4
$\begingroup$

I have a conjecture that seems rather obvious but the proof seems elusive.

Consider a diffusion given by, $dX_t = \mu(X_t) dt + \sigma(X_t) db_t$

where $b_t$ is a standard Brownian motion.

$\mu,\sigma$ are piecewise Lipschitz, bounded, of locally bounded variation. If there are any other nice properties required, I would assume I have them.

Suppose we are interested in computing the following -

$v(x) = \mathbb{E}^x \int_0^\infty e^{-rt} f(X_t) dt$

where $f$ is the flow payoff that our agent collects. $f$ is also piecewise Lipschitz, bounded, locally bdd variation etc. Let $f$ be between $m$ and $M$.

I want to prove that $v$ is continuous. The reason why I think it sounds obviously true is the following -

Suppose we want to prove continuity at $x$. Fix some point $a$ to the left of $x$ and consider a sequence $\{x_n\} \in [a,x]$ such that $x_n \rightarrow x$.

Let $\tau_n := \inf \{t: X^{x_n}_t \notin (a,x)\}$

Then,

$v(x_n) \le \mathbb{E}^{x_n} \left[ e^{-r\tau_n} [ P(X^{x_n}_{\tau_n} = a) v(a) + P(X^{x_n}_{\tau_n} = x) v(x)] + (1-e^{-r\tau_n})M\right]$

$M$ is because that's the upper bound of $f$.

Now, if, $x_n \rightarrow x$ $\Rightarrow$ $\tau_n \rightarrow 0$, and $P(X^{x_n}_{\tau_n} = x) =1$.

then we can get continuity. But I do not know how to prove these two things which sound so obviously true. The only approach I am aware of is the time change formula but that seems like an overkill for this problem. Any help would be immensely appreciated. Thanks.

Improve this question
$\endgroup$
1
  • $\begingroup$ Three comments: (1) You obviously need some lower bound on $|\sigma(x)|$. (2) Strictly speaking, your argument only gives upper semi-continuity, you also need a simpler bound $\nu(x_n) \ge \mathbb{E}^{x_n}[e^{-r\tau_n}[P(X_{\tau_n}^{x_n}=a)]]\nu(a)$. (3) Your process is "locally Feller" except at a discrete set of points, so you only need to worry about points where any of the coefficients is discontinuous. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 28, 2017 at 8:49

1 Answer 1

Reset to default
1
$\begingroup$

The key thing is that your diffusion is Feller under the stated conditions, so $x \mapsto \mathbf E_x f(X_t)$ is continuous. Therefore $x \mapsto \int_0^{\infty} \mathbf E_x f(X_t) e^{-rt} \ d t$ is continuous and using Fubini gives the required result. For reference material on continuous dependence on the initial condition see e.g. Karatzas and Shreve, Rogers & Williams, or Kallenberg.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thanks for the answer. From what I know, the diffusion is not Feller continuous because the drift and volatility are piecewise Lipschitz and not Lipschitz everywhere. Also, $f$ is piecewise Lipschitz. Do you think there are some standard results like the one you are referring to that can do our job? $\endgroup$
    – avk255
    Commented Jun 3, 2016 at 13:44
  • $\begingroup$ Are $\mu$ and $\sigma$ locally Lipschitz and in particular continuous? In this case, further assuming you have uniqueness of the martingale solution, Kallenberg Thm 21.11 gives the Feller property. $\endgroup$
    – Joris Bierkens
    Commented Jun 3, 2016 at 13:46
  • $\begingroup$ No, $\mu$ and $\sigma$ are piecewise continuous. We can assume that the number of discontinuities is finite. They are bounded, $\sigma$ is bounded from below by a strictly positive number. Hence, by Nakao, I will have a strong solution. But, I must work with piecewise continuous $\mu$ and $\sigma$. $\endgroup$
    – avk255
    Commented Jun 3, 2016 at 13:55
  • $\begingroup$ How about $f$, is it not continuous either? $\endgroup$
    – Joris Bierkens
    Commented Jun 3, 2016 at 14:16
  • $\begingroup$ $f$ is not continuous either. It is piecewise continuous, bounded, finite points of discontinuity. $\endgroup$
    – avk255
    Commented Jun 3, 2016 at 17:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .