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What do you mean by "so we can price as usual"? What you showed is that for every $c \in \mathbb R$ we can find a probability measure such that the drift of $S$ is $c$. But that does not really say anything about pricing. You can easily see that $V_t = E^Q_t[e^{-r(T-t)} \Phi_T]$ does not give arbitrage free prices with your choice of $Q$. Indeed if $\Phi_T ...


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The first consideration is to set prices which do not generate arbitrage opportunities. The existence of a risk-neutral probability measure ensures that the model is arbitrage-free. In the Black-Scholes setting, as you mentioned, the market is complete and there as a unique martingale measure, hence only one possible price for each derivative. In a ...


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I would consider the financial applications of the Greeks: hedging. The "main" greeks, viz. Delta, Gamma, Theta, Vega and Rho, all have intuitive financial meanings. Gamma is the rate of change of your Delta (how many shares of stock to own) with respect to the stock price, so a high Gamma implies you will be rebalancing in large quantities (often ...


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My guess is that you may have different seasonalities in the data as per each agricultural commodity characteristics. My guess is that there may be seasonality within the year as per the specific commodity depending if one or more harvestings occur a year. Also due to the specific unceirtanty due to planting, growth season and harvesting times. In addition ...


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The clue was to establish that there are typos in the script. Hence, I should have aimed to prove $\frac{Ke^{-r\tau}}{S^2\sigma\sqrt{\tau}}\Phi'(d_2)=\frac{\Phi'(d_1)}{S\sigma \sqrt{\tau}}$. Therefore, we have \begin{equation*} \frac{\partial^2 C}{\partial S^2}=Ke^{-r\tau}\mathbb{E}[\frac{\delta(S-U)}{U}] = Ke^{-r\tau} \int^{\infty}_0 \frac{\delta(S-u)}{u} ...


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Since $Y=e^{(r-\frac{\sigma^2}{2})\tau + \sigma \sqrt{\tau}Z}$, then \begin{align*} xY > K \Leftrightarrow Z > -d_2, \end{align*} where \begin{align*} d_2 = \frac{\ln \frac{x}{K} + (r-\frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}. \end{align*} Consequently, \begin{align*} e^{-r\tau}\mathbb{E}\big(Y \mathbb{1}_{\{xY >K\}} \big) &= ...


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Try to give David Spiegelhalter a read/listen to David Spiegelhalter's work and research. He is a statistician and a Professor of the Public Understanding of Risk at Cambridge England. Rather than new ways of calculating risk, he looks at ways of communicating risk to a general public that doesn't have any knowledge of stats. I Linked an interesting ...


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Your derivation is incorrect because the option does not have a fixed maturity $T$. Instead it ends whenever the stock reaches 100. Mathematically this means that we have a stopping time which is a random variable $\tau(\omega) = \inf \{t \ge 0 | S_t(\omega) = 100\}$. It is usually assumed that the stock price eventually reaches 100 i.e. that $\tau$ is ...


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if only one person can make a choice, it strikes me as unlikely that it can reduce value. Ultimately, a choice means that the holder can choose between one of a number of portfolios on a given date. They will choose the one of maximal value. As long as the without choice portfolio was one of the ones they could have chosen, value can only go up.


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Maybe something like the option to lend money at negative interest would bring about a lower optionality value...



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