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 $H_a < S_a$.
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)