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 agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.