Suppose $u(\cdot)$ and $v(\cdot)$ are two differentiable, strictly increasing, and strictly concave real functions. Specifically, $v(\cdot)$ is "more concave" than $u(\cdot)$ in the sense that there exists an increasing and strictly concave function $\phi(\cdot)$ such that $v(x)=\phi(u(x))$ at all $x$. It is also equivalent to \begin{equation} \frac{v''(x)}{v'(x)}<\frac{u''(x)}{u'(x)} \textrm{ for any }x\,. \end{equation}

Let $p_i\in(0,1), \sum_{i\in I}p_i=1$ be probabilities and $|I|>2$. Let $x_i$ and $y_i$ be strictly positive for all $i\in I$. Assume \begin{equation} \sum_{i\in I}p_ix_i<\sum_{i\in I}p_iy_i, \end{equation} and \begin{equation} \sum_{i\in I}p_iu(x_i)=\sum_{i\in I}p_iu(y_i). \end{equation}

Conjecture: \begin{equation} \sum_{i\in I}p_iv(x_i)>\sum_{i\in I}p_iv(y_i). \end{equation} I believe this is right (after trying many numerical examples) and I think a clever use of Jensen's inequality (or its variants) will do this. But I'm stuck on doing it formally. Any hints/thoughts on providing a formal proof?

Remark: this is related to my other post: Proving an inequality of the expectation of concave functions?

Update: after some more attempts, I believe some techniques in convex analysis would be helpful. Geometrically, the middle equation represents a hyperplane in the $R^{|I|}$ space, and the desired result (very roughly) says that a concave transformation of that hyperplane should be separated from a convex transformation of it.

To be clear, I wasn't saying the conjecture should be generally true. Any thoughts on finding any sufficient conditions to make it work would be very helpful.

  • $\begingroup$ Should it really be $|I| > 2$? $\endgroup$ – fourierwho Dec 27 '17 at 3:14
  • $\begingroup$ Yes, I've already managed to prove the case of $|I|=2$, and I believe it's true for $|I|>2$ as well. $\endgroup$ – Tuzi Dec 27 '17 at 3:29
  • $\begingroup$ Then it's induction. $\endgroup$ – fourierwho Dec 27 '17 at 3:30
  • $\begingroup$ Could you be more specific? I don't think a usual induction argument works here because starting for $|I|=3$ one couldn't write the inequalities in a way that the $|I|=2$ case can handle, due to the non-linearity of $u(\cdot)$ and $v(\cdot)$. $\endgroup$ – Tuzi Dec 27 '17 at 3:38
  • $\begingroup$ The cardinality of $I$ is finite, infinite countable or uncountable? $\endgroup$ – induction601 Dec 27 '17 at 5:45

The reverse inequality holds. A differentiable function of one variable is concave on an interval $J$ if and only if the function lies below all of its tangent, i.e., $$f(x) \leq f(y) + f'(y)(x-y)$$ for all $x,y\in J$. In particular, if $f′(c) =0$, then $c$ is a global maximum of $f(x)$.

Applying this for $v$, we have $$p_i(v(x_i) - v(y_i)) \le p_i v'(y_i)(x_i-y_i)$$ for all $p_i\in(0,1)$ and all $i\in I$.

Summing up over $i$ \begin{align} \sum_{i\in I}p_i(v(x_i) - v(y_i)) &\le \sum_{i\in I} p_i v'(y_i)(x_i-y_i) \\ &\le \sum_{i\in I} p_i |v'(y_i)|(x_i-y_i) \qquad \text{since} \,\, w\le |w|, \forall w \in \mathbb{R} \\ &\le \min_i |v'(y_i)|\sum_{i\in I} p_i (x_i-y_i) <0. \end{align} The second inequality holds since $v$ is increasing so that $|v'|=v'$ iff $v'>0$.

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    $\begingroup$ Could you elaborate more on the second inequality in the last part? if $x_i-y_i < 0$, we can't conclude that $w(x_i-y_i) \le |w|(x_i-y_i)$. $\endgroup$ – induction601 Dec 27 '17 at 16:47
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    $\begingroup$ Sorry but I'm not sure if the second inequality in your proposed answer is correct, as the above comment also suggested. More generally, as I stated in the conditions of my initial question, we can surely have $\sum_{i}p_iu(x_i)=\sum_{i}p_iu(y_i)$ for a concave function $u(\cdot)$. This means one can't generally prove ``the reserve inequality holds.'' $\endgroup$ – Tuzi Dec 27 '17 at 17:03
  • $\begingroup$ @induction601 Right, the second inequality holds in one and only one case $v'>0$ which is already happen since $v$ is increasing i,e, $v'>0$ and this implies that $|v'|\ge v'\ge0$ $\endgroup$ – Mohammad W. Alomari Dec 27 '17 at 18:27
  • $\begingroup$ @mwomath But still the second inequality in your proposed answer is not true because $x_i-y_i$ is negative for some $i$'s... $\endgroup$ – Tuzi Dec 28 '17 at 1:53
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    $\begingroup$ @mwomath the conjecture may be right or wrong but I still didn't see your answer to be correct. As I said and other people also pointed out the second inequality in your proposed answer is clearly not correct. Could you provide a different argument to support your claim. Thanks. $\endgroup$ – Tuzi Dec 28 '17 at 3:07

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