Starting with a single stick of unit length, a point $p \in (0, 1)$ is picked uniformly at random along the stick and the stick is snapped, producing two sticks of length $p$ and $1-p$.


At each next stage, a stick is picked uniformly at random, and a point is picked uniformly at random along the length of that stick, and it is snapped.


**Question:** After n snaps, what is the expected length of the longest remaining stick?


**Remarks:**


Myself and a friend of mine did some simulations and found some pretty unexpected results. The expected value after $500$ splits is approximately $0.2098$, which is massive for that many splits, at least intuitively.


On the other hand, it can be proven rather easily that the expected value does go to $0$ as $n \to \infty$. But the decay seems extremely slow.



Expected length of longest stick in a stick snapping process
It seems that the length of the longest stick is of order $n^{2\sqrt{2}-3} = n^{-0.171\ldots}$ as $n\to\infty$. This follows from a discrete-time analogue of the homogeneous fragmentation process, see chapter 1.5 of J. Bertoin, *Random fragmentation and coagulation processes*. Vol. 102. Cambridge University Press, 2006.


Let us denote by $X\_{n,0} \geq X\_{n,1} \geq \cdots \geq X\_{n,n}$ the ordered sizes of the sticks after $n$ snaps. The first result we need is the following.


**Lemma:** $\chi\_{n}(p) := \mathbb{E}\left[\sum\_{i=0}^n X\_{n,i}^p\right] = \frac{1}{n!}\left(\frac{2}{1+p}\right)\_n,$ where $(a)\_n=a(a+1)\cdots (a+n-1)$ is the rising Pochhammer symbol.


**Proof:** We have
\begin{align\*}
\mathbb{E}\left[\sum\_{i=0}^n X\_{n,i}^p\middle| X\_{n-1,0},\ldots\right] &= \frac{1}{n}\sum\_{k=0}^{n-1}\Big(\mathbb{E}\left[x^p + (X\_{n-1,k}-x)^p\middle|X\_{n-1,k}\right]+\sum\_{\substack{i=0\\i\neq k}}^{n-1}X\_{n-1,i}^p\Big)\\
&= \frac{1}{n}\left(\frac{2}{p+1}+n-1\right)\sum\_{i=0}^{n-1}X\_{n-1,i}^p,
\end{align\*}
where $x$ is uniform in $(0,X\_{n-1,k})$. Hence $\chi\_n(p) = \frac{1}{n}\left(\frac{2}{p+1}+n-1\right) \chi\_{n-1}(p)$.
Together with $\chi\_0(p)=1$, this gives the claimed formula for $\chi\_n(p)$. $\square$


Following Corollary 1.4 in Bertoin's book, we note that
\begin{equation}
n^{\frac{p-1}{p+1}} \chi\_n(p) = \frac{1}{\Gamma\left(\frac{2}{p+1}\right)} + O(n^{-1}).
\end{equation}
In particular it is bounded for any $p>-1$. Since $X\_{n,0}^p < \sum\_{i=0}^n X\_{n,i}^p$, we deduce that $n^{\frac{p-1}{p+1}}X\_{n,0}^p$ is bounded as $n\to\infty$. Hence
\begin{equation}
\limsup\_{n\to\infty} \frac{\log X\_{n,0}}{\log n} \leq -\frac{1}{p}\frac{p-1}{p+1} \leq -\frac{1}{\bar{p}}\frac{\bar{p}-1}{\bar{p}+1} = 2\sqrt{2}-3,
\end{equation}
because the maximum $\bar{p}$ of $-\frac{1}{p}\frac{p-1}{p+1}$ is achieved at $\bar{p} = 1+\sqrt{2}$.
Similarly one can derive a matching lower bound by noting that
\begin{equation}
X\_{n,0}^\epsilon \geq \frac{\sum\_{i=0}^{n}X\_{n,i}^p}{\sum\_{i=0}^{n}X\_{n,i}^{p-\epsilon}}
\end{equation}
for any $\epsilon>0$, which implies
\begin{equation}
\liminf\_{n\to\infty} \frac{\log X\_{n,0}}{\log n} \geq - \frac{\frac{p-1}{p+1} - \frac{p-\epsilon-1}{p-\epsilon+1}}{\epsilon} 
\end{equation}
for any $\epsilon>0$ and $0<p<\bar{p}$. Letting $\epsilon$ approach $0$ and $p$ approach $\bar{p}$, we thus have
\begin{equation}
\liminf\_{n\to\infty} \frac{\log X\_{n,0}}{\log n} \geq -\frac{2}{(1-\bar{p})^2}=2\sqrt{2}-3.
\end{equation}
We may therefore conclude that
\begin{equation}
\lim\_{n\to\infty} \frac{\log X\_{n,0}}{\log n} =2\sqrt{2}-3.
\end{equation}


