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If $\mu$ is large, then it is more likely for the call to finish in the money. Your and my intuitions suggest that this means that the option is more valuable. But this is wrong. A call option is an insurance policy. A call option is useful because it protects you in the case that the value of the stock goes down. That is why call options are valuable for ...


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Girsanov'Theorem let $\theta_t$ be an adapted procee such that the solution of SDE $$dL_t=-L_t\, \theta_t \,dW_t , \, L_0=1$$ is a Martingale.We set $Q{{|}_{\mathcal{F}_t}}=L_t\,P{{|}_{\mathcal{F}_t}}$,then $$W_{t}^{Q}=W_{t}^{P}+\int_{0}^{t} \theta_s\,ds$$ is a standard wiener process under Q measure. Result Now we assume $\{S_t\}_{t\geq0}$ be a ...


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$r-\frac{\sigma^2}{2}$ for the drift only applies to the log-returns. The Euler discretisation simply discretises the SDE directly. You'd use the risk-free rate for you drift under the risk-neutral measure for your question. For your reference: Please read the wikipedia for more details.


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The Feynman-Kac theorem can be used in both directions. That is, If we know that $r_t$ follows the Ito process as described by the following stochastic differential equation \begin{align} d{{r}_{t}}=\mu ({{r}_{t}},t)dt+\sigma ...


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EDIT I think I figured it out. Under the $\mathbb{Q}$ measure, $\begin{equation} S_t \sim LN(ln(S_0) + r - 0.5\sigma^2, \ \sigma\sqrt{t}) \end{equation}$ Under the $\mathbb{P}$ measure $\begin{equation} S_t \sim LN(ln(S_0) + \alpha - 0.5\sigma^2, \ \sigma\sqrt{t}) \end{equation}$ Suppose we simulate the following stock prices Si = 96.33, 69.04, ...



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