Probability of going broke

[jason1990]

Well, it depends on what you're used to as to which numerical method is easier. I would be more comfortable with the simulation, because I would be less worried about round off errors. And you're right about needing to be careful about how long to run the simulation, but also about how many times. I would be more comfortable dealing with these technicalities than dealing with the technicalities of round off errors. I suppose it's just a matter of what your background/taste is.

[pzhon]

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I'm a little surprised by these asymptotics. Thinking in terms of the Central Limit Theorem, when n is large, the path of the bankroll looks more and more like a Brownian motion with drift. Using the hitting time probabilities for a Brownian motion with drift, the probability of going broke is approximately

e^{-2mb/s^2}
= e^{-100000n/1105000}
= (0.913476)^n.

This does not agree with your asymptotics.

[/ QUOTE ]

The disagreement is smaller than it looks. We are using different definitions of n. I was talking about a bankroll of 100n. I believe you are talking about a bankroll of 1000n. To convert between the asymptotics, 0.9909571^10 = 0.913163.

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Have you computed c? Do you know it's not zero? Also, there seems to be a uniqueness issue in the difference equation. There doesn't seem to be enough boundary conditions to uniquely determine the solution.

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I haven't computed c in this example. I will, later. I think there should be some principle that allows you to bound c over all similar problems, which I think is more interesting than the particular case.

I think there are enough boundary conditions. Ten roots of the characteristic polynomial have magnitude greater than 1, and the coefficients of the exponential terms corresponding to these roots must be 0. (They would correspond to sequences satisfying the recurrence bounded in the opposite direction.) I'm a bit unsatisfied by this, since I don't know why the number of roots with magnitudes greater than 1 is right. I think there must be a general result about the roots of this type of polynomial.

[jason1990]

[ QUOTE ]
The disagreement is smaller than it looks. We are using different definitions of n. I was talking about a bankroll of 100n. I believe you are talking about a bankroll of 1000n. To convert between the asymptotics, 0.9909571^10 = 0.913163.

[/ QUOTE ]
Aha! Okay, then. My guess is that the remaining discrepancy is due to round off errors. So c=1, right? No need to compute?

As for what you said about boundary conditions, it reminded me of the heat equation. If you solve the heat equation on the entire real line (no boundary conditions) and all you specify is initial conditions, then you don't get a unique solution. But if you require the solution to have some bounded growth rate, then you do. (I'm a little fuzzy on the details, but I think this is generally correct.) It seems like something similar is happening here. I don't know if this is conceptually relevant, but I thought it was interesting.

[pzhon]

[ QUOTE ]
[ QUOTE ]
The disagreement is smaller than it looks...
0.9909571^10 = 0.913163. [vs. 0.913476]

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Aha! Okay, then. My guess is that the remaining discrepancy is due to round off errors. So c=1, right? No need to compute?

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I don't see how to get c=1. I still haven't computed c, though.

I think the remaining difference is not just a round-off error. The bases have different algebraic degrees. I'm not sure what the source of the difference is. Perhaps it is that the Central Limit Theorem does not work rapidly enough, particularly at the tails, and that is where we would like to use it for this problem.

[ QUOTE ]

As for what you said about boundary conditions, it reminded me of the heat equation. If you solve the heat equation on the entire real line (no boundary conditions) and all you specify is initial conditions, then you don't get a unique solution. But if you require the solution to have some bounded growth rate, then you do. (I'm a little fuzzy on the details, but I think this is generally correct.) It seems like something similar is happening here. I don't know if this is conceptually relevant, but I thought it was interesting.

[/ QUOTE ]
Yes, this is close to what I was thinking. In both cases, there is some sort of spectrum, and for the answer to be "physical" there must be no energy (zero coefficients) in half of the spectrum, and with this restriction the solution may be unique.

If you vary the polynomial, the roots may change magnitude. Can a root have magnitude 1, and if so, what happens?

[jason1990]

[ QUOTE ]
I don't see how to get c=1. I still haven't computed c, though.

I think the remaining difference is not just a round-off error. The bases have different algebraic degrees. I'm not sure what the source of the difference is. Perhaps it is that the Central Limit Theorem does not work rapidly enough, particularly at the tails, and that is where we would like to use it for this problem.

[/ QUOTE ]
Well, I was naively thinking our asymptotics must agree and this trivially gives c=1. But of course you're right about algebraic degree. Even in the simple case where the bets are +2 and -1 with equal probability, the true asymptotics are given by the difference equation and are

[(-1+sqrt{5})/2]^n

and the (incorrect) asymptotics given by the Brownian approximation are

[e^{-.4}]^n.

One problem which I originally thought would "disappear" in the limit is this: the player's wealth consists of two components, a simple random walk (SRW) which moves +1.5 and -1.5 with equal probability and a drift which moves at constant speed of +.5. "Scaling" the SRW gives Brownian motion. But this scaling means scaling both time and space *and* scaling them in different ways. However, in order to achieve a scaling limit for the drift, we must scale time and space in the *same* way.

So saying that the player's wealth looks a lot like a Brownian motion with drift when his bankroll is large is true to some extent, but it does not literally mean that his wealth converges to a Brownian motion with drift under some suitable scaling. This is, of course, because there is no way to consistently scale both components.

Obviously, this problem doesn't "disappear" in the limit. I believe it does "disappear" in many interesting applications, but certainly not here. I wonder also how they deal with this in mathematical finance. There they often model stocks and other risky commodities as diffusions with drift. But, in certain circumstances, just like here, these diffusions will have different asymptotic behaviors than their discrete counterparts. This is definitely something I'll be thinking more about.

Also, thinking more about boundary conditions, I think we simply need to add the one additional "boundary" condition that p(n)-->0 as n-->infty. I suppose this was obvious and probably what you were thinking all along, but I thought I'd say it explicitly.