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Standard O-U Formulas: Take the Ornstein–Uhlenbeck process defined by the SDE $$ dX_t = -\frac{\mu}{\theta} X_t dt + \frac{\sigma}{\theta^{1/2}} dW_t $$

where $\mu > 0, \theta > 0, $ and $\sigma > 0$ and $W_t$ is standard Brownian Motion

We know that the conditional variance is, $$ Var(X_t | X_0 = x_0) = \frac{\sigma ^2 \left(1-e^{-\frac{2 \mu t}{\theta }}\right)}{2 \mu },\quad \forall t > 0 $$ This can be derived by using the Kolmogorov Forward Equation for the pdf $p(x,t)$ with a boundary value at $x_0$, $$ \frac{\partial p(x,t)}{\partial t} = -\frac{\mu}{\theta}\frac{\partial (x p(x,t))}{\partial x} + \frac{1}{2}\frac{\sigma^2}{\theta}\frac{\partial^2 p(x,t)}{\partial x^2} $$

We also know that we can find conditional expectations, for some measurable $f(\cdot)$ with typical properties, as $$ u(x,t) = E_t\left[\int_{t}^{T} e^{- \rho \tau}f(X_\tau)d \tau | X_t = x\right] $$ being equivalent to the solution to the PDE: $$ 0 = \frac{\partial u(x,t)}{\partial t} - \frac{\mu}{\theta}x\frac{\partial u(x,t)}{\partial x} + \frac{1}{2}\frac{\sigma^2}{\theta}\frac{\partial^2 u(x,t)}{\partial x^2}-\rho u(x,t) + f(x) $$ subject to a boundary value $u(x,T) = U(x)$


Taking the limit: Define the stochastic process $X^*_t$ as the limit of this process as $\theta \to 0$. Naively taking the limit, (if it limit exists?), $$ Var(X^*_t | X^*_0 = x^*_0) = \frac{\sigma^2}{2 \mu},\quad \forall t > 0 $$

This appears to show that the process is IID in the sense of $X_t \perp X_{\tau}$ for any $t \neq \tau$. This also appears to coincide with multiplying the KFE by $\theta$ and then taking the limit, which shows the pdf is asymptotically an ODE and independent of time. $$ \frac{d (x p(x))}{d x } = -\frac{1}{2}\frac{\sigma^2}{\mu}\frac{d^2 p(x)}{d x^2 } $$

What happens to the conditional expectation with the Feynman-Kac formula is less clear. Multiplying by $\theta$ and taking the limit, it appears that most of the terms drop out, leaving us with an ODE not connected to the discounting or $f(\cdot)$ $$ \frac{d u(x)}{d x } = \frac{1}{2}\frac{\sigma^2}{\mu}\frac{d^2 u(x)}{d x^2 } $$ which doesn't seem to be meaningful. Applying the boundary value, doesn't this say that $u(x,t) = U(x)$ for all $t$?


Question 1: Does the limit of this stochastic process exist?

Question 2: Is this stochastic process measurable? Does the strange form that the naive application of the Feynman-Kac formula takes suggest that it is only defined for trivial functions?

Question 3: It sure looks like we can't really take the limit directly and solve expectations. Is there a different(and better) way to take this limit, maybe along the lines of a rapidly mean reverting stochastic process and singular perturbations for small $\theta$? https://www.princeton.edu/~sircar/Public/ARTICLES/fpss3.9.pdf


ADDED: It seems like the answer to the latter part of the problem may be that the $X^*_t$ process cannot be measurable. To see this: let $P(x,t | x_0)$ be the CDF of $X^*_t$ given $X_0$ for $t \geq 0$. (Note: I am pretty sure taking the limit that this becomes normal, and $P(x,t | x_0) \sim N(0, \frac{\sigma^2}{2 \mu}),\quad \forall t > 0,\, \forall x_0$

As this is independent of $t$, this means that (1) $\sup\{X_{t}\} = \sup\{N(0, \frac{\sigma^2}{2 \mu})\} = \infty$ for any $t > 0$ and (2) $\inf\{X_{t}\} = \inf\{N(0, \frac{\sigma^2}{2 \mu})\} = -\infty$ for any $t > 0$. Since these are true for arbitrarily small $t$, there is no way to take an integral over time and have the $\inf$ and $\sup$ converge (or the Lebesgue equivalent). I believe this rough description generalizes to show the non-existence of any continuous time IID process that is both measureable and non-deterministic

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  • $\begingroup$ Generally you should rescale time in these situations. Here if you take $t=\theta s$ then $dt=\theta ds$ and $dW_t=\theta^{1/2} dW_s$, so you wind up with $dX_s = -\mu ds + \sigma dW_s$ considered on the (large) interval $[0,T/\theta]$. $\endgroup$
    – Ian
    Jan 8 '15 at 18:44
  • $\begingroup$ Thanks @Ian. Do you suspect that there is any way to setup a stochastic process rescaling time that is asymptotically IID and still measureable? $\endgroup$
    – jlperla
    Jan 8 '15 at 18:50
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Consider the Langevin system of equations: $$ \begin{cases} X_t = \varepsilon^{-1} U_t \,dt\\ U_t = - \varepsilon^{-2}\mu U_t\,dt + \varepsilon^{-1} \sigma \,dW_t \end{cases} $$ You will recognize that the process $U$ is the same as your process $X$.

In Pardoux and Veretennikov's article On Poisson equation and diffusion approximation 2, it is show that $X_t$ converges in distribution to a brownian motion (with diffusion coefficient $\sigma^2 /\mu^2$ - give or take a constant factor). In fact the article gives a much more general result that you should definitely check out.

It can be considered that $U$ represents the velocity of a particle, so that $X$ then represents its position. In the limit of large timescales (the one given to you by @Ian) then the position process $X$ goes to a brownian motion (in distribution but I'm sure even in probability). The velocity process $U$ being therefore a white noise of sorts.

You can also check Zeev Schuss's book Theory and Applications of Stochastic Processes, chapter 8 Diffusion Approximations to Langevin’s Equation. He has another book on diffusion limits that you should check but I don't know anything about its contents.

If you play around with the powers of $\varepsilon$ you obtain different limits (overdamped Langevin, Smoluchowski-Kramers limit).

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  • $\begingroup$ Thank you very much, this is extremely valuable. I have ordered the Schuss book. From the referenced Pardoux and Veretennikov, for those who care about the solution to this problem, it seems that Theorem 4 is the key to the answer, as it gives a practical infinitesimal generator, from what I can read. $\endgroup$
    – jlperla
    Mar 26 '15 at 16:44
  • $\begingroup$ @jlperla I should mention that Schuss doesn't spend a huge amount of time on his mathematical demonstrations (i.e. some of the things he says in this chapter make sense but you're not sure if they're true since they lack a proof), other parts on the general theory of SDEs seems very good to me. I can send you some screen shots to see if the book has what you really need. $\endgroup$ Mar 27 '15 at 10:33
  • $\begingroup$ Thank you for the offer but grant money makes costs less unimportant. One last thing: the motivation for my question was to try to construct a stochastic process that is asymptotically IID (in a loose sense) but is still measureable for expectations over time (i.e. can calculate things like Kolmogorov Backward Equations). It sounds that while the velocity process $U_t$ looks like white noise, but it isn't necessarily measureable. The $X_t$ process above converges to brownian motion, which is measurable but is markov rather than IID. Hence, this construction doesn't deliver what I wanted. $\endgroup$
    – jlperla
    Mar 27 '15 at 16:14

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