# Bounding Brownian motion and an Ito process simultaneously

Let $$(W_t)_{t\geq0}$$ be a standard Brownian motion in $$\mathbb{R}^n$$ and $$(A_t)_{t\geq0}$$ be an adapted matrix-valued process such that $$A_t$$ is a positive symmetric matrix with bounded operator norm : for all $$t>0$$, $$\vert\vert A_t \vert\vert_{op} \leq b$$ for some constant $$b$$. That is $$0\preceq A \preceq bI_n$$ Define, for all time $$t\geq0$$ the martingale $$M_t=\int_0^sA_sdW_s$$.

Is it true that both processes can be simultaneously confined ? That is for all $$T>0$$ and all $$R>0$$, $$\mathbb{P}(\forall t \in [0,T], W_t \in B(0,R) \text{ and }\forall t \in [0,T], M_t \in B(0,R)) > 0 \quad?$$ For my purpose I would actually need the slightly weaker :

$$\mathbb{P}(W_T \in B(0,R) \text{ and }\forall t \in [0,T], M_t \in B(0,R)) > 0$$ Fix some $$T>0$$. I tried to work on the event $$E_R = \{\forall t \in [0,T], W_t \in B(0,R)\}$$ but it is unclear to me how $$M_t$$ conditionned to that event behaves, as there could be cancellations assuring that $$W_t$$ is small that might be destroyed in the stochastic integral right ? On the other hand it looks plausible to me that $$\omega \in E_R \implies M_t(\omega) \in B(0,\lambda R)$$ for some $$\lambda$$ potentially large but I might be wrong ?

EDIT : The general answer is no, as shown in an answer below. What kind of hypothesis on $$A_t$$ could fix the issue ? If $$A_t$$ is an homotethy for instance it is of course true. Is :

$$aI_n\preceq A \preceq bI_n \quad \text{ almost surely}$$ for some $$a,b>0$$ enough for instance ?

• With the new assumption of low bounded for A, It seems to me that you want is a kind of Harnack's inequality for $(W_t,M_t)$. (so that the probability density of is lower bounded.) May 17, 2021 at 8:12

If I am not mistaken, if you allow for degenerate $$A_t$$, then the answer is negative, and the reason is not really related to cancellations of any kind. In fact, $$M_t$$ alone can escape a given ball at a given time with probability one.

Consider the following 2-D case. Let $$O = (-1, 0)$$ be an arbitrarily chosen point, and let $$M_t$$ be the process that behaves like a 1-D Brownian motion in direction perpendicular to $$M_t - P$$. Then $$M_t$$ aligns to circles centered at $$O$$, but due to non-zero curvature of these circles, it will move away from $$P$$ with positive and deterministic velocity.

To be specific: let $$B_t = (X_t, Y_t)$$, and define $$M_t = O + (U_t, V_t)$$ to be the solution of the system of SDEs $$dU = \frac{V^2 dX - U V dY}{U^2 + V^2} , \qquad dV = \frac{-U V dX + U^2 dY}{U^2 + V^2} ,$$ with initial values $$(U_0, V_0) = -O = (1, 0)$$. The coefficients of this system of SDEs form a projection matrix in direction orthogonal to $$(U, V) = M - O$$. Then it is elementary to check that $$d\langle U \rangle = \frac{V^2}{U^2 + V^2} dt , \qquad d\langle V \rangle = \frac{U^2}{U^2 + V^2} dt ,$$ and $$U dU + V dV = 0 .$$ Therefore, $$d(U^2 + V^2) = 2 U dU + 2 V dV + d\langle U \rangle + d\langle V \rangle = dt .$$ In other words, $$|M - O| = \sqrt{1 + t}$$, and hence the probability that $$M_t$$ is in the unit ball is zero for $$t \geqslant 3$$.

• Thanks for this illumaniting example. Now what if I assume that $A_t$ is non-degenerate, or stronger $$aI_n\preceq A_t \preceq bI_n$$ almost surely for some $a,b >0$. I guess it is still not enough ? Or more generally, what kind of hypothesis on $A_t$ would be required to obtain a statement in the spirit of what I'm asking ? May 10, 2021 at 12:02
• In the "uniformly elliptic" case, the result should be true, at least without the constraint on $W_t$. One should be able to show that for a suitable function $\phi$ and constant $\lambda$ the process $e^{\lambda t} \phi(M_t) \mathbb{1}_{\{t < \tau_R\}}$ is a sub-martingale (here $\tau_R$ is the first exit time from the ball $B(0, R)$), and so with positive probability $M_t$ does not leave the ball $B(0, R)$ until any given time. The function $\phi(x) = J_\nu(\mu |x|)$ for sufficiently large $\nu$ and appropriate $\mu$ should do, but I have no time to check the details. May 10, 2021 at 18:36
• A similar method might acutally work for $(W_t, M_t)$, with an appropriate $\phi$, but off the top of my head I do not have a good candidate. May 10, 2021 at 18:38
• This is a beautiful answer @MateuszKwaśnicki. Can I ask how you created the picture of the process? Jul 24, 2021 at 5:46
• Thanks. The image shows a very naive approximation by a discrete-time random walk with appropriately small step size. I do not think I still have the Mathematica code, but there's nothing fancy here. Jul 24, 2021 at 20:42

In one dimension, with T=1, I am going to represent a function $$f(X_1)$$ satisfying $$E(f(X_1)) = 0$$ as a martingale. I believe it is the case that when represented as $$M_t = \int^t A(s) dW(s)$$ that $$\min f^\prime < A < \max f^ \prime$$, (quick proof: $$M_t = E(f(X_1|F_t)$$ but $$E(f(X_1| X_t = x) = \int \phi(y-x)f(y) dy = \int \phi(y)f(y - x ) dy$$ for an appropriate normal distribution $$\phi$$, and just take $$\partial_y$$. One is using Ito, but there is no need to calculate $$\partial^ 2_y$$ or $$\partial _t$$.. They must cancel out. ) so as long as $$\min f^\prime < 0 <,\max f^ \prime < \infty$$. It satisfies your conditions.  Furthermore, as long as $$f(0) > 0$$, there is some neighborhood of 0 where if $$X_1$$ is in it, then $$f(X_1)$$ defanitely is not. So take $$f(x) = ax + 1, x <0, bx + 1, x \geq 0$$. a and b must satisfy $$a > 0, b>0$$  $$\frac {(-a + b)} {\sqrt(2 \pi)} = -1$$ in order to be a mean 0 martingale, but that is easy to do.