16

I don't think that the reason given in the paper by Bailleul and Bernicot is a good one. Basically, they treat an example which is simple enough so that it is still manageable to describe the various bits and pieces needed to control their solutions "by hand" instead of combining them into a single object in a more coherent way. This being said, there are ...


15

First, note that the right comparison is not with the Riemann integral but rather with the Riemann-Stieltjes integral. To be concrete, consider $\int_0^1 X_s dW_s$ where $W$ is Brownian motion and $X_s$ is an adapted, not differentiable process (for example you can take $X_s=W_s$). Now replace $X_s$ by $X_s=X_{t_i}+\Delta_i$ where $\Delta_i$ is a random ...


12

Actually, it is quite possible to condition Brownian motion to hit a given point at a given time. The process is a called a Brownian bridge and the distribution of the winding number of the Brownian bridge is even known explicitly. It has been computed by Marc Yor in the paper Marc Yor: Loi de l'indice du lacet brownien, et distribution de Hartman-Watson. ...


11

A continuous strong Markov process is a semimartingale if and only if it can be represented as a time-changed Ito diffusion. More generally, a Hunt process is a semimartingale if and only if it is a time-changed jump diffusion. This is a result of Cinlar, Jacod, Protter, and Sharpe, from their truly amazing paper Semimartingales and Markov processes.


10

You might want to look into the Wong-Zakai theorem. Essentially, it states that if $\xi_\epsilon$ is a sequence of smooth approximations to white noise, then the solutions to the random ODE $$ {dx \over dt} = f(x) + g(x) \xi_\epsilon\;, $$ converge to the solutions to the SDE (in Stratonovich form) $$ dx = f(x)\,dt + g(x)\circ dW\;. $$ There are of course ...


9

The process $\|B(t)\|$ is called $n$-dimensional Bessel process (or Bessel process with parameter $\nu=\frac{n}{2}-1$). I think formula $\bf 4$.1.1.4 of Borodin-Salminen "Handbook of Brownian Motion -Facts and Formulae" is what you're looking for.


9

Unfortunately, calling the Skorohod integral an "integral" is a bit of a misnomer, as it doesn't really have many of the properties which you would naturally associate with integrals, except for the fact that it "magically" coincides with the Itô integral when its integrand is adapted. The best way (I think) to get some kind of intuition for the Skorohod ...


9

Your intutive reasoning is leading you astray because you are thinking of Brownian motion as behaving like a smooth curve, for which there is a well-defined "direction" in which it is heading. Brownian motion isn't like that. (As you probably know, it's almost surely nowhere differentiable, so its "velocity vector" is undefined.) Let $\tau$ be the time at ...


9

A good way to measure the set of maxima is the Hausdorff dimension of the set of records, which for BM is a.s. 1/2. Because of time/scale invariance, the dimension is the same for $\alpha B_\cdot$, and if the drift is nice enough to have absolute continuity, the same holds for your drifted diffusion. This however changes when you move from BM to fractional ...


8

I found this explanation somewhere and wrote it down in my personal notes. I will explain with an example that I think exemplify why Riemann-Stieltjes will provide the wrong answer. First, let's remember how we can define the Riemann-Stieltjes integral below \begin{equation}\int_0^t Z(x)dZ(x)=\lim_{n\rightarrow\infty}\sum_{k=1}^n(Z(t_k)-Z(t_{k-1}))Z(t_{k-1}...


8

Yes, you do need the integrand $f(X_t)$ to be predictable. If it is merely adapted you may not get a martingale. Intuitively, the stochastic integral $\int Y_t\,dM_t$ tells you the profit from a stock trading strategy. $M_t$ is the share price at time $t$, and $Y_t$ is the number of shares you hold at time $t$. Requiring that $Y$ be predictable means ...


8

Aspects of Brownian Motion (Mansuy & Yor) give an expression for the joint Laplace transform of $B_t$ and $\int_0^t B_s^2 ds$ (section 2.1). For $\delta$-dimensional Brownian motion, $$ \mathbb{E}\left[\exp\left(-\alpha|B_t|-\frac{b^2}{2}\int_0^t |B_s|^2 ds \right) \right] $$ is equal to $$\left(\text{ch}(bt)+2\frac{\alpha}{b}\text{sh}(bt)\right)^{-\...


8

You can expand the integrand in powers of z. The coefficient of $z^k$ is $$e^{-ky^2}\sum_{j=1}^{k-1} {1\over j!(k-j)!}=e^{-ky^2}(2^k-2)/k!.$$ This yields the following series expression for the integral: $$\sum_{k=1}^\infty \sqrt{\pi\over k}(-2+2^k)z^k/k!.$$


8

Yes. Unexpected weak solutions to the SDE $$ d Y = - \Phi'(Y) dt + \sqrt{2} dW \quad Y(0) \in \mathbb{R} $$ are available. To see this, transform the associated Fokker-Planck equation into a Schrödinger equation following e.g. Chapter 5 of H. Risken, The Fokker-Planck Equation, Springer-Verlag, 1989. Loosely speaking, if the resulting Schrö...


8

One way to define the "enclosed area" for a curve $\mathbf{r}(t)$ in the $x$-$y$ plane of duration $T$, with $\mathbf{r}(0)=\mathbf{r}(T)$, is via the socalled algebraic area $A=\tfrac{1}{2}\int_0^T (\mathbf{r}\times\dot{\mathbf{r}})\cdot\hat{z}\,dt$. (The dot indicates the time derivative and $\hat{z}$ is a unit vector perpendicular to the plane.) A loop ...


8

One can apply a deterministic result, called Garsia--Rodemich--Rumsey inequality, to estimate $\mathrm{E}[||X||^\alpha_{\gamma;[0,T]}]$. Here is a particular form of this result, which is most convenient for us. For any $\alpha >1$, $\delta> 1/\alpha$, there is a constant $C(\delta,\alpha)$ such that for any $f\in C([0,T],S)$ and $t,s\in[0,T]$ $$ ...


7

I know a version which exactly gives the constant $O(p^{1/2})$ for $p\ge 2$. It is contained in a lecture note by D. Khoshnevisan on SPDE.


7

Hi it is possible to get some Feynman-Kac formula in this case. The proof only use the martingale property and Itô's formula for jump-diffusion processes. So let's have $X$ s.t. (I took the compensated version of your sde): $dX_t=[\mu(t,X_t)+\lambda(t)\gamma(t,X_t)]dt + \sigma(t,X_t)dW_t+ \gamma(t,X_{t-})d\tilde{N}_t$ where $\tilde{N}_t$ is a compensated ...


7

Here's a proof of the statement for $f=0$, so that $X=W$ is a Wiener process. (The proof with general $f$ is a bit more involved, and I give this further below). I'll base the proof on the following simple result. Here, I am using $B_\epsilon=\left\{\omega\in\mathcal{W}^n\colon\sup_{t\in[0,1]}\lVert\omega(t)\rVert\le\epsilon\right\}$ for the $\epsilon$-ball ...


7

The time change described in the question may be handled as follows. Recall that if $W(t)$ is a standard Brownian motion then $$ W(\tau(b))-W(\tau(a)) $$ has the same distribution as $$ \int_a^b \sqrt{\tau'(t)} dW(t) $$ where $a \le b$. Therefore the time-changed process $\tilde X_t = X_{\tau(t)}$ satisfies: $$ d \tilde X_{t} = \delta \tau'(t) dt + 2 \...


7

I do not think Hsu's book is a good place to start with, although it has some strong holds-like details in calculation, neither its depth nor its clarity is comparable to Stroock, Daniel W. An introduction to the analysis of paths on a Riemannian manifold. No. 74. American Mathematical Soc., 2005. Don't be deceived by the title of this book, this is ...


7

First, a martingale is always only specified with respect to a filtration, and so is thus a local martingale. You do not specify any filtration in your problem, so I assume you mean the natural filtration of the local martingale (i.e., the smallest filtration w.r.t. which $X$ is a local martingale). Second, your prove/disprove statement does not consist of ...


7

Percolation 'noise' is generated by a perfectly good family of random variables, the 'quad-crossing' events. Basically, for every diffeomorphic image of the unit square, this random variable is $1$ if one can cross from the left to the right face traversing only open bonds and $0$ otherwise. Let's call this random variable $X_\phi$ where $\phi$ is the ...


7

To complement the excellent answer by Ofer Zeitouni, let me offer a functional analysis perspective. We want to define an integral of the following form: $\int F(W_t)dW_t=\int F(W_t)W'_tdt$, say, for a nice $F$. We can ask, generally, when is the integral $\int G(t)H(t)dt$ naturally defined? An obvious answer is: whenever $G$ belongs to some function space ...


7

Here's an approach that comes from Li, Xue-Mei, Strict local martingales: examples, Stat. Probab. Lett. 129, 65-68 (2017). ZBL1386.60159, https://arxiv.org/abs/1609.00935. Indeed, she mentions this very example after Corollary 4 (bottom of page 4 in the arXiv version). Lemma. Suppose $X_t$ is a continuous martingale, and let $\langle X \rangle_t$ be its ...


6

The estimate you are interested in has already been studied. The tail bound $ \mathbb{P} \left( \sup_{t \in [0,1]} | J_n^f (t)| \ge K \right) \le C_1 \exp( -C_2 K^{2/n}) $ was proved by C. Borell. The main tools are infinite dimensional isoperimetric inequalities. It was later proved by M. Ledoux that we even have $ \lim_{K \to \infty} \frac{1}{K^{2/n}}\...


6

Actually your SDE may be solved explicitly. Look at the more general SDE, \begin{align} dX_{t} = (a X_{t}^{n} + b X_{t}) dt + c X_{t} dW_{t} \end{align} where $n > 1$ and $a,b,c \in \mathbb{R}$. It has a solution given by \begin{align} dX_{t} = \Theta_{t} \Bigl( X_{0}^{1-n} + a(1-n)\int_{0}^{t} \Theta^{n-1}_{s} ds \Bigr)^{\frac{1}{1-n}} \end{align} with \...


6

Some of Malliavin's ideas are well-explained in his own book: Stochastic Analysis. I think the main geometric idea behind his proof of Hormander's theorem is the idea of submersion. More precisely, if we consider a stochastic differential equation in Stratonovitch form $dX_t=V_0(X_t) dt +\sum_{i=1}^n V_i(X_t) \circ dW^i_t $ where $W$ is a Wiener process ...


6

Let $(W_t)_{t\geq 0}$ be a Brownian motion. By the Skorokhod embedding theorem there exist a stopping time $T$ such that $W_T \,{\buildrel d \over =}\ X_1-X_0$, see e.g. "The Skorokhod embedding problem and its offspring", Obłój, (link) (Actually, we may choose $T$ in such a way, that $W^T = (W_{T \wedge t})$ is a uniformly integrable martingale). ...


6

OK, since I slipped again (there is no way I will believe nowadays that after 40 one's abilities do not decline sharply, though when I was younger I was more optimistic about the lifetime of a mathematician), let's try to figure it out. We are going to use the standard trick of considering the truncated stopping times $T_\tau=\min(\tau,T)$. Everything is ...


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