Please consider a random walk on a finite N-dimensional lattice with vectors $(x\_1, ..., x\_N)$. We define the origin to be $(0, ..., 0)$ and the target to be at the point in the lattice furthest away from the origin - i.e. $(||x\_1||, ..., ||x\_N||)$ where $||x\_k||$ is the integer length of the lattice in the $x\_{k}$ dimension. Here, each step of the random walk is a uniformly distributed, strictly positive random integer in each of the N-dimensions with an upper-bound value defined by the requirement that one cannot exceed the dimensions of the lattice. 


Is there a nice method, aside from explicit path-counting, to derive the probability density for hitting times provided an arbitrary lattice as defined above? 


Some computational results: For the $N=1$ case I expected the target hitting time (defined as the number of steps to reach the target) to fit well with a logarithmic growth function of the form $A\*ln(S)$ where A is a positive real number and $"S"$ is the number of integer steps one takes to reach the target from the origin. Running simulations (averaging 10,000 times) this yielded a decent fit with the value of $A$ ~ 1.146 for $||x|| = 100$, but $A$ decreases to ~1.095 for $||x|| = 1,000$ and decreased further ~1.069 for $||x||=10,000$. 



Hitting times for an N-dimensional random walk on a lattice with (strictly positive) random integer steps
Assuming you mean Leonid Kovalev's interpretation, the distribution of the hitting time in the $N = 1$ case is the same as the distribution of the number of cycles of a random permutation of $[n]$.


To be more specific, I'll change coordinates. Let $X\_0 = (x\_0^1, \ldots, x\_0^N) = (S, S, \ldots, S)$. Let $X\_1 = (x\_1^1, \ldots, x\_1^N)$, where $x\_1^j$ is chosen uniformly at random from $0, 1, \ldots, x\_0^j-1$. Define $X\_2$ similarly in terms of $X\_1$, and so on. 


Then the sequence $(x\_0^1, x\_1^1, x\_2^1, x\_3^1, \ldots)$ are as follows:


* $x\_0^1 = L$, of course.
* $x\_1^1$ is uniformly distributed on $\{ 0, \ldots, S-1 \}$.
* $x\_2^1$ is uniformly distributed on $\{ 0, \ldots, x\_1^1-1 \}$.


and so on...
In particular the distribution of $x\_1^1$ is the distribution of number of elements in a random permutation on $S$ elements which are {\it not} in the cycle containing 1; In particular the distribution of $x\_1^1$ is the distribution of number of elements in a random permutation on $S$ elements which are {\it not} in any of the $k$ cycles with the smallest minima.


So the distribution of the smallest $k$ for which $x\_k^1 = 0$ is exactly the distribution of the number of cycles of a random permutation of $\{ 1, \ldots, S \}$; this is $1 + 1/2 + \cdots + 1/S \sim \log S + \gamma$, where $\gamma = 0.577\ldots$ is the Euler-Mascheroni constant.


In the higher-dimensional cases, the time to reach $(0, \ldots, 0)$ is just the {\it maximum} of the number of cycles of $N$ permutations chosen uniformly at random; this should still have expectation $\log S$ plus a constant depending on $N$.


