I've recently come across the problem where, given $X_i\in \text{Uniform}(0,1)$ (wlog. ordered by value here), $1\leq i<n$, these samples divide the interval, and you are looking for the expected length of the longest segment. Ie. letting $X_0=0, X_{n}=1$, and $I_i=|X_i-X_{i-1}|$. As noted here, we have that $\mathbb{E}[\max\{I_i\}]=\frac{1}{n}H_n$ with $H_n$ being the $n$-th harmonic number.
I was interested in finding the cumulative distribution function of $Y=\max\{I_1,\ldots,I_n\}$ for some $n$, denote it by $F_n(y)$. The subscript $n$ determines the number of intervals, so $$F_1(y)=\left\lbrace\begin{matrix}0\text{ if }x<1\\ 1\text{ else}\end{matrix}\right..$$
Considering the location of the $n$-th dividing point, I've been able to find a recursion for these cdf's $$F_n(y)=\int^x_0F_{k}\left(\frac{x}{t}\right)F_{n-k}\left(\frac{x}{1-t}\right)dt=\int^x_0F_{n-1}\left(\frac{x}{1-t}\right)dt$$ setting $k=1$. The results seem okay for $n=1,2,3$ with a correct expectation and the cdf plots agree with the simulations I ran. I was wondering if there is a general solution for this cdf, or any ideas on how to approach solving such a functional recursion relation?
