Yuval Peres: Welcome. We're delighted to start the theory day. And the first speaker is Noam Nisan,
who will tell us about the borders of Border's Theorem.
Noam Nisan: Okay. So hi. So Border with a capital B is actually a person who has a nice theorem
called Border's Theorem, that is very useful in auction theory. And the theorem, the point of this talk is
that this theorem cannot be generalized as you would want, and that's why it has borders. So let me
start with a very simple problem on marginal probabilities.
So what I have here is some probability space that has separated into four regions maybe according to
maybe two random variables, [indiscernible] random variable is the column 1. And what I have here is
the probability in each one of these pieces. So assume it's all conditioned, and being in this row in this
column, the probability of some event is 0.1. Condition on this one you get 0.5, and what I'm
interested really in the marginal probabilities here instead of writing marginal probabilities, I wrote 2
times the marginal probabilities. So I'm assuming everything is uniform in these four things. So the
marginal probability of the event, just given that we are in the first row, is 0.3. So I'm writing 0.3 so we
can see that it's just the sum of these two numbers. And similarly has the marginal probabilities relative
to the second guy. And the total sum, of course, should probably be the same; the sum of the average
of these marginal probabilities should be the total probability of the event, and this should work out
both this way and this way. Okay?
and the question is what kind of numbers can you have in the blue rows? Can you have any kind of
numbers? If I give you a set of numbers in the rows can you tell me whether it's possible or not?
That's the question. So it might seem completely true; surely you can get any number between 0 and 1,
but the answer is apparently not. Can you fill in the matrix here so that the average of these two
marginal [indiscernible] this marginal probability is 0.2, and this marginal probability is 1.6, and so on.
Is that possible or not? So that's sort of an interesting actual question.
And it's not completely clear. I will prove shortly that it is not. But what kind of numbers can you
have there? So in general, if you wish to use natural computational problem, will be something like
that. Suppose you're given -- now we have n different random variables, each one taking m different
values. And I'm giving you the marginal probabilities for every value of every random variable. So
you get XI of j, the marginal probability of some event that we're thinking about when a certain random
variable has a value j.
And the question, is that possible? If there are any probability space. So I'm assuming that my random
variables are uniform. Just for simplicity you could have the same thing with other probabilities. Is
there any way to set up a probability space such that the probability -- so what does it mean to set up a
probability space? You need for every vector, for every profile of values of random variables, to know
what is the probability of the event that we're talking about. Is there any way to set it up so that you get
these marginal probabilities? So that's a natural question. It would seem to be very simple.
And one thing that I should point out, that obviously the set of possible values of these marginal
probabilities is the polytope. Because it's really just a projection of the linear -- so if you wish in higher
dimensions, the dimension of taking into account, oh, the little XI is here, each one of them is just
bounded between 0 and 1, and we're just projecting on to their averages. So this is a polytope. And the
question is, is this a polytope you can recognize efficiently? Seems like a very interesting natural
question.
So why was I interested in this question in variables? Because so -- so why was I? Because it
somehow came up as an auction-like problem. So let me continue to try to say where am I trying to
go? I'm going to try to give you this from an auction perspective. How do you get this kind of problem
from an auction? Once we get that, then I can tell about Border's Theorem, which is very nice.
So I'm going to tell you what it is and I'm going really to tell you why it's really so cool and why lately
in the algorithmic mechanism design community people have really been working it out. And how I
would like to generalize it. And then I would fix -- one way that I would like to generalize it, the very
simple more general economic scenario that I would like to get a Border-like theorem for, and I would
like to describe it. And then I will start playing with this problem and seeing what can I do anything
with that problem.
And when I start playing with it, it turns out that it's going to be translated into a function
[indiscernible] into a problem in Boolean function, and the Chow parameter is basically on the low
Fourier Coefficient of the Boolean function. And this reduction will be very simple, apparently.
Surprising but simple. And once we get it that way, then we can actually show it's going to be a #Phard problem. And that implies that we're not going to be able to actually get a generalization of our
theorem because it's too complicated in complexity terms. If you had the nice [indiscernible] to be like
an [indiscernible] something like that, and you won't be able to do that. And that already gives us our
bad result. And then also we will close the loop and go back to our original problem saying why the
problem that I showed you, whether you can recognize these marginal probabilities are also #P-hard.
So it's a difficult computational problem. So that's the plan.
>>: So just to understand, you're claiming that there is a polytope such that membership in this
polytope is a #P-hard?
>>: So I understand, will it be as hard if every variable [indiscernible]?
Noam Nisan: Yes, we're going to actually show it to everybody.
>>: So basically the problem is you are in two-dimensional -- you are projecting this two-dimension
the vector of probabilities -Noam Nisan: Two n values.
>>: And now you're asking -Noam Nisan: If there is a way to -- [inaudible]. A projection of this original polytope is size although
it's exponential in the variables and the projection is huge variables, but yet it's complicated.
Everything is going to be very easy, but I hope to spin your head a bit when going between model to
model.
Yes?
>>: So it is a suggestion now that if you're not restricted to [inaudible]?
Noam Nisan: Then everything very easy. Yes. That's the whole point; it's a probability between 0 and
1. That's really the sad thing.
>>: Okay.
Noam Nisan: Okay. So let me cast that -- really cast that in an option. So this is the original example
that I have here, you have the marginal probability of an event. And here you have the probability
theorem, the marginal probabilities. I'm going to recast that as an option where I'm going to treat this
number as the probability that there's an auction between two players, where this is the probability is
the first player wins, and then the second player wins with probability of 0.9. So for this example we
always allocate the item.
So let me just switch that to that situation. Now I have an auction. And so the question is, is there an
auction, if you wish, where the first player made the probability 0.4, if it has this type value in this
situation, and with the probability 1.4 in this situation. And the second player wins with probability
1.4, if it's in this situation; probability 0.[indiscernible] if it's in this situation. So we're assuming that
there are two players, each one has two possible values, and these two values are IID, so you get each
one of the four combinations of probability 1.
So again, what I write here is this is the probability if the second player wins, this is the probability if
the first player wins. That's two times the marginal probability. In the world of auctions, this is called
the interim allocation of this player.
So back to this situation. So we ask the question, can you fill up the marginal probabilities? Can you
fill up the probabilities to get these marginal probabilities? And in the auction setting, can you find
where the probability that the first player wins under this type. We're now going to call these types of
the players; okay? And the second player wins under this type and so on. So now we translated the
same problem into these question marks.
And here is the proof that you cannot do that. So the proof is by coloring the different question marks.
And it goes like this. Well, in this case this player wins with probability 1.6, which means that the sum
of the red value has to be 1.6. The sum of the green values has to be 1.6. Now, the sum of the purple
value -- so now I've already taken a little bit of the -- some of the question marks in these big cells. I
didn't touch this cell at all. And this cell has probability exactly one quarter, or in terms of actual
numbers, relative probabilities are 1. So the sum of the purple question marks is 1. And then you see
you cannot have, because all the question marks together have to be 4. And even without the black
question marks, which are positive numbers, we're already beyond 4. So that's the proof that you
cannot do that.
So generalizing this is exactly Border's Theorem. The Border asks the following kind of question. And
now let me tell you I'm doing a little cheat here. For two players, my original problem with probability
space is the auction problem is exactly the same, because anything that the first player didn't take the
second player took. Now we're going to have more players, more dimensions; then it's different
because in the auction I have a bunch of different parameters. The probabilities of the first player wins,
the second player wins, the third player wins, and that's not exactly the same as the original problem.
And originally my thinking was that that's a very tiny complication that we should get rid of, but
apparently that's not the case. But I'm not continuing in the auction world.
So Border asks the following problem. You have n vectors, each one of them having m probabilities.
These are the interim allocation probabilities, the marginal probabilities. And the question, does there
exist an auction with these marginal winning probabilities? Very simple.
And in a second I'll talk about the various incentives and things like that. But for now really the
question is, the content of Border's Theorem is just -- because there's the probabilities work out. And
that's the question.
Now Border prove this theorem twice. In 1991 he proved it, and then sixteen years later he wrote
another paper saying, "Oh, by the way, it's just an MP duality." So this is that. So once you have the
correct thing you can prove it with MP duality. You can even actually cast it as a [indiscernible],
though it's tricky.
And then there is the following thing. So he says the following are actually clearly necessary
conditions for this to work. And the necessary conditions are the following. You take some subset of
values for the first player, some subset of values for the second player, and so on. These are the things
that I'm going to basically color, if you wish. And I'm going to look at the sum of the winning
probabilities, assuming that the first player had the value here. Okay? So the first set, it's a big X for a
subset of values is simply the sum of the winning probabilities that he should get.
So I'm not going to look at the sum of the probabilities that he should get if each one of the players
should get when he has a value in these subset of items. And then I counted part of the question marks,
and the question marks I never touched are the ones that are in the compliment of all of these sets that
I'm now counting. So I'm actually adding the probability in terms of just -- if we're talking about
uniform over these values, it's just the size of the subset in terms of probability. And of course you
need all these probabilities to be at most 1 so things would work out. And the interesting theorem is
that this is also sufficient.
Let me give an example in a two-dimensional kind of thing. So you're getting these XIs, capital XIs
here, and capital XIs here. And the question is, is that feasible? Can you find an option in each one of
these cells, figuring out the probabilities of the column player who gets the item and the probabilities of
the role player who gets the item. Can you fill it up such that you get these marginal probabilities,
these interim allocation probabilities.
And the point is, here is the necessary condition. Let's look at these two rows as the probability role
player has one of these two values, and look at the probabilities of one of the column players has these
two values. So you need the sum of these three probabilities, plus the sum of these two probabilities,
plus all this sum; it better be less than 1. Otherwise you're certainly not going to be able to do that.
And the statement says that these are the only constraints that you have. It's a very nice
characterization of this polytope. That's Border's Theorem.
Okay, so now let me just say a few words how to make this completely into an auction kind of thing.
So if you're not interested in auctions, you can probably zone out. So really in auction you may have -so first of all, I only talked about like a uniform distribution over these possible values of ->>: If I was just hearing right, the probabilities then there's not much you can say between the case of
1 marginals and, say, 5 marginals, you can kind of think about [indiscernible].
Noam Nisan: So it turns out that there's a big difference.
>>: There is a difference?
Noam Nisan: There is a big difference. So that's exactly -- I want to go back to the 1 marginal,
because that's my original problem. And I will give motivation from auction theory that I want to go
back to 1. And it will turn out to make a difference. So for very long time I thought it didn't make a
difference, and it didn't work out exactly and I didn't know why.
So basically now it's not going to be just abstract values. There are going to be actually real values, V1
to VM for each one of the values of the players. And these are really the dollar amounts -- this is really
the value that they get for getting the item. So now these are actual values, and they're not just place
holders, first row, second row, third row.
And we're going to assume -- so everything here in the auction world that we're talking about in the
Bayesian setting, where you are living with this prior probability distribution and the values of the
players. And all the discussion that I'm doing is only an expectation. We're assuming that all the
players are just Bayesian probability maximizers and doing all the calculations this way. So each
player basically knows the distribution of the others but knows it's own value.
And by the way, Border didn't do it only in this finite case that I'm thinking about. It works also for
infinite. But if you want to think about computational issues, you'd better make it final, so I'm making
everything final, finite, and the Border will not be happy. But it's the same really. And I'm assuming
everything is uniform. And these values, again it doesn't matter; Border's Theorem extends naturally
and there's nothing new there.
So beyond these values there's one other interesting thing that an auction needs to specify; how much
do you pay when you win? That's another thing that an auction really needs to specify. And it turns out
that since we're really talking only an expectation, it's enough to talk about the expected payment that I
have if I have certain value. And because all the considerations are an expectation, I can just take that
expected value and say give it everywhere, something like that.
So there are going to be -- there is going to be PI[j], the expected payment of player I when it's value is
V[j]. So that's something that an auction would need to specify to really become an auction.
And then you have two really interesting important constraints for whatever you're doing. One of them
is called individual rationality, means the players should not lose. That is, what they pay should be less
than what they gain. What is it that they gain? Well, with probability XI[j], they win. That's the
probability, is they win the item over everything assuming they have value V[j]. And in that case
they're making V[j] dollars. So this is their expected gain and it better not pay more than that.
And the second, which is incentive one, says that if I have value J I'd better act like a guy with value J,
and I'm not going to gain by misrepresenting my value. And that basically translates to this very simple
inequality, saying that if I have value J and I act like a guy J, I'm going to get better utility, which is
what I gain from getting the item minus what I pay, than if I act according to some other value J prime.
So for us, it's none of these -- it's very important, but it's all simple linear inequalities in the Xs and in
the Ps, in the capital X and the marginal probabilities, in the interim location probabilities, and in the
expected payments.
So why is this Border's Theorem so cool? Basically because it allows us to optimize auctions to do all
calculations on auctions only using the interim location rules, rather which are only a polynomial in
many of them, rather than the whole exponential size matrix of probabilities.
So the kind of things we really are interested in -- okay, so by the way, why Border didn't give a
polynomial algorithm, just gave the characterization, actually it's not trivial. You have to be clever.
But you can actually get a polynomial algorithm for that. And then you really can just use linear
programming to do whatever you want to, optimize most anything that you're interested in in the
auction space.
So suppose that we want to optimize -- for some reason we want to optimize [indiscernible], the
payment, 0.7 times the payment that the first player gives and the futility of the second player times
0.9, then maybe the probability is the third player wins, and the revenue of the auctioneer and the
wealth that we get in the system, all of these are just quantities that are very natural in terms of the Ps
and Xs. They are not just linear quantities in the Ps and Xs. And then since we can now recognize the
set of -- recognize the set of possible interim allocation rules, we just need to do optimization of linear
function over very simple polytope, that we have all the constraints, incentive controls, the constraints.
The constraints are the feasibility ones the Border gives us. Plus the incentive constraints and the
individual rationale constraints that come from the auction world which are just linear constraint, and
the few of them, and everything is very simple. So you can just run your ellipsoid algorithm and you're
happy, you optimize whatever you want. That's really cool.
So now when you look at it from an auction point of view, it's not that -- there are other things beyond
the auction of a single item. There are more complicated things like a public project, there are multiitem auctions; there are lots things you would like to get the same kind of results for. And for that you
will just need to be able to generalize the Border's Theorem to these other scenarios. You need to
specify them in the same way. And you can get the same optimization for anything that you would
want to do.
And that's been the program that these people have been doing for the last few years. And there's a
situation as follows. If you really want nice mathematical, clean, exact extensions of Border's
Theorem, very slight generalizations. On the other hand, if you're allowing yourself to be a computer
scientist and allowing yourself approximation, allowing yourself not clean mathematically-nice
descriptions of the constraints but just the computational efficient ones, then there are completely,
almost extremely -- almost completely general generalizations work. So that is a situation, which is
really amazing results, because now we can optimize anything that you want really for any algorithmic
mechanism design problem if you're happy to be a computational approximate guy.
So I wanted at the beginning to not be a computational approximate guy but to be an exact
mathematical guy and get the same kind of approximations, but only very little was known. And the
question is can we get exact result, exact optimization result, and just generalize the theorem? And like
a mathematician would and get the exact theorem, and which first of all would have a nice
mathematical structure that maybe we can understand and work with, not just this completely
ununderstandable computational issues. And do it exactly.
>>: Maybe an algorithm that we learn to love and think that [indiscernible] thinks is nice.
Noam Nisan: So computer scientists have learned to love and think that other things are nice; right?
But we would like to get something non-approximate if possible. And it seems there are worse
problems that are simpler than the auctions should be able to do that. So our point is that, no, this
basically, they did what's possible. So now we really should love it. You can't do much better than
that.
So basically unless the complexity collapses on our head, there is not going to be a nice
characterization, a nice characterization of the constraints for the Border-like polytope for public
project problems for multi-optimizations even for unit-demand bidder for essentially any more
complicated problems that we can think of that was not dealt by the tiny extensions that we said, oh,
cannot be done exactly.
And the way we're going to show that, I'm going to show it basically for the public projects, and I'm
going to show the following interesting art. A public project setting, this is a very simple mechanism
design setting, for if you want to compute just optimal revenue, which is definitely one of the many
things you can definitely compute exactly if you had the nice Border-like theorem. I'm going to show
the computing revenue is #P-hard. So let me give the problem.
Here is the problem. So this government need to decide whether to build a bridge or not. And the
government is nice, just going to ask each person in the world -- in the country whether he values the
bridge or not. So it turns out that the people are completely independent of this country, and they can
each one of them have either two values. Half of the people have no value whatsoever for the
government. And half of the people have a value Wi which depends on the nature of the person, and
the government exactly knows what this WI is, how much if I like the bridge they would know how
much I like the bridge according to my demographic. But they don't know whether I like it or not.
So usually the usual story of this is called the public project. Then it's a Boolean public project,
because there are only two possible values here. And usually the story is the government wants to
make everyone as happy as possible, which in this case is very easy. Definitely builds the bridge,
because the value is 0 even for those that don't like it.
But our government really wants to maximize revenue, so they're going to ask you how much -whether you want it or not and they're going to charge you something if you say, yeah, build the bridge.
And how much are they going to charge you? As much as possible. So they want to design the best
mechanism to actually extract value from the players, from all the people in the country, and then get to
a decision of building the bridge or not. So they don't care about build the bridge, if people are happy
or not, but they just are happy to get the maximum possible revenue.
>>: They can charge different prices for [inaudible].
Noam Nisan: They can charge different prices from different people. And that's the problem. What is
the best revenue you can have? So one thing, which is a little strange for people coming from this
field. This is a single parameter problem, Myerson kills this problem. But there's a tiny little caveat in
Myerson theorem. He tells you what the optional thing is, so it's very easy to figure out using Myerson
what is the optimal mechanism to get most revenue. The only tiny little difficulty is that revenue itself
is going to be some kind of expectation at this point. And it turns out that computing that expectation is
going to be difficult.
Now if we can do the Border kind of thing of optimizing, we would know also the value of the
expectation, and we could do that. So even though Myerson tells us what is the optimal revenue, what
the optimal mechanism is not able to tell us what is the optimal revenue is, except by a formula that is
not computable efficiently.
>>: So is that parameter needed [inaudible]?
Noam Nisan: No, no, no. It's not. It's not, but the point -- so the mechanism -- so that's interesting.
We know what the best mechanism is and I'm going to tell you in a second also. We're going to derive
it without Myerson. We're going to derive it from first person principles. So I'm going to have the best
mechanism, but I'm not going to be able to tell you what is its revenue.
>>: But you can [indiscernible].
Noam Nisan: You can definitely approximate. By the way, so you can approximate everything as we
said.
>>: [indiscernible].
Noam Nisan: Right, yes. And so that by itself may not be that critical, but the point is what you learn
from it is that we were not able to characterize the polytope. Because if we could characterize the
polytope, what are the feasible interim allocation rules here, then we would also directly be able to
figure out exactly the revenue, which we can't. So that's really the point here.
So now I'm going to do the switch and try to figure out from first principles what is optimal
mechanisms here. And it's going to be the following thing. So for every possible -- so first of all, let's
find our indicator valuables, XI, which specifies whether my value is VI=WI or public known constant,
or it's 0. So XI, now I'm going move toward 01 situation rather than a 0 WI situation. Which is good;
we're getting closer to Boolean function. And for every possible profile of values of the citizen, of the
bidders, there's going to be a probability that the bridge is going to be built. It doesn't have to be a
decision yes or no, it could be some probability. We're allowing randomized mechanism to the design.
So I'm going to have this Boolean function Fi, which of course is a probability. So it unfortunately has
to be between 0 and 1.
And now let's see what is my revenue going to be? So let's try to think how much money am I going to
take from a player whose value is 0; how much am I going to take from a player from value WI.
So a player with value 0 has 0 value if the bridge is built. So we better pay 0. Expected payment
should be 0 when his value is 0, because I could give the money. But that's not really going to help me.
It's easy to see. So I definitely cannot take more than 0.
So now suppose that my value is WI, how much money can be taken from me if I'm expected to be
truthful? Well, let's see what happens if I say [indiscernible] rather than 0. So then I'm going to maybe
increase the probability that the bridge is built from the probability of F given that I said 0, to the
probability that F is going to be built given that I'm going to say 1; whereas the expectation is over
what I don't know what the other players are saying. So this is the extra probability that I'm winning by
lying and by saying 1 over 0. And in that case my value from it is WI. So if I'm charged more than WI
times the delta of these probabilities, I'm not going to want to say 1 because I'd rather say 0, I don't
want the bridge if you're charging me so much. So the amount of money that I can be charged the most
if I say 1 is exactly this number.
So basically we've really reduced -- so a tiny bit is missing, but it really is the proof. We completely
reduced the revenue maximization problem to the following interesting problem. It's almost a Boolean
function, like a randomized Boolean function. It's bounded between 0 and 1. We want to find the
Boolean function or the Fourier Boolean function that maximizes the weighted sum with WI weights of
the probability of 1 minus the probability of 0.
>>: Shouldn't it be a Boolean function? Because -Noam Nisan: Would turn out to be a Boolean function. But it's not clear at this point. So how can we
figure this out? How can we find this thing?
So furthermore, let's look at this thing. So now it really looks like a Boolean function, so Parikshit and
other -- [indiscernible] this guy is going to say, oh, this is something we know how to do. And this is
how they do it. So if you look at the problem XI equals 1 minus XI equals 0, it's really -- I'm just
summing up F of (x) times either negative 1 or 1, according to rather (x) is positive -- is 0 or 1, okay?
So instead of this difference, I'm really having here f(x) times si I of (x), si I is negative 1 to the power
of XI, just turning 01 to negative 1/1.
So why is this so useful? Oh, and by the way, this thing is really the Fourier coefficient of I, of the
singleton I of this function. Also called the Chow parameter.
Yes?
>>: Why did you say that you don't know in advance that it is Boolean? Aren't you maximizing some
Boolean function?
>>: Looks like a vertex of the [indiscernible] maximizing the Boolean function over the ->>: Yes, that's what it is.
>>: [inaudible].
>>: In the expectation, if you want to add the randomness of the algorithm [indiscernible].
>>: [inaudible].
>>: Yeah, the maximum must be the vertex of the [indiscernible].
>>: Maximums have no random [indiscernible].
Noam Nisan: So these are the Fourier Coefficient. So really what are we asking? We're really asking
what is the Boolean function or Fourier Boolean function to maximize the weighted sum of the Chow
parameters, of the single [indiscernible] coefficient.
>>: So you're trying to find x even though [indiscernible].
Noam Nisan: That's it. Given WI's, that's a mathematical problem because maybe it will be difficult to
tell what F is. But it will not be difficult.
So we've taken what we want to optimize, we wrote it now in this [indiscernible], this notation, which
is so very nice, because we can just move the expectation to the outside. And now we take the
expectation outside. And in its place we now need to optimize this thing. But this is very nice to
optimize. We can do it term by term. Because F has to be between 0 and 1. So how can I maximize
this? Well, if this number is negative, I want to take 0. If this number is positive I want to take 1.
Completely obvious. So basically what I'm taking is I'm taking either 0 or 1 according to the sign of
this. So the optimal -- the function that maximizes this is really the sign or the 01 sign of your thing,
which is really the weighted threshold function. So look at sum of the weight of the XI that are 1 and
compare them to the sum of the weight of the XIs that are 0 and see which one is bigger and according
to that you build or not build. And that's the best thing.
So by the way, Myerson would also tell you the same thing surprisingly.
And as you said, this is a Boolean function, which is really nice because it means that the best option is
deterministic. It's also a monotone Boolean function, which actually means you can make it into not
just Bayesian incentive compared off of Boolean dominate strategy [indiscernible], so it's all very nice.
And now the only problem that we have, so we know what's the optimal function. Can we figure out
the sum of the weight [indiscernible] of influences [indiscernible]. And it turns out that's very easy to
solve. That's #P-hard. That's basically known that playing with sharp parameter and figuring out it's
#P-hard. So our problem is given these Ws, and you just want to compute not what the function is, we
know what the function is, but just what the weighted sum of these things is. And you can see that if
you just add another bit to it and see when does it play. So suppose I added another [indiscernible]
item with a very tiny weight. Then when can it play? It can only come into play when the other guys
are exactly balancing out to 0. And if they're balancing out to 0, that's basically the probability. So his
influence, basically his sum of weight will be the probability that they balance to 0, and you can see
that's almost a knapsack problem. So this is #-knapsack. And you can fill the details. It's very easy.
So that's #P-hard. So let's see what did we get or Boolean functions, and then as I say go back to what
do we get for auctions. And I hopefully finish on time.
So we show that it's #P-hard to compute the W weighted sum of influences. So influences are these
[indiscernible], the difference between the probability with 1 and 0, these are really [indiscernible]
coefficient of not just the arbitrary thing, exactly of the W threshold function, because we know that
already is optimal. From this I want to say -- but that also means that you're not able to optimize
arbitrary Boolean function, arbitrary linear functions over valid Chow parameters. Because if you
could do that, you could run the same ellipsoid and get the optimal thing. So also even recognizing, at
least under Tourig reductions, recognizing whether you're given a set of possible free coefficient for the
first level, realizing whether there's any Boolean suitable semi-Boolean functions that gives you these
is going to be #P-hard.
And the same thing what happens with the auction world. So in the auction world we already know
that the optimal revenue is exactly the answer to this question, which is #P-hard. And in fact it means
that because we can't figure out the revenue for the Boolean public project, you can't characterize -you can do a Border-kind of generalization, because if you did the Border type generalization where
there's inequalities of the new polytope were reasonably recognizable, then you would do #P, and
reasonable and that's that.
And back to marginal probabilities. So I started with the question of the problem that I started with
was -- or this is a slightly special case of problem. You get the input P1 to Pn. These are the marginal
probabilities of the different end players. And I'm going to assume the special case that the total
probability is going to be exactly one-half, so I'm going to call that the balance thing, the total
probability is one-half. So the situation for the other one of the two types is going to be exactly one-
half minus P1. So this is a special case.
So I'm giving these things. And the question is does this exist a probability space with a [indiscernible]
that is the probability exactly one-half of the event such as that for some other events which are also
IID and balanced, so these are the exactly the [indiscernible] different possibilities. We have the
marginal probabilities given this way. Okay? So that's the problem we started with, for the special
case of exactly probability one-half.
And now that's really equivalent to the question that we already answered in Boolean function.
Because in Boolean function we have the Qs, and the Qs were not the probabilities themselves but the
gap between the probability for 1, the probability for 0. So that really translating back and forth is
completely trivial and that's the story. So the original problem #P-hard. So even though it's only a
single dimensional variant of the Border thing, it's harder, even though the Border thing is easy.
And I'm done. Thank you. [applause]
Yuval Peres: Any questions?
>>: So is the positive results sort of like the negative result [indiscernible] computations seems to
allow the polynomial approximations maybe?
Noam Nisan: Yes. So we do have a partial -- we do have like matching. You can do it approximately.
Which anyway, sort of known in the auction world. So it's not nothing new so far. We're trying to get
it nicer and more generally. So basically what you get from the auction world you can do
approximation extremely well, [indiscernible].
>>: So you said before that it's 2 marginal, 3 marginal problems are significantly harder than in what
sense? [indiscernible].
Noam Nisan: So in the auction thing, when you have a different number for each player basically, then
it becomes easy. When you have less than that one, there's a single player number that you get different
marginal than the same number, then it becomes hard apparently.
Yuval Peres: Anything else? Okay. We have a coffee break until the next talk is at 1:30. [applause]