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If you want a really intuitive answer, I thought of two things to explain the key idea: From a Masters student point of view: Assume $\xi \sim N(0, \sigma^2)$, prove that: $\mathbb{E}[f(\xi)\xi] = \sigma^2 \mathbb{E}[f'(\xi)] = \mathbb{E}[\xi^2]\mathbb{E}[f'(\xi)]$ A more intuitive financial explanation may go like this: Consider you have a binary ...


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Different measures have different properties. Using a particular measure may make it easy to derive an analytic formula since a rate is driftless. When performing Monte Carlo, the sign of the drifts changes with measure which affects convergence. There is also the problem in the terminal measure that the numeraire can get very small and so some paths can ...


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If I am not mistaken, the Feynman-Kac formula is related to the Kolmogorov's backward equation, so I would expect it to be available only for Markov processes. Diffusions are usually of Markovian type, in contrast to general Ito processes or more to say, general semimartinagales. Intuitively, the PDE/PIDE/... will describe the dynamics of ...


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I can clarify 100% that $(dw)^2$= $dt$ and recommend you to accept it as a fact. Like any other differential, this differential is defined in terms of its integral: $$ \int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}[W(t_{k+1})-W(t_{k})]^{2} $$ Where $t_{k}=t_{0}+k(t_{1}-t_{0})/n$. Since $$ ...


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You derivation here is flawed because you are deriving with respect to two processes and you do not take into account that the variable $W_t$ is stochastic and hence $S_t$ is as well. So, to derive $S_t$ from $dS_t$, you have to apply Ito's Lemma, see this question for details. This is the "classic" way you see it. If you want to do it the other way ...


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A martingale must have constant expectation, such that adding a deterministic finite variation process $(b-r)dt$ would break the martingale property (except for when its a constant, which it is not by multiplication with $dt$). Hence the finite variation process must be eliminated under $Q$ for LRS to be an (equivalent) martingale measure, and as shown the ...


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I saw a quote from Brigo & Mercurio "IR models" (page 26, 2.1 No-Arbitrage in Continuous Time) . May be it will help you to find answer: Harrison and Pliska (1983) proved the following fundamental result. A financial market is (arbitrage free and) complete if and only if there exists a unique equivalent martingale measure.



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