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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.

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Are you sure about your result? Assuming geometric Brownian motion, the probability of the stock price reaching $H>0$ is equal to $1$ (with infinite time). So the price of the option should be $1$ dollar, if interest rates are zero. – FKaria Apr 6 '14 at 14:53
Two period binomial? If stock goes to H or D, then price of the replicating portfolio is (1 - D) / (H - D). – Richard Herron Apr 6 '14 at 20:05
up vote 4 down vote accepted

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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