Let $X$ be a Brownian motion and let $$H_a = \inf\{ s \ge 0 \mid X_s = a \} \;\ \text{and} \;\ S_a = \inf\{ s \ge 0 \mid X_s > a \}.$$

Now, I've shown that $H_a$ and $S_a$ are equal almost surely (pretty straightforward). Importantly (I'm assuming), I've shown that $H = (H_a)_{a\ge0}$ and $S = (S_a)_{a\ge0}$ are not almost surely equal (in fact, I'm pretty sure that I could extend this to 'almost surely not equal').

I want to show that $S$ is a Levy process but that $H$. I think I have a proof for $S$.

$$S_b - S_a = \inf\{s \ge 0 \mid X_s > b\} - \inf\{s \ge 0 \mid X_s > b\}$$ $$ = \inf\{s \ge S_a \mid X_s > b\} \sim \inf\{s \ge 0 \mid X_s > b-a\} = S_{b-a}$$ by the strong Markov property (of $X$). So we have stationary increments. Further, the strong Markov property again says that we have independence of the past.

Note that $X_{S_a} = a = X_{H_a}$. Unfortunately, I don't see why this doesn't apply in exactly the same way with $S$ replaced by $H$! Any advice would be most appreciated!

Note that there are also these two SE questions.

Unfortunately, they don't clarify it for me either! Indeed, in the one with the answer, the answerer appears to strongly imply that $H$ is a Levy process. My question is taken from a past exam paper, so I'm guessing that it's right (not always the case though!).

Here's the exact question.


I've actually thought of a better argument than I gave for the second part of (b). There's no non-trivial interval on which BM is non-decreasing. In particular, on any non-trivial interval it must attain a maximum (which is finite almost surely). Let this be attained at $a$. Then $H_a < S_a$.

  • $\begingroup$ I am afraid that when you write $X_{S_a} = a = X_{H_a}$ this is wrong. Take for example a Poisson Process started at 0. Then $S_0=0$ and $H_0 =\tau$ where $\tau$ is the first time $X$ jumps, then $X_{S_0}=0\not=X_{H_0}=1$. Best regards $\endgroup$ – TheBridge Apr 27 '16 at 15:02
  • $\begingroup$ Don't understand your claim that $H = (H_a)_{a\ge0}$ and $S = (S_a)_{a\ge0}$ are not almost surely equal. Do you assume the Brownian motion $X$ has continuous sample path? If so, $H$ and $S$ are the same. $\endgroup$ – Jay.H Apr 27 '16 at 15:15
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    $\begingroup$ @SmileySam: Concerning point b): Does your definition of "Levy process" require a process that is right continuous with left limits? (I ask this because $H$ does have stationary independent increments.) $\endgroup$ – John Dawkins Apr 27 '16 at 19:13
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    $\begingroup$ Assertion d) is false. The process $S$ is right continuous with left limits, and has (almost surely) countable many jumps. But $a\mapsto S_a$ is continuous for all $a$ outside that countable set of jump times. $\endgroup$ – John Dawkins Apr 27 '16 at 19:15
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    $\begingroup$ For (d), perhaps the author meant to say: the set of discontinuous points are dense, a.e. $\endgroup$ – Jay.H Apr 27 '16 at 20:07

The question was incorrect. See comments for a (short) discussion and suggestion of what it should be. This answer is being posted merely so that the question is removed from the unanswered pile. (Answer written by OP)


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