Monday, April 1, 2019

no arbitrage theory - What is the fair price of this option?


Without having to use Black-Scholes, how do I price this option using a basic no-arbitrage argument?


Question


Assume zero interest rate and a stock with current price at $1 that pays no dividend. When the price hits level $H (H>1) for the first time you can exercise the option and receive $1. What is the fair price C of the option today?


My thoughts so far


According to my book, the answer is 1H. I'm stuck on the reasoning.


Clearly I'm not going to pay more than $1H for this option. If C>1H then I would simply sell an option and buy C shares with 0 initial investment. Then:



  • If the stock reaches H I pay off the option which costs $1 but I have $CH>1 worth of shares.

  • If the stock does not reach H I don't owe the option owner anything but I still have CH>0 shares.



What if C<1H? Then CH<1 and I could buy 1 option at $C by borrowing C shares at $1 each. Then:



  • If the stock reaches H then I receive 1CH>0 once I pay back the C shares at $H each.

  • But if the stock does not reach H, then I do not get to exercise my option and I still owe CSt where St is the current price of the stock. This is where I am stuck.



Answer



This option is a perpetual one touch option. Its price depends on the model used; additional assumptions are required to get a model-independent price.


Let us first consider 3 important example models for stock price S.


Constant: S(t)1. There is 0 probability that the perpetual one touch pays off, so its price is 0.



Black-Scholes: S follows geometric Brownian motion with volatility σ>0. Option price C(S,t) satisfies a PDE Ct+1/2σ2S2Css=0. Since it is perpetual, C(S,t) cannot depend on t and so Ct=0. Then the PDE reduces to an ODE Css=0. With boundary values C(0)=0 and C(H)=1 the solution is C(S)=S/H. With S(0)=1 option value is 1/H.


Bachelier: S follows arithmetic Brownian motion with volatility σ>0 and no drift. Since Brownian motion is recurrent, with probability one S will reach the level H. Thus the perpetual one touch has value 1.


Note: Geometric Brownian motion is not guaranteed to reach the level H. When we take log of GBM, it is an arithmetic Brownian motion with drift 1/2σ2dt. This negative drift is enough to allow some paths of log-spot to stay below the barrier level at log(H). The probability of hitting the barrier is the option price C(S,t) we calculated by PDE above.


Now lets return to the original question about making a model-independent no arbitrage price. Clearly from the examples it is impossible; different models give different prices.


We can get a little farther by assuming that S(t)0. In this case the original poster correctly argues the fair value has C1/H. But we still get a range of prices. The Black-Scholes model with zero rates and positive volatility gives C=1/H. But for the constant model the fair value is 0. Any value 0C1/H is possible: consider the model where at time 0 with risk-neutral probability HC the stock follows a GBM with volatility σ>0 and with probability 1HC it remains fixed at 1 forever. The expected value under the risk neutral measure is HC1/H+(1HC)0=C.


There is no obvious choice for a hidden assumption to rule out these other models. So there is not a model-free fair value of this option.


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