Tag Info

Hot answers tagged

3

Hints Brownian motion: Apply Itô's formula to $f(t,x) := \exp(\beta x - t \beta^2/2).$ Poisson process: Use that $(N_t)_{t \geq 0}$ has independent increments, i.e. $N_t-N_s$ is independent of $F_s^N$. Start with the easier one: $$\mathbb{E}(N_t-\lambda t \mid F_s^N) = \mathbb{E}((N_t-N_s)+N_s \mid F_s^N) - \lambda t = \ldots$$


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

Re 1., the expanded definition (maybe clearer) of the random process $\tilde B$ is that, for every $\omega$ in $\Omega$, $\tilde B_{U(\omega)}=0$ and $\tilde B_t(\omega)=B_t(\omega)$ if $t\ne U(\omega)$. In short, $$ \tilde B_t=B_t\,\mathbf 1_{U\ne t}. $$ Re 2., the almost sure path discontinuity of $\tilde B$ should be obvious. To wit, the event that ...


2

One is the probability that at least one path will hit the ball. Actually, no (but "the probability that at least one path" is difficult to interpret anyway). (1) is the probability of all the paths that hit the boundary ever. The other is the probability of all the paths that start from $x$ and hit $\partial B_{0,r}$ at time $t$. Actually, no. ...


2

They are functions of the independent vectors $\bigl(X(s),X(t)\bigr)$ and $\bigl(Y(s),Y(t)\bigr)$, QED.


2

Yes--and this is quite general: for every collection $(X_s)_{0\leqslant s\leqslant t}$ of random variables, $$\sigma(X_s;0\leqslant s\leqslant t)=\sigma\left(\bigcup_{0\leqslant s\leqslant t}\sigma(X_s)\right).$$ Proof: Double inclusion.


2

For each fixed $t$, the random variable $X_t=W_{2t}-W_t$ is centered normal with variance $t$ hence indeed distributed like $W_t$. But the process $(X_t)$ is not a Brownian motion. To see this, note that $X_2=W_4-W_2$ and $X_1=W_2-W_1$ are the increments of a Brownian motion on some disjoint time intervals hence they are independent while $W_2$ and $W_1$ ...


1

Here is a more general statement: Consider some Markov process $(Y_t)$ and define, for every $t$, $Z_t=F(t,Y_t)$, where $F$ is measurable and, for every $t$, $F(t,\ )$ has a measurable inverse. Then, $\sigma(Z_s;s\leqslant t)=\sigma(Y_s;s\leqslant t)$ for every $t$ and the process $(Z_t)$ is Markov. Can you prove this and apply it to your setting?


1

$\{W_0>0\} = \{\tau_+=0\}$ is not correct. Note that $\tau_+(\omega)=0$ if, and only if, there exists a sequence of numbers $(t_n)_n \subseteq [0,\infty)$ such that $t_n \to 0$ and $W_{t_n}(\omega)>0$. Show the following equalities: $$\begin{align*} \{\tau_+=0\} &= \{\omega; \exists (t_n)_n \subseteq [0,\infty), t_n \downarrow 0: ...


1

For every nonnegative $u$, the distribution of $W(u)$ is $N(0,u)$ hence, for any parameter $u(x)$, the assertion that $T(x)\stackrel{d}{\to}N(0,u(x))$ and the assertion that $T(x)\stackrel{d}{\to}W(u(x))$ are logically equivalent.


1

It follows straight from the definition of convergence in distribution that $$(X_j^n-X_{j-1}^n)_{j=1}^t \stackrel{d}{\to} (W(y_j)-W(y_{j-1}))_{j=1}^t.$$ Define a mapping $g: \mathbb{R}^t \to \mathbb{R}^t$ by $$g(x) :=(x_1,x_1+x_2,\ldots,x_1+\ldots+x_t).$$ where $x := (x_1,\ldots,x_t)$. Then $g$ is continuous and $$g((y_j-y_{j-1})_{j=1}^t) = ...


1

$U$ is a random variable, but its value is determined before the Brownian motion begins to evolve, we can tell if t equals $U$ or not. Look at the finite dimensional distribution $(\tilde{B}_{t_1}, \cdots, \tilde{B}_{t_n})$ of $\tilde{B}_t$, we have $$(\tilde{B}_{t_1}, \cdots, \tilde{B}_{t_n}) = ({B}_{t_1}, \cdots, {B}_{t_n})$$ on the event $I = \{U\neq ...


1

Assume that $X$ solves $$ dX_t = \mu(t,X_t) dt + \sigma(t, X_t) dW_t. $$ In integral form, this means $$ X_t = X_0 + \int_0^t \mu(s,X_s) ds + \int_0^t \sigma(s,X_s) dW_s. $$ Now, the quadratic variation of a stochastic integral process $H\cdot Y$, where $(H\cdot Y)_t = \int_0^t H_s dY_s$, is $$ [H\cdot Y]_t = \int_0^t H_s^2 d[Y]_s. $$ You can find this ...


1

The conditions that a process $X$ is has independent increments and that these increments are normal uniquely determines the finite-dimensional distributions of the process, and so, in law, there exists only a single Levy process with normal increments, and that Levy process is the Brownian motion. However, stochatic processes can exist relative to ...


1

The line B2 <- t*rev(B1) produces $t\mapsto t\cdot B(1-t)$, not $Y:t\mapsto t\cdot B(1/t)$. Two signs of what is going wrong are that B2 produces graphs that are (i) too flat near $t=0$ and (ii) quite similar near $t=1$ to a mirrored version of the B1 graphs near $t=0$.


1

Let $(t_i)_{i \in \mathbb{N}}$ be a dense subset of $[0,T]$ such that $t_k = T$ for some $k \geq 1$. If we set $\mathcal{H}_n := \sigma(B_{t_1},\ldots,B_{t_n})$ and $$\mathcal{H}_{\infty}:= \sigma \left( \bigcup_{n \in \mathbb{N}} \mathcal{H}_n \right) \qquad \quad \mathcal{F}_T := \sigma(B_s; s \leq T),$$ then $\mathcal{H}_{\infty} = \mathcal{F}_T$. ...



Only top voted, non community-wiki answers of a minimum length are eligible