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Important assumptions: - we have zero interest rate, - option is perpetual, EDIT: with probability 1, share price will hit the barrier $H$ (in fact this is a hidden assumption that price changes continuously or we can at least trade at the very moment when $S_t = H$). No, we can't assume that, because , as @q.t.f noted, it would imply arbitrage. ...


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Unfortunately I cannot upvote user2142's answer because I lack the reputation, but his reasoning makes sense to me: the price is $\$1/H$ because as the seller of the option you buy $1/H$ shares for the premium. You sell them when the $S_t$ hits $H$ to obtain the $\$1$ you have to pay to the option buyer. I think the price is model free for any model with ...


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Let $T= \inf\{t>0: S_t = H\}$. Then the option payoff is given by $\mathbb{1}_{\{T < \infty\}}$, and the value of the option is given by $\mathbb{P}(T< \infty)$. We assume that the stock price process is a geometric Brownian motion, that is, for $t>0$ $$ S_t = \exp\big(-\frac{1}{2}\sigma^2 t + \sigma W_t\big),$$ where $\{W_t, t \geq 0\}$ is a ...


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Look at the B-S parameters for the dynamics of the stock. $\frac{dS}{S} = \mu dt + \sigma dt$ $\sigma$ is independent of strike in the B-S model, which means all derivatives priced assuming these dynamics should have the same volatility. This clearly is not the case given the existence of smile and skew. You can't assume the BS model produces the "fair" ...


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The claim payoff you describe, $g(M)$, looks to me like a tight butterfly spread that pays off only in one state of the world. Can't you just replicate that by short two calls with strike $K_0$ and long two calls, with strikes one either side at $K_0\pm 1$? Then the price of your option would be $C(K_0+1)+C(K_0-1)-2\cdot C(K_0)$. This is effectively the ...



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