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Have you come across Ito lemma / Ito calculus?. As Gordon suggests: divide the top equation by St, so you get dSt/St which is same as differential of ln(St),which is $dln(St)=\frac{dSt}{St}$ by chain rule. Ito states: $df(W_t,t)=\frac{df}{dWt}\times dW_t+\frac{df}{dt}\times dt+\frac{1}{2}\times \frac{d^{2}f}{dW_t^{2}}\times dW_t^{2}$ Now from Ito: $d ln(... 2 I try to be precise. There is no dependence in$\omega$for the$t_i$sequence. simple process Let$t_0<t_1<\dots<t_n<t_{n+1}$be a increasing sequence of real-numbers then$f$is said to be a simple process on a filtered space$(\Omega,\mathcal{F})$if: $$f(t,\omega) = \sum_{i\geq 0}\mathbb{1}_{t\in[t_i,t_{i+1})}\xi_i(\omega)\text{ where }\... 0 Are you talking about something like this?$$dx(t)=\ldots\ dt+[x(t)]^\gamma\ dW(t)$$If \gamma is zero then you've got BM, if it's one you get GBM, inbetween you have a 'mix'. 0 Idea Let B be a standard brownian motion starting from x_0=0, m_T = \inf_{u\leq T}B_u and M_T =\sup_{u\leq T}B_u. Let's define if it exists for A\in\sigma(B_u,u\leq T), \mathbb{P}(A | B_T=x_T)\stackrel{\rm def}{=}\lim_{\varepsilon\to 0}\mathbb{P}(A|B_T\in(x_T-\varepsilon,x_T+\varepsilon))$$\begin{split} \mathbb{P}(\tau_U\leq T \cap \tau_U\leq ... 2 Assuming$\theta>0$(take$\tilde{X}=\mu-X$if it is not the case) Let us denote$\text{erfi}(x)$the imaginary error function Let us denote$\tau_L$,resp.$\tau_U$the hitting time of$L$resp.$U$where$L<U$1) Using Ito's lemma, prove that : $$Y_t = \text{erfi}\left(\sqrt{\frac{\theta}{\sigma^2}}\left(X_t-\mu\right)\right) \text{ is a martingale}$$ ... 2 :D Is it a joke? $$\underset{\kappa \to 0 }{\mathop{\lim }}\,\frac{1-e^{-\kappa(T-t)}}{\kappa}=\underset{\kappa \to 0 }{\mathop{\lim }}\,\frac{\frac{d}{d\kappa}\left(1-e^{-\kappa(T-t)}\right)}{\frac{d}{d\kappa}\kappa}=\underset{\kappa \to 0 }{\mathop{\lim }}\,(T-t) e^{-\kappa(T-t)}=T-t$$ 5 IMHO the problem isn't stated correctly indeed, in the sense that the Radon-Nikodym derivative provided as the "solution" is not the unique way to define a measure$\mathbb{Q}$equivalent to$\mathbb{P}$and under which$X_t$is a martingale. Just take $$\frac {d\mathbb{Q}}{d\mathbb{P}} =\mathcal{E}\left(-\int_0^t \cos(s) dW_s + a\right)$$ for any$a \in \...

2

By the usual integrating factor method, \begin{align*} r_t = r_0e^{-\beta t} + \int_0^t \alpha(s) e^{-\beta(t-s)}ds +\sigma \int_0^t e^{-\beta(t-s)}dW_s. \end{align*} Let \begin{align*} x_t &=\sigma \int_0^t e^{-\beta(t-s)}dW_s, \textrm { and}\\ y_t &=r_0e^{-\beta t} + \int_0^t \alpha(s) e^{-\beta(t-s)}ds. \end{align*} Then $r_t = x_t + y_t$, ...

2

\begin{align*} \int_0^T W(t)\, dt &{}= \int_0^T\!\!\int_0^t dW(u)\,dt \\ &{}= \int_0^T\!\!\int_u^T dt\, dW(u) \\&{}= \int_0^T (T - u)\,dW(u) \\&{}= TW(T) - \int_0^T u\, dW(u) \end{align*} however i am not sure if it is what you are asking for

8

First, note $$\mathbb{E^Q}\left[\int_0^t e^{-a(t-s)}dW_s\right]=0$$ and $$\mathbb{Var^Q}\left[\int_0^t e^{-a(t-s)}dW_s\right]=\mathbb{E^Q}\left[\int_{0}^{t} e^{-2a(t-s)}ds\right]=\frac{1}{2a}(1-e^{-2at})$$ therefore $$\mathbb{E^Q}[r_t]=r_0 e^{-at} + \frac{b}{a}(1 - e^{-at})$$ $$\mathbb{Var^Q}(r_t)=\frac{\sigma^2}{2a}(1-e^{-2at})$$ second The Itô integral ...

0

Hint Let $\,H_0(x,t)=1$ , $H_1(x,t)=x$ and for every $n\ge 2$ set $${{H}_{n}}(x,t)=x {{H}_{n -1}}(x,t)-(n-1)\,t\,{{H}_{n-2}}(x,t)$$ then ${{H}_{n }}(W_t ,t)$ is a Martingale. For exapmple $$H_1(W_t,t)=W_t$$ $$\qquad H_2(W_t,t)=W_t^2-t$$ $$\qquad\qquad H_3(W_t,t)=W_t^3-3tW_t$$ $$\vdots$$

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