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As a follow-up to this question, for a Markov process $X_t$ with drift $u>0$ and variance $\sigma^2$, the Laplace transform of the stopping time $T=inf\{t: X_t = a >0\}$ is

$$ \mathbb{E}[\exp(-\lambda T)] = \exp\left(-a* \frac{\sqrt{u^2 + 2\sigma^2\lambda} - u}{\sigma^2}\right) $$

If taking derivative and let $\lambda = 0$, I will get $E[T] = a/u$, which is independent of $\sigma^2$. Is this right?

gt6989b
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Lei Hao
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  • What do you mean general Markov process like this? Do you mean a diffusion process given by $dX_t=\mu dt + \sigma dB_t$? – Ian Nov 09 '21 at 03:38
  • Also what is this $u$, is it supposed to be $\mu$? – Ian Nov 09 '21 at 03:41
  • What I have in mind is $X_t = ut + \sigma B_t$ where $B_t$ is a standard Brownian motion. Is this the same process as your notation? – Lei Hao Nov 09 '21 at 03:42
  • Yes. I would not describe this as a "general Markov process". It is a diffusion process. – Ian Nov 09 '21 at 03:42
  • Ah, okay. Is the expectation right? – Lei Hao Nov 09 '21 at 03:43
  • The end result is right (I checked with a different method). I don't know about the Laplace transform result offhand. – Ian Nov 09 '21 at 12:10

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