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A stick of length $1$ is broken into $n$ pieces, with the cut points chosen as $n-1$ iid Uniform points. If $L_1, \ldots, L_n$ are the lengths of the pieces, are $L_1, \ldots, L_n$ exchangeable?

I can prove that $L_1, \ldots, L_n$ aren't independent and are identically distributed, but is there a formal way to show that $L_1, \ldots, L_n$ is exchangeable (that the joint distribution doesn't change if the indices are permuted)? Thanks.

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    Are you sure that you can prove they are independent? it doesn't seem like they would be. And if you could, then any collection of iid random variables is also exchangeable. – 2017-01-10
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    Thanks, I meant to say they aren't independent. – 2017-01-11
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    Other than $\Sigma L_i = 1$, are there other reasons why they are not independent? I mean, are $L_1, \ldots, L_{n-1}$ independent? (This isn't a coy hint, I just don't know and I'm curious to know what you showed.) – 2017-01-11

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Exchangeable means that the distribution of $(L_1,\cdots,L_n)$ is the same as $(L_{\pi(1)},\cdots,L_{\pi(n)})$ where $\pi$ is any permutation of $[n]$. Equivalently, the distribution is the same to any single transposition of $L_i,L_j$. More technically, $L_i$ is the length of the $i$'th piece, if pieces are tabulated left to right.

The proof is as follows. The pieces $L_i$ are determined by $U_0:=0,U_1,\cdots,U_{n-1},U_n:=1$, the cut points. Then $L_i=U_i-U_{i-1}$. Although the $L_i$ are not independent, you can now see that any pair $i\neq j$ is exchangeable since $U_i$ are independent and $U_i-U_{i-1}$ and $U_j-U_{j-1}$ can be swapped without changing the distribution.

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Let's experiment with a simulation in R statistical software, using $n = 4,$ so that there are $n-1 = 3$ uniformly distributed cuts, and four lengths. The experiment is iterated $m = 100,000$ times. The four lengths from the $i$th iteration are stored in the $i$th for of the $m \times n$ matrix $L.$ Lengths $L_j,\, j=1,2,3,4$ are the columns of $L.$ [Some facts related to those derived in the Answer by @AlexR. (+1), appearing just recently, are illustrated.]

We show a scatterplot of $L_2$ vs. $L_1$ showing the strong association between the first and second lengths. A correlation matrix shows the linear components of the associations of the four lengths (due to the restriction that they must sum to 1). Notice that $P(L_1 < L_2) \approx 1/2,$ and the same seems to be true for other pairs $(L_i, L_k).$ Also, $E(L_i) \approx 1/4,\, i = 1,2,3,4.$

All correlations $r_{ik} \approx -1/3,$ for $i \ne k,$ regardless of $|i-k|.$

m = 10^5; L = matrix(0, nrow=m, ncol=4)
for (i in 1:m) {
  x = sort(runif(3))
  L[i,] = diff(c(0,x,1)) }
plot(L[,1], L[,2], pch=".")
cor(L)
##            [,1]       [,2]       [,3]       [,4]
## [1,]  1.0000000 -0.3323449 -0.3353112 -0.3316366
## [2,] -0.3323449  1.0000000 -0.3347688 -0.3329160
## [3,] -0.3353112 -0.3347688  1.0000000 -0.3330164
## [4,] -0.3316366 -0.3329160 -0.3330164  1.0000000

View of the first six rows of $L:$

 head(L)
 ##            [,1]       [,2]       [,3]       [,4]
 ## [1,] 0.03374415 0.32182321 0.60317138 0.04126126
 ## [2,] 0.72121079 0.02778852 0.05219687 0.19880382
 ## [3,] 0.15893803 0.16255597 0.16551246 0.51299354
 ## [4,] 0.09773046 0.07345969 0.23131406 0.59749578
 ## [5,] 0.66563371 0.09203583 0.18899067 0.05333980
 ## [6,] 0.22231089 0.07900022 0.07720272 0.62148617

 colMeans(L)
 ## 0.2495914 0.2495984 0.2511221 0.2496881

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