Bounding Brownian motion and an Ito process simultaneously

Question

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 ?

Answer 1

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 $t \geqslant 3$.

Answer 2

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.