Poly math geniuses--Jules needs your help!

[Jules]

The problem below is excerpted from "Money is Memory" by Narayana Kocherlakota (Journal of Economic Theory 81, 232-251, 1998).

"Consider a world with an infinite number of trading posts located at the integer points along the real line. In period 1 and in every period thereafter, at each trading post there are J "stayers" and J "movers." At the end of each period t, the movers move 2^(t-1) trading posts to the right; the stayers stay at their current trading posts....

Suppose the matching process pairs stayer j with mover j at each trading post. Consider the mover j who started life at trading post 0; in period t, he arrives at trading post 2^(t-1) - 1. It is tedious but simple to show that for this agent, the set of trading histories (Qt) [I will explain] contains all stayers and movers labled j who begin life at posts {0, 1, ..., 2^(t-1) - 1}. At the same time, the set Qt for mover j at trading post (2^t - 1) contains all stayers and movers labelled j who began life at posts {2^(t-1), ..., 2^t - 1}."

Now some remarks about Qt, the set of trading histories. Qt consists of mover j himself and his trading partners in period t, all of their trading partners in period (t-1), all of those people's trading partners in (t-2), and so on.

So for example, if mover j begins at trading post 0, in t=5 he arrives at trading post 15 (0-->1-->3-->7-->15). Remember that the mover is matched with a stayer at each trading post. So his trading partners at t=5 includes the stayers at 0, 1, 3, 7, and 15. But the set Qt also contains their trading partners at period t=4. For instance, the stayer at trading post 15 had trading partners who were movers who began life at trading posts 8, 12, 14, and 15. At t=3, each of these movers had trading partners who were stayers. And so on until you work all the way back to t=1. You should then find that Qt for mover j contains all movers and stayers who begin life at posts {0, 1, 2, 3, ..., 15}.

Now Kocherlakota may think it is simple to show this in general for any t, but this is where I need some help.

[JohnT]

I think the only "official" Polymath genius on these boards is MtG.

[Japher]

Consider the mover j who started life at trading post 0; in period t, he arrives at trading post 2^(t-1) - 1

I don't think that equation is right for the position of the mover j. With that you get that j's current position would be p(t)=2^(t-1)+(p-1); where p(t) is position at time t and p is position at time=0. With that you get that at t=0, p(t=0)=-p-0.5 where it should be p(t=0)=p...

Yet, I graphed the position versus the step increase and fitted a line with an R value of 1 to it and got

y=2x-(p-1) or p(t)=2m-(p-1)

or p(t)-p=2m+1 where m=2^(t-1)

really just 2*(2^(t-1))-1 for j who started at 0

I don't really know what you are trying to prove, but I just thought I would point that out.

[Jules]

Originally posted by Japher
I don't think that equation is right for the position of the mover j. With that you get that j's current position would be p(t)=2^(t-1)+(p-1); where p(t) is position at time t and p is position at time=0. With that you get that at t=0, p(t=0)=-p-0.5 where it should be p(t=0)=p...

Ahh, I see where you're confused. There is no t=0. The game is assumed to start at t=1; nothing exists before that. So t can be all the integers strictly greater than zero. The Excel table I posted was for the purposes of assisting people with the numerical example I gave: suppose mover j begins life at trading post zero (t=1)and we forward to period t=5. Sorry I didn't explain it earlier, but I had to rush to class. The row above the dotted line marks the time periods (in this case t = 1, 2, 3, 4, 5, 6). Below each time period is just a segment of trading post positions (which will hypothetically extend all the way back to negative infinity and all the way forward to positive infinity). So the example we have is mover j is at TP zero at t=1 (go down the t=1 column to find zero). At the end of t=1, mover j moves 2^(1-1) = 1 TPs to the right and ends up at TP one at the beginning of period t=2, 2^(2-1) - 1 = 1 (move to the cell directly to the right). Then at the end of t=2, j moves 2^(2-1) = 2 TPs to the right and is at TP three at the beginning of t=3, 2^(3-1) - 1 = 3. And so on until at t=6, j arrives at TP 31.

btw, I quoted this directly out of a published paper, so the formula better be right. It would be rather embarrassing to both the author and the Journal of Economic Theory if there was a mistake.

[Japher]

You should then find that Qt for mover j contains all movers and stayers who begin life at posts

I don't think I understand the pairing. Let me see if I do..

A mover from TP 0 moves to TP 31 via the equation 2^(t-1)... This gives the intervals at which he moves the 1, 2, 4, 8, etc... thus getting to posts 1, 3, 7, 15, etc... At each post he makes a connection [Q(t)] with the stayer, the stayer has connections will all the subsequent movers that have "visited" this stayer. Since the stayer has connections with these other movers this gives the current mover a connection with those stayers...

I.e. Bob starts at 0 goes to 1 where he meets Greg. Greg has just got done visiting with Henry who, at t=1, went to 2 (because of the equation). Now, Bob has a connection with Greg who has a connection with Henry thus Bob has a connection with Henry and Greg... This continues.

Eventually, when Bob gets to post 31 he will have had a connection with everyone from post 1-31 whether they are direct connections or indirect... Correct?

If so then to include the stayer at TP 0 you would have to start at t=0... Right? Also, this produce an exponential number of connections which is intuitively right. Though that really gets you know where.

I read some of Kocherlakota stuff today, and I don't understand it, I am no economist.

[Japher]

edit: gibberish

[Jules]

Originally posted by Japher
I don't really know what you are trying to prove, but I just thought I would point that out.

And now the punchline....

"Good thing I didn't tell them about the horny economist--!"

No, wait!


The purpose of this paper is to provide a logic for the essentiality of money in conducting transactions. Why do we need this tangible thing called money if it's possible to keep a perfect record of all a person's transactions, on a computer say? So when you receive income, the computer will just credit your account and when you purchase goods your account will automatically be debited and the appropriate seller credited. And the computer will do this over your entire lifetime and for all agents you transact with, directly or indirectly.

One possibility is that in the real world we don't have this kind of perfect recordkeeping device, so agents may be hesitant to enter into commitments. Hence the existence of tangible money may be welfare improving. This motivates the title of the paper, "Money is Memory." The author shows that under certain assumptions, the existence of money produces the same outcome as when there exists perfect memory.

The trading post game I described above is one possible application of a general framework which the author goes on to describe. One crucial assumption we need is that the agent has no direct or indirect information about his future trading partners. So for the exercise, we are supposed to show why we need this odd 2^(t-1) movement rule in order to satisfy the above assumption.

So I need to show that for mover j who begins the game at TP zero, at any time t his set Qt of all direct and indirect trading partners is all movers and stayers who begin at TPs {0, 1, ..., 2^(t-1) - 1}. Also at time t there is a mover at TP (2^t - 1), which is where our mover j is going to be at time (t + 1), whose set Qt consists of all movers and stayers who begin life at TPs {2^(t-1), ..., 2^t - 1}. Notice that the intersection of these two sets is empty, which implies that the assumption is satisfied: at any time t, mover j has no direct or indirect information about his future trading partners.

Sorry I didn't explain any of this earlier, but I was a bit rushed.

[Japher]

edit: more gibberish

[Jules]

Originally posted by Japher
I don't think I understand the pairing. Let me see if I do..

A mover from TP 0 moves to TP 31 via the equation 2^(t-1)... This gives the intervals at which he moves the 1, 2, 4, 8, etc... thus getting to posts 1, 3, 7, 15, etc... At each post he makes a connection [Q(t)] with the stayer, the stayer has connections will all the subsequent movers that have "visited" this stayer. Since the stayer has connections with these other movers this gives the current mover a connection with those stayers...

I.e. Bob starts at 0 goes to 1 where he meets Greg. Greg has just got done visiting with Henry who, at t=1, went to 2 (because of the equation). Now, Bob has a connection with Greg who has a connection with Henry thus Bob has a connection with Henry and Greg... This continues.

Eventually, when Bob gets to post 31 he will have had a connection with everyone from post 1-31 whether they are direct connections or indirect... Correct?

Yes. You've got it right.

If so then to include the stayer at TP 0 you would have to start at t=0... Right?

No, because the set Qt also includes the person you meet today. So at t=1, mover j at TP 0 is matched with a stayer at TP 0.

[Urban Ranger]

Originally posted by JohnT
I think the only "official" Polymath genius on these boards is MtG.

What about Chowlett, the math major at Cambridge, or the physicists?

[Urban Ranger]

Re: Poly math geniuses--Jules needs your help!

Originally posted by Jules
Consider the mover j who started life at trading post 0; in period t, he arrives at trading post 2^(t-1) - 1.

That doesn't seem right.

As t starts at 1, 2^(t-1) would be the series 1, 2, 4, 8, ...

So, trader j should arrive at post 2^(t-1) + t₀ -1 at time t, which would be post 1 at t=1.

Edit: fixed equation

[Japher]

Yeah, I find this confussing as for j to move from t=1 to t=2 you move 2^(1-1) not 2^(2-1), or they move at the t they WERE at not the t the ARE at.

t=1, j=0

to get to t=2, j moves 2^(1-1)=1, to 2 => t=2, j=1

to get to t=3, j moves 2^(2-1)=2, to 3 => t=3, j=3

to get to t=4, j moves 2^(3-1)=4, to 7 => t=4, j=7

.
.
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which is not how most people write an equation (maybe this is how they do it in economics?).

[JohnT]

Originally posted by Urban Ranger


What about Chowlett, the math major at Cambridge, or the physicists?

The fact that you felt you had to specify what field they are knowledgable in automatically disqualifies them from Polymath status. Good going, UR!

[KrazyHorse]

Proof is by a ridiculously simple induction.

[JohnT]

Proof by definition.