Perturbation of a stochastic differential equation

Question

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 $\epsilon\rightarrow0$?

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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.

Answer 1

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

Answer 2

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?

Answer 3

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 $\epsilon\rightarrow0,\,\forall \omega\in\Omega$.