Is it easy to write down the large deviations rate for the maximizer of a random walk with negative drift?


Let $X\_i$ be the (iid, mean $-\mu$, variance $\sigma$, arbitrarily nice tails) jumps of a random walk $S\_i$. I am interested in the location of the maximum of $S\_i$, i.e. in $\arg\max\_k S\_k$. (If $X\_i$ are continuous, the maximizer is almost surely unique.)


There's a trivial exponential upper bound
$$
\mathbb{P}(\arg\max\_k(S\_k) > n) \leq \mathbb{P}(S\_{n+1} > 0) = \mathbb{P}(\sum\_{i=1}^n X\_i > 0)
$$
and the trivial exponential lower bound
$$
\mathbb{P}(\arg\max\_k(S\_k) > n) \geq \mathbb{P}(X\_i > 0\ \forall i \leq n)
$$
and they (of course) don't match. Is there a limit
$$
\lim\_{n \to \infty}n^{-1}\log\mathbb{P}(\arg\max\_k(S\_k) > n),
$$
and is it possible to write it down in terms of the distribution of the jumps $X\_i$?



Large deviations for maximizer of random walk with drift
The upper bound gives the correct large deviations; that is, the answer is
$$n^{-1}\log P(\mbox{ arg max}\_k S\_k>n)\to -I(0)\,,$$
where $I(0)=-\inf\_\lambda \log E(e^{\lambda X\_1})$.


The upper bound appears in the OP. For the lower bound, just consider the event 
$$ {\cal A}\_n:=\{S\_n\geq \epsilon n \}$$
and note that 
$P(\mbox{ argmax}\_k S\_k <n(1-\delta)|{\cal A}\_n)$ goes to $0$ as $n\to\infty$ if $\delta>>\epsilon$
(but both small). But $n^{-1}\log P({\cal A}\_n)\to-I(\epsilon)$. Now use the continuity of $I$ in $\epsilon$ (here you need to assume something on the tails; finite logarithmic mgf is enough).


