28 Jan 09
Originally posted by aethsilgneWhen starting economics students usually get the feeling that many things are just assumed with a sleight of hand. The thing is that most of those assumptions can be derived from much weaker (and sensible) axioms.
Nah I study Political Science, I take evening classes in economics but the lecturer is so dull that I think of quitting 🙁
28 Jan 09
1 edit
I'm not blaming economics, I'm blaming the lecturer! 🙂
To get back to the original post of this thread:
Given we have perfect information (preferences and indifference functions) on an individual, we can predict all his rational (minimal cost) behaviour. Given that no man is equal, everyone has different preferences. This may lead to some actions being regarded as irrational in one others point of view. Hence all actions of an individual can be predicted mathematically, altho this mathematical formula will be different for every single person (and assumes perfect information and set budgets).
I'll leave you to do the numbers 🙂.Thanks for the interesting thread!
28 Jan 09
2 edits
Originally posted by PalynkaSomebody had to give in to your demands, because I'm very intrigued to see what happens next 🙂.
I knew it would be straightforward for you. 🙂
Give me a method for assigning a real number to an infinite, but countable set. What happens if X is uncountable?
Edit - Adam, did his proof made it clear what I meant by that?
There is rather more to this than I thought before -- we will have to POP and LOAD*! I'm going to try a topological solution**.
We consider X/I, totally ordered by P, as a topological space with the order topology, i.e. the topology generated by sets of the form {x: xPy and zPx}.
Suppose we have a function u:X/I --> R with the desired property, namely that xPy if and only if u(x)<u(y). Then let Q denote the image of X/I in R. Note that if u(x) = u(y), then x~Py and y~Px, so x and y represent the same I-class, i.e. u is injective. Thus u:X/I-->Q is a bijection.
Q inherits a topology from R as a subspace, namely the topology generated by the intersections of open intervals, and the order-preservingness of u guarantees that u and its inverse (on Q) are continuous with respect to this topology.
Thus a function of the type Palynka asked for is exactly an embedding (wrt the order topologies) of X/I into R. Now Palynka's question is: Under what conditions is such an embedding possible and how is it done explicitly when X/I is countable?
For the case when X/I is countable, pick your favourite real number r > 1. Pick some arbitrary enumeration of X/I = {x(i) : i in N}. For any x in X/I, let A(x) = {i in N : x(i)Px}. Let u(x) be the sum over A(x) of r^(-i). Then if xPy, A(x) is contained in A(y), so that u(x)<u(y). On the other hand, u(x) < u(y) means that the sum determining u(y) is taken over a set of natural numbers than that for u(x), since all terms are positive, i.e. A(y) contains A(x), so xPy. This is a bijection since X/I is totally ordered by P, and it has the Palynka Ordering Property (POP).
The question of when such an embedding exists is a harder. Since we can do it for countable ordered sets, the natural next step is to do it for sets with countable subsets which are dense in the order topology. I feel like the converse is rather nastier, i.e. there are embeddable X/I which are not separable.
In any case, the idea is to take a countable subset D of X/I and construct u: D-->R as above, and then find v: X/I-->R such that v restricts to u on D, and then show that this v has the POP. Given x in X/I, choose a sequence of x(n)s in D converging to x (by density). Then since u is an embedding, the sequence {u(x(n))} has a limit v in R (this can be had by pulling back the metric on R and observing that the x(n)s are a Cauchy sequence, and so must be their images, and applying completeness of R, among other ways). We have to check that this gives the same value when we choose a different sequence y(n) converging to x, but this is immediate since u is an open map and any neighbourhood of x contains all but finitely many of the x(n) and y(n). Thus we define v(x) to be the pointwise limit of u(x(n)) for any sequence x(n) --> x. v restricts to u on D.
Now take xPy in X/I. Now we actually have to assume that X/I is connected, which gives some w with xPwPy, since otherwise the stuff larger than y and smaller than x would be a separation of X/I by open sets. By sticking an element of D between x and w and between w and z, and applying the definition of v, we get that v(x)<v(y).
Thus: u exists when X/I is connected and has a countable dense subset, with respect to the order topology.
I don't think the converse is true; As I said, I think there is something a little weaker than separability that does the trick, and I am going to call it Linear Order Approximately Dense, or LOAD, and I'll try to figure out what it might be tomorrow evening.
*This is somewhere between popping and locking and locking and loading.
**When the only tool you have is a hammer etc.
28 Jan 09
What about this:
Consider the following ordering on R^2. We say (x,y)P(s,t) if x<s or, if x=s, s<t. In other words, compare the first coordinate first, and, if they are the same, compare the second. Thus:
(1,1)P(2,-1)
(1,1)P(1,2)
P-intervals are easy to visualise, and they generate an order topology on R^2. This is the simplest example I can think of where separability fails -- any dense set must have the cardinality of R. Can anyone construct a function u:R^2 --> R with the POP property for this ordering?
28 Jan 09
Originally posted by ChronicLeakyThat is the classical counter-example for proving (by contradiction) that the existence of such a function is not guaranteed for uncountable sets. They're called lexicographical preferences.
What about this:
Consider the following ordering on R^2. We say (x,y)P(s,t) if x<s or, if x=s, s<t. In other words, compare the first coordinate first, and, if they are the same, compare the second. Thus:
(1,1)P(2,-1)
(1,1)P(1,2)
P-intervals are easy to visualise, and they generate an order topology on R^2. This is the simplest example I c ...[text shortened]... lity of R. Can anyone construct a function u:R^2 --> R with the POP property for this ordering?
Great work, I'm VERY impressed! 🙂
Indeed, there needs to be a countable order-dense subset or no such u exists.
28 Jan 09
Originally posted by PalynkaThanks 🙂. People often lexicographically order things over here in Grouptheoristan as well, so it's not a real rabbit-out-of a hat. Maybe a practical example of a lexicographic preference-ordering on a subset of R^2 could be made from the complaint that "it's not the heat, it's the humidity".
That is the classical counter-example for proving (by contradiction) that the existence of such a function is not guaranteed for uncountable sets. They're called lexicographical preferences.
Great work, I'm VERY impressed! 🙂
Indeed, there needs to be a countable order-dense subset or no such u exists.
I'm eager to see where this goes next. Will there be many different preferences on the same set, like if everyone has a different favourite temperature/humidity combination?
28 Jan 09
I was thinking first going the classical route into uncertainty and prove under which axioms expected utility holds. But you're so fast that you've probably scared everybody away!
Anyway, I'll see if anybody is following my other simpler explanation and we'll see how/where/if this goes anywhere. I don't have a clear objective here...
29 Jan 09
Originally posted by PalynkaI'll be reading with interest wherever this goes 🙂.
I was thinking first going the classical route into uncertainty and prove under which axioms expected utility holds. But you're so fast that you've probably scared everybody away!
Anyway, I'll see if anybody is following my other simpler explanation and we'll see how/where/if this goes anywhere. I don't have a clear objective here...
As penitence for scaring people and probably getting us booted to the science forum, I could try to give real examples of the different topologies that can come from different kinds of preferences. They might involve chess, and they might help to integrate that proof into your exposition.
29 Jan 09
Originally posted by ChronicLeakyWhat are you waiting for?? 😠
I'll be reading with interest wherever this goes 🙂.
As penitence for scaring people and probably getting us booted to the science forum, I could try to give real examples of the different topologies that can come from different kinds of preferences. They might involve chess, and they might help to integrate that proof into your exposition.
29 Jan 09
Originally posted by TyrannosauruschexIt already is in the process of being mathametized: Functional MRI's are mapping emotional responses and able to pick out who is depressed, clinically depressed, and who is normal at a success rate of over 86%. That will extend to other mental problems like schistphrenia (sp)? and more. It will lead to a situation where psychologists and the rest will be able to be independently judged in their diagnoses.
Could human behavior be modelled mathematically?