Breaking sticks, not throwing them

[jason1990]

[ QUOTE ]
[ QUOTE ]
Was that just a device designed to work in the (k!)^2? My use of the two independent Poisson sequences was exactly such a device. Was Aaron doing the same thing? It will be interesting to find out. I hope he explains how he derived the continued fraction from that table of data.

[/ QUOTE ]
The requirement that two players both have good break games could be considered a way to force the 1/(k!)^2 into the infinite series. Similarly, the device of selecting the same number of players on each round could be considered a device to get the 2^(-n) out of it.

I was working from a different point of view, in which I was concerned about the ratio of the partial sums of the infinite series. But it comes to the same thing. The infinite series I need to get my recurrence relation

[/ QUOTE ]
Wait! Stop right there. What recurrence relation? Tell us what the connection is between

(\sum_{k=0}^n k/(k!)^2) / (\sum_{k=0}^n 1/(k!)^2)

and the continued fraction. Why did you want to have terms that look like this, and how did you then derive the continued fraction from them? This is what we are excited to know.

[AaronBrown]

[ QUOTE ]
Since you think that the Black-Scholes PDE is an SPDE, you apparently do not even know what an SPDE is. . . .

There are many examples in physics and finance, for instance, where a lack of rigor will lead to the wrong answer. If you already know the right answer to the problem, then you do not need to worry about such a trap. But a blind application of known tools without regard for the often subtle hypotheses involved, can and does lead to blatantly incorrect conclusions in real-life applications. I understand that you are not at all concerned with rigor. You are of course entitled to your opinion on the matter. I am just glad that you are not designing the airplane I am about to fly on.

[/ QUOTE ]
I really, really wish you would stop saying things like the first and last sentence I quoted.

I suspect we do not disagree too much on the middle. Yes, people make mistakes by not being rigorous enough. When rigor is needed, it's a great thing, and beautiful. But lots of people waste everyone's time insisting on rigor for its own sake. As one more or less random example, the long refusal to accept Cantor's work on set theory because it was "not constructive." I lose patience with people who don't have to solve real problems every day, but want to load down useful tools with unnecessary baggage.

[ QUOTE ]
Sorry if I was unclear. I was hoping for a derivation, not a calculation. I had already calculated the ratio of sums. I also already proved that the limit is the continued fraction.

[/ QUOTE ]
But the integer parts of the ratios are the coefficients of the continued fraction. That's the whole point. What more is there to prove? Why the integer parts of the of the ratios are successive integers? I don't know of a meaningful way to answer that, they just are.

Remember, I didn't generate those numbers, look at them and say "aha! continued fraction" (although I probably would have if I'd seen them). I had a continued fraction solution to the original puzzle that was wrong, so I tried to figure out how to change the problem to make the solution right. I was trying to demonstrate to you that a continued fraction solution could be used to solve a problem like the broken stick one.

I had two series that did the trick from my mistaken original solution to the broken stick problem. So I tried to set up a related problem (the lottery ticket problem) in which these would be the correct series. I did not solve the lottery ticket problem with continued fractions (although I probably would have, if I didn't already know the answer by construction).

[ QUOTE ]
I am not trying to be unfair. I believe I genuinely misunderstood you. I thought you could derive the terms in the ln(2) continued fraction by using the original puzzle and without using the fact that you had already solved it by some other method.

[/ QUOTE ]
I can derive the terms in ln(2) directly from the broken stick problem, without knowing ln(2) is the answer. For example, I can derive the terms if the stick does not start at unit length, in which case the answer is not simple.

However, there's nothing illuminating about the method. It's no different in principle from generating the decimal expression term by term. It requires a computer program (but not a simulation, a deterministic program), it's too tedious by hand.

In a good continued fraction solution, I can solve for the kth term without solving for all the terms up to k first. I can do it from general principles, not by expressing the answer as an infinite series first. I thought I could do that for the broken stick problem, I was wrong. I can do it for the lottery ticket problem.

[ QUOTE ]
Wait! Stop right there. What recurrence relation? Tell us what the connection is between

(\sum_{k=0}^n k/(k!)^2) / (\sum_{k=0}^n 1/(k!)^2)

and the continued fraction. Why did you want to have terms that look like this, and how did you then derive the continued fraction from them? This is what we are excited to know.

[/ QUOTE ]
I really think I've done this, although I don't think about it your way. Let N_k be the sum from k to infinity of i/(i!)^2 and D_k be the sum from k to infinity of 1/(i!)^2. You ask for N_0/D_0.

I also wanted N_0/D_0, but I didn't have derive the terms of the infinite series. I didn't need to for the continued fraction solution. I did have all the information I needed to solve for the terms, but I didn't do it. If I had, in my formulation it would have been more natural to use N_0 = sum from 0 to infinity of 1/{(k!)^2*(k+1)}, which has the same value as your numerator. I did know that the numerator terms were i times the corresponding denominator terms.

For a continued fraction solution, you want some kind of recurrence relation, like the one I used to solve the quadratic equation. In this case it's N_k/D_k = k + D_k+1/N_k+1. Note that although D_0 > N_0, for all i > 0 D_i < N_i so the last term is a fractional part.

This means, as I tried to say in my original post on this solution, that the kth term in the continued fraction expansion is k, as it is the integer floor function of N_k/D_k. When you subtract k and invert the residual, you get N_k+1/D_k+1.

I really think this is pretty simple, and it's getting loaded up with a lot of unnecessary formalism. The wildy non-rigorous solution, from the beginning is:

Notice that the probability of losing, given that you finish on round k, is 1/(k+1). Therefore the probability of losing, given that you make it to round k, is the sum of the probabilities of finishing on round i divided by (k+1) from i = k to infinity, divided by the same sum without the division by k+1.

Wave hands and guess that the leading terms dominate so the inverse ratio of the partial sums might be k, therefore that the continued fraction solution might be 0 + 1/(1 + 1/(2 + 1/(3 +. . .

Look for a recurrence relation between the probability of winning if the game is not over by the round i and the probability of winning if the is not over by round i+1.

Notice (incorrectly) the recurrance relation between the probabilities of winning given that you get to successive rounds, N_k/D_k = k + D_k+1/N_k+1. Think that you have demonstrated the continued fraction solution.

Check it and notice it is wrong, but remain excited because it was so pretty and surprising when it worked, and it is close to the correct answer. Despise the actual correct answer, ln(2) based on summing a simple infinite series. It seems pedestrian.

Come up with a problem for which this is the correct answer. Succeed, until you realize it is really a different kind of problem.

Work out all the details of the solution, with help from others, and realize the continued fraction solution is more flawed than you first thought. It is only close to the correct numerical answer by coincidence.

Apologize, get on with life. Hope everyone gained something from the experience.

[jason1990]

[ QUOTE ]
[ QUOTE ]
Since you think that the Black-Scholes PDE is an SPDE, you apparently do not even know what an SPDE is. . . .

There are many examples in physics and finance, for instance, where a lack of rigor will lead to the wrong answer. If you already know the right answer to the problem, then you do not need to worry about such a trap. But a blind application of known tools without regard for the often subtle hypotheses involved, can and does lead to blatantly incorrect conclusions in real-life applications. I understand that you are not at all concerned with rigor. You are of course entitled to your opinion on the matter. I am just glad that you are not designing the airplane I am about to fly on.

[/ QUOTE ]
I really, really wish you would stop saying things like the first and last sentence I quoted.

[/ QUOTE ]
The last sentence is true. In my experience, among all applied scientists, engineers are the most concerned with rigor. They do not want to take any chances that their results are not correct. You do not seem to have that attitude, so I am indeed glad you do not work in such a field.

The first sentence is also true. This may be a case of an unknown unknown, as Rumsfeld might say. You do not know what an SPDE is, but you do not know that you do not know it. There are no SPDEs in Black-Scholes. At all. The Black-Scholes PDE is a parabolic PDE with no noise term. It is impossible to mistake it for an SPDE, when you know what one is. But since you have probably never seen an SPDE, you probably do not know what I am talking about.

The post I was responding to displayed an ignorance of and a disrespect for my profession. You speak presumptuously as though you know what I am talking about. It may not be evident to a lay person, but it is quite clear to me that you do not. You want people to say the "magic word" to you when they ask you to demonstrate your skills with a puzzle. But this time it is you who should be asking me what an SPDE is and how it differs from what you thought it was. But you cannot even bring yourself to admit that you do not know.

PairTheBoard is right. This place would not be the same without you. (Or without me too, probably.) We are providing a nice bit of entertainment for the masses. But do not get annoyed with me when you know you do your fair share of annoying as well.

[jason1990]

[ QUOTE ]
I really think I've done this, although I don't think about it your way. Let N_k be the sum from k to infinity of i/(i!)^2 and D_k be the sum from k to infinity of 1/(i!)^2. You ask for N_0/D_0.

...

For a continued fraction solution, you want some kind of recurrence relation, like the one I used to solve the quadratic equation. In this case it's N_k/D_k = k + D_k+1/N_k+1.

[/ QUOTE ]
Is this right? Taking 28 terms, I calculate N_0 = 1.590637, D_0 = 2.279585, N_1 = 1.590637, and D_1 = 1.279585. This gives

N_0/D_0 = 0.697775 and D_1/N_1 = 0.804448.

But according to your formula, these should be equal, right? Am I missing something simple here?

[jason1990]

Can anyone comment on this? I have been trying to understand Aaron's derivation of the consecutive integer continued fraction for the last hundred or so posts. I realize that it was based on a mistake and does not solve the original puzzle. But he posted an elaborate lottery puzzle which he says it does solve. He gave the numbers for that lottery puzzle. The solution is N_0/D_0, where

N_k = \sum_{i=k}^\infty i/(i!)^2, and
D_k = \sum_{i=k}^\infty 1/(i!)^2.

He says that he derived the continued fraction for N_0/D_0 from the recurrence relation

N_k/D_k = k + D_{k+1}/N_{k+1}.

I think I can see how the continued fraction would follow from this recurrence relation. That part seems pretty straightforward. But my calculations indicate that this recurrence relation is false.

Have I done my calculations wrong? Have I misunderstood Aaron's derivation? If this recurrence relation is false, then did Aaron simply arrive at the correct continued fraction by accident? That would seem like an awfully big coincidence. In short, is there anyone still following this thread who understands enough about what Aaron is trying to do to point me in the right direction? If there is a slick way to get from N_0/D_0 to the continued fraction, I think it would be a very nice connection between probability (e.g. the Poisson example I gave earlier) and the continued fraction constant.

[jay_shark]

I stopped following it after I read Aarons first solution and Pzhons previous solution .

There is no way this problem can be solved using continued fractions .

[PairTheBoard]

[ QUOTE ]
N_k/D_k = k + D_{k+1}/N_{k+1}.

[/ QUOTE ]

Aaron has probably been called away to consult with the NSA on some matter of vital national security and will be back to answer when he is done with the matters of state.

I'm wondering, does that recurrence relation have to be exact? It seems to me from reading Aaron's previous posts that it might have to just satisfy some Interger Part thingy. Do your calculations show it to be close?

PairTheBoard

[jason1990]

[ QUOTE ]
[ QUOTE ]
N_k/D_k = k + D_{k+1}/N_{k+1}.

[/ QUOTE ]

Aaron has probably been called away to consult with the NSA on some matter of vital national security and will be back to answer when he is done with the matters of state.

I'm wondering, does that recurrence relation have to be exact? It seems to me from reading Aaron's previous posts that it might have to just satisfy some Interger Part thingy. Do your calculations show it to be close?

PairTheBoard

[/ QUOTE ]
Well, I suppose it must be close enough, since it has already been proven that N_0/D_0 is equal to the continued fraction. But how could one know that a priori? And doesn't it get quite complicated when you start using those integer part thingies? I mean, don't you end up layering them on with each step, resulting in integer parts of integer parts of ...

Is it the case that he mistakenly thought this recurrence relation was true, used it to derive the continued fraction, and then happened to be right by accident?

[PairTheBoard]

I don't know. If I can work up the ambition to plow into the algebra maybe I can look at the equation myself. I'm getting a bit weary of it though. Aaron should provide his math for the equation.

I think This Post by Aaron best describes his thinking for working with continued fractions. It's an alien approach to me.

I'm still not clear on how he could look to a continued fraction to calculate an acumulating probabilty when the partials of the continued fraction oscillate.

I'm also curious about this statement by him in that post.

[ QUOTE ]
Aaron -
There is even a theorem due to Euler that says something like this (that is, if successive terms in an infinite sum have integer ratios, those integers are the coefficients of the continued fraction representation of the sum).

[/ QUOTE ]

Did he say that right? Wouldn't that mean that,

1/2 + 1/4 +1/8 + ...

has a continued fraction representation of something like,

[0;2,2,2,2,2,...] ?

Can that be right? You certainly don't have the partials of the continued fraction equal to the partial sums because the partial sums are increasing while the partials for the continued fraction oscillate. Maybe it is right though. I don't know. Maybe the partials don't have to match, just the whole thing.

If this is what he meant, it seems like it would be easy to construct probabilty problems with continued fractions as natural solutions.

I suppose I could google Euler. But I'll probably wait for Aaron to come back instead.

PairTheBoard

[jason1990]

[ QUOTE ]
I'm also curious about this statement by him in that post.

[ QUOTE ]
Aaron -
There is even a theorem due to Euler that says something like this (that is, if successive terms in an infinite sum have integer ratios, those integers are the coefficients of the continued fraction representation of the sum).

[/ QUOTE ]

Did he say that right? Wouldn't that mean that,

1/2 + 1/4 +1/8 + ...

has a continued fraction representation of something like,

[0;2,2,2,2,2,...] ?

Can that be right?

[/ QUOTE ]
Isn't this 1/(2 + 1/(2 + ... ? If we call that x, then x = 1/(2 + x), so x^2 + 2x - 1 = 0. So no, I guess that can't be right.

I thought maybe he meant this. Suppose

N = a_0b_0 + a_1b_1 + a_2b_2 + ...
D = b_0 + b_1 + b_2 + ...

Then N/D = a_0 + 1/(a_1 + 1/(a_2 + ...

This would prove that (*) in my Poisson proof is the continued fraction constant. Unfortunately, it looks like it would also prove that the expected value of every non-negative integer-valued random variable is 1/(1 + 1/(2 + 1/(3 + ... So maybe that's not what he meant.

And now that I have looked again at that post you linked to, I see that Aaron gives a different recurrence relation there:

[ QUOTE ]
D_i = N_i *(i + 1+ N_i+1/D_i+1)

[/ QUOTE ]
I suspect that D_i and N_i may not be defined in exactly the same way as they are in his last post. Now I have to sort through everything to see if this is in fact a different recurrence relation and, if it is, whether or not this one is right.

Maybe it's time to give up the quest. But I still think a "natural" proof of my Poisson claim would be something special. pzhon, are you reading this? Have you seen the Poisson example before?