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To simplify the question, we start with standard Brownian motion(BM) $B_t$. Then, $$\lim_{\epsilon \to 0} \frac{1}{\epsilon}\mathbb P( 1- \epsilon <B_1 < 1 ) = \phi(1),$$ where $\phi$ is the density function of the standard normal distribution.

Next, let $B^1_t$ be the killed BM at $1$, i.e. $$B^1_t = B_{t\wedge \tau}, \hbox{ for } \tau = \inf\{t>0: B_t \ge 1\}.$$ [Q1] Does the following identity hold: $$\lim_{\epsilon \to 0} \frac{1}{\epsilon}\mathbb P( 1- \epsilon <B^1_1 < 1 ) = 0?$$ [Q2] If [Q1] is yes, can we replace $B_t$ by any other diffusion process starting from zero?

Remark: This question is related to the reason why the fokker-planck equation of a killed diffusion imposes zero boundary condition. See Fokker-Planck equation for a truncated process

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    $\begingroup$ of course you can do this explicitly for Brownian motion absorbed at 1. Also for the two sided there is a series expansion along the lines you are suggesting. Some sources to look at are karlin and taylor vol. 2, Ito and McKean, and Port & Stone, classical potential theory and ... , where I believe I dug out the analagous fact for multi dimensional Brownian motion for something I was doing about 1982, now lost to time. $\endgroup$
    – user83457
    Commented Dec 18, 2017 at 8:52
  • $\begingroup$ @michael, Thanks for your suggestions and it's helpful. For the Brownian motion, yes, I can find the Fourier series expansion, which indicates the density vanishing at the boundary. I guess it may be feasible to use Girsanov theorem for a drifted Brownian motion with bounded drift. I do not know if the diffusion is given in a more broad scope. $\endgroup$
    – kenneth
    Commented Dec 18, 2017 at 13:30
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    $\begingroup$ It is, but also, with girsanov you can go immediatele from the density under driftless to the density with drift, because the likehood ratio only depends on the value and time. The fourier expansion comes in as the eigenfunctions for the laplacian, and I think you just replace that operator with the generator of the diffusion. $\endgroup$
    – user83457
    Commented Dec 18, 2017 at 13:39
  • $\begingroup$ @michael Sounds great, I will try. $\endgroup$
    – kenneth
    Commented Dec 19, 2017 at 4:40
  • $\begingroup$ Did you by any chance find a reference/answer for this question ? Thanks ! $\endgroup$
    – kantadou
    Commented Apr 4, 2019 at 14:51

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