5
$\begingroup$

Suppose we have the following two stochastic differential equations for $x_0$ and $x$ respectively \begin{align} dx_0 &= -k_0(t)(x_0-1)dt+\eta_0(t) x_0\,dB \tag1\\ dx &= -(k_0(t)+\epsilon k_1(t))(x-1)dt+(\eta_0(t)+\epsilon \eta_1(t)) x\,dB \tag2 \end{align} with initial condition $x(t=0)=x_0(t=0)$, where $\epsilon>0$ is a constant parameter and $k_i(t)$ and $\eta_i(t)$ for $i\in\{0,1\}$ are $t$-dependent functions. Each SDE thus has a unique solution for any given initial value.

Let $$x=x_0+\epsilon y$$ and substitute it into Eq. (2) and collect up-to 1'st power the same power terms of $\epsilon$. $$(dx_0+k_0(t)(x_0-1)-\eta_0(t) x_0 dB)+\epsilon(dy+(k_0(t)y+k_1(t)(x_0-1))\,dt-(\eta_0(t)y+\eta_1(t)x_0)dB)+O(\epsilon^2)=0. \tag3$$ The term in the first parenthesis vanishes due to Eq. (1). We set $$dx_1=-(k_0(t)x_1+k_1(t)(x_0-1))dt+(\eta_0(t)x_1+\eta_1(t)x_0)dB. \tag4$$ with initial condition $x_1(t=0)=0$.

$(\text{Eq}.(3)-\text{Eq}.(1))/\epsilon$ gives $$dy=-(k_0y+k_1(x_0-1)+\epsilon k_1y)\,dt+(\eta_0y+\eta_1x_0+\epsilon \eta_1y)\,dB. \tag5$$ $\text{Eq}.(5)-\text{Eq}.(4)$ gives $$dz = -(k_0z+\epsilon k_1y)\,dt+(\eta_0z+\epsilon \eta_1y)\,dB \tag6$$ where $z=y-x_1$.

Question: Does $z\rightarrow0$ as path/trajectory or function of time, in some sense, e.g. in distribution or probability, as $\epsilon\rightarrow0$?


We can use the Duhamel's principle to obtain an explicit Ito integral solution for Eq. (2) and convergence is clear and it is pathwise. But I would like to use this problem as a model for techniques that can be generalized to the case where the factors in front of $dt$ and $dB$ are Lipschitz continuous.

$\endgroup$
3
$\begingroup$

Proposition 1: Given $T>0$. $\mathbf E\big[z(t,\epsilon)^2\big]<\epsilon^2Be^{AT},\,\forall t\in[0,T]$ for some positive constants $\epsilon_0,\,M,\,A,\,B$, and $y(t,\epsilon)\rightarrow x(t)$ in probability as $\epsilon\rightarrow0$ uniformly for $t\in[0,T]$

Proof: Here I can use Duhamel's principle or the linearity of Eq.(6) to obtain $z$ as an integral of $\epsilon y$. The convergence is pathwise. However, I would like to find a technique that is generalizable to any SDE with factors that are Lipschitz continuous $C^1$ functions. So I will proceed with the following approach.

I am going to prove for any given $T>0$, \begin{align} \mathbf E\big[y(t,\epsilon)^2\big]&<M, \tag{7.1} \\ \mathbf E\big[z(t,\epsilon)^2\big]&<\epsilon^2Be^{AT}, \tag{7.2}\\ \forall\epsilon\in(0,\epsilon_0),&\ t\in[0,T] \end{align} for some positive constants $\epsilon_0,\,M,\,A,\,B$. Therefore by the Chebyshev-Markov inequality, we have $z(t,\omega,\epsilon)\rightarrow0$ in probability as $\epsilon\rightarrow0$ uniformly for $t\in[0,T]$.

We will show the derivation of Eq.(7.2) given Eq.(7.1). A similar technique applies to Eq.(7.1) without premising on Eq.(7.2).

Take the integral form of Eq. (6), square it and apply the Cauchy-Schwartz inequality, $$\frac14z(t)^2\le\Big(\int_0^t k_0z\,ds\Big)^2+\Big(\int_0^t\epsilon k_1y\,ds\Big)^2+\Big(\int_0^t\eta_0z\,dB_s\Big)^2+\Big(\int_0^t\epsilon \eta_1y\,dB_s\Big)^2. \tag8 $$

We make use of the following inequality. For a deterministic function $a(t)$ and a stochastic function $u(t,\omega)$ where $\omega$ is an element of the sample space, \begin{align} &\mathbf E\Big[\Big(\int_0^t a(s)u(s,\omega)dB(s,\omega)\Big)^2\Big] \\ =&\mathbf E\Big[\int_0^t (a(s)u(s,\omega))^2\,ds\Big]\quad \text{(Ito isometry)} \\ \le&\int_0^t a(s)^2ds\int_0^t \mathbf E[u(s,\omega)^2]ds.\quad\text{(Cauchy-Schwartz inequality)}\tag9 \end{align}

Take expectation on Eq.(8) and apply Eq.(9) and again Cauchy-Schwartz, we have $$\mathbf E[z(t)^2]\le \alpha(t)\int_0^t\mathbf E[z(s)^2]\,ds +\epsilon^2\beta(t)\int_0^t\mathbf E[y(s)^2]\,ds$$ for some positive nondecreasing continuous deterministic funtions $\alpha(t)$ and $\beta(t)$. Let $v(t):=\int_0^t\mathbf E[z(s)^2]\,ds$, from Eq.(7.1), we have $$v'(t)\le \alpha(t)v(t)+\epsilon^2\beta(t)< Av(t)+\epsilon^2B,\ \forall t\in[0,T]$$ for some positive $A, B$ as increasing functions of $T$. That implies $$\mathbf E[z(t,\epsilon)^2]=v'(t)< \epsilon^2 Be^{At},\ \forall t\in[0,T].$$

In the following, we will further prove that $\forall \delta>0,\,\exists\epsilon \ni$ $$P\Big[\sup_{t\in[0,T]}z(t,\epsilon)^2>\delta\Big]<\epsilon^2 B_1e^{A_1t}.$$

(to be continued)

QED

$\endgroup$
2
$\begingroup$

Isn't by your definition $x_1 = \frac{x-x_0}{\epsilon}$ ?

I suppose you want to understand the convergence of $\frac{x-x_0}{\epsilon}$ as $\epsilon$ converges to $0$ ? Am I correct?

$\endgroup$
  • $\begingroup$ Good point. I have edited my question accordingly. Please check. It appears you are new to the site. The above is not an answer. By the convention here, you should not write a non-answer comment in the answer space but should click "add a comment" to post your comment. $\endgroup$ – Hans May 22 '18 at 5:03
  • $\begingroup$ unless mistaken hans, you can't comment unless you have a mininum reputation which is why I upvoted Taro NGUYEN response. $\endgroup$ – The Bridge May 22 '18 at 9:41
  • $\begingroup$ @TheBridge: Good point. $\endgroup$ – Hans May 22 '18 at 22:54
  • $\begingroup$ Taro: Have you had time to check my new formulation? $\endgroup$ – Hans May 22 '18 at 22:58
  • 1
    $\begingroup$ Taro and @TheBridge: I have obtained a solution and its proof. Please review my answer as I am filling in the details. Please let me know if you see anything wrong. Thank you. $\endgroup$ – Hans May 23 '18 at 22:31
0
$\begingroup$

As stated in the previous answer of Hans, the convergence for this question is pathwise, by virtue of it the SDE per se having an explicit solution. Therefor this approach is not generalizable. Still, I will state the proposition and its proof.

Proposition: Let $\Omega$ be the sample space, $y(t,\epsilon,\omega)\rightarrow x_1(t.\omega)$ as $\epsilon\rightarrow0,\,\forall \omega\in\Omega$.

$\endgroup$

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.

Not the answer you're looking for? Browse other questions tagged or ask your own question.