Poisson Martingales and Gambler's Ruin

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In summary, the conversation discusses using a Poisson martingale to find the probability of an outcome in Gambler's Ruin. The solution involves defining various parameters and using the martingale principle. However, the conversation also touches on finding upper and lower bounds for the probability, which is where the use of the optional stopping theorem and the Poisson jump comes in.
  • #1
RedZone2k2
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I posted this in the HW help section, but I had no responses. I figure that this place may be better to answer this question. If this is against the rules or anything, mods please remove it!


Homework Statement


Find the probability of an outcome of Gambler's Ruin using a Poisson martingale.

Let m be the parameter of a Poisson Process (ie the lambda)
Let N(t) be a continuous Poisson process at time t>=0
Let M(t) = N(t) - mt

Homework Equations


Now, define s = inf{t | M(t) <= -a or M(t) >= b} (ie the stopping time)
Let r = P(M(s)>=b)

The Attempt at a Solution


I have shown that M(t) is a martingale with E[M(t)] = 0 for all t
Using the martingale principle, I have:
E[M(t)] = 0 = a*(1-r) + b*r
r = a/(a+b).

This is the solution for discrete gambler's ruin. However, my professor says that this value should be an upper bound for r, and the lower bound should be a/(a+b+1) <= r. I'm not sure how to get this.

My guess should be that it has something to do with the Poisson distribution, and the only difference is that b is bigger by 1. But, if there is a Poisson jump at some episilon of time, its just as if we have a+1 dollars, and our opponent has b-1 dollars; hence I should see a change in the numerator of the bound (and not the denominator). Is there any intuition for this? I'm not necessarily just looking for the answer (or equation), but more of understanding why there should be two bounds to my probability.

Thanks!
 
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  • #2
Can you describe parameters a and b?
 
  • #3
Sure,

I think of it like this:

You start out with a dollars, and your opponent starts out with b dollars. The variable M(t) is the sum of all your winnings and losses. You play until one person beats out the other, so you play until M(t) is either -a (where you are bust) or +b (where you bust your opponent). You want to find the probability that you win.
 
  • #4
If I'm starting out with $a, and M is a martingale, isn't E[M(t)] = E[M(0)] = a for any t > 0 ?
 
  • #5
M(t) represents the net change from your starting amount, so E[(M(t)] = 0.
 
  • #6
You're making an arithmetical error: 0 = a*(1 - r) + b*r implies r = a/(a - b), which raises a further issue: a = b implies r = a/0.

[I am guessing you had a typo in the equation, you meant 0 = -a*(1-r) + b*r. That makes sense. See this, where h = a in your problem and N = a + b in your problem. With [itex]\alpha[/itex] = 1/2, you do obtain the symmetrical solution h/N = a/(a+b). However, you'd get a different probability if [itex]\alpha\ne[/itex] 1/2.]
 
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  • #7
Yeah, sorry, that was a typo; what you have written above is correct.
 
  • #8
See my edits.
 
  • #9
RedZone2k2 said:
Find the probability of an outcome of Gambler's Ruin using a Poisson martingale.

Let m be the parameter of a Poisson Process (ie the lambda)
Let N(t) be a continuous Poisson process at time t>=0
Let M(t) = N(t) - mt

Now, define s = inf{t | M(t) <= -a or M(t) >= b} (ie the stopping time)
Let r = P(M(s)>=b)

First you'll need the optional stopping theorem to get E[M(s)]=M(0)=0.

Also you know the process first crosses the upper barrier via a jump, or the lower barrier via a non-jump, i.e. M(s)<=-a -> M(s)=a, and M(s)>=b -> M(s)<b+1.

Apply those to E[M(s)] to get two inequalities involving r.

HTH
 
  • #10
I figured it had to do with a Poisson jump. Thanks!
 

FAQ: Poisson Martingales and Gambler's Ruin

What is a Poisson martingale?

A Poisson martingale is a type of stochastic process that models the random behavior of a Poisson process. It is a sequence of random variables that satisfies certain mathematical properties, including being a martingale. This means that the expected value of the next value in the sequence is equal to the current value, given the information available at the current time.

How is a Poisson martingale used in gambling?

In gambling, a Poisson martingale is often used to model the probability of winning or losing a bet. It can be used in games such as roulette or coin tosses, where the outcome of each round is independent of previous rounds. The martingale property allows for the calculation of the expected value of the player's winnings over time.

What is Gambler's Ruin?

Gambler's Ruin is a mathematical concept that describes the inevitable loss of a gambler who continues to bet on a game with a negative expected value. It is based on the idea that, over time, the player will eventually lose all of their money. This concept is often used to analyze the risks involved in gambling and to determine optimal betting strategies.

How are Poisson martingales and Gambler's Ruin related?

Poisson martingales are often used to model the progression of a gambler's winnings or losses over time. As the gambler continues to bet, their winnings will follow a Poisson martingale until they eventually reach Gambler's Ruin. This means that the expected value of their winnings will eventually reach zero, and they will have lost all of their money.

Can Poisson martingales and Gambler's Ruin be applied to real-life situations?

Yes, these mathematical concepts can be applied to real-life situations involving risk and probability. They can be used to analyze the risks involved in gambling or in other situations where outcomes are uncertain. However, it is important to note that these models are simplifications and may not accurately reflect the complexities of real-life situations.

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