Suppose that we have the following inequality:

$u( y- d) - u(y-d') \ge u(y' - d) - u( y' -d')$.

The concavity of $u$ together with $y\le y'$ then implies that $d \le d'$.

I sometimes come across this problem in economic papers, but I don't understand the logic behind this. Can anyone explain this?

  • $\begingroup$ Please consider a more concise title, e.g. Implication of the concavity of $u$ $\endgroup$ – emeryville Jan 6 at 9:54

I think you have the inequality sign wrong. The inequality as you write it cannot hold. To see this, consider the numerical case where $y=4 > y'=3, d=1 < d'=2$ and $u= \sqrt(x)$. I let you do the calculation yourself.

I suppose the correct inequality is: \begin{equation} u(y−d)−u(y−d') \leq u(y'−d)−u(y′−d′) \end{equation}

This equation holds true due to the curvature of a concave function. If you're on the upward sloping part of the function, the closer you are to the summit, the flatter the slope. So the vertical distance between the images of the two points in the domain on the left side of the equation is smaller than the same on the right side if $y>y'$ and $d \leq d'$, and equal if $y=y'$. On the downward sloping part it's the other way around. The set of inequality conditions imply each other. Changing one inequality requires changing the others. You can check this for yourself by picking points on a concave curve that satisfy the inequalities.

It is really just an elaborate version of $|x| \leq |y|$, where if $x \leq 0, y \leq x$ and if $x \geq 0, y \geq x$. I hope this answers your question.

  • $\begingroup$ Thanks. I realized that I wrote the inequality of $y$ wrong. I edited it. $\endgroup$ – Daeseon Jan 6 at 21:21

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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