Consider a random walk on $\mathbb{Z}$ with triangular drift and jumps that are standard normals. That is, $$ \begin{cases} RW_{t+1} = RW_t - d + \epsilon_t, \quad t \geq 0,\\ RW_{t-1} = RW_t - d + \epsilon_{t-1}, \quad t \leq 0, \end{cases} $$ where $\epsilon_t$ are iid standard normals.

Using Skorohod embedding, we can think of $W_t$ as the values of a Brownian motion with triangular drift at integer time points: $$ RW_t = B_t - d |t|, \quad \text{for $t \in \mathbb{Z}$} $$ where $B_t$ is a standard Brownian motion (and equality is in distribution). This tells me that $$ \max_{t \in \mathbb{Z}} RW_t \leq \max_{t \in \mathbb{R}} (B_t - d|t|) $$ (in the sense of stochastic domination), and the difference between them is not too large (~max of a Brownian bridge with different endpoints?).

Question: is something similar true for $\arg\max$, the location of the maximum? That is, is it true that $$ \arg\max_{t \in \mathbb{Z}} RW_t \quad \text{ is not too far from } \quad \arg\max_{t \in \mathbb{R}} (B_t - d|t|) ? $$ (Stochastic domination seems unlikely.)

Update: as Martin and Ofer point out, on a fixed sample path the maximizers can be arbitrarily far from each other. Nonetheless, is there a true statement of the form $$ \arg\max_{t \in \mathbb{Z}} RW_t \leq 2 \cdot \arg\max_{t \in \mathbb{R}} (B_t - d|t|) + 3, $$ for some values of 2 and 3? (In the sense of stochastic domination, of course.)

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    $\begingroup$ If $d$ is of order one, then both arg maxes will be of order one and at distance of order one of each other. In that sense, they will not be "close". Do you have some limiting case in mind? In the limit $d \to 0$, one would expect the statement to make sense and to be true. $\endgroup$ – Martin Hairer Sep 22 '13 at 19:00
  • $\begingroup$ The end goal is constructing conservative confidence intervals for the maximizer of $RW_t$ using $B_t - d|t|$ (whose maximizer has a known distribution). So a statement of the form "one is at most twice the other" (for some value of 2), or "their difference is at most blah" (for some nice blah), or any combination of the two, would work. $\endgroup$ – Elena Yudovina Sep 23 '13 at 16:33
  • $\begingroup$ To elaborate on Martin's reply: if $d$ is fixed, then with positive probability the argmax's can be far from each other (a picture should convince you of that, so it is a question about probability, not about regularity of the argmax). As $d\to 0$, the probability tends to $0$. So the only sense of the question is if you make it quantitative. $\endgroup$ – ofer zeitouni Sep 24 '13 at 20:05
  • $\begingroup$ True enough. I edited the question to clarify the sort of statement I would like to be true. $\endgroup$ – Elena Yudovina Sep 25 '13 at 21:11
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    $\begingroup$ The answer is I believe yes, if $2$ and $3$ are allowed to depend on $d$. The reason is that with the statement as you wrote it, it is enough to consider $t>0$ (due to symmetry) and then all you need to check is the tail estimate $P(argmax(B_t-d t)>T)$ as $T\to\infty$ (you need a precise asymptotic). It does decay at the same rate for the RW and BM, and your factor of $2$ takes care of the rest. $\endgroup$ – ofer zeitouni Sep 25 '13 at 22:07

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