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 $\frac{1}{H}$. I'm stuck on the reasoning.

Clearly I'm not going to pay more than \$$\frac{1}{H}$ for this option. If $C > \frac{1}{H}$ 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 $C<\frac{1}{H}$? 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 $1-CH > 0$ 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 $C S_t $ where $S_t$ 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: $S(t) \equiv 1.$ 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 $\sigma > 0.$ Option price $C(S,t)$ satisfies a PDE $C_t + 1/2 \sigma^2 S^2 C_{ss} = 0.$ Since it is perpetual, $C(S,t)$ cannot depend on $t$ and so $C_t = 0.$ Then the PDE reduces to an ODE $C_{ss}=0.$ 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 $\sigma > 0$ 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 $-1/2 \sigma^2 dt.$ 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 $S(t) \ge 0.$ In this case the original poster correctly argues the fair value has $C \le 1/H.$ 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 $0 \le C \le 1/H$ is possible: consider the model where at time $0$ with risk-neutral probability $HC$ the stock follows a GBM with volatility $\sigma > 0$ and with probability $1-HC$ it remains fixed at 1 forever. The expected value under the risk neutral measure is $HC \cdot 1/H + (1-HC) \cdot 0 = C.$

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.