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Historical returns are not to be used 'untreated' for the calculation of option prices. The expectation that you will be using in Monte Carlo will take the form $$C(K,T) = E^Q\{D(T)\ \max[0, S_T-K, 0]\}$$ where $T$ is the maturity, $K$ is the strike price, $S$ is the stock price and $D$ is the discount factor. But the expectation is taken under the 'risk ...

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Most of the time, when you have a simple SDE without a drift, it's a martingale because the Wiener process itself is a martingale. In your example, you have a constant with the Wiener process, therefore the whole process must also be a martingale because the expectation is clearly X(t). However, we can't conclude a driftless SDE is always a martingale. ...

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You can use the "Merton Jump Diffusion Model" to price European Options with jumps. The other points of your question are rather of practical relevance only. The negative drift of the underlying is usually not important, because the pricing goes under the riskneutral measure $Q$.

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I think that for any $q>0$ it becomes optimal to exercise an American call for a sufficiently high spot price $S$: if the spot increases enough, the dividend yield corresponds to sufficient cash dividend to render exercise optimal. This would happen irrespective of the value of $r$ or the sign of $r-q$. What matters is that, for a given strike $K$, the ...

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If the option is fairly priced, as under the Black-Scholes Model, one cannot gain by selling it early, because the money you get just reflects the fairly expected value. If you think you have some inside-information to be bullish on the underlying, you would not sell it early, but this is then just your assumption, so as you may aswell assume to be bearish. ...

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Note that, for a smooth function and constant a $$f(S_t) = f(a) + f'(a) (S_t-a) + \int_a^{\infty}(S_t-x)^+f^{''}(x)dx + \int_{0}^a(x - S_t)^+f^{''}(x)dx.$$ Then, the payoff $1/S_t$ can be approximately hedged by call and put options: \frac{1}{S_t} = \frac{1}{a} -\frac{1}{a^2}(S_t-a)+ 2\bigg[\int_a^{\infty}\frac{(S_t-x)^+}{x^3}dx + \int_{0}^a\frac{(x - ...

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