Suppose IBM costs $75 today.Suppose you live forever.Suppose interest rates are zero.How much should you pay today for a derivative that pays you $1 if and when IBM hits $100 for the first time?I don't know the answer but I think it's $0.75. Because from now to the end of time IBM will at some point either go out of business (derivative pays nothing) are hit $100 (derivative pays $1), so it is just a one-step binomial tree. What do you guys think?

this sounds like a perpertual barrier optiontake a look at the thread http://www.wilmott.com/310/messageview. ... d1=paradox

reza,Thanks very much, that answered my question. I was just concerned that my solution was way too simple, but I guess not. However that thread confused me because I don't see the paradox. I don't get why S approaches the barrier with probability 1. Doesn't S approach the barrier with risk-neutral probability S(t)/barrier?

Piranha,The paradox arises because under the standard lognormal stock price model, the stock price can never actually reach 0, it just gets lower and lower. So the stock will wander around, and as it never reaches 0, the only barrier it can actually cross is 100 - so that will be the barrier that it crosses, eventually.Another way of looking at it is to assume that the stock goes up by a factor of u with probability p and down by a factor of d with probability q. And 100/75=u^3So the probability of it getting to 100 eventually is:ppp + x1 qpppp + x2 qqppppp + x3 qqqpppppp + ...(where xi is the number of ways in which the stock can get to 100 at time 2i+3 but not before. )I used to know how to prove it (but have forgotten how to!), but it is reasonably easy to show that where p>=q this series works out to 1.

Got it. Thanks for the explanation gjlipman.

A quick answer might be like this, let tau be the first time S hit 100 or 0, tau1 is the first time S hit 100 before it hit 0, tau2 be the first time S hits 0 before it hits 100. Now the price should be given by the probability of S hits 100 before it hits 0, i.e. tau=tau1. Let the a=prob(tau=tau1), b=prob(tau=tau2). We have 100*a+0*b=75, so a=0.75 is the answer.

Martingale - that proof works well, however it assumes that it is an aritmetic r.w. and not a geometric random walk - in a geometric random walk it will never get to zero.

should we also incorporate the credit risk? i.e. when the stock price fall below a certain level, the firm will go bankrupt. In that case, it will not reach 100 with probability 1. Does that make sense?

I think we need to ask ourselves what we are trying to work out - the true value of the instrument, or the value that the model should spit out. For example, we can use a lognormal b.m. (as in B-S), which will mean that the stock will never go to zero (and never go broke). We can fix this up by using a lognormal b.m., but consider that the company is broke if the stock price goes below a value k. Or we can use a normal b.m. (arithmetic rather than geometric). Each of these will have different implications. And will imply different values and probabilities of exercise for the perpetual binary barrier discussed.Also, I think (from reading the thread linked below) that if you are using a lognormal b.m. and building a binomial tree then while the upwards movement u will equal 1/d (where d is the downward movement), in the case of r=0 then the probability that it goes up = (1-d)/(u-d)<50%. This means that the probability that it reaches the top barrier in a finite amount of time will be less than 1. This is the reason that the option won't have value $1.

What I used in my argument does not assume any model, except it is a martingale, and it might hit zero. We can assume that instead of zero, taking a small number epsilon and let it goes to zero and we still get the answer, or under a default risk, the equity does have probability to hits zero. Johnny can give us a model here... Johnny? (something like the creditgrade model)

Thanks for the reference Martingale - I'll do my best to live up to it! The short answer is that you may as well follow the CreditGrades approach and assume a zero share price in default. This obviously means that you also need to assume a share price distribution that will allow for a zero share price. CreditGrades assume a mixture of GBM and Arithmetic BM (ABM). Here's the longer answer:If the legal system gives shareholders control of the "corporate reorganisation" you can bet that when the dust settles the shareholders will get something, even if it's not much. This is relevent to the US as Chapter XI does indeed give control to the shareholders (known as "debtor in possession"). This means that for US shares it's appropriate to assume some low (but non-zero) share price in default as (almost without exception) shareholders never agree to a bankruptcy settlement that gives them nothing.At the other end of the scale is the English bankruptcy process of liquidation in which the senior creditors appoint a liquidator to sell assets until they have got their money back. The obligations of the liquidator are to the creditors, not to the shareholders, which means that this process can turn into a firesale. In this circumstance its often better to assume a zero share price in default.CreditGrades assumes a zero share price in default regardless of jurisdiction. This raises the issue that a share price cannot reach zero under GBM. CreditGrades get around this issue by modelling the share price volatility as a mixture of GBM and ABM, giving a function like this:Vol (SharePrice) = A + B/SharePriceand a shareprice process of:dS = m.S dt + (A.S + B) dWFinal conclusion: if you want to assume zero share price, choose a share price distribution that allows it. If you want to assume a non-zero share price, don't spend too much effort trying to model the low share price, because it's the outcome of a game between creditors, shareholders, the tax authorities and various other parties. Choosing an arbitrary low share price is likely to be just as accurate as trying to model this process!

QuoteOriginally posted by: rezathis sounds like a perpertual barrier optiontake a look at the thread <a target=new class=ftalternatingbarlinklarge href="http://www.wilmott.com/310/messageview. ... ox</a>Well mighty Piranha this is a perpetual binary option , this can also be priced using a simple arbitrage argument with vanilla calls and the underlying asset suppose you wrote 100 of those calls and pocketd the premiums , and if the premiums were bigger than 0.75$ then by simply buying the asset with the $75 and pocketing the extra change you are in for sure profit, either which way (if IBM hits 100 or not)If the premium was lesser than $0.75 then you could do the reverse and so by simple arbitrage the price has to be $0.75<hehe>

Well done quantie - I like that proof! And it works regardless of the interest rate or the model assumed by the stock (eg geometric B.M or a.b.m or credit risk inclusive).Just extending on from my ramble below - a proof that under the geometric brownian motion the probability of reaching the barrier of $X is equal to the stock price/k.We choose u and d from the volatility such that d=1/u.We choose p and 1-p from u, d, mu (which I am taking to be e^r). p=(mu-d)/(u-d)Let S represent the current stock price.Let P(s) represent the probability that the stock ever reaches barrier of $k given it is at s.ThenP(s)=p P(s u) + (1-p) P (s d)subject to boundary conditions:P(k)=100%0<=P(s)<=1 for all sThis has the solution (let me know if you want the details):P(s)=1 if p>= .5implies that P(S)=1 for any S.or if p<.5 (as it will be if mu=1)P(s)=[p/(1-p)]^k where s=X d^k or k=log base d of (s/k).and we can show pretty easily that in the second case P(s)=[(mu-d)/(u-mu)]^k = [d (mu-d)/(1-d mu)]^kIf S=X d^j then P(S)=[d (mu-d)/(1-d mu)]^jand if mu=1 then P(S)=d^j = S/X.Giving us the desired result.

I also like Quantie's proof, but it depends on IBM not paying dividends. This is true of all the pricing arguments. If IBM is planning to pay a liquidating dividend of $75 tomorrow, the option is worthless.My objection to this quiz is the assumption that you live forever and know for certain interest rates will be zero forever; yet you further assume that IBM must either go to zero or cross $100. Why cannot IBM simply continue to exist with a positive stock price under $100? You can't use the argument that something would drive it up or down eventually, because you're assuming certainty with respect to horizon and interest rates.If you stick in any limit on the horizon, then you will get some probability that IBM stays between $0 and $100 for the period. If you stick in any small uncertainty or interest rate, you get some probability that IBM will stay under $100 long enough that your payment is not worth much.In other words, while I agree that $0.75 is the answer to the quiz as posed, I assert that $0.75 is not the limit of the price as horizon goes to infinity, or as interest rate goes to zero.

Aaron,I agree that it is kind of a poor problem. A professor posed the problem in class and then afterword, as a hint, gave the assumption that IBM must go to 0 or 100. I wasn't sure if it was a valid hint, which is why I didn't include it in the problem statement (it was only part of my guess at a solution). However even without the assumption that IBM goes to 0 or 100 the option must be worth 0.75 as was shown by quantie.Oh, and the problem does assume no dividends (I should have stated that).