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Edited: Just realised my first post was somewhat misleading and not precise. Thanks to the two commetators that pointed it out.

I am working on an article and ended up wondering for which values of $\gamma>0$ does the following inequality hold:

$$\frac{\sum_iN_ic_{i,t}^\gamma}{\left(\sum_iN_ic_{i,t}\right)^\gamma}<\frac{\sum_iN_ic_{i,0}^\gamma}{\left(\sum_iN_ic_{i,0}\right)^\gamma}$$

For $N_i \in [0,1]$ and $\sum_i N_i=1$. This can be rewritten as:

$$\frac{\mathbb{E}_i \left[ c_{i,t}^\gamma\right]}{\mathbb{E}_i \left[c_{i,t}\right]^\gamma}<\frac{\mathbb{E}_i \left[ c_{i,0}^\gamma\right]}{\mathbb{E}_i \left[c_{i,0}\right]^\gamma}$$

Where $c_{i,t}\geq c_{i,0}$ for every $i$. Of course I already noticed that the expresion holds with equality when $\gamma =1$. Somewhat inspired in Jensen´s inequality, my intuition is that it should hold for $\gamma \in (0,1)$ but I haven´t been able to prove it. Any suggestions?

Note: Jensen´s inequality states that $\mathbb{E}\left( \varphi(x) \right) > \varphi \left(\mathbb{E}x \right)$ if $\varphi(x)$ is a convex function.

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  • $\begingroup$ If $c_0$ is a known value why do you even bother with expectations on the right hand side? $\endgroup$ – hejseb Jun 9 '16 at 18:08
  • $\begingroup$ hejseb, you were right. Just edited my post so it is not trivial, I was confused about the expectation. $\endgroup$ – user344188 Jun 9 '16 at 19:37
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For $0<\gamma <1$, $x^{\gamma}$ is a concave function. Then, by Jensen's inequality \begin{align*} \frac{E(c_t^{\gamma})}{(E(c_t))^{\gamma}} &= E\left(\left(\frac{c_t}{E(c_t)} \right)^{\gamma} \right)\\ &< \left(E\left(\frac{c_t}{E(c_t)}\right) \right)^{\gamma}\\ &=1. \end{align*}

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  • $\begingroup$ That works, but I would actually expect $\mathbb{E}(c_t)>c_0$. As it is consumption, which grows over time (sorry for not providing enough context). $\endgroup$ – user344188 Jun 9 '16 at 19:02
  • $\begingroup$ @user344188: I revised. In fact, we do not have to assume that $E(c_t)=c_0$. Note that the right hand side, in your inequality, equals to 1, as $c_0$ is a known quantity. $\endgroup$ – Gordon Jun 9 '16 at 19:10
  • $\begingroup$ You are right. And your answer is in fact correct. But I was confused, $c_0$ shouldn´t be deterministic. Just edited my post in order to provide enough context and correct my mistake. Thanks. If you can check it out that would be great. $\endgroup$ – user344188 Jun 9 '16 at 19:39

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