullrich Aug 28 '16 at 15:00

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