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rickyzhang
Posts: 17
Joined: September 4th, 2008, 9:02 am

a gambler's ruin problem

April 23rd, 2009, 3:11 am

Just finish a homework exercise in my textbook. Let's denote A's fortune at time n as X_n. two Stopping time T_0=inf{n>0:X_n=0} and T_20=inf{n>0:X_n=20}Q: calculate P(T_20>T_0 | X_0=7)Apply gambler's ruin solution in Durrett text book (3rd edition) on page 293. Then you are done.In my exercise, I need to compute E(min(T_20,T_0)). It use martingale stopping time theorem. But it is very interesting...
 
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alexrem
Posts: 12
Joined: April 3rd, 2009, 4:57 pm

a gambler's ruin problem

April 23rd, 2009, 10:01 pm

QuoteOriginally posted by: cdmurray80Thats a random walk on a bridge right? e.g. a drunk man stands on position 7 of a bridge that has length 20, and takes one step left or right every turn. If he falls of the closer side A wins.I've seen this problem but I don't know an easy solution technique...maybe someone could point one out?The only solution technique I know is this: considering the gambling problem:Let P(1) be the chance for A to win when B has 1 dollarLet P(2) be the chance for A to win when B has 2 dollars...Let P(N) be the chance for A to win when B has N dollars, which is the total,so P(N) = 0Let P(0) is the chance for A to win when B has 0 dollars, which means A won, so P(0) = 1Then we can get N equations:P(0) = 1P(1) = 0.5*1 + 0.5*P(2)P(2) = 0.5( P(1) + P(3) )P(3) = 0.5( P(2) + P(4) )...P(N) = 0Then you have P(k)=0.5(P(k-1)+P(k+1)) so P(k+1)-P(k) = P(k)-P(k-1) for k = 1, 2,...,N-1. Let P(k+1)-P(k) = d.Then P(N)-P(0) = (P(N)-P(N-1))+(P(N-1)-P(N-2))+...+(P(1)-P(0)) = NdP(N)=1, P(0) = 0 so d = 1/NAnd so for every k=1,2,...,N P(k)-P(0)=kd=k/NSo when k=7, N=20 then P(7)=7/20.
 
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SixSigma
Posts: 64
Joined: September 19th, 2002, 8:56 pm

a gambler's ruin problem

July 13th, 2009, 9:20 pm

The bridge analogy clarified this one perfectly, this is the "motion" of a single pot of money under Brownian motion. The game begins with "A" holding seven dollars. The game stops when "A" has zero dollars (A loss) or twenty dollars (B loss). I did a Monte Carlo with 1,000,000 games to find that "A" lost 666,283 times while "B" lost 333,717 times.Cheers
 
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Wan
Posts: 1
Joined: February 14th, 2006, 12:21 am

a gambler's ruin problem

July 14th, 2009, 2:42 am

The stopping time of a martingale is a martingale. Use the martingale property.
 
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aaron81
Posts: 3
Joined: June 12th, 2009, 3:50 pm

a gambler's ruin problem

July 15th, 2009, 2:50 am

Last edited by aaron81 on July 14th, 2009, 10:00 pm, edited 1 time in total.
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