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Given function $h_1(s)\ge0, h_1'(s)\ge0$, $h_2(s)\ge0, h_2'(s)\ge0$ for $s\in[0,\bar s]$ and $g(s)$ is a density function on $[0,\bar s]$. I already shown that $$ \int_0^{\bar s}h_1(s)g(s)ds\ge \int_0^{\bar s}h_2(s)g(s)ds $$ and $$ \int_0^{\bar s}h_1(s)^\sigma g(s)ds\ge \int_0^{\bar s}h_2(s)^\sigma g(s)ds $$ for $\sigma\ge1$. It seems the inequality is also valid for $\sigma\in[0,1]$ (I also used numerical values to plot the integrals). But I am not able to prove it formally. Any hints?

Does it help if the following condition is satisfied: there is a unique $s^*\in[0,\bar s]$ such that $$ h_1(s)>(<)h_2(s), for\; s>(<)s^*. $$

Glenn
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  • The result is doubtful in full generality. Which $(g,h_1,h_2)$ did you check it on? – Did Jul 28 '16 at 09:27
  • @Did, Actually, my belief of generality comes from the original question with specific and complicated $h_1, h_2$ funcitions. It seems obvious that the conjecture should be true in that setup. But you are right that in general it is doubtful. Could you give some hints of possible additional conditions that can prove the conjecture? Thanks anyway. – Glenn Jul 28 '16 at 09:40
  • An obvious sufficient condition would be $h_1(s)\geq h_2(s)$ for all $s\in[0,\bar{s}]$. By the way, how did you show the first inequality? (What where the conditions on $h_1$ and $h_2$. Or is the first inequality an assumption?) – smcc Jul 28 '16 at 12:10
  • @smcc, The first inequality is not an assumption and it is proved. As said above, the inequalities are established under specific and complicated funtional form of $h_1, h_2$ and $g$ which are all functions of a distribution function $F(s)$ and $s$. Do you think it will help if the following condition is satisfied: there is a unique $s^\in[0,\bar s]$ such that $ h_1(s)>(<)h_2(s), for; s>(<)s^$. – Glenn Jul 28 '16 at 12:20
  • I don't understand the votes to close. The question is very clear to me. The inequality has been proven for $\sigma \ge 1$ and OP wants hints towards proving whether it's also true or not for $0 \le \sigma \le 1$. What's unclear? –  Jul 28 '16 at 12:56

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