Why No-Arbitrage Alone Cannot Price a Barrier Option

Summary

The document considers an option that pays one dollar if a non-dividend-paying stock, initially priced at one dollar, reaches an upper barrier. A geometric Brownian motion assumption under risk-neutral pricing gives the questioner a value of the reciprocal of the barrier. The accepted answer explains that absence of arbitrage does not by itself determine this derivative’s price: the underlying stock’s dynamics affect the probability of reaching the barrier and therefore the payoff value.

As a counterexample, the answer considers a constant stock price with zero interest. The discounted stock remains a martingale, so the example is arbitrage-free under the stated argument. If the barrier is above that constant price, the stock never reaches it and the option is worthless. This illustrates why different admissible dynamics can imply different prices. The text does not develop a broader pricing framework or specify market completeness; its point is that no-arbitrage alone is insufficient to recover the geometric-Brownian-motion valuation.

Key ideas

• The value of a barrier option depends on the underlying price process and its chance of reaching the barrier.
• No-arbitrage does not uniquely determine the option value without further assumptions about dynamics.
• A constant stock process with zero interest can be arbitrage-free while never reaching a higher barrier.
• Under that constant-process example, the barrier payoff is never triggered and the option is worthless.

Tags

# How to price this option without using BS framework


# How to price this option without using BS framework












We have a stock at price 1 dollar which pays no dividend. Also we assume zero interest rate. When the price hits $H$ dollars for the first time where $H>1$, we can exercise the option and receive 1 dollar. What is the price of the option?

I can price this option assuming the stock price follows a geometric brownian motion under risk neutral measure which gives me the price as $\frac{1}{H}$. I am curious if we can price this option without this assumption, using only no arbitrage principle.

## Answer by Probilitator (score 4, accepted)

https://quant.stackexchange.com/a/10845

The dynamics of the underlying stock process are obviously crucial to the derivative's price. Thus if you don't necessarily assume $S_t$ to be log normally distributed (B&S-Model) you won't get the same price even if the market is arbitrage free.

Example: Assume $S_t=C$ $ \forall t \in \mathbb{R}^+$ and $r=0$. Thus $S_t$ is constant and the interest rate equals zero. In this setting $S_t$ will be a martingale under the bank account numeraire. To be more precise $$E\left[e^{-\int^t_0 r_s ds}S_t |\mathcal{F}_u\right]=E\left[e^{-\int^t_0 0 ds}C|\mathcal{F}_u\right]=C$$. Now with $S_t$ being contant this mmeans $$E\left[e^{-\int^t_0 r_s ds}S_t |\mathcal{F}_u\right]=S_u$$ By the first fundamental theorem of asset pricing our market is thus free of arbitrage.

Now let us assume we are pricing the instrument described in your question. And let $H>C$. Obviosuly it's price is obiously equals zero for the proces is constant and will never hit the "knock-in-boundary" $H$

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