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I got one question about the Nash-SWF. Typically it is defined as the product of individual utilities, ie. $$ NSWF:=u_1(x_1) \cdot u_2(x_2) \cdot u_3(x_3) \cdot ... $$ For this to make sense, individual utilities are restricted to always being positive. Is there a way to adjust the Nash-SWF to work for utility fcts that are always negative, like $-e^{-ax}$? Meaning all individuals have the same utility fct. which is $-e^{-ax}$.

Thanks a lot!

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  • $\begingroup$ As you said, it makes sense that the utility functions are defined positive. You have to justify, why it is make sense to you, that you can assume negative individual utilities. But you will have another problem. If the number of individuals are odd, then you will always have a negative NSWF. If the number of individuals are even, then you will always have a positive NSWF. $\endgroup$
    – callculus42
    Oct 6, 2015 at 12:09
  • $\begingroup$ There could be negaitve utility functions but in most cases with this kind of function, the marginal utility is positive and decreasing. For exemple ramsey model without discount rate is an exemple. you can find and example by this link : uhero.hawaii.edu/assets/WP_2013-9.pdf (by the way, the paper is published in Resource and Energy Economics, which is a high ranked journal.) $\endgroup$
    – optimal control
    Oct 6, 2015 at 12:24
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    $\begingroup$ I disagree with @calculus: I don't see any particular reason why utility should be positive. In macroeconomics, most utility functions are negative-valued values for a set of (attainable) points. Just think of $log(c)$ $\endgroup$
    – FooBar
    Oct 6, 2015 at 12:29
  • $\begingroup$ @FooBar You are right, it is often used. But the problem of changing sign is still there. $\endgroup$
    – callculus42
    Oct 6, 2015 at 12:46
  • $\begingroup$ @calculus which makes the question (on how to solve the problem) all the more interesting! :) $\endgroup$
    – FooBar
    Oct 6, 2015 at 12:50

1 Answer 1

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I completely agree with @FooBar. Of course the utility function can be negative in some of its range. Conceptually, one could invoke the cardinality vs. the ordinality argument as a response to your question. I would think that your problem is as follows:

$$ \ max\, SWF=u(x_{1})\times u(x_{2})\times...\times u(x_{N}) \ s.t \ f(x_{1}....x_{N})\leq g \ $$

The constraint(s) can be some resource constraint. All that matters here is that the SWF satisfies the normal properties associated with the existence of a maximum on the constraint set (You can invoke the assumptions of the Weierstrass Theorem here). Moreover, you should check Second Order Sufficient Conditions (in this case the negative definiteness of the Bordered Hessian) for a maximum. The value of the welfare function by itself is irrelevant as long as it is a maximum on the constraint set.

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  • $\begingroup$ Ok, so some more details: I got a mass X of individuals with utility -exp(-af(p)), a mass Y of individuals with utility -exp(-ag(p)) and a mass Z of individuals with utility -exp(-ah(p)). Now i want to do the Nash SWF and maximize wrt p. But as already said, maximizing the NSWF only makes sense if all individual utilities are always positive, which is not the case here, they are all always negative. There are no constraints. "a" is a risk constant, h(), g(), f() are functions. $\endgroup$
    – Vincent
    Oct 6, 2015 at 14:18
  • $\begingroup$ Unless the utility function has bliss point, the lack of constraint is problematic.. If your individual utilities were always positive, then you would set each utility as being infinite and if your utilities were always negative, you would set them at 0. Here, the latter is not possible because for the utility function to attain 0, the object being exponentiated will have to be negative infinity. If you had a constraint, the problem could be both solved and interesting, independent of the range of the utility function $\endgroup$
    – ChinG
    Oct 6, 2015 at 14:31
  • $\begingroup$ Cannot set each utility to infinite (or zero). Each utility depends on p, but while one may rise in p the other may fall in p. They are also non-linear in p, so might first increase for small p and then fall for large p or the other way round. For completeness, there actually is another constraint, which is 0<p<1, but i think it's not needed/interesting here. $\endgroup$
    – Vincent
    Oct 6, 2015 at 14:35
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    $\begingroup$ I think this answer is actually a very good hint and should point you in right direction. Sit down and read what was written and think carefully, I think the constraint you just mentioned is going to be important. $\endgroup$
    – cc7768
    Oct 6, 2015 at 15:55

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