1
$\begingroup$

This it taken from "Heard on the Street", Section B.

Consider a market with $0$ risk-free rate, no transactions costs etc. The IBM stock costs \$75 and does not pay dividends. Design a security which pays \$1 if IBM stock reaches \$100. What does it cost?

The answer is \$.75 which can be proved by no-arbitrary considerations. At the same time, risk-neutral pricing suggests that the price is $$ \mathsf P\{\text{IBM has hit \$100 at least once}\} = 1. $$

| improve this question | | | | |
$\endgroup$
2
$\begingroup$

As I see it the question does not enforce that the market is free of arbitrage. This is why you can get to contradicting prices. Thus you can't actually apply a risk-neutral argument here without making additional assumptions.

You yourself provide the example of such an arbitrage. If the underlying process had a B&S dynamics you could just borrow money for free, buy the stock and wait until it hits the $100 mark.

Seeing how one does not know the dynamics of the underlying process it makes sense to enter into the static hedge of holding 0.75 of the stock.

Related Question: Price Option and find replicating portfolio

| improve this answer | | | | |
$\endgroup$
  • $\begingroup$ I think I see your point: when we are making an equivalent change of measure, we have to restrict ourselves to finite intervals of time, otherwise changing the drift changes the null events. Thanks $\endgroup$ – Ilya Apr 7 '14 at 13:48
  • $\begingroup$ Price this derivative in the digital framework. This will allow the same logic to be implied such as cash-or-nothing options. You could price in the barrier framework as well specifying a few extra parameters within the BS methodology. These methods will allow the security to be priced under the risk neutral measure implying a martingale. $\endgroup$ – Andrew Mann Apr 7 '14 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.