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4

First, simulate the paths of the process $(X_t)$ defined by $$X_t=\sigma W_t+\left(r-\tfrac12\sigma^2\right)t,$$ using Euler's scheme $X^\varepsilon_0=0$ and$$X_{n+1}^{\varepsilon}=X_n^{\varepsilon}+\sigma\sqrt\varepsilon Z_n+\left(r-\tfrac12\sigma^2\right)\varepsilon,$$ for every positive $\varepsilon$ and every $n$, where the process $(Z_n)$ is i.i.d. ...


3

The main goal of the exercise is to make you realize (and use the fact) that the random variable $X$ defined as $$ X=\int_0^tY_s\mathrm ds, \qquad Y_s=W_s-\frac{s}{t}W_t, $$ is a linear combination of the gaussian family $(W_s)_{0\leqslant s\leqslant t}$ and that, as such, $X$ is itself gaussian. Hence, to fully determine the distribution of $X$, all ...


3

Hint: Apply Itô's formula to $$Z_t := \text{arsinh} \, X_t = \int_0^{X_t} \frac{1}{\sqrt{1+y^2}} \, dy.$$ General approach: The given SDE is an autonomous, i.e. the coefficients do not depend explicitly on the time: $$dX_t = b(X_t) \, dt + \sigma(X_t) \, dW_t. \tag{1}$$ In some cases, these SDEs can be transformed into linear SDEs - and since linear SDEs ...


3

Let $(X_t,\mathcal{F}_t)_{t \geq 0}$ be an (càdlàg) $L^2$-martingale, i.e. a martingale which satisfies $$\sup_{t < \infty} \mathbb{E}(X_t^2)<\infty.$$ Then it follows from the Doob-Meyer decomposition that there exists a unique increasing previsible process $(A_t)_{t \geq 0}$ such that $A_0=0$ and $$(X_t^2-A_t,\mathcal{F}_t)_{t \geq 0} \,\, ...


2

It is exactly the other way round: We define $(1)$ to be the same thing as $(2)$; it's just a more convenient way to write $(2)$. That this notation makes sense can be seen by integrating both sides of $(1)$ from $0$ to $T$: $$\underbrace{\int_0^T dX_t}_{X_T-X_0 = X_T-x} = \int_0^T b(t,X_t) \, dt + \int_0^T \sigma(t,X_t) \,dW_t.$$ A similar situation pops ...


2

Expand the inequality $\langle X-Y\rangle\geqslant0$.


1

Hint: the SDE verified by $r(s+t)$ is the same as $r(t)$, hence $$ r(t+s) = r(t)e^{-\alpha s} + \mu (1-e^{-\alpha s}) +\sigma \int_0^se^{-\alpha (s-u)}dW_{t+u} $$as in your first expression, with initial condition $r(t)$.


1

Let $Y_t=\displaystyle\int_0^t\mathrm e^{au}\mathrm dW_u$ then, for every $t$, $X_t=b\mathrm e^{-at}Y_t+z(t)$ where $z(\ )$ is deterministic hence $$\mathrm{cov}(X_t,X_s)=(b\mathrm e^{-at})(b\mathrm e^{-as})\mathrm{cov}(Y_t,Y_s),$$ and, for every $t$, $Y_t=\displaystyle\int_0^\infty g_t(u)dW_u$ where $g_t$ is deterministic since $g_t(u)=\mathrm e^{au}\mathbf ...


1

For simplicity, I'll assume that $(M_t)_{t \geq 0}$ is a continuous martingale. The argumentation is similar for a local martingale, but more technical. Set $$I_n(t) := \int_0^t I_{n-1}(s) \, dM_s, \qquad n \in \mathbb{N}, t \geq 0, \tag{1}$$ and suppose that $$(n-1) I_{n-2}(t) = I_{n-2}(t) M_t - I_{n-3}(t) \langle M \rangle_t, \qquad t \geq 0. \tag{2}$$ ...


1

Complementing the answer above to make explicit use of Ito's isometry as you requested. The appropriate version of ito's isometry to use in this case is the following: $$ \mathbb{E} \int_0^T f(r)dW(r) \int_0^T g(r) dW(r) = \mathbb{E} \int_0^T f(s)g(s)ds,$$ where in your case $f(r) = e^{ar} \mathbf{1}_{(0,s)}(r)$, and $f(r) = e^{ar} \mathbf{1}_{(0,t)}(r)$. I ...


1

Quite generally, if the canonical filtration $(\mathcal F_t)$ is generated by the process $N=(N_t)$, then, for every stopping time $T$, the sigma-algebra $\mathcal F_T$ is generated by the process $N^T=(N^T_t)_t$ obtained by stopping $N$ at time $T$. That is, $N^T_t=N_{\min\{T,t\}}$ for every $t\geqslant0$. In your case $N^T$ is $\sigma(T)$-measurable since ...



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