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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 $\epsilon$.
$$(dx_0+k_0(t)(x_0-1)-\eta_0(t) x_0 dB)+\epsilon(dy+k_0(t)y+k_1(t)(x_0-1)-(\eta_0(t)y+\eta_1(t)x_0)dB)+O(\epsilon^2)=0. \tag3$$
The term in the first parenthesis vanishes due to Eq. (1). We set 
$$dx_1=-k_0(t)x_1-k_1(t)(x_0-1)+(\eta_0(t)x_1+\eta_1(t)x_0)dB. \tag4$$
with initial condition $x_1(t=0)=0$.

Question: Does [$]y\rightarrow x_1[$] with respect to the trajectories in some sense, e.g. in probability or distribution, as [$]\epsilon\rightarrow0[$]?

It depends on what are the distributions of x_0 and x_1. What are you trying to prove?

The distributions of [$]x_0, x, x_1[$] are determined completely by the respective SDE's [$](1), (2), (3)[$]. I am trying to see if [$]\frac{x-x_0}{\epsilon}[$] approaches, in some sense, perhaps in distribution or probability, [$]x_1[$] determined by Eq. (3) as [$]\epsilon\rightarrow0[$].

"determined completely by the respective SDE's"


You have to be more specific than that to answer your question.

Are you talking about the initial condition? If so, this is a good point. I have added [$]x_0(t=0)=x(t=0)[$], and thus [$]x_1(t=0)=0[$] to my first post to reflect that. I have also edited my first post to make it more rigorous and clearer.

No, I mean: what is the distribution of x0(t) and x1(t)?

Like I said before, the distribution at any [$]t[$], and in fact the whole lot of trajectories, were uniquely determined by the SDE and its initial condition so long as the SDE satisfy some regularity condition. That is the whole point of any differential equation, getting the state at any other coordinate point from any given coordinate point.

Exactly and of course. The trajectory generates more than the instantaneous distribution at time [$]t[$] but the probability distribution/density of the whole function [$]x: R^+ \rightarrow R[$]. The well behaved SDE together with its initial condition uniquely determines the trajectory or the whole aforementioned function in time and its associated distribution (probability density) for the whole trajectory/path/function. Therefore, like I said before, presenting an SDE is equivalent to presenting a distribution for the whole trajectory/path/function, and as a corollary, it gives the distribution of say [$]x[$] at any given time [$]t[$] of your choice.  What are you saying is missing?

I have changed the notation in the first post a bit for clarity.

Yes, I studied mathematical physics. You guessed it from my use of the word "trajectory"? What about you?

Ha, quite observant of you for my ease of use of the perturbation theory. What did you study?

However, "blind faith" is an erroneous choice of words. Something is a blind faith, only when it is a sure belief without reason. The very act of posing the question is a refutation of that phrase. Moreover, the question is well posed in the mathematically rigorous fashion. That further contradicts the word "blind".

Now, does my answer resolve your technical issue?

Gosh, you have one fragile ego. Why don't you point out the specific statements demonstrating my "arrogant attitude" and "again"? Only you have the privilege to accuse me of blind faith and I am not even allowed to defend myself? That is some weird entitlement complex you have harboured from I-don't-know-where. 

I think I have earned the allowance to make an observation. In the two encounters we have had, all your comments are vague or erroneous or emotional. I always answer your questions head-on and with, perhaps brutal, honesty. You either evade, or make vague or erroneous statements, or snap/snipe emotionally. If you think my statement is unclear, you should ask for proof or details. If you think it is wrong, point out the logical error or give a counterexample. If you want to make a statement, make one that is concrete and logical. You never do.

Your question is naive and you're too slow and balky to understand why.

I wanted to study physical mathematics, but they didn't offer that at Leiden University. 

Ho, I was just wondering where troll katastrofa with a wounded ego is.   You are stalking me with a vendetta, aren't you? You had zilch to say after the question had been posed for several days and when the mathematics was discussed but simply could not bear passing up the opportunity to talk trash once the opportunity presented itself. Haha, so pathetic. I understand you just cannot bear to see your twin go down alone. "Slow and balky"? Apt self-portrait of your ineptitude in "high school" linear algebra demonstrated to the hilt a mere few weeks ago. Do I need to exhibit the links here for all to see? 
 
If you have anything mathematically meaningful to say, say it with mathematical rigor and some semblance of professionalism. Stop behaving like an infant throwing temper tantrum. Act like an adult in case you are one. Don’t make people think your psychological age is stunted at three. You don’t see yourself, but the immaturity you have displayed here is stunningly embarrassing.

Critter noise notwithstanding, I think I have found the answer to be affirmative and the convergence is pathwise almost surely. I have constructed a proof which will be written up later. The method can be extended to a more general setting.

Pathwise and in distribution are two different things, genius.

He understands only himself. And only himself understands him.

1. I meant to type "the answer to be affirmative and the convergence is in probability, possibly pathwise almost surely". In fact, the convergence in probability is uniform in every time interval.

2. The original question is:  

Question: Does [$]y\rightarrow x_1[$] with respect to the trajectories in some sense, e.g. in probability or distribution, as [$]\epsilon\rightarrow0[$]?

It is asking whether there is a convergence in a sense as yet to be determined. It is an open question. What are your objections to the question?

3. Your, ISayMoo's, previous posts demand to know the distribution of [$]x(t)[$]'s and [$]y(t)[$] at every [$]t[$]. It is not clear what your objection is. Are you asking for the explicitly written expression of the distribution of [$]x(t)[$]'s and [$]y(t)[$] at every [$]t[$]? What is it for?

Do you object to what I said as follows? Given the initial conditions [$]x(t=0), x_0(t=0), y(t=0), x_1(t=0)[$], [$]x(t), x_0(t), y(t), x_1(t)[$] are all uniquely determined for every sample point and so the distributions of those variables at every [$]t[$].

4. 

The problem with pathwise convergence is the same as with pointwise convergence of functions - it's a weak form of convergence. 

Weak compared to what form? The pointwise convergence implies convergence in probability. The gist of the proof is that the set of the sample point whose distance to the target exceeds the positive [$]\epsilon[$] infinitely many times in the sequence is a subset of the point-wise divergent set which is of measure zero. Therefore the former is stronger than the latter which in turn is stronger than convergence in distribution.  I can write out the proof if you want to see it.

More importantly, how is this an objection to my original question, which asks for if there is a convergence and if there is, in what sense the convergence is. How can you object to a question. My upcoming answer provides one solution. What is the objection in that? 

The rate of convergence will depend on what path you're on.

This is true. Nevertheless the rate of convergence is irrelevant. However varied the rate is, the random variables at those sample points will eventually converge. The set of all those converging sample points can not be in the set [$]U(\epsilon)[$] the point of which remains further than any given positive [$]\epsilon[$] away from the target infinitely many times in the sequence. Since the measure of point-wise convergent set is [$]1[$], the measure of  [$]U(\epsilon)[$] has to be zero. Therefore the almost sure point-wise convergence implies convergence in probability.

So you don't know if the prices (or other expectations) computed for one process will converge to the prices computed for the other process, even if the processes convergence pathwise.

This is a non sequitur. The convergence of distribution is equivalent to the convergence of the expectation of all bounded continuous functions.

Pathwise and in distribution are two different things, genius.

Good job for retaining something from the lesson. But do you have a point to make? Did your elementary school teach you how to read?