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5

It has to do with the Girsanov theorem that relates the equivalent measures $\mathbb Q$ and $\mathbb P\,.$ To make intuitively clear what happens I like to give the following "baby Girsanov" example: Let $X$ be standard normal having probability density $$p(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$ under $\mathbb P\,.$ If the equivalent ...

4

Intro: I think intuition is super important for this one, so my answer below focuses on the intuition here. Short answer The volatility parameter is meant to describe the behaviour of $S(t)$, whilst the drift only describes the probabilities of the instrument $S(t)$ moving up or down: changing the drift (and therefore the probabilities of "up" and &...

3

Extract from my answer about what the VIX measures (more details on the notation and the conventions I am using can be found in the preceding sections from that answer): About changing the measure (This section is based on currently non-public lecture notes by Dylan Possamaï for a course on mathematical finance. I will update it with precise references if ...

2

As you have already noted, $W_t$ is normally distributed with $$W_t\sim \mathrm{N}(0,t)$$ Then, $$\varphi(z)\equiv\mathrm{E}\left(e^{izW_t}\right)=e^{-\frac{1}{2}z^2t}$$ is the characteristic function of the Normal distribution.

1

Alternatively, you can solve the ODE and recover the result: With regards to the SDE for $X_t$, I got the same ODE as described above. We can derive the solution, $m(t)$, to the ODE as follows: \begin{align} \frac{dm(t)}{dt} &= \frac{1}{2} (iz)^2\cdot m(t)\\ &\Updownarrow\\ \frac{2\cdot\frac{dm(t)}{dt}}{m(t)} &= -z^2\\ &\Updownarrow\\ \int \...

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