Distribution of the maximum of a "bridge" random walk

Question

I found a card problem in an old book that essentially boiled down to this question: Suppose we have a random walk $S_n = X_1 + X_2 + \ldots + X_n$ that starts at $0$, where $X_i \hspace{0.1cm}$ is $1$ or $ -1$ with equal probability of $0.5$

We know that $S_{n} = 0$ and from that point on the walk no longer changes. (i.e it is only "walking" for n steps, and stops at $0$). The walk is n steps long

• Let $Z = \max(S_1,S_2,\ldots S_n)$. What is $E[Z]$? What is the distribution of Z?

Here is what I've managed so far: My first instinct was to use the reflection principle to count the number of walks that hit a height $k$ but do not surpass it. The only way I could manage this was in the following way: We know that there are $\binom{i}{\frac{i+k}{2}}$ paths between $(0,0)$ and a point $(i,k)$. Using the reflection principle, $\binom{i}{\frac{i+k+2}{2}}$ of these paths touch $k+1$ (meaning k would not be the maximum). So altogether there are $\binom{i}{\frac{i+k}{2}} - \binom{i}{\frac{i+k+2}{2}} = \frac{2k+2}{i+k+2} \cdot \binom{i}{\frac{i+k}{2}}$ paths from $(0,0)$ to $(i,k)$ that don't go higher than k.

Applying the same logic you can find that the number of paths from $(i,k)$ to $(n,0)$ that don't go higher than $k$ is $\frac{2k+2}{n-i+k+2} \cdot \binom{n-i}{\frac{n-i+k}{2}}$.

Now the way to go forward would be to multiply these two quantities to get the total number of paths for a given i, sum over all possible i, and divide by $\binom{n}{\frac{n}{2}}$ (total number of paths) to get $P(Z = k)$

But this seems to be extremely ugly and I can't figure out any way to work out this "sum over i's". I feel like there's got to be a simpler approach to this problem, and I can't figure out a way to simplify this sum. I also haven't found any MSE posts that answer this question for a "bridge"-like walk like in my case (I'm aware of the infinite case)

Any ideas?

Answer 1

I will use the notation $M_{n}=\max(S_1,\dots,S_{n}$). I can prove that $$ E[M_n]\sim \sqrt{\pi n/8}\qquad \text{as }n\to\infty. $$ First, write $$ E[M_n]=\sum_{m=1}^{n/2}P(M_n\ge m). $$ We can compute $P(M_n\ge m)$ using the reflection principle. Random walks which start at zero, eventually reach $+m$, and then end at zero, are in bijection with random walks starting at zero and ending at $+2m$. To count the number of paths where $M_n\ge m$, we apply this reflection bijection, and note the reflected paths must have $n/2+m$ up-steps and $n/2-m$ down-steps, so $$ P(M_n\ge m)=\binom{n}{n/2+m}\Big/\binom{n}{n/2} . $$ Therefore, we just need to compute $\sum_{m=1}^{n/2}\binom{n}{n/2+m}\big/\binom{n}{n/2} $. For this, the following asymptotic lemma is useful.

Lemma: As $n,m\to\infty$ in such a way that $m\le n^{2/3}$, then $$\binom{n}{n/2+m}\Big/\binom{n}{n/2} \sim e^{-2m^2/n}\cdot \left(1+O\left(n^{-1/3}\right)\right)$$If instead $m,n\to\infty$ with $m\ge n^{2/3}$, then $$\binom{n}{n/2+m}\Big/\binom{n}{n/2}=O(\exp(-n^{1/3}))$$

Before I prove the lemma, let me show why it is useful. To compute $\sum_{m=1}^{n/2}\binom{n}{n/2+m}\big/\binom{n}{n/2}$, we split the sum into two regimes, one where $m\le n^{2/3}$, and one where $m>n^{2/3}$. The first regime becomes a Riemman sum: $$ \begin{align} \sum_{m=1}^{n^{2/3}} \binom{n}{n/2+m}\big/\binom{n}{n/2} &\sim \sum_{m=1}^{n^{2/3}} e^{-2m^2/n} \\&=\sqrt{n}\sum_{m=1}^{n^{2/3}} \frac1{\sqrt n}e^{-2(m/\sqrt{n})^2} \\&\sim \sqrt n\int_0^\infty dx \,e^{-2x^2} \\&=\sqrt{\frac{\pi n}{8}}. \end{align} $$ The second sum can then shown to be negligible compared to the first.

Proof of Lemma: Write $$\binom{n}{n/2+m}\big/\binom{n}{n/2}=\left(\frac{(n/2)!}{(n/2-m)!(n/2)^m}\right)\Big/ \left(\frac{(n/2+m)!}{(n/2)!(n/2)^m}\right).\tag1$$ Using this earlier answer of mine, I showed that $$ \frac{n!}{(n-k)!n^k}= \begin{cases} e^{-k^2/2n}(1+O(n^{-1/3})) & \text{if }k\le n^{2/3} \\ O\big(\exp(-n^{-1/3})\big) & \text{if }k\ge n^{2/3} \end{cases}\tag2 $$ Using the exact same method, you can additionally prove $$ \frac{(n+k)!}{n!n^k}= \begin{cases} e^{k^2/2n}(1+O(n^{-1/3})) & \text{if }k\le n^{2/3} \\ O\big(\exp(-n^{-1/3})\big) & \text{if }k\ge n^{2/3}\tag3 \end{cases} $$ Apply $(2)$ to the numerator of $(1)$, and apply $(3)$ to the denominator of $(1)$, and the lemma is proved. $\tag*{$\blacksquare$}$