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Suppose $U_1(x,y) = y - 0.5x$ and $U_2(x,y) = x - 0.5y$ where $U_i$ is the pay-off function of player $P_i$. What are all the pareto optimal solutions for $x,y \in [0,1]$?

I can't think of a way to do this holistically. Examples like $(0,1)$ and $(1,0)$ work (and can be verified), but is there a general approach to solve similar problems?

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  • $\begingroup$ You use Lagrange's method, which requires calculus. It still involves checking 1,0 and 0,1 manually though! $\endgroup$
    – RegressForward
    Commented Mar 7, 2022 at 14:34
  • $\begingroup$ @RegressForward How so? What do I maximise? $\endgroup$
    – not tdm's twin
    Commented Mar 7, 2022 at 15:28
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    $\begingroup$ Following videos might be of some help: m.youtube.com/playlist?list=PLUJGfL_499TKsujAH6aeObLCw5VvSjzAx $\endgroup$
    – Amit
    Commented Mar 7, 2022 at 15:37
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    $\begingroup$ For this particular case, (1, 0) and (0, 1) are the only solutions. In general, you would maximise one player's payoff subject to the constraint that the other player's payoff is constant. For example, maximise $y-0.5x$ subject to $x-0.5y=c$, where x and y are nonnegative. Different values of the parameter (c, and others if relevant) would give you different pareto optimal allocations. The set of all solutions corresponding to permissible values of the parameter would give you all the solutions. $\endgroup$
    – Ishan Kashyap Hazarika
    Commented Mar 8, 2022 at 3:45
  • $\begingroup$ @IshanKashyapHazarika Thanks for the method (+1). But there are more solutions. Going by your method, if we fix a particular $c$ for $U_2$, then the $U_1$ is maxxed at $x=1$ as $\max [1.5x - 2c] = 1.5 - 2c$. Similarly, for the other case. So $(x,1)$ and $(1,y)$ are all the possible solutions. $\endgroup$
    – not tdm's twin
    Commented Mar 8, 2022 at 14:43

1 Answer 1

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The linearity guarantees that interior solutions make no sense, and these are the basis of most analytical recipes. Still, one can solve this problem by thinking a bit.

You can first show that in every Pareto optimum, one has $x=1$ or $y=1$. Otherwise, one can increase both values by the same amount and that will make both agents better off.

The condition $x=1$ or $y=1$ is also sufficient for a Pareto optimum. For example, you can take the case $x=1$ and calculate that every allocation that makes $1$ better off must make $2$ worse of and vice versa. Similarly, for $x=2$.

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  • $\begingroup$ I wasn't looking for this. I do understand that if a point $(x,y)$ lies on $[0,1] \times [0,1]$ (such that neither of $x,y$ is $1$), then you choose a point that lies between the two lines and every such point corresponds to a pareto improvement. This tells that the top and right edges of $[0,1] \times [0,1]$ are the solutions. But I was looking for a more "general" answer. Like, if there is generally a way to find pareto optimal solutions. $\endgroup$
    – not tdm's twin
    Commented Mar 7, 2022 at 15:23
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    $\begingroup$ @Gang'sBigBoss Than you should edit your question and make it more general, e.g., don't use linear utility functions! $\endgroup$
    – Giskard
    Commented Mar 7, 2022 at 20:39
  • $\begingroup$ @Gang'sBigBoss, 1. Draw the indifference curves of $P_1$. 2. Draw the indifference curves of $P_2$. 3. For a given point in $[0,1]^2$ sketch the set of Pareto improvements by intersecting the corresponding better-sets of the players. 4. Find a point such that its set of Pareto improvements has empty intersection with $[0,1]^2$. 5. Find all such points. 6. Use this graphical approach to describe the set of Pareto optima algebraically. 7. Prove necessity and sufficiency. $\endgroup$
    – VARulle
    Commented Mar 10, 2022 at 9:56

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