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Ralph Boles
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Jun2805, 04:49 PM (EST) 

"Recomb. Binom. Tree"

Hello everyone; I am wondering about the following issue. I am working with a binomial tree model for a valuation. Basically v1=1, and in each period, with probability p the valuation moves "up" with probability 1p the valuation moves "down"; moving up means that the valuation is multiplied by a factor u, moving down means it is multiplied by a factor "d" I am interested in passing to the continuous limit. So baiscally chop the time period T into N pieces and take the limit as N>infinity. For that we need functions u(N), d(N) It is easy to verify that when p=1/2 using functions e^mDT sroot(DT), e^mDTsroot(DT) leads to a geometric brownian motion, but not for p<>1/2 are there some functional forms for u(N),d(N) that would lead to a nice SDE having p as a parameter for any possible value of p? 

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Graham C
Member since Feb503

Jun3005, 07:13 AM (EST) 

1. "RE: Recomb. Binom. Tree"
In response to message #0

I don't understand this, though it may well be my fault. If you say that after a period T there is a probability p the valuation will be multipled by u and a probability (1p) that it will be multiplied by d, you are saying that at T there are only two possible states. If you break T into N 'pieces' then if it may go up or down at each of the N, then there are multiple (N plus 1) possible outcomes with differing probabilities. So your initial statement is now invalid. If you want there to be only two possible outcomes, then at each N it must either always go 'up' or always go 'down'. Only the first period can then be probabilistic (probability p of up): subsequent moves would have to be certain. In that case (assuming periods are identical apart from the first) the change in valuation at each step must simply be the Nth root of u  or the Nth root of d. The valuation after m steps is then u^(m/N) or d^(m/N). But I'm sure you must have meant something different. 

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Ralph Boles
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Jun3005, 00:05 AM (EST) 

2. "Recomb. Binom. Tree"
In response to message #1

At time 1 the state is 1. At time 2 the state is u w/prob p, d w/ prob 1p At time 3 the state is u^2 w/ prob p^2, d^2 w/prob (1p)^2, ud w/ prob 2p(1p)but conditional on being in state k at time t, at time t+1 it is in state uk w/ prob p, and in state dk w/ prob 1p and 0 on all others. hope that makes more sense 

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Graham C
Member since Feb503

Jul0405, 07:52 AM (EST) 

3. "RE: Recomb. Binom. Tree"
In response to message #2

>At time 1 the state is 1. >At time 2 the state is u w/prob p, d w/ prob 1p >At time 3 the state is u^2 w/ prob p^2, d^2 w/prob (1p)^2, >ud w/ prob 2p(1p) > >but conditional on being in state k at time t, at time t+1 >it is in state uk w/ prob p, and in state dk w/ prob 1p and >0 on all others. > >hope that makes more sense It's a bit clearer. However, at time t it will be in one of the states (u^n*d^(tn)), n=0..t, all with nonzero probability. At t+1 it will be in one of the states (u^(n+1)*d^(tn)) with probability p and (u^n*d^(tn+1)) with probability (1p). If by 'uk' and 'dk' you mean (u^(n+1)*d^(tn)) and (u^n*d^(tn+1)) respectively then what you're saying is always true. 

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mpdlc
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Jul3105, 10:51 PM (EST) 

4. "RE: Recomb. Binom. Tree"
In response to message #0

Hello Ralph, I do not know if you still pursuing the answer, The problem as you stated, I understand belong to Markov nature. So I suggest you check in the Web for the two classical ones this type: The Gambler's Ruin Problem and the Random Walk Problem. I do believe must be a large number of papers on the subject. If you cannot find I will search them for you. Good luck 

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