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Exam tactics

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Say you have 20 essay topics. You know that there will be 9 exam questions and you’ll be expected to answer 4 questions. How many topics ought you revise? The simplest answer is 15. If you study 15 topics, even if all 5 topics you didn’t study get picked, there’ll be 4 left that you did study (because 9 get picked, remember). But that’s still a lot of topics! If you studied 14 topics, what are the odds that you’d only have 3 revised topics on the exam?

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Written by Seamus

April 4, 2011 at 3:42 pm

Posted in maths

Bets and groups

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I’ve been thinking about bets recently. As you do. I’m interested in the Dutch book theorem, and central to the proof is having a method to formally represent what a bet is. Borrowing from this paper by Frank Döring, we can think of a bet as a list of ordered pairs of an event and a number: [E_1,S_1;E_2,S_2;\ldots ;E_n,S_n] . The E_i are the events and the S_i are the stakes. So if Alice and Bob are betting on whether a coin will land heads or tails, with the winner gaining €1 from the loser the bets they are accepting are as follows: (let’s say Alice bets the coin will land heads) [H,1;T,-1] . Bob is taking the other side of the bet, so he is accepting the bet: [H,-1;T,1]. Flipping the sign on each stake turns one bet into its “complement” if you like. Taking both bets (i.e. taking Alice’s bet and Bob’s bet) would mean a net profit of 0, however the world turns out. This is a kind of “neutral bet”. We can stipulate that the E_i form a partition of the space of possible outcomes. That is, they are mutually exclusive and exhaustive. A bet has to say what happens in all eventualities. Say you were betting on the roll of a die, and the deal was that you won if a 1,2, or 3 came up and that your opponent won if a 5 or 6 came up. What would happen if a 4 was rolled? The bet as it stands doesn’t say… So for simplicity let’s say bets specify what happens in all eventualities.

Perhaps a better formulation of Alice’s and Bob’s bets is as follows: [H,1;T,-1;X,0] for Alice and the “signs reversed” version for Bob. The X here is supposed to cover all remaining possibilities, and stipulates that in the event that the coin lands on its edge, say, then “all bets are off”.

I like Döring’s framework for bets. It’s wonderfully general. J.Y. Halpern’s book Reasoning about uncertainty has a different framework which is less general, but still sufficient for proving the Dutch book theorem. Döring’s framework allows him to show that no (nontrivial) kinds of conditional probability measure can be justified by a Dutch book argument.A Dutch book in Döring’s framework is a bet where all the stakes are negative.

I was thinking about this framework for talking about bets, and I realised a couple of things. Döring discusses combining bets by taking pairwise intersection of the events, and summing the corresponding stakes, to get you a new bet. The events will still partition the event space, and the net profit will be the same for all outcomes, whether you take each bet individually or take the combined bet. We’ve already seen how there’s a neutral bet, and it should be obvious that combining any bet with the neutral bet doesn’t change anything. And for each bet, flipping the signs on all the stakes gives you a bet which, when combined with the original bet, gives you the neutral bet. Basically, the set of all bets with the rule of combination form a group. (Associativity of the group action follows from associativity of taking intersections and addition of numbers). An abelian group, in fact.

For each event we can define a function that maps a bet to the net gain of the bet if that event occurs is a homomorphism from the group of bets to the additive reals. For each probability function over the events, there is a homomorphism from bets to additive reals that returns the bet’s expected value given that probability distribution.

I don’t know if there is any actual point to this, but I found it neat that you can do this stuff. I wonder if you can put a Dutch book in terms of the existence or otherwise of some group theoretic stucture. A Dutch book in this framework is a bet with all stakes being the same sign (that is, whatever happens, it’s always the same party that benefits). So perhaps one can exploit the fact that this is the same as the maximum stake being negative or the minimum stake being positive, and neither of these functions from bets to additive reals are homomorphisms.

Written by Seamus

September 13, 2010 at 11:54 am

Posted in maths, philosophy

White’s coin puzzle for imprecise probabilities

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[Caveat lector: I use a whole bunch of different labels for people who prefer sharp credences versus people who prefer imprecise credences. I hope the context makes it obvious which group I’m referring to in each instance. Also, this was all written rather quickly as a way for me to get my ideas straight. So I might well have overlooked something that defuses the problems I discuss. Please do tell me if this is the case.]

On two occasions now people have told me that there’s this paper by Roger White that gives a pretty strong argument against having imprecise degrees of belief. Now, I like imprecise credence, so I felt I needed to read and refute this paper. So I sat out on my tiny balcony in a rare spell of London sunshine and I read the paper. I feel slightly uneasy about it for two different reasons. Reasons that seem to pull in different directions. First, I do think the argument is pretty good, but I don’t like the conclusion. So that’s one reason to be uneasy. The other reason is that it feels like this argument can be turned against sharp probabilists as well…

The puzzle goes like this. You don’t know whether or not the proposition “P” is true or false. Indeed, you don’t even know what proposition “P” is, but you know that either P is true or ¬P is true. I write whichever of those propositions is true on the “Heads” side of a coin, after having painted over the coin such that you can’t tell which side is heads. I write the false proposition on the tails side. I am going to flip the coin, and show you whichever side lands upwards. You know the coin is fair. Now we want to know what sort of degrees of belief it is reasonable to have in various propositions.

It seems clear that your degree of belief in the proposition “The coin will land heads” should be a half. I’m not in the business of arguing why this is so. If you disagree with that, I take that to be a reductio of your view of probability. Whatever else your degrees of belief ought to do, they ought (ceteris paribus) to make your credence in a fair coin’s landing heads 1/2.

What ought you believe about P? Well, the set up is such that you have no idea whether P. So your belief regarding P should be maximally non-committal. That is, your representor should be such that C(P)=[0,1], the whole interval. This is, I think, the strength of imprecise probabilities over point probabilities: they do better at representing total ignorance. Your information regarding P and regarding ¬P is identical, symmetric. So, if you wanted sharp probabilities, the Principle of Indifference (PI, sometimes called the Principle of Insufficient Reason) suggests that you ought to consider those propositions equally likely. That is, if you have no more reason to favour one outcome over any other, all the outcomes ought to be considered equally likely. In this case C(P)=1/2=C(¬P). In sharp probabilities, you can’t distinguish total ignorance from strong statistical evidence that the two propositions are equally likely. Consider proposition M: “the 1000th child born in the UK since 2000 is male”. We have strong statistical evidence that supports assigning this proposition equal weight to the proposition F (that that child is female). I’ll come back to that later.

So what’s the problem with imprecise probabilities according to White? Imagine that I flip the coin and the “P” side is facing upward. What degrees of belief ought you have now in the coin’s being heads up? You can’t tell whether the heads or tails face is face up, so it seems like your degree of belief should remain unchanged: 1/2. Given that you can’t see whether it’s heads or tails, you’ve learned nothing that bears on whether P is the true proposition. So it seems that your degree of belief in P should remain the same full unit interval: [0,1].

But: you know that the coin landed heads IF AND ONLY IF P is true. This suggests that your degree of belief in heads should be the same as your belief in P. But they are thoroughly different: 1/2 and [0,1]. So what should you do? Dilate your degree of belief in heads to [0,1]? Squish your degree of belief in P to 1/2? Neither proposal seems particularly appetising. So this is a major problem, right?*

What I want to do now is modify the problem, and try and explore intuitions about what sharp credencers should do in similar situations. First I should note that the original problem is no problem for them, since PI tells them to have C(P)=1/2 anyway, so the credences match up. But I worry about this escape clause for sharp people, since it is still the case that the reasons for their having 1/2 in each case are quite different, and it seems almost an accident or a coincidence that they escape…

Consider a more general game. I have N cards that have a number 1..N on one side of each. The reverse sides are identical. Now, on those reverse sides I write a proposition as before. On card number 1 I write the true proposition out of P and ¬P. On the remaining 2..N I write the false one. Again you don’t know which is which etc. Now I shuffle the cards well, pick one and place it “proposition side up” on the table. For simplicity, let’s say it says “P”. I take it as obvious that your credence in the proposition “This is card number 1” should be 1/N. What credence should you have in P? Well, PI says it should be 1/2. But it should also be 1/N, since we know that P is true IF AND ONLY IF this is card number 1.

Imagine the case where N is large, 1 million, say. I get the feeling that in this case, you would want to say that it is overwhelmingly likely that P is false: 999,999 cards have the false proposition on, so it’s really likely that one of them has been picked. So my credence in P being true should be something like 1/1,000,000. Put it the other way round. Say there’s only 1 card. Then if you see the card says “P”, that as good as tells you that P is true, so your credence should move to 1.

On the other hand, if we’re thinking about a proposition we are very confident in, say A: “Audrey Hepburn was born in Belgium” (it’s true, look it up.). Let’s say C(A)=0.999 (not 1 because of residual uncertainty regarding wikipedia’s accuracy.). Now, if we have 2 cards and the one drawn says A, that’s good reason to believe that that card is the number 1 card. So in this case, it’s the belief regarding the card that moves, not the belief regarding the proposition.

What about the same game but with a million cards? Despite my strong conviction that Audrey Heburn was, if briefly an Ixelloise (?), the chance of the card drawn being number 1 is so small that maybe that should trump my original confidence and cause me to revise down my belief in A.

Here’s another trickier case. Now imagine playing the same game with some proposition I have strong reason to believe should have credence 1/2, like M defined above. And let’s say we’re playing with only 3 cards. For simplicity, let’s imagine that M is shown. How should your credences change in this situation? Again, it seems that C(M)=C(1) is required by the set up of the game. But I’m less sure which credence should move.

In any case, is there a principled way to decide whether it’s your belief about the card or your belief about the proposition that should change. And if there isn’t, doesn’t this tell against the sharp credentist as much as against the imprecise one?

On objection you might have is that this is all going to be cleared up by applying some Bayes’ theorem in these circumstances, since in these cases (as opposed to White’s original one) see which proposition is drawn really does count as learning something. I don’t buy this, since the set up requires that your degrees of belief be identical in the two propositions. Updating on one given the other is going to shift the two closer together, but I don’t think that’s going to solve the problem.


* The third option, a little squish and a little dilate to make them match seems unappealing, and I ignore it for now, since it seems to have BOTH problems that the above approaches do…

Written by Seamus

July 19, 2010 at 1:04 pm

Stupid size comparisons

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Today’s Guardian contains this spectacular piece of idiocy. Speaking of a chunk of ice that fell off a glacier and caused a huge wave, they said:

The chunk of ice [was] estimated to be the size of four football pitches

First, a minor quibble. Not all football pitches are the same size. Four football pitches could be anything between 20,000 and 52,000 square yards. (Thats 16,000-43,000 m^2). That’s a pretty big margin of error, even for an estimate.

Second, much more major gripe. Chunks of ice are 3-dimensional. What does comparing it to an area, a 2-dimensional thing, even mean? Does that mean the area of glacier that was lost, as seen from above? If so, how thick was the ice below it? That makes a huge difference to how much stuff we are talking about.

Now, even if the ice were only a centimetre thick and we take the lower estimate for the area, that’s still 160,000 kg of ice. Which sounds a lot, but is an order of magnitude less than the weight of water in an olympic sized swimming pool (2.5m kg).

I bet there are whole websites dedicated to this kind of stupid size comparison thing. It sometimes comes up in New Scientist’s Feedback column…

This post brought to you by pedantry, my boundless capacity for procrastination and WolframAlpha.

Written by Seamus

April 14, 2010 at 1:17 pm

Geek poetry

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I’ve not posted for a while: I’ve been busy doing things other than procrastinating! OK that’s a lie, but the procrastination hasn’t taken the form of blog posts for a while. My website looks much the same as ever, but lots has changed under the hood, as it were. It now validates as XHTML and the columns are the same height and extend to accommodate as much text as needed. Hoorah. I also have an essay I’m rather proud of. (Actual work, shock horror!) It is probably over long and not all that much of it can be adapted to fit into my literature review, but I’m still happy with the (almost) finished product. On an unrelated note, here are some poems that appealed to the geek in me.

Here is the halting problem proven in poem form.

A poem composed entirely of punctuation. (I may have linked to this before.)

I remember a maths lecturer at Warwick starting a lesson by telling us:
Integral t-squared dt
from 1 to the cube root of 3
times the cosine
of three pi over 9
equals log of the cube root of ‘e’.

More maths limericks here.

A history of Western philosophy in limerick form you say? Well why not?

And of course there’s limerickdb. The marked geeky charm of the top 150 indicates that this project is from the chap behind xkcd.

And while it’s not a poem, it’s certainly the same ballpark.

And finally my own contribution thanks to getting bored during measure theory lectures. I give you a haiku about basic measure spaces:

A finite union
of disjoint rectangles is

I have tons more of these on some scrap of paper in my old notes folder. I also wrote a limerick about Rene Magritte once… (I rhymed “Rene Magritte” with “ceci n’est pas une pipe”)

Written by Seamus

August 6, 2009 at 12:13 am

Posted in internet, logic, maths

Tagged with , , , ,

Majority logic I

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When I went for an interview at Oxford for my undergrad degree one of the things we had to do was a kind of logic test. It had things like “All men wear hats, some men wear ties” and you were asked what it was possible to conclude from that. Presumably they were looking for something like “Some men wear ties and hats”. One of the questions was “Most men wear ties, most men wear hats”. I said that it wasn’t possible to conclude anything from this. But it was then explained to me that if hat-wearers and tie-wearers are both in a majority, then there must be some who are both: the proper conclusion is that “Some men wear ties and hats”.

“All men” is standing in for the universal quantifier (x) and “Some men” stands for the existential quantifier (Ex). But “Most men” is awkward to put in similar terms without resorting to some sort of logic enriched with predicates relating to numerical proportions. But that sort of “statistical logic” is too complex. The fact that some predicate applies not to all but to a majority of some domain is an easily understood fact that does not need a whole logic of proportions behind it. Let’s make “Most of” a new kind of quantifier: (Mx). What properties does (Mx) have?

  • (Mx)Px \wedge (Mx)Qx \rightarrow (Ex)Px \wedge Qx

This is the property my interviewer at Oxford was exploiting. If most x’s are P’s and most x’s are Q’s, then some x’s must be both.

  • (x)Px \rightarrow (Mx) Px
  • (Mx) Px \rightarrow (Ex)Px

These two properties simply say that if all men wear hats then most men wear hats and if most men wear hats then some men wear hats.

Do we really need a third quantifier here? Is there some way to express “most of” in terms of universal and existential quantifiers? I’m not sure there is. Here’s a first stab. When you say “(Mx)Px” what you are really saying is:

  • [(Ex)Px ] \wedge [(Mx)Qx \rightarrow (Ex)Px \wedge Qx]

But that’s no good, because that expression still contains an “Mx”. We are kind of going beyond standard predicate logic, but only a little bit. Perhaps Mx shouldn’t be a quantifier, but a property of predicates? But I don’t think that is a particularly satisfying suggestion. A property of predicates would be a third-order entity, and that seems extravagant given the modest goal of formalising the easily understood concept of a majority.

Soon I’ll post again and discuss the “dual” of (Mx): “Only a few men wear hats” “Hat-wearers are in a minority”.

A note on the symbolism: I wanted to use \forall and \exists but I couldn’t think of a way to “invert” the letter “M” in a similar fashion. It shares the vertical symmetry of “A” but when reflected around a horizontal axis as \forall is, you get a “W”, which is no good if the idea is to come up with a new symbol. So I reverted to old-style quantifiers. Suggestions regarding how to “symbolise” majority are most welcome.

Written by Seamus

May 26, 2009 at 10:34 am

Posted in logic, maths, philosophy

Tagged with

Dutch books

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I’ve been thinking about the Dutch Book Argument (DBA) recently. I think the constraints on rational betting preference that underpin the force of the argument are unreasonable. At least they are not always reasonable. I think a better way to think of the argument is as a conditional argument: “Given these rational betting preference conditions, this is how rational degrees of belief should be constrained.” Then you can have a whole series of different DBAs with different betting preference conditions leading to different constraints on credence. It would be interesting to see how one would have to constrain betting preferences in order to have you beliefs behave like upper and lower probabilities or Dempster-Shafer belief functions

I think this isn’t to undermine the force of the DBA, but to reinforce it. With this wider framework we can understand why people often fail to reason probabilistically. We can understand what aspects of rational betting preference are “non-probabilistic”.

That’s not to say that the DBA isn’t without its flaws. Some elements that concern me are:

  • Using betting behaviour as a proxy for belief
  • Existence of exactly specific numerical credence (and utility)
  • Reasoning as calculating expected utilities
  • The idealisations involved in discussing “ideal rational agents”: utility maximising, purely self interested, perfect calculating agents…

Written by Seamus

February 22, 2009 at 7:27 pm