Question about random walk with fixed endpoint, and a reference request We have a random walk of length $n$, starting at $0$ and ending at $-6\,\sqrt{n}$. Can we give any sort of high probability bound on the number of steps before we first reach the value $-2\, \sqrt{n}$? I'm really looking for anything related to the random variable defined by the first time we reach the value $-2\,\sqrt{n}$, like probability distribution, expected value, etc. 
As for the reference request, are there any books which discuss random walks with a given endpoint in more detail? The unfortunate thing is that knowing the endpoint loses the martingale property, so I don't really know of many ways to analyze this type of walk.
 A: This follows the basic ideas I laid out in the comments.
Let $n = m^2$ be the total number of steps taken and $S_i$ be the value of the random walk at time $i$. We are interested in the probability distribution of the stopping time $\tau = \inf\{i: S_i = 2m\}$ conditional on the event $\{S_n = 6m\}$.
The key is to break things down according to the possible values that $\tau$ can take on and then consider how they could occur. First, note that $\tau$ must be an even integer in the range $[2m,n-4m]$ since we are considering a stopping time that stops when the height of the process reaches an even integer.
Suppose $\tau = k$. Then $S_i < 2 m$ for all $i < k$. 

Question: How many paths are there from $(0,0)$ to $(k,2m)$ that never hit $2m$ prior to time $k$? A: We find all the paths from $(0,0)$ to $(k-1,2m-1)$ and subtract away every path from $(0,0)$ to $(k-1,2m-1)$ which hits $2m$ at some point.

The number of paths from $(0,0)$ to $(k-1,2m-1)$ that hit $2m$ can be computed by the reflection principle since this is the same as the number of paths that go from $(1,1)$ to $(k,2m)$ without ever crossing zero (just rotate 180 degrees). By the reflection principle, this is ${k-1 \choose a}$ where $a = m + k/2$. Hence, the total number of paths from $(0,0)$ to $(k-1,2m-1)$ that never hit $2m$ prior to time $k$ is ${k-1 \choose a-1} - {k-1 \choose a} = \frac{2 m}{k} {k \choose a}$.
The total number of paths from $(k,2m)$ to $(n, 6m)$ is simply ${n-k \choose b}$ where $b = (n + 4m - k)/2$.
Finally, the total number of paths from $(0,0)$ to $(n, 6m)$ is ${n \choose a+b}$.
Hence, 
$$
\mathbb{P}(\tau = k \mid S_n = 6m) = \frac{2m}{k} \frac{{k \choose a}{n-k \choose b}}{{n \choose a+b}} = \frac{2m}{k} \frac{{k \choose m+k/2}{n-k \choose (n+4m-k)/2}}{{n \choose (n+6m)/2}}
$$ for each even $k \in [2m, n-4m]
$. Otherwise the probability is zero.

Here is an example with $n = 400$. One sample path is shown with the two horizontal barriers corresponding to $2m = 40$ and $6m = 120$. The vertical line indicates the value of $\tau$ for this sample path at which the lower barrier was hit for the first time. The histogram in the background is the distribution of $\tau$ with a height of 50 corresponding to a probability of 0.01. Some skew to the distribution is evident.

Aside: Note that it is easy to generate sample paths that always end at $6m$. Simply take the cumulative sum of the coordinates of a permutation of the vector $(+1,\ldots,+1,-1,\ldots,-1)$ which contains $(n+6m)/2$ entries that are $+1$ with the rest being $-1$.
