# Tag Info

16

Computable, absolutely normal numbers do actually exist. See V. Becher, S. Figueira: An example of a computable absolutely normal number, Theoretical Computer Science 270 (2002), 947-958.

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 ...

14

Geometric rough paths have the property that if you want to solve an equation with values in a manifold, choose a coordinate chart, and write in local coordinates $$dY^i = V_0^i(Y)\,dt + \sum_j V_j^i(Y)\,dX_j$$ for some vector fields $V_i$ (with the obvious abuse of notation that the solution actually depends on the choice of $\mathbb{X}$, not just on $X$),...

11

As to question 2: Planar Brownian motion started at $y_0$ will almost surely loop around $y_0$, i.e., disconnect $y_0$ from $\infty$ immediately, so it has to hit $A$ and $B$ immediately, too, and $\tau_A = \tau_B = 0$ a.s., no matter what kind of Jordan curve $C$ is. This should also imply that the answer to the first question is yes, by the strong Markov ...

10

In the definition, "diagonal" does not make much sense, you probably mean "self-adjoint", although this can be relaxed to "m-sectorial" (powers of $A$ still make sense then). Also, it is irrelevant for the definition whether or not $A$ has some essential spectrum. Anyway, the main reason why these interpolation spaces are useful for the analysis of SPDEs (...

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 ...

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 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 $t,s\in[0,T]$ $$... 8 First let us check that T exists and is unique. Let \mathrm{Sym}_n be the space of symmetric matrices (with real coefficients), \mathrm{M}_n the space of all matrices and \mathrm{Alt}_n the space of antisymmetric matrices. Claim: the map \mathrm{Alt}_n \rightarrow \mathrm{M}_n / \mathrm{Sym}_n, T \mapsto BT is injective. Let T be in the kernel, ... 7 The time change described in the question may be handled as follows. Recall that if W(t) is a standard Brownian motion then$$ 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

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

What I mean is that $$X_\epsilon^\tau(t) = \int_{-\infty}^t P_{t-s} \Pi_0^\perp (\partial_x X_\epsilon^{\tau_1}(s)\, \partial_x X_\epsilon^{\tau_2}(s))\,ds\;,$$ where $P_t$ denotes convolution with the heat kernel. The product appearing on the right is the usual pointwise product of two random variables with values in the space of continuous functions. The ...

7

There are several relevant papers: Path Integral Approach to Relativistic Quantum Mechanics: Two-Dimensional Dirac Equation (1987) Path Integral for Relativistic Equations of Motion (1997) Path Integral for the Dirac Equation (1998) On the Feynman Path Integral for the Dirac Equation in the General Dimensional Spacetime (2014) On the construction of the ...

6

The argument in JKO (short for Jordan-Kinderlehrer-Otto-98) is based on a Lyapunov function for the Fokker-Planck equation. As such, it requires that this Lyapunov function evaluated at $\rho^0$ be finite. (The main result of JKO assumes this condition.) This is a strong constraint on $\rho^0$: it seems to require, e.g., that the support of $\rho^0$ be $\... 6 If the function$f : \mathbb{R} \to \mathbb{R}$is discontinuous, then$u(t,x)= \mathbb{E}_xf(X(t))$may not satisfy the initial condition, in the sense that the limit statement: $$\lim_{(t,s) \to (0^+,x)} u(t,s) = f(x) \quad \forall x \in \mathbb{R} \tag{\star}$$ may not hold. To be concrete, consider $$d X(t) = d B(t) \;, \quad X(0) = x$$ where$B(...

6

Yes. A systematic study of stochastic (differential) algebra could be found in Grenander, Ulf. Probabilities on algebraic structures. Dover Books, 1981. Grenander studied the operation of integration on what are called "stochastic semi-groups". More specifically the Lie group representing the probability measures equipped with covariate derivatives(Lie ...

6

Using the notation of the OP, let $I(t)=\int_0^t X_s^2 ds$ where $X$ solves the above SDE. By Chebyshev's inequality, we have that $$P(I(t) > \alpha \mid X_0 = x) \le \frac{E\left\{ I(t) \mid X_0 = x \right\}}{\alpha} \;.$$ Fortunately, thanks to a Feynmann-Kac formula, the function $$u(t,x)=E\left\{ I(t) \mid X_0 = x \right\}$$ appearing in the ...

6

As Kwaśnicki remarked, the velocity process $v_t$ is a Brownian bridge, which can be represented as: $$v_t = v_0 (1 - \frac{t}{T}) + v_T \frac{t}{T} + (T - t) \int_0^t \frac{1}{T-s} d B_s \;.$$ (For an intro to this representation, see the first exercise of the following exercise sheet on Brownian bridges). As before, the position process $x_T$ is ...

6

There is no English translation of El Karoui's lecture notes, however her work on Snell envelopes is described in Reflected Solutions of Backward SDE'S, and Related Obstacle Problems for PDE's. For a text book treatment of Snell envelopes see Methods of Mathematical Finance by Karatzas and Shreve.

6

The inequality is not true in general — additional assumptions are needed. I think some kind of monotonicity of $a_1$ and $a_2$ should help, but this is merely a guess. Here is a counterexample. Consider $a_2 = 1$, so that $X_2(1) = W(1)$. Let $a_1(s, x) = 1$ when $|x| < 1$ and $a_1(s, x) = 0$ otherwise. Then $X_2(1) = W(1 \wedge \tau)$, where $\tau$ is ...

5

I think the chain rule $d[f(Z_t)]=f'(Z_t)\circ dZ_t$ is valid when the product $\circ$ is defined as in Stratonovich stochastic integral (while the SDE uses Ito's). Note that $g(Z_t)\circ dt$ doesn't differ from the ordinary product $g(Z_t)\ dt$, but $g(W_t)\circ dW_t$ is Ito's $g(W_t)\ dW_t$ plus $\frac12 g'(W_t)\ dt$.

5

It looks to me like nothing's wrong; the local time $\Lambda^Z_t(0)$ of $Z$ at zero is identically zero. This makes a certain amount of intuitive sense, because $Z$ should have "bounded variation at zero". If we follow the definitions in "The Pedestrian's Guide to Local Time" by Björk, local time $\Lambda^Z_t(x)$ is cadlag in $x$ and satisfies \begin{...

5

Loops don't get canceled out in the signature. (You might like to compute the signature of a circle. The second iterated integral is nonzero; in fact, by Green's theorem, it gives you the area inside the circle, maybe up to a factor of $\frac{1}{2}$ or something.) What does get cancelled out is any segment where the path retraces or doubles back on itself. ...

5

As indicated in the comments, the field is very wide, but I understand from the comment of the OP to zab's answer that there is a specific interest in the more narrow subtopic of applications of fractional Brownian motion to quantitative finance. Here are some overviews: Fractional Brownian Motion in Finance (2003) Fractional Brownian Motion and ...

5

It is not true that the bound $dM/dt = -\beta M^2$ implies that $M$ blows up almost surely. For example, with $b = 2$, there is a non-zero probability that $\beta < ce^{-t}$ for all $t>0$, for any given $c > 1$. Therefore, if $M(0) > -1/c$, no blow-up occurs.

5

I believe we can. Let $\ell_n \to \ell$ in your topology, and fix $t_0 \in \mathbb R_+$. We show that $\Gamma(\ell_n)(t_0) \to \Gamma(\ell)(t_0)$. We work over the interval $[0, T]$ with $T > t_0$. Step 1: We first note that $\ell_n$ converges to $\ell$ in measure. Indeed, let $\varepsilon > 0$ be arbitrary. As $\ell$ is nonincreasing, $\ell$ contains ...

4

For a probabilistic and analytic treatment to your questions check out: Chapter 1 of Second Order PDE's in Finite and Infinite Dimension. S. Cerrai. Springer, 2001 Chapter 2 of Analytical Methods for Markov Semigroups. L. Lorenzi, M. Bertoldi, CRC Press, 2006. respectively. I believe both references treat the Kolmogorov equation where time flows forward,...

4

The distribution of the process $Y$ is locally equivalent to the distribution of $X$ if and only if the flow $\Theta$ preserves the principal symbol of $L$. If $\Theta$ does not preserve the principal symbol of $L$, then the two processes $X$ and $Y$ do not have the same quadratic variation and thus their distributions can not be locally equivalent. If $\... 4 Well, if there are regions on the real line where the processes never get (which can be the case, e.g. if$f\equiv 1$and$\sigma$has zeroes with fast enough decay), then, clearly, you cannot say anything about the$\sigma$'s there. Otherwise the quadratic variation is the invariant that ensures$\sigma=\pm\tilde{\sigma}$. Suppose that$|\sigma(x)| <a&...

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