>>: Hi everyone. It's good to have Yash back here. He'll give his talk.
He's at New England now and headed to Columbia in a few months, I believe.
>> Yashodhan Kanoria:
>>:
Yes.
And he'll tell us something about matchings, yeah.
>> Yashodhan Kanoria: Thanks Malek. Thank you for coming. So this is joint
work with Itai Ashlagi and Jacob D. Leshno. So we're going to talk about large
random matching markets, and tied to matching market, two-sided matching
markets is what we're going to consider, and we're going to look at, try to
understand sort of which side of the market is able to exercise its choice and
which side is not able to exercise its choice.
So what determines this? Okay. So here's a setting. So we will call the two
sides of the market men and women. So this column M is a set of men, column W
is a set of women. And each agent has strict preferences over the other side
of the market.
Okay. We'll denote matchings by mu. And we will call the matching stable if
there is no man/woman pair who both prefer each other over their partner in the
match. Okay. So such a pair would be sort of a source of instability.
If there is no such pair, then we call the matching stable. Now, the set of
stable matching, it has a nice structure. In fact, it's a lattice. And it's
non-empty. And the two extreme points of the lattice, we usually call them the
men optimal stable match and the woman optimal stable match.
So now what happens in the men optimal stable match is that for every man, his
partner is his best possible stable partner. This is happening simultaneously
for all the men. Consequence of the lattice structure. And so again under the
MOSM for every woman her husband is her worst possible stable husband. And the
woman optimal stable match is a semantic counterpart where the women are doing
as well as they possibly can, the men are doing as badly as they possibly can
within the set of stable matches.
>>: How the lattice --
>> Yashodhan Kanoria: So formally, so a set has -- so to define a lattice you
should have a set with a partial order.
>>: [indiscernible].
>> Yashodhan Kanoria: Partial order you would define here is that, for
instance, you could say that -- so matching mu is bigger than mu prime. If
under mu, the partner for every man is at least better, at least as good as the
partner in the mu prime, and for the women, it's at most as good on the mu,
compared to mu prime.
>>: [inaudible].
>> Yashodhan Kanoria:
So --
>>: [inaudible].
>> Yashodhan Kanoria: So it's just one lattice. So you would say that,
simultaneously say the men are doing better off and the women -- men are better
off and the women are worse off. Okay?
So mu.
And the women are weakly worse off.
>>: [inaudible].
>> Yashodhan Kanoria:
So the ordering requires stability.
This is just --
>>: [inaudible].
>> Yashodhan Kanoria: It's an ordering over matches. If you look at the set
of stable matches, then that subset also has this lattice structure, which
means that given any two elements you can find one that -- so the least one
that's larger than both of them and the smallest one that is -- no, the largest
one that's smaller than both of them.
>>: Stable matching, every man is weakly better off than [inaudible] for women,
then it's the opposite?
>> Yashodhan Kanoria:
I think that is true, yes.
So we now are going to impose some stochastic structure on the preferences of
the two sides of the market. So we are going to look at these random matching
markets with N man and N plus K women. K will be something larger than one for
all theoretical results. Complete preference list, drawn uniformly at random
and [inaudible] okay?
So it's a time model where the agent's preferences have no correlation at all.
They're independent, identically distributed and uniformly random. So each
reference list is a permutation on the other side, complete permutation. And
there's no correlation.
Okay. So here's a definition, a couple of definitions. So under a match mu,
we'll denote by little RM a man's rank of his wife. Okay. So a man M has a
preference list. How low on the list is his wife? That's here little R. For
instance, if the man has his top choice, little R will be one. If he has a
last choice, it's going to be N plus K. Otherwise, it's something in between.
Now, here's a metric for how well the men are doing. So for all the men who
are matched, we take the average of the ranks of their wives. And in fact all
men will be matched in this kind of a setting.
So this condition is not required. And this is the average rank of the wives
for the men. Okay. This is a metric of how well the men are doing. So this
average rank is small, they're doing well.
And similarly, we define the women's average rank of their husbands.
>>: So women do the rank by M plus K.
>> Yashodhan Kanoria: M plus K. So sorry.
matched women which again is going to be N.
So we'll divide by the number of
We'll divide by it. Thank you.
So we're not the first ones to look at random matching markets. These in fact
have been quite heavily studied. Notably by Pittel and this work by Knuth,
Motwani and Pittel. But it's been very typical to look at balanced random
matching markets where you have an equal number of men and women and what they
found there is that under the man optimal stable match, the men are doing
really well.
So the average rank of the men for their wives is just log N and in fact the
factor in front is 1. So it's 1 times log N. And the women's average rank of
husbands is N by log N. So these are asymptotic results as N tends to
infinity. This is what the average ranks look like. Okay?
So in the man optimal stable match, the men are doing really well.
have to go down their list. But the women are doing really badly.
to go down almost a linearly amount in their list.
They hardly
They have
The woman optimal stable match is the semantic counterpart, because this is a
balanced symmetric setting. Roth and Parenson, they sort of did an empirical
investigation of the national residency matching program. And there was sort
of this debate of whether the residents should be the ones proposing in the.
Gale–Shapley algorithm or the matching, the residency program should be the
ones proposing. So who proposes determines which endpoint of the lattice you
read.
And what they argue there in that paper is that in fact it doesn't matter,
because empirically if you take the preference list, both these extreme points
look almost the same. The two extreme points look almost the same.
So now there is this large gap between theoretical understanding of these
random matching markets where the man optimal stable match and the woman
optimal stable match are very, very far apart from each other. They look very
different. Which side proposes is extremely important in determining the
outcome. And in practice you have this NRMP data, and in that data you see
these two extreme points are almost the same.
>>: How many people are in this market?
that different.
N over log N and log N might not be
>>: If you have N being like a thousand, log N is going to be something,
something like 10, and N over log N might be 100 or more. So this kicks in
fairly early.
I think -- so I don't know the number for NRMP but it's of that order. I mean,
it should be 10 to the 3 or something of that order. Maybe even larger.
And then you can simulate these things. And the gap is -- I'm going to show
you some simulations. The gap is evident even for quite smaller user effect.
So Immorlica, Mahdian and also follow-up by William Patak look at random
matching markets with a twist. So they have balanced matching markets where
one side has short preference list. So they have a short list and they say if
we don't get one of these partners, then we are happy to be unmatched.
Okay. This, for instance, could be because market imposes a rule.
residency program you can't submit arbitrarily long list.
Like the
And what they find is that most agents have only one stable partner in the
setting. But this is somehow a way to make the market imbalanced, because one
side is not able to specify full preference list. And a lot of the agents on
that side remain unmatched with random preferences.
Okay. So they're showing that this core is small. They're not studying how -whether the outcome is good for one side or the other.
They're showing the code is small with respect to individual agents and most
agents don't have multiple stable matches. We have some understanding now of
what Roth Parenson observed in the data. And there's also this recent work by
Lee. I can talk about it later, it's not very closely related.
So here is another punchline for our work. So we look at random matching
markets with different numbers of men and women, and what I mean by different
is that it's different by one or more. It need not -- we don't need some large
difference in the size of the two sides of the market. It can be just that
there's one extra woman. No more than that. And under this very sort of weak
requirement, what we find is that if you look at the set of stable matchings,
the short side of the market is much better off.
So roughly the short side is able to choose, be able to exercise choice, get
somebody very high on their list, and the long side is roughly chosen. So it's
like they assign roughly a random partner.
And this is true under all stable matchings, including the two endpoints of the
stable matching lattice.
>>: So that's saying here are these thousand men, thousand women random
preferences, then there is a massive difference between them?
>> Yashodhan Kanoria:
One extra woman.
>>: Then you just add one extra men and then the entire structure is like
completely changed?
>> Yashodhan Kanoria: Yeah. Yeah, so we were also surprise d and we think
it's cool. So let's not start with that theorem because you might not believe
it. So let's look at some simulations.
>>: [inaudible] [laughter].
>>: [inaudible] like end up like small degrees, such that like they have
preferences on the set of number of others and the others are -- so they are
just completely not interested, isn't that ->> Yashodhan Kanoria: Sure. So we are not -- so we are saying that in this
model with random preferences, complete reference list. So we can talk about
sort of how to interpret the results.
And you wouldn't say that, okay, this is the reason why you observe X, Y, Z in
the real world. So there are many steps before where you can say that.
But it has been sort of common to study stable matches as a proxy for expected
outcomes even in decentralized real world market, and definitely for
decentralized markets like NRMP where you're running an algorithm and
explicitly finding stable matches.
So one interpretation of this kind of result is that you need to be careful
especially when you're studying these decentralized markets of using the stable
marriage concept because it's kind of fragile. You just add a couple of agents
and the set of stable matches can completely transform itself.
>>: One should expect that is the real data, right?
>> Yashodhan Kanoria:
You would expect --
>>: But that is basically because you're doing random matching, right, it's
because of the model, one should expect this thing --
>> Yashodhan Kanoria: You should not, right. So in a decentralized market,
for instance, if you look at a dating market or a marriage market in a certain
geographical region, you don't expect adding or introducing like two extra
agents on one side of the market is going to transform outcomes.
>>: [inaudible] transcend ->> Yashodhan Kanoria:
Yeah.
>>: [inaudible].
>> Yashodhan Kanoria: It's obviously in those situations you have [inaudible]
with dealing with preferences and effects are smoothed out over some window.
So you don't expect something so stark. And so one sort of implication of a
result is that, okay, be careful while using something as naive as a stable
match definition that is commonly used.
>>: All the other could be, assuming you have a perfect, everybody has a random
preference is not a good assumption.
>> Yashodhan Kanoria: That of course. But so -- so based on this analysis,
which has been -- so it happened a long time back and it's been followed up by
a bunch of work, it seemed like these random preferences are causing a large
core. Okay?
And at least one thing that we found is that is sort of -- that is a
[inaudible] phenomenon and in fact with random preferences, unless there's an
exact match on the sides of the two sides of the market, the core is not large.
So this is something -- this is of interest. Now, of course, there are -- so
there's reason to believe that there are going to be correlations between
preferences and this might affect other conclusions that you can draw.
So, for instance, which side chooses will definitely be something that you
would want to check whether continues to hold on the correlation. In fact we
talk about it a bit later it does not in the same way we find. But so one
method is, okay, if you're looking at these random markets, this is a knife
edge and you might as well sort of take a broader view and say that, okay, this
is not really what we are concerned with. And many of the properties we expect
the whole also for these random markets.
So here is a result of simulations. So we are holding a number of women
constant at 40. And we are weighing the number of men. And for each number of
men we are going to draw a number of samples of matching markets and then take
the empirical average of the man's rank of their wives, the two extreme points
of the lattice.
The man optimal stable match and the woman optimal stable match. And what we
see is that these two extreme points, they're very close to each other for most
points on the X axis, except for the one where the number of men is exactly
equal to the number of women, which is here when it's 40. So they're very
close to each other. The men are doing really well when there are fewer men
than the number of women. So they're getting the top choice or the second
preferred woman.
And suddenly when the number of men exceeds 40 they start doing much worse, and
it sort of approaches the case where they match to somebody randomly which
would amount to a rank of 20.5 or some such thing.
>>: Does each one know where she is in the ranking?
>> Yashodhan Kanoria:
>>: Number one.
So we're --
[laughter].
>> Yashodhan Kanoria:
So that's -- want an answer to that?
So we notice this large gap when there are 40 men.
for like 40, for a market of size 40.
[laughter].
But, see, this is happening
>>: So again, the expectation only when they're masked, the condition of what?
You don't say it's 40 when they're not [inaudible] right?
>> Yashodhan Kanoria:
No.
Yeah.
So this is for those who are matched.
>>: [inaudible] that is 40 you have a jump, two ->> Yashodhan Kanoria:
Yes.
So this is sort of illustrative of the same phenomenon. But slightly different
plot. So now we're varying the number of men but keeping the number of women
constant at two different options: Either the number of men and women are the
same both are N or number of women is one more. So let's now first look at the
case where the number of men and women is the same.
So those are the dark blue plots. So one is hidden here. This is the average
rank of wives under the man optimal stable match. But under the woman optimal
stable match the men are doing significantly worse. Okay. So in the balanced
case there is this large gap between the man optimal and the woman optimal.
But the moment we add one extra woman, the man and woman optimal both are
strongly favored -- so they strongly favor the men and they're very close to
each other.
>>: So [inaudible] log N and N over log N.
>> Yashodhan Kanoria: Yes, exactly. Exactly. And these numbers are really
quite small. So we are going to very large N. And this is when there is one
less woman and now again the man and woman optimal become close to each other
but the men are doing badly anywhere in the set of stable matches. So the
average rank of the wives is large. So here is our main theorem. So we look
at a random matching market, with excess women. This number of excess women
can be anything between one and N by two. So this two is coming for some
technical reasons.
I mean, we can change it if required. And what we prove is that with high
probability, in any stable match, the men are doing really well. So their
average rank is no more than this quantity here, basically logarithmic.
The women are doing really badly. So the average rank of, the women average
rank of husbands is quite large. It's almost linearly large. It's like N over
log N. And for the -- so the rank -- so men's rank of their wives under the
man optimal stable match obviously smaller than that under the woman optimal
stable match. But what we are proving here is that it's not much smaller.
It's very, very close. Okay. So the ratio of them informally converges to
one.
And all this happens with high probability.
>>: [inaudible] at the other expense?
>> Yashodhan Kanoria:
.3.
So this stuff works out for epsilon being N to the minus
>>: Okay.
>> Yashodhan Kanoria: Okay. So this one is a little bit hard to parse, and
I'll give you a bunch of corollaries that might be easier to parse. They may
not be corollaries, but they come from our proof. Here's corollary one. If
there are N men and one extra woman, then with high probability the average
rank of wives for the men is no more than one plus epsilon log N. The average
rank of husband for the women is at least N over 2 log N.
And this is true in all stable matches. So this is a surprising thing.
under the woman optimal stable match, wherein in the balanced case --
Even
>>: You have five before and now you have one. So it seems like you had weaker
hypothesis before -- you had stronger hypothesis ->>: Have to control all K.
>> Yashodhan Kanoria: Like I said, this is not strictly corollary but we
proved it. The constant factors got a little better. So, yeah, I mean good
point. So this is not a corollary of that theorem, but we proved this. Okay.
So even under the woman optimal stable match, when the women were doing much
better in the balanced case, the moment you introduce one extra, they're doing
much worse and the men are doing [indiscernible]. What about if you have a
linear number of extra women.
So here what we show there is that in fact the average rank of wives is bounded
by a constant, which you can bound in terms of this extra number of women,
extra fraction of women. And the average rank of women is in fact linearly
large. Again, in all stable matches.
This third corollary really doesn't follow from the theorem, but it's part of
the proof. Now, what we've shown here is that all but a sublinear fraction of
agents have a unique stable match with high probability. So not only are they
doing almost as well, in fact, they don't have multiple stable matches. In
other words, they're matched to the same person under the man optimal stable
match and the woman optimal stable match.
Okay.
>>: [inaudible].
>> Yashodhan Kanoria:
Yeah.
>>: So this is even if K is equal to 1?
>> Yashodhan Kanoria: Yeah. With high probability. Yeah. So the proof is
not terribly hard. So let me give you some sense of how we did it. So our
first step was to write down -- so let me first give you sort of a heuristic
reason for to believe that this happened, okay.
So think of the man optimal stable match. This can be thought of as being
computed by the Gale-Shapley stable algorithm. What does Gale-Shapley do. You
bring men in one by one. The man proposes to the first woman in his list. Or
let's do the parallel version. So all the men come in. They propose to their
top reference woman. The women look at the proposals they have received and
reject all but the best proposal. All the rejected men then go to the next
woman in their list, and they propose. And this iteratively keep doing this
until for each man they're matched to a woman or they have reached the end of
their list.
Okay. Now, in the balanced case, one can show that -- so the analysis is
basically like coupon collector. And one can show that the number of proposals
that occurs is like N log N. So number of proposals is the same as the sum of
ranks of wives. So now what if we introduce at this point an extra man? Okay.
Now Gale-Shapley can be run in any order it can be asynchronous. You can think
of introducing this man after the first N have been proposed and done their
thing. So this last guy comes in.
He proposes. And gets rejected a bunch of times. At some point he gets
accepted. But then there's another man who is now unmatched. That guy starts
proposing. And this goes on until some man has reached the end of his list.
Okay. Now, there is a part that I'm going to sort of gloss over, and that is
that these proposals are accepted quite frequently. So the man who is
proposing is actually switching around quite often. So now you have these N
plus one men. So roughly N men. And you're sort of throwing proposals at
these. So think of them as bins. And think of this as throwing proposals into
these N bins. And we want to wait until one of these bins is completely full.
So it has N proposals thrown into it. Now, if these proposals are just thrown,
these balls are thrown into these bins you would need to wait until you throw
an average of N by log N balls per bin. In order for the max to hit N. So
this then is the average rank of wives and the moment you introduce one extra
man.
Now, we could not prove things this way, because it's like -- when you have to
basically analyze order N squared steps and there's a bunch of correlations.
So we had a different way of doing it. That's what I'm going to describe. So
this is the intuition.
Okay. So now so what I described here was for adding one extra man now we're
going to switch back to the setting for the theorem. There is one extra woman.
Okay. So now you would have to do this with the woman proposing, and it has a
large number already squared proposals. This was difficult to analyze. So
instead we are going to continue with the man proposing and then add an extra
algorithm here, which will somehow take the man optimal stable results from the
man proposing and transform it into the woman optimal stable match in nearly
linear number of steps instead of quadratic number of steps.
Okay. Okay. So the first algorithm consists of two steps, the first step is
just the regular Gale Shapley algorithm. So I just describe it for
completeness.
So think of the asynchronous version of Gale Shapley, the main one by one.
Each man proposes to his next most preferred woman in his list until he gets
accepted.
At that point you sort of have a new man doing the proposals. You have this
chain will finally terminate in an unmatched woman. You do this for all the
men. At the end all the men are matched, and you have the man optimal stable
match.
Now, here is our key algorithm for going from the man optimal stable match to
the woman optimal stable match. Okay. So what are we going to do? Set the
set S to be the woman who are unmatched. Okay. So there is this rural
hospital theorem, ruled by Roth and a couple of others. And what it says is
that under all stable matches you have the same set of unmatched data.
a nice intuition to carry with us.
This is
If under the man optimal stable match we find certain women are unmatched, in
fact these women are going to be unmatched under all stable matches.
Okay. So we set S to be the set of unmatched women. And we will choose W had
to be somebody who is not in this current set S. So it's a woman who is
matched. And what we are going to do is to make that woman reject her current
partner, make him sort of propose to his next most preferred woman. Run this
chain. And see if we can somehow find the stable match as a result of this
chain.
So the only modification to the chain that we had for Gale Shapley is that this
W had herself will only accept a proposal if that proposal is preferable to M
whom she started with. Okay? So if this chain terminates with W hat accepting
a proposal, it's not hard to see that the match at that point is a stable
match, which in fact is more favorable towards the women than the one you
started with. Okay. So we're going to try to do this, should do these run
these chains on top of each other.
At the end we'll get the woman stable optimal match. Okay. So I'm waving my
hand here but we have a proof that this algorithm, when you run it sequentially
over the women who are matched will actually give you the woman optimal stable
match at the end. So a couple of details here. Suppose the chain does not end
with the proposal being accepted with W hat by W hat. Then it ends by, with
acceptance by a woman who is in the unmatched set.
So what we infer from that is that W hat to begin with at the beginning of the
phase was already matched to her best stable woman and our attempt to find her
best stable husband and an attempt to find an even better husband as we fail.
So we will roll our match back to the match at the beginning of the phase, and
we call that a terminal phase. On the other hand, if W hat does end up
accepting a proposal and that's how the phase ends, we'll call it an
improvement phase, because women actually are better off at the end.
So we have an algorithm. The problem is that we are sort of not -- we're
willing proposals women on men preference list multiple times. And this kind
of thing introduces correlation and makes it difficult for us to analyze. So
we are going to modify this algorithm.
So the first thing we notice here is that we don't have to wait for W
unmatched. So for an unmatched woman to accept a proposal, if the woman who
accepts the proposal has already previously sort of been through this process,
she's already matched to her best stable man. And if you now, if now she
accepts the proposal it's definitely going to end up in a terminal chain that
will go to W unmatched. So we don't have to wait until we reach the unmatched
woman. It's sufficient to wait until we have a proposal accepted by any member
of S.
Any member of S already has her best stable husband and is going to lead to a
termination. So we make this modification. And what we also do is to add not
just W hat who triggered terminal phase to S, we'll also add all women who
accepted proposals in this terminal phase to S. Because, again, we can infer
that for all these women they've now been matched to their best stable husband.
Okay. So these modifications, what they buy us
match preference list is revealed at most once,
basically a uniformly random entry. Subject to
interviewed, but that number is small. So it's
is that now each entry in a
and in a random model it's
some women already having been
possible to control it.
So how do we analyze our algorithm? The part one is just the algorithm one,
which is we run Gale Shapley. So that has been analyzed before. And so we
know this bound. It comes from COOP and collection. And at the end of Gale
Shapley we have the man optimal stable match.
The second algorithm we analyze in two parts. So one choice we make is to
order the women in a convenient way. Our first woman who we consider and try
to find a better stable husband for is the one who has received the maximum
number of proposals in part one. Turns out that is convenient for us.
So what do we show? We show that firstly the probability that first phase that
we trigger in algorithm two is an improvement phase is small. So we show that
it's much more likely that that phase ends with acceptance by an unmatched
woman then acceptance by the W hat that we started with. Okay.
So here formally is the bound, but basically it's very likely to be a terminal
phase and not only that it's likely to be fairly long terminal phase, unless
the number of unmatched women we started with -- so unless the imbalance in the
market is large, in which case it's likely that you'll propose quite quickly to
an unmatched woman, but in any case the set S after the first terminal phase
already becomes large with high probability.
Okay. So this part two just considers the first woman for whom you're trying
to find a better stable husband shows that most likely she does not have a
better stable husband but moreover, along with her we're able to find a bunch
of others who don't have better stable husbands, and this then helps us to
handle all the rest of the women, who will come up when we try to look for
better stable husband for them.
Okay. So we show that the number of proposals in algorithm two itself is
small, because acceptance is quite likely. And most acceptances actually end
up adding a woman to S because they're typically of terminal phases. We're
able to bound the number of acceptances in improvement phases, because the set
S is large. So, again, typically a phase ends with termination rather than
improvement. Okay. So only agents who were involved in improvement phases
actually have multiple stable matches. And we're able to show that with high
probability that it is a sublinear number of such agents. So this is basically
the proof.
So the million-dollar question is, of course, if we can allow correlation in
the preferences. And we don't yet know how to do it. It definitely sounds
difficult. Now, one thing that we notice is that the question of which side
chooses in the market actually has a different answer than the perfect
correlations, the model that we study. So, for instance, if the short side of
the market has perfectly correlated preferences, then there is in fact a unique
stable match in that setting, and that unique stable match corresponds to,
let's say the men have perfectly correlated preferences the unique stable match
corresponds to the women coming in that order and picking the best available
man.
Okay. So there are N women who are going to be matched, and if they have
random preferences, then say the first 80 percent of them are going to be able
to basically exercise their choice, because there's still a linear fraction of
men who are unmatched and somebody quite high on the list is going to be
unmatched.
>>: What do you mean by [inaudible].
>> Yashodhan Kanoria:
So everybody has the same preference list.
>>: [inaudible] then of course you don't need any randomness?
>> Yashodhan Kanoria: Yeah, yeah. So this is something that you can just talk
through the example. If this were a correlation from one site of the market,
that implies that there is a unique stable match for the market. And there is
a description for that, which is that the other side comes in that order and
takes their pick from who remains in the market.
So what we're saying is even if the market is imbalanced and there's perfect
correlation from the short side, with random preferences from the other side,
the long side actually chooses here, basically. So this result is not robust
to introducing correlations in this way. But the other -- so the other
implication of the result was that the core is remarkably small for random
matching markets the moment you have any imbalance. And the other extreme, if
you have perfect correlations from one side, the core has a unique element.
There is a unique stable match. Now what happens in between? We have two
extremes for full completely random preferences. The core is very small like
we proved. Perfect correlation from one side and again you have a unique core.
What happens in between?
So this is a very interesting and relevant for programs like the NRMP and
understanding the result.
Another question is whether we can extend the many-to-one matchings and this we
do have some leads on. So hopefully the next time I give this talk I'll have
something on that. Yeah. So some more numerical results. So here again so we
are fixing the number of -- here's exactly 100 men, and these are the number of
extra women.
So if there is less women, the men are doing badly. So this is the average
ranks for the men. The men are doing badly until the number of women exceeds
the number of men and then they're doing really well.
How many men have multiple stable matches? So, again, if the number of men and
women is the same, this fraction of agents with multiple stable matches seems
to converge to 100 percent. But if it's different it seems to converge to
zero.
Yeah? If you keep the number of men fixed, vary the number of women, so this
is similar to the plot I showed right in the beginning. The average rank,
man's average rank of wife, there's 1,000 men, it's in the hundreds. And all
of a sudden the number of men -- so in the number of women exceeds the number
of men it comes down to something very close to zero. It's really a sharp
phenomenon. This is with the larger size of the market and it becomes even
sharper than it was for a small size of the market.
So in conclusion, in random matching markets being on the short side is much
more important than the benefit of proposing. And we get a core convergence
result decision. So most agents have a single stable partner. That's it,
thank you.
[applause]
>> Yashodhan Kanoria:
Yeah.
>>: You have 12 men and no women in a room, then what results do you get?
>> Yashodhan Kanoria:
so if you have a large
anyway, right? So now
matched, they have bad
Yeah, so that's a good point. So there is obviously -imbalance, that many agents are going to be unmatched
we are saying that not only that, the ones who do get
outcomes.
Does that answer your question?
>>: My question was not a serious one.
that there were no women. So ->> Yashodhan Kanoria:
I was just noting that in this room
Here you looked like it was a serious question.
>>: No, it was not a serious question.
>> Yashodhan Kanoria:
All right.
>>: But like I guess you mentioned at times people have shorter lists. Well,
everybody has a shorter list. And but then the problem with that is you don't
get complete matches. Most to perfect matching.
>> Yashodhan Kanoria: Analyzed the -- how good the outcomes are for the two
sides. They just said that the core is small. So an agent, the probability
that that agent has multiple stable matches is small. So now -- so we've
analyzed sort of which side of the market gets to choose. So that is a new
thing. But also the other kind of result is stronger. So now we're saying
with high probability over the market, almost all agents will have a unique
stable match. There they're doing it this way individually so we're also
stronger in that core. And the most odd thing is you need a one extra woman.
If you're cutting the preference list to a constant size, and there are equal
number of agents to begin with, it's like having a linear fraction of
imbalance, basically. So that result is significantly easier to prove.
>>: Okay.
[applause]
No other questions.