Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $W$ be a standard Brownian Motion. Let $\epsilon>0$ be given. Let $X^\epsilon$ be the process which diffuses like $W$ on $(-\epsilon,\infty)$, but when $X^\epsilon$ reaches the level $-\epsilon$, it is immediately brought back to the value zero. It then diffuses again according to $W$ until hitting $-\epsilon$, and then is brought back to zero, and so forth. Let $X^0$ be a reflected Brownian Motion (reflected at zero). Then, as $\epsilon \rightarrow 0$, in what sense does $X^\epsilon \rightarrow X^0$ Are there any references for this? I'm also interested in when $W$ is a diffusion.

share|improve this question
    
I'd be interested in knowing some (easily) readable references on this sort of thing too. One "classic" paper on diffusions with boundary conditions is Stroock and Varadhan's "Diffusions with boundary conditions." –  ShawnD Oct 23 '11 at 5:48
    
One thought: the reflected BM is a standard BM plus a local time term at zero. Using the interpretation of local time in terms of downcrossings should possibly do the trick. –  weakstar Oct 23 '11 at 14:28
    
You're saying that it's the value of the Brownian motion plus the minimal value attained rounded up to the nearest multiple of $\epsilon$. That really makes me want to take the limit by deleting the "rounded up" bit. Is that a thing? –  Will Sawin Oct 23 '11 at 18:49
1  
The comments above have the correct idea. We have $X^\epsilon_t=W_t+\epsilon\lfloor\epsilon^{-1}\max\_{s\le t}(-W_s)\rfloor$ which converges uniformly to $W_t-\min\_{s\le t}W_s$. Convergence for this particular process is especially simple. –  George Lowther Oct 23 '11 at 21:02
add comment

1 Answer

It looks like it should converge in distribution in Skorokhod space D. Martingale problem approach (see the book by Ethier & Kurtz on Markov processes) should work.

share|improve this answer
    
Thanks for the reference, will take a look. –  weakstar Oct 23 '11 at 14:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.