# Tag Info

1

We have: $$d(X_tY_t)=X_tdY_t + Y_tdX_t+d\langle X,Y\rangle_t={X_t\cos(X_t+Y_t)dW_t}+{Y_t\sin(X_t+Y_t)dW_t}+{\sin(X_t+Y_t)\cos(X_t+Y_t)dt}$$ Can you take it from here or do you need me to continue? EDIT: the two Brownian motions are in fact independent so the last term is 0. As said in a comment, you have to proceed differently.

0

Suppose $(\mu,\sigma)$ obeys the linear growth condition $$\left|\mu(t,x)\right|+\left|\sigma(t,x)\right|<C(1+\left|x\right|),\ \forall t\in[0,T],\, x\in\mathbf R$$ for some positive constant $C$. By the Cauchy-Schwarz inequality and the Gronwall inequality, $$\mathbf E[r^2]<3\mathbf E[r(0)^2]e^{a(1+T)t}, \forall t\in[0,T]$$ for some positive ...

1

Express $Y$ as an Ito process: $$dY_t=X_t\,dt = X_t\,dt +\ 0\,dB_t.$$

1

This might be helpful: Theorem 29 on page 173 of Protter's "Stochastic integration and differential equations": Let $M$ be a local martingale and let $H\in \mathcal P$ be locally bounded. then the stochastic integral $H\cdot M$ is a local martingale. I'm reading that book, but haven't got that far yet, will be interested to see if someone can provide a ...

2

4

The distribution of $\int_0^t W^2_s\,ds$ was found by Cameron and Martin in the 1940s. Mark Kac (1949, Transactions of the AMS) applied his method to find the Laplace transform of this integral: $$\Bbb E\left[\exp(-u\int_0^t W^2_s\,ds)\right]={1\over\sqrt{\cosh(t\sqrt{2u})}}.$$

1

It's enough that $f$ be Lebesgue measurable and $\int_0^t [f(s)]^2\,ds<\infty$ for each $t$. Indeed, in this case your stochastic integral is well defined. Moreover, if we set $A_t:=\int_0^t[f(s)]^2\,ds$, then $Y$ is a (mean zero) continuous martingale with quadratic variation $\langle Y\rangle_t=A_t$. Let $\tau(t):=\inf\{s:A_s>t\}$. Then ...

0

If a function is continuous, you can connect its Riemann integral with a limit of a sum of elements. In your case, they are normally distributed. EDIT : let n an integer, and we define the a subdivision of the interval $[0,t]$ , as follows $t_i=\frac{t}{n}i$ with $i \in${ ${0,...,n}$} Since $W$ is a.s continuous, as well as h, we have $t \rightarrow ... 0 Ok i forgot to change the boundaries, so we have: $$\frac{\lambda^n}{\Gamma(n)} \int_0^{\infty} \exp(-\frac{\lambda}{y}) \frac{1}{y^n} dy = \star$$ so i substitute$x = \frac{1}{y}, \; dy = -\frac{dx}{x^2} $and$\phi(y) = \frac{1}{y}$, so$\phi(\infty) = 0, \; \phi(0) = \infty$. $$\star =- \frac{\lambda^n}{\Gamma(n)} \int_\infty^{0} \exp(-\lambda x) ... 6 That should not be true. The problem is that you're not considering the square of the stochastic integral, i.e. the random variable J= (\int_0^1 {W_s} ds)^2, which would indeed be the square of a normally distributed random variable, but the r.v. S= (\int_0^1 {W_s}^2 ds). A note in the direction of finding a solution would be a discretization attempt, ... 3 \text{Cov} \, (X_t,X_r)=\text{Cov} \, (\int_0^ts^3W_sdW_s,\int_0^rs^3W_sdW_s)=\int_0^t s^6 W^2_s ds This identity does not hold true. Note that the left-hand side is a fixed real number whereas the right-hand side is a random variable. If you apply Itô's isometry correctly, you find$$\text{cov} \, (X_t,X_r) = \color{red}{\mathbb{E} \bigg( }\int_0^t ... 2 Let$X_{t}$and$W_{t}$be defined as above. Let's apply Ito with$f(x,y)=xy^{2}$to get the semimartingale decomposition of$f(X_{t},W_{t})=X_{t}W_{t}^{2}. \begin{align*} X_{t}W_{t}^{2}&=\int_{0}^{t}W_{s}^{2}\text{sgn}(W_{s})dW_{s}+2\int_{0}^{t}X_{s}W_{s}dW_{s}\\ ... 1 I would like to add something to the excellent answer of Siron: although it is good to know the Ito Integral of a deterministic function is Gaussian, we can solve my problem without that property. Let\quad g \colon [0,T] \longmapsto \mathbb{R}$a deterministic function. Then if we define:$\quad Y_t = \int_0^t g(s) \mathrm{d} W_s \iff \mathrm{d} Y_t = ...

1

Your calculation is almost good. However, your application of Ito's lemma is not entirely correct. I will outline the details for you: Since $X(t) = \int_{0}^{t} \sigma(s) dW_s$, it follows that $dX(t) = \sigma(t)dW_t$. Define $Z(t) = \exp(iuX(t)) = f(t,X(t))$. Now, an easy calculation shows $$\begin{cases} \displaystyle\frac{\partial f}{\partial t} = 0, \\ ... 2 Given that \sigma(s) is a deterministic function, then the process (X(t))_{t \geq 0}, where$$X(t) = \int_{0}^{t} \sigma(s) dW_s,$$is a Gaussian process with zero mean and covariance function \rho(s,t) = \displaystyle \int_{0}^{\min(s,t)} \sigma(u)^2 du. A proof of this theorem can be found in Schreve's stochastic calculus for Finance II. Hence, ... 3 Being predictable (as in being measurable with respect to the predictable \sigma-algebra) always implies being progressive (as in being measurable with respect to the progressive \sigma-algebra), but the other implication only holds in special cases, such as when e.g. the underlying filtration is the usual completion of the filtration generated by a ... 1 To answer your first question, stochastic integrals of this type are Ito integrals. They are defined as a specialized limit of a Riemann-like sum with respect to a partition. Under appropriate conditions on processes X_s and B_s, we can define the integral as a limit of left-hand (non-anticipatory) sums:$$\int_0^t X_s dB_s = \lim_{n \to \infty} ...

1

We have $\mathbb{P}(B_s=0)=0$ for all $s>0$, and therefore it follows directly from Tonelli's theorem that $$\mathbb{E} \left( \int_0^t 1_{\{B_s=0\}} \, ds \right) = \int_0^t \underbrace{\mathbb{E}(1_{\{B_s=0\}})}_{=\mathbb{P}(B_s=0)} \, ds = 0.$$ Since the integrand is non-negative this implies $$\int_0^t 1_{\{B_s=0\}} \, ds = 0 \quad \text{a.s.}$$

2

I'll rewrite the proof in terms of the process $\tilde{S}_t$, rather than using stochastic integrals. Well, let me first say that the idea of Girsanov's theorem is to 'eliminate' the drift term of $\tilde{S}_t$ by changing the probability measure (to $\mathbb{Q}$). We have, $$d\tilde{S}_t = \tilde{S}_t(\mu-r)dt + \tilde{S}_t \sigma dB_t.$$ Since $\tilde{B}_t ... 1 For Brownian motion it is true that$B=C$. This means that if$f\in B$there exists$g\in C$such that$f(\omega,t)=g(\omega,t)$for$\Bbb R\times\mu$-a.e.$(\omega,t)\in\Omega\times[0,T]$. This does not mean that$t\mapsto f(\cdot,t)$is a predictable process. The matter is discussed in some detail in Chapter 3 of Introduction to Stochastic Integration by ... 0 Here is the idea (you need to fill in some details)$R_j = o(|\Delta t_j|^2 + |\Delta X_j|^2)$means $$R_j \le r(|\Delta t_j|^2 + |\Delta X_j|^2)(|\Delta t_j|^2 + |\Delta X_j|^2)$$, where$r(y)\to 0 \ as \ y\to 0$(using the condition that$g$is twice continuously differentiable, you can show such$r$exists and does not depends on$j$). Now$\$|\sum_j R_j| ...

Top 50 recent answers are included