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3

For the last question. We assume that \begin{align*} S_t = S_0 e^{(r-q-\frac{1}{2}\sigma^2)t + \sigma W_t}, \end{align*} where $W$ is a standard Brownian motion, $r$ is the interest rate, $q$ is the dividend yield, and $\sigma$ is the volatility. Then, \begin{align*} X_{u+a}-X_a &= (r-q-\frac{1}{2}\sigma^2)a + \sigma(W_{u+a}-W_u)\\ &\sim ...

3

For the first question, since by definition, \begin{align*} \varepsilon_t^{if} = e^{i \int_0^{t}f\big(\frac{1}{\xi}\langle M\rangle_s\big)\frac{dM_s}{\sqrt{\xi}} + \frac{1}{2}\int_0^t f\big(\frac{1}{\xi}\langle M\rangle_s\big)\frac{d\langle M\rangle_s}{\xi}}, \end{align*} then, \begin{align*} d\varepsilon_t^{if} = i \varepsilon_t^{if} ...

3

Note that, for $0 \leq s < t$, \begin{align*} W_t^3 &= (W_t-W_s+W_s)^3\\ &= (W_t-W_s)^3 + 3(W_t-W_s)^2 W_s + 3 (W_t-W_s) W_s^2 + W_s^3. \end{align*} Moreover, \begin{align*} E\big( (W_t-W_s)^3 \mid \mathcal{F}_s\big) &= E\big( (W_t-W_s)^3\big)\\ &= 0,\\ E\big((W_t-W_s)^2 W_s \mid \mathcal{F}_s\big) &= W_s E\big( (W_t-W_s)^2\big)\\ ...

2

Note that $X$ is a continuous martingale. Moreover, the quadratic variation is given by \begin{align*} \langle X_t, \, X_t\rangle = \int_0^t |\sigma_u|^2 du = c^2 t. \end{align*} That is, \begin{align*} \langle X_t/c, \, X_t/c\rangle = t. \end{align*} From Levy's characterization, $X/c$ is by law a Brownian motion, which we denote by $\beta$. Then, by law, ...

2

It appears that we need only to observe the following: \begin{align*} \lim_{\lambda\rightarrow 0}\frac{1}{\lambda}\int_0^{\lambda t}\sigma^2_u du &= \lim_{\lambda\rightarrow 0}\int_0^{ t}\sigma^2_{\lambda u} du\\ &= \int_0^{ t}\sigma^2_{0} du \\ &=\sigma^2_{0} t. \end{align*}

2

In order to apply Ito's lemma, your function needs to be a twice-differentiable function. There is no issue with the non-differentiability of the Wiener process. $\frac{dF}{dX}$ involves differentiating F, not the Wiener process X. Using a simple analogy: instantaneous velocity ($\frac{dD}{dt}$) is the derivative of position (D) over time; what is ...

2

We write the differential form of Ito formula for simplification. Actually, the differential form for Ito formula $$dF(W(t)) = 2W(t)dW(t) + dt$$ means the integral form for Ito formula, $$\int{dF} = \int{2W(t)dW(t)} + \int{dt}$$ which make sense in mathemaitcs.

1

You are right that a Wiener process can not be differenciated in the conventional way since the derivative in respect to time does not exist. For this reason Ito lemma should be used to integrate and differenciate Brownian or Wiener processes as these are considered ito processes.

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