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3

Providing only a sketch here, using Ito integral definition (and commuting limit, summations and expectation), the result boils down to studying the expectation term: $$ E\left[ f_{t_{i-1}} (W_{t_i}-W_{t_{i-1}}) \cdot g_{t_{j-1}} (W_{t_j} - W_{t_{j-1}}) \right]. $$ If the intervals don't overlap, $i\not= j$, and say $t_i \leq t_{j-1}$, then $f_{t_{i-1}} g_{...


0

I think you can't definitely solve the integral, but there should be some way to solve for a distribution on the integral. Provided the integral is smooth enough for all values the random variable could take on. Good luck!


5

No. Itō’s formula helps you derive the dynamics of $f (S_\cdot )$ given the SDE followed by $S$. Here this is not the case. You simply have: $$ \mathrm{d} \left[\int{g(S_t)\mathrm{d}S_t}\right] = g(S_t) dS_t $$


5

By the definition of the quadratic covariation $$ \int_0^t dD_u dS_u = [D,S]_t = \lim_{\Vert P\Vert \to 0}\sum_{k=1}^{n}\left(D_{t_k}-D_{t_{k-1}}\right)\left(S_{t_k}-S_{t_{k-1}}\right). $$ We note that: $$|\sum_{k=1}^{n}\left(D_{t_k}-D_{t_{k-1}}\right)\left(S_{t_k}-S_{t_{k-1}}\right)|\leq \max_{1\leq k\leq n} |S_{t_k}-S_{t_{k-1}}| \left( \sum_{k=1}^{n}|D_{...


2

I don't think that you can apply Girsanov's theorem in this way, noting that I don't understand your (short) argument in the comments. This is how I would proceed, which makes sense mathematically, but the economic interpretation is then a bit strange. Let us write the SDE's a bit different, noting that it still preservers the same correlation structure \...


1

A Lévy process is defined as (Lévy process and Stochastic Calculus, David Applebaum): Suppose that we are given a probability space $(\Omega, \mathcal{F}, P)$. A Lévy process $X = (X (t), t \geq 0)$ taking values in $\mathbb{R}^d$ is essentially a stochastic process having stationary and independent increments; we always assume that $X (0) = 0$ with ...


1

Consider the generalised geometric Brownian motion $$\text{d}S_t = \mu(t)S_t \text{d}t+\sigma(t)S_t \text{d}W_t.$$ Using Itô's Lemma, you get $$\text{d}\ln(S_t) = \left(\mu(t)-\frac{1}{2}\sigma^2(t)\right)\text{d}t+\sigma(t) \text{d}W_t.$$ Thus, by definition of an SDE, $$\ln(S_t) =\ln(S_0)+\int_0^t \left(\mu(s)-\frac{1}{2}\sigma^2(s)\right)\text{d}s+\int_0^...


6

Consider the Heston (1993) model under the real world measure ($\mathbb{P}$) \begin{align*} \mathrm{d}S_t&=\mu^\mathbb{P} S_t\mathrm{d}t+\sqrt{v_t}S_t\mathrm{d}B_{S,t}^\mathbb{P}, \\ \mathrm{d}v_t&=\kappa^\mathbb{P} (\bar{v}^\mathbb{P}-v_t)\mathrm{d}t+\sigma^\mathbb{P}\sqrt{v_t}\mathrm{d}B_{v,t}^\mathbb{P}, \end{align*} where $\mathrm{d}B_{S,t}^\...


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