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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.
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" $H$
