Maximum of dependent random variables Consider the following process: Pick $N$ numbers uniformly at random from $U[0,1]$. Suppose that they are numbered such that $w_1 \geq w_2 \geq \cdots \geq w_N$. Denote $X_i=w_i-w_{i+1}$. Find $\mathbb{E}\left[ \max_i X_i \right]$.
Here's what I tried:
\begin{aligned}
\mathbb{E}\left( \max_i X_i \right) = & \int_{1/N}^1 1 - \mathbb{P}(\max_i X_i \leq x) dx \\
= & \int_{1/N}^1 \int_{1/N}^1 \cdots \int_{1/N}^1 \mathbb{P} \left( X_1\leq x ∧ X_2\leq x ∧ \cdots X_N\leq x \right)
\end{aligned}
Well, things got quickly out of hand here. I am aware that integral limits above are not correct since $\sum_i X_i = 1$.
Is there a different way around for dependent random variables?
 A: One way to think about this is to define the joint pdf:
$$f(w_1,w_2,...,w_N)=f(w_1)f(w_2|w_1)f(w_3|w_2,w_1)...f(w_N|w_N-1,...w_1)$$
which I believe can be written as:
$$f(w_1,w_2,...,w_N)=f_N(w_1|1)f_{N-1}(w_2|w_1)f_{N-2}(w_3|w_2)...f_1(w_N|w_N-1)$$
such that $f_{N-2}(w_3|w_2)$ is the pdf for $w_3$ which represents the maximum of $N-2$ uniform random variables over the range $[0,w_2]$.
Using: https://stats.stackexchange.com/questions/18433/how-do-you-calculate-the-probability-density-function-of-the-maximum-of-a-sample
$$f_n(w|b)={n w^{n-1}\over b^n}$$
Interestingly enough, many terms cancel, leaving:
$$f(w_1,w_2,...,w_N) = N!$$
provided $w_1>w_2>...>w_N$ and zero otherwise. This supports the symmetry comment - it is uniform within the slice of space in which the ordering holds. Maybe this should have been obvious.
As a first attempt, with the erroneous assumption that the $X_i$ could be treated as independent. The next step is to consider $X_i=W_i-W_{i+1}$. Its pdf can be constructed as a convolution. Doing this by hand for $i=1$ and for $N=2$ and $N=3$ (work not shown here), suggests again a very simple form:
$$h_N(x_1) = N(1-x_1)^{N-1}$$
As stated in the comment the pdf for $X_i$ is the same for all $i$. The cdf for each is therefore,
$$H_N(x_i) = 1-(1-x_i)^N$$
Finally consider $Z_N=\max(X_i)$. Its cdf can be constructed through the product of the $(N-1)$ cdf's for $X_i$: 
$$P_N(Z_N<z)=\left(H_N(z)\right)^{N-1}$$
So the pdf is:
$$g_N(z)=N(N-1)(1-z)^{N-1}\left[1-(1-z)^N\right]^{N-2}$$
and the expectation value is:
$$E[Z_N]=\int_0^1 z\,g_N(z)\,dz$$
If we consider $U_N=1-Z_N$, the expectation works out easily (using WolframALPHA):
$$E[U_N] = {N-1\over N}{\Gamma(N-1)\Gamma({1\over N})\over\Gamma(N+{1\over N})}$$
so finally,
$$E[Z_N] = 1 - E[U_N] = 1 - {N-1\over N}{\Gamma(N-1)\Gamma({1\over N})\over\Gamma(N+{1\over N})}$$
A plot of $E[Z_N]$ vs $N$ is below (also Wolfram Alpha)

A correct approach treats the joint pdf for $x_i$, without the mistaken assumption above that the $X_i$ are independent. Transform from $w_i, i=1...n$ to $x_i = w_i - w_{i+1}, i = 1...N-1$ by the relations:
$$w_2 = w_1 - x_1, w_3 = w_2 - x_2 = w_1 - x_1 - x_2, ...$$
so
$$h(x_1,...,x_{N-1}) = \int_s^1 f(w_1, w_1-x_1, w_1 - x_1 - x_2, ...) \,dw_1$$
where the integral is over all possible $w_1$, from $s=\sum_{i=1}^{N-1}x_i$ to 1. This integral is easy, since the integrand is constant:
$$h(x_1,...,x_{N-1}) = N! (1-s)$$
provided $s<1$ and 0 otherwise. To find the pdf for $Z_N=\max(X_i)$, evaluate:
$$g_N(z) = \int h(z,x_2,x_3,...)\,dx_2\,dx_3... + \int h(x_1,z,x_2,...)\,dx_1\,dx_3... + ...$$
over the space $x_i<z$. Given the symmetry of $h()$, each integral is the same, so
$$g_N(z) = (N-1)N!\int \left(1-z-\sum_{i=2}^{N-1}x_i\right)\,dx_2\,dx_3...\,dx_{N-1}$$
where the boundaries for integration are given by the conditions:
$$0<x_i<z \mathrm{\ \ and\ \ } z+\sum_{i=2}^{N-1}x_i<1$$
I don't have tools to do this analytically for general N, so evaluated these for $N=2,3,4$. The pdfs are not simple forms (piecewise polynomials). The expectations for $Z_N$ turn out to be simple fractions:
$$\begin{align} E[Z_2] & = 1/3 \\ E[Z_3] & = 3/8 \\ E[Z_4] & = 11/30 \end{align} $$
A numerical approach was used to find that,
$$\begin{align} E[Z_5] & = 25/72 \\ E[Z_6] & = 137/420 \end{align} $$
With these values it became apparent that there is a very simple relation
$$E[Z_N] = {1\over N+1}\sum_{i=1}^{N-1}{1\over i}$$
which correctly predicted the numerical result for $E[Z_7]$. No doubt there is a way to solve this problem analytically, since the result has such a simple form. Can anyone explain the approach to take?
