Oct 10 JDN 2459498
When this post goes live, I will be married. We already had a long engagement, but it was made even longer by the pandemic: We originally planned to be married in October 2020, but then rescheduled for October 2021. Back then, we naively thought that the pandemic would be under control by now and we could have a wedding without COVID testing and masks. As it turns out, all we really accomplished was having a wedding where everyone is vaccinated—and the venue still required testing and masks. Still, it should at least be safer than it was last year, because everyone is vaccinated.
Since marriage is on my mind, I thought I would at least say a few things about the behavioral economics of marriage.
Now when I say the “economics of marriage” you likely have in mind things like tax laws that advantage (or disadvantage) marriage at different incomes, or the efficiency gains from living together that allow you to save money relative to each having your own place. That isn’t what I’m interested in.
What I want to talk about today is something a bit less economic, but more directly about marriage: the matching process by which one finds a spouse.
Economists would refer to marriage as a matching market. Unlike a conventional market where you can buy and sell arbitrary quantities, marriage is (usually; polygamy notwithstanding) a one-to-one arrangement. And unlike even the job market (which is also a one-to-one matching market), marriage usually doesn’t involve direct monetary payments (though in cultures with dowries it arguably does).
The usual model of a matching market has two separate pools: Employers and employees, for example. Typical heteronormative analyses of marriage have done likewise, separating men and women into different pools. But it turns out that sometimes men marry men and women marry women.
So what happens to our matching theory if we allow the pools to overlap?
I think the most sensible way to do it, actually, is to have only one pool: people who want to get married. Then, the way we capture the fact that most—but not all—men only want to marry women, and most—but not all—women only want to marry men is through the utililty function: Heterosexuals are simply those for whom a same-sex match would have very low utility. This would actually mean modeling marriage as a form of the stable roommates problem. (Oh my god, they were roommates!)
The stable roommates problem actually turns out to be harder than the conventional (heteronormative) stable marriage problem; in fact, while the hetero marriage problem (as I’ll henceforth call it) guarantees at least one stable matching for any preference ordering, the queer marriage problem can fail to have any stable solutions. While the hetero marriage problem ensures that everyone will eventually be matched to someone (if the number of men is equal to the number of women), sadly, the queer marriage problem can result in some people being forever rejected and forever alone. (There. Now you can blame the gays for ruining something: We ruined marriage matching.)
The queer marriage problem is actually more general than the hetero marriage problem: The hetero marriage problem is just the queer marriage problem with a particular utility function that assigns everyone strictly gendered preferences.
The best known algorithm for the queer marriage problem is an extension of the standard Gale-Shapley algorithm for the hetero marriage problem, with the same O(n^2) complexity in theory but a considerably more complicated implementation in practice. Honestly, while I can clearly grok the standard algorithm well enough to explain it to someone, I’m not sure I completely follow this one.
Then again, maybe preference orderings aren’t such a great approach after all. There has been a movement in economics toward what is called ordinal utility, where we speak only of preference orderings: You can like A more than B, but there’s no way to say how much more. But I for one am much more inclined toward cardinal utility, where differences have magnitudes: I like Coke more than Pepsi, and I like getting massaged more than being stabbed—and the difference between Coke and Pepsi is a lot smaller than the difference between getting massaged and being stabbed. (Many economists make much of the notion that even cardinal utility is “equivalent up to an affine transformation”, but I’ve got some news for you: So are temperature and time. All you are really doing by making an “affine transformation” is assigning a starting point and a unit of measurement. Temperature has a sensible absolute zero to use as a starting point, you say? Well, so does utility—not existing. )
With cardinal utility, I can offer you a very simple naive algorithm for finding an optimal match: Just try out every possible set of matchings and pick the one that has the highest total utility.
There are up to n!/((n/2)! 2^n) possible matchings to check, so this could take a long time—but it should work. I’m sure there’s a more efficient algorithm out there, but I don’t have the mental energy to figure it out at the moment. It might still be NP-hard, but I doubt it’s that hard.
Moreover, even once we find a utility-maximizing matching, that doesn’t guarantee a stable matching: Some people might still prefer to change even if it would end up reducing total utility.
Here’s a simple set of preferences for which that becomes an issue. In this table, the row is the person making the evaluation, and the columns are how much utility they assign to a match with each person. The total utility of a match is just the sum of utility from the two partners. The utility of “matching with yourself” is the utility of not being matched at all.
A | B | C | D | |
A | 0 | 3 | 2 | 1 |
B | 2 | 0 | 3 | 1 |
C | 3 | 2 | 0 | 1 |
D | 3 | 2 | 1 | 0 |
Since everyone prefers every other person to not being matched at all (likely not true in real life!), the optimal matchings will always match everyone with someone. Thus, there are actually only 3 matchings to compare:
AB, CD: (3+2)+(1+1) = 7
AC, BD: (2+3)+(1+2) = 8
AD, BC: (1+3)+(3+2) = 9
The optimal matching, in utilitarian terms, is to match A with D and B with C. This yields total utility of 9.
But that’s not stable, because A prefers C over D, and C prefers A over B. So A and C would choose to pair up instead.
In fact, this set of preferences yields no stable matching at all. For anyone who is partnered with D, another member will rate them highest, and D’s partner will prefer that person over D (because D is everyone’s last choice).
There is always a nonempty set of utility-maximizing matchings. (There must be at least one, and could in principle have as many as there are possible matchings.) This actually just follows from the well-ordering property of the real numbers: Any finite set of reals has a maximum.
As this counterexample shows, there isn’t always a stable matching.
So here are a couple of interesting theoretical questions that this gives rise to:
1. If there is a stable matching, must it be in the set of utility-maximizing matchings?
2. If there is a stable matching, must all utility-maximizing matchings be stable?
Question 1 asks whether being stable implies being utility-maximizing.
Question 2 asks whether being utility-maximizing implies being stable—conditional on there being at least one stable possibility.
So, what is the answer to these questions? I don’t know! I’m actually not sure anyone does! We may have stumbled onto cutting-edge research!
I found a paper showing that these properties do not hold when you are doing the hetero marriage problem and you use multiplicative utility for matchings, but this is the queer marriage problem, and moreover I think multiplicative utility is the wrong approach. It doesn’t make sense to me to say that a marriage where one person is extremely happy and the other is indifferent to leaving is equivalent to a marriage where both partners are indifferent to leaving, but that’s what you’d get if you multiply 1*0 = 0. And if you allow negative utility from matchings (i.e. some people would prefer to remain single than to be in a particular match—which seems sensible enough, right?), since -1*-1 = 1, multiplicative utility yields the incredibly perverse result that two people who despise each other constitute a great match. Additive utility solves both problems: 1+0 = 1 and -1+-1 = -2, so, as we would hope, like + indifferent = like, and hate + hate = even more hate.
There is something to be said for the idea that two people who kind of like each other is better than one person ecstatic and the other miserable, but (1) that’s actually debatable, isn’t it? And (2) I think that would be better captured by somehow penalizing inequality in matches, not by using multiplicative utility.
Of course, I haven’t done a really thorough literature search, so other papers may exist. Nor have I spent a lot of time just trying to puzzle through this problem myself. Perhaps I should; this is sort of my job, after all. But even if I had the spare energy to invest heavily in research at the moment (which I sadly do not), I’ve been warned many times that pure theory papers are hard to publish, and I have enough trouble getting published as it is… so perhaps not.
My intuition is telling me that 2 is probably true but 1 is probably false. That is, I would guess that the set of stable matchings, when it’s not empty, is actually larger than the set of utility-maximizing matchings.
I think where I’m getting that intuition is from the properties of Pareto-efficient allocations: Any utility-maximizing allocation is necessarily Pareto-efficient, but many Pareto-efficient allocations are not utility-maximizing. A stable matching is sort of a strengthening of the notion of a Pareto-efficient allocation (though the problem of finding a Pareto-efficient matching for the general queer marriage problem has been solved).
But it is interesting to note that while a Pareto-efficient allocation must exist (typically there are many, but there must be at least one, because it’s impossible to have a cycle of Pareto improvements as long as preferences are transitive), it’s entirely possible to have no stable matchings at all.