1
$\begingroup$

Let function $F(x,y,z)$ defined on $[0,1]\times[0,1]\times[0,1]$ be increasing in $x$ and $y$. By increasing I mean $\frac{\partial F}{\partial x}\geq0$ and $\frac{\partial F}{\partial y}\geq0$.

By quasi-concave I mean $$F(x',y',z)\geq F(x,y,z)\Rightarrow[\frac{\partial F}{\partial x},\ \frac{\partial F}{\partial y}][x'-x,\ y'-y ]^T\geq0.$$

Define $$H(x,y)=\min\limits_{z\in[x,1]}F(x,y,z).$$

My question: Is function $H(x,y)$ quasi-concave in $x$ and $y$? Why?

EDIT 1: I think $H(x,y)$ is still increasing in $x$ and $y$, because for any $x'\geq x$ and $y'\geq y$, we have $\min\limits_{z\in[x',1]}F(x',y,z)\geq \min\limits_{z\in[x,1]}F(x,y,z)$ and $\min\limits_{z\in[x,1]}F(x,y',z)\geq \min\limits_{z\in[x,1]}F(x,y,z)$. Is it right?

EDIT 2: Correction of the quasi-concavity.

$\endgroup$
5
  • $\begingroup$ Well. If $F$ is increasing I can also say it's quasi convex right? I believe by increasing you mean increasing in any variable if others are fixed. For e.g. increasing in $z$ if $x$ and $y$ are fixed. $\endgroup$ Jun 25, 2020 at 11:52
  • $\begingroup$ @Shiv Tavker Yes, but I'm confused about the minimized function. I want to see if function H is quasi-concave or not... $\endgroup$
    – HXW
    Jun 25, 2020 at 12:08
  • $\begingroup$ Isn't the argmin just $z^* = x$, given the way you are setting up the problem? $\endgroup$
    – user762914
    Jun 25, 2020 at 12:26
  • $\begingroup$ @Renard But function F is not necessarily increasing in z... $\endgroup$
    – HXW
    Jun 25, 2020 at 13:26
  • $\begingroup$ @Huaixin Now the First inequality in Edit 1 may not be correct. $\endgroup$ Jun 25, 2020 at 14:07

2 Answers 2

0
$\begingroup$

An increasing function is only necessarily quasiconcave if it is a map from $\mathbb{R}$ to $\mathbb{R}$. The function $x^2+y^2+z^2$ is strictly convex on $[0,1]^3$, and increasing for any meaningful definition of increasing.

$\endgroup$
2
  • $\begingroup$ True. I guess calling quasi convex is safer? $\endgroup$ Jun 25, 2020 at 12:31
  • $\begingroup$ Thanks! I corrected my formulation. $\endgroup$
    – HXW
    Jun 25, 2020 at 13:52
0
$\begingroup$

Since, $F(x, y, z)$ is increasing, $$\ H(x, y) = \min_{z\in[x, 1]}F(x, y, z) = F(x, y, x) $$ Can you finish it now?

$\endgroup$
3
  • $\begingroup$ F is increasing in x and y, but it is not necessarily increasing in z. $\endgroup$
    – HXW
    Jun 25, 2020 at 13:23
  • $\begingroup$ Oh you edited the question! $\endgroup$ Jun 25, 2020 at 14:03
  • $\begingroup$ I just corrected the formulation, sorry... $\endgroup$
    – HXW
    Jun 25, 2020 at 14:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.