1
$\begingroup$

Let $\epsilon > 0$ be a small parameter and consider the following lemma.

Lemma. Let $B(t)$ be a bounded, continuous, $R^{n \times n}$-valued function defined on a time interval $[0,T]$ such that all real parts of the eigenvalues of $B(t)$ are bounded from above by $-2\kappa$, where $\kappa$ is a positive constant. Let $\Phi^\epsilon(t,s)$ be the fundamental matrix that solves the initial value problem (IVP): \begin{equation} \label{ivp} \frac{\partial \Phi^\epsilon(t,s)}{\partial t} = \frac{B(t)}{\epsilon} \Phi^\epsilon(t,s), \ \ \Phi^\epsilon(s,s)=I, \ \ 0 \leq s \leq t \leq T. \end{equation} Then there exists a constant $C$ and a constant $\epsilon_{0} > 0$ such that \begin{equation}\|\Phi^\epsilon(t,s)\| \leq C e^{-\kappa(t-s)/\epsilon} \end{equation} for all $\epsilon \leq \epsilon_{0}$ and for all $s,t \in [0,T]$.

Let us work with the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and let $E$ denotes expectation with respect to $\mathbb{P}$. If $B(t)$ is bounded random process the constant $\epsilon_0$ (related to the modulus of continuity of the function $B$) depends, in general, on $\omega \in \Omega$, i.e. $\epsilon_0 = \epsilon_0(\omega)$.

Now consider $$A^\epsilon(t) = \int_0^t \Phi^\epsilon(t,s) F(s) ds,$$ where $F(s)$ is a bounded continuous stochastic process and $\Phi^\epsilon(t,s)$ solves the IVP above (with $B(t)$ satisfying the assumptions in the lemma but is random).

I would like to derive an estimate for $A^\epsilon(t)$ in the limit as $\epsilon \to 0$, in particular, a bound on $$E\sup_{t \in [0,T]} |A^\epsilon(t)|^p$$ for $p>0$. Here is what I have so far (setting $p =1$ for simplicity for now). For $\epsilon < \epsilon_0(\omega)$,

$$\left|\int_0^t \Phi^\epsilon(t,s) F(s) ds \right| \leq \sup_{s \in [0,t]}|F(s)| C \int_0^t e^{-\kappa (t-s)/\epsilon} ds =: D \frac{\epsilon}{\kappa }(1-e^{-\kappa t/\epsilon}).$$ The above is a $\mathbb{P}$-a.s. bound. Now, $$E \sup_{t \in [0,T]} \left|\int_0^t \Phi^\epsilon(t,s) F(s) ds \right| \leq \frac{2D}{\kappa } E \epsilon_0(\omega).$$ This estimate is not helpful as I would like to obtain its asymptotic behavior as $\epsilon \to 0$. Moreover, $\epsilon_0(\omega)$ is a random variable whose expectation is generally not known. I am aiming for the above quantity to be of $O(\epsilon)$ in the limit $\epsilon \to 0$ but I am not sure how I could get this (had the $\epsilon_0$ been a constant rather than a random variable, it would be easy to get the asymptotic behavior).

$\endgroup$
  • $\begingroup$ It is not even clear where you got your last displayed formula from. As to the question, it isn't clear why a tiny portion of the probability space won't create a huge blow-up in general as $\varepsilon\to 0$. Notice that when you divide by $\varepsilon$, you improve the decay if it is there but I do not see how you can control the possible growth for $\varepsilon>\varepsilon_0(\omega)$ at all without additional assumptions. $\endgroup$ – fedja Apr 19 '18 at 20:35
  • $\begingroup$ There is a missing factor of 2 in the last formula. I used the fact $E \epsilon < E \epsilon_0$ since $\epsilon < \epsilon_0$. Do you mind elaborating on the huge blow-up that you mentioned? $\endgroup$ – randomg Apr 19 '18 at 20:44
  • $\begingroup$ Note that here we are only interested in the behavior as $\epsilon \to 0$, so we only care about controlling things for $\epsilon < \epsilon_0(\omega)$. $\endgroup$ – randomg Apr 19 '18 at 20:55
  • $\begingroup$ Nope. You also need to show that there are no huge blow-ups on tiny sets where $\varepsilon_0(\omega)$ is very small and I do not see how you are going to do it in such a generality. $\endgroup$ – fedja Apr 19 '18 at 22:13

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.