5

I am looking at the hitting time of a two sided Gaussian random walk i.e.

$S_{n}=\sum_{i=1}^{n}X_{i}$

where $X_{i}$ are i.i.d normally distributed random variables. The hitting time is $\tau=\inf\{n:S_{n}\notin [a,b]\}$ where a and b are constants. Most of the literature I have come across deals with expectations of $\tau$. I have an upper bound for $P(\tau > t)$. I am trying to find a lower bound or rather the exact distribution for the hitting time. Really appreciate any help you can provide. Thanks in advance!

aks3010
  • 51
  • 1
    This thread discusses the result for brownian motion: http://math.stackexchange.com/questions/32302/first-exit-time-for-brownian-motion-without-drift – muaddib May 31 '15 at 03:22
  • 1
    Sorry, I forgot to add that $X_{i}$'s have non-zero means. – aks3010 May 31 '15 at 14:55
  • When you say the exact distribution, are you looking for a closed form solution, or how to go about setting up an equation to calculate the distribution? – muaddib May 31 '15 at 14:57
  • I would like to have a closed form solution which I can analyze. A setting of an equation which would give me a tractable distribution would also be fine. – aks3010 May 31 '15 at 18:15
  • Using the optional stopping theorem on the martingale $M_n(r) = \frac{e^{rS_n}}{\mathbb{E} e^{rS_n}}$ and doing some analysis should give the exact distribution. – J Richey May 18 '17 at 01:12
  • @JRichey I am very interested in this approach, would you mind providing some more details? What would be the stopping time we use when applying the optional stopping theorem in this case? – ttb Apr 25 '18 at 01:10
  • @ttb I think this is the approach I had in mind: first, check that $M_n(r)$ is a martingale for each $r \in \mathbb{R}$, and use the OST on $M_n(r)$ with the stopping time $\tau$ that OP defines. This should give (after some algebra, and -- I think this holds! -- that $\mathbb{P}(S_\tau = a) = \frac{b}{b-a}$), an exact expression for $\mathbb{E}e^{r \tau}$ as a function of $r$. This is just the PGF (Laplace transform) of $\tau$ (change variables). The distribution of $\tau$ is hiding in the coefficients of the power expansion of that function, so you can use some derivatives to tease them out. – J Richey Apr 26 '18 at 02:54
  • @ttb I am curious to know what happens here. Maybe I will try it myself! Do post a solution if you figure something out. – J Richey Apr 26 '18 at 03:00
  • I later figured out, there's a result by A. Wald and other results in Renewal Theory which specify the distribution neglecting the overshoot over the thresholds. Without using OST, I could derive a closed form which can be computed but is analytically intractable using Owen's T-function. https://en.wikipedia.org/wiki/Owen's_T_function – aks3010 Apr 26 '18 at 06:11

0 Answers0