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6

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 ...


5

Let's skip to the stochastic differential equation (SDE): $$ dF=\left[\frac{\partial F}{\partial t}+\mu \frac{\partial F}{\partial x}+\frac{1}{2}\sigma^2 \frac{\partial^2 F}{\partial x^2} \right]dt + \sigma \frac{\partial F}{\partial x}dW $$ What does this equation actually represent? It suggests that a change in $F$ (represented by $\Delta F$) equals a ...


2

I'd like to give an alternative derivation not involving the clever (mystifying?) transformation to the heat equation and thus present a more general technique for solving constant coefficeint advection-diffusion PDEs. All we need is the Fourier transform: \begin{align*} \mathcal{F}[f] & = \int_{-\infty}^\infty e^{-i \omega y} f(y) dy, \end{align*} ...


1

As you have guessed correctly, these type of questions can be answered using Ito's Lemma.We have: \begin{equation} d(M_t)= d(Z_t e^{\int_0^tF(Z_u)du})=d(Z_t) e^{\int_0^tF(Z_u)du}+Z_t d(e^{\int_0^tF(Z_u)du})+d(Z_t)d(e^{\int_0^tF(Z_u)du}) \end{equation} For the first two terms on R.H.S, we have: \begin{equation} d(Z_t) e^{\int_0^tF(Z_u)du} = (f(W_t)dW_t + ...



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