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Let $f_1, f_2$ be two smooth strictly-quasiconcave functions. Do there always exist monotone transformations $g_1,g_2$ such that the sum $g_1\circ f_1 + g_2 \circ f_2$ is ​a strictly-​quasiconcave function?

While this question is mathematic, its motivation is economic. Given two people with strictly-convex preferences, we would like to represent their common preferences by a social welfare function. One natural way to do this is to represent each person's preferences by some strictly-quasiconcave function and take their sum. However, the sum of strictly-quasiconcave functions is not necessarily quasiconcave. So it is interesting whether we can always find specific representative functions such that the sum is quasiconcave?

I already know that, without the "strictly", the answer is no. Here is an example: enter image description here

Both functions are weakly-quasiconcave. When one function is increasing, the other is flat, and when the other is decreasing, the first is flat. Therefore, regardless of what transformations we apply to them, the sum will look like a wave and will not be quasaiconcave.

Note: I asked this question on MathOverflow some time ago. Taneli Huuskonen came up with the negative example for weakly-quasiconcave function, but the question for strictly-quasiconcave function is still open. Due to the economic importance of quasiconcave functions, I thought someone here might have an answer.

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  • $\begingroup$ This is similar in flavor to a recent paper, rasmusen.org/papers/quasi-connell-rasmusen.pdf I suspect the answer to your question is generally no $\endgroup$ – user11305 Mar 23 '18 at 22:25
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    $\begingroup$ @WaitakereCity Indeed, their paper was the first thing I tried. if $f_1$ and $f_2$ are "concavifiable" by their definition, then they are "quasiconcavifiable" by my definition. Pick $g_1$ and $g_2$ such that $g_i\circ f_i$ is concave. Then, the sum is also concave, hence the sum is quasiconcave. However, some functions are not concavifiable, so this approach does not work. My condition is weaker, so there is stil hope it can always be done. $\endgroup$ – Erel Segal-Halevi Mar 25 '18 at 4:45
  • $\begingroup$ Right, yes it is an interesting problem. I've been thinking about it a little bit. The counterexample for the weak case was nice, and it makes me question my conjecture/suspicion above. I tried approaching it in terms of upper contour sets as well, and that approach may be more fruitful. $\endgroup$ – user11305 Mar 25 '18 at 4:52
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In one dimension if the functions $f_i$ are bounded, e.g. defined on a compact interval, then yes. Take $g_1$ such that it makes $f_1$ almost constant, e.g. $g_1(x)=\frac{x}{N(\max f_1-\min f_1)}$ where $N>0$ is large. Take $g_2(x)=x$ Then $g_1\circ f_1+g_2\circ f_2\approx f_2$, which is strictly quasiconcave by assumption.

In one dimension a strictly quasiconcave function is either strictly increasing, strictly decreasing or single-peaked.

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  • $\begingroup$ Interesting. Why does this work only in one dimension? $\endgroup$ – Erel Segal-Halevi Sep 30 '18 at 6:38
  • $\begingroup$ I do not claim it only works in one dimension, just that one dimension is a sufficient condition for it working. Probably the same `flattening argument' can be used in any number of dimensions if the functions are bounded. With unbounded functions, there may be problems. $\endgroup$ – Sander Heinsalu Oct 1 '18 at 0:32

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