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Consider $n$ random variables $t_1$ through $t_n$ each of which is uniformly randomly chosen on $[0,1]$ and labelled left to right so that $0\leq t_1 \leq t_2 \ldots \leq t_n \leq 1$. We readily see that their joint density is $f_t(t_1,\ldots,t_n) = n!$

Now introduce $n+1$ new variables corresponding to the distances between successive rv's as follows: \begin{align} s_1:= t_1\\ s_i:= t_{i} - t_{i-1},~ 1\leq i\leq n \\ s_{n+1} :=1-\sum_{1\leq i\leq n} s_i = 1-t_n \end{align}

Note that $0\leq s_i\leq 1$ with the added constraint that $\sum_{1\leq i\leq {n+1}}s_i =1$.

Now I'd like to determine the joint density $f_s(s_1,\ldots,s_n)$, where $s_{n+1}$ is omitted as it is functionally determined by the preceding $s_i$'s. From these notes (p. 7) we have that:

$f_s (s_1,\ldots, s_n) = \det(M)f_t = \det [\frac{\partial{(t_1,\ldots,t_n)}}{\partial{(s_1,\ldots,s_n)}}] f_t(t_1,\ldots,t_n)$

where $M$ is the matrix corresponding to the Jacobian. From the definitions, it's easy to see that $M_{i-1,i} = -1$ and $M_{i,i} = +1$. This causes the Jacobian $\det(M)$ to evaluate to +1 for small values of $n$ (and very possibly this pattern is true for all $n$). Thus, $f_s = f_t = n!$. However, for the function $f_s$ to be a (uniform) pdf , it can only be defined as $$f_s(s_1,\ldots,s_n)= 1/n!$$ so that it integrates to unity over its domain.

Where is the error in computing the Jacobian $\det(M)$?

In other words, how do I incorporate the constraint $t_i \leq t_{i+1}$ into the transformation?

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The domain of $f_s$ is $ D=\{(s_i)\mid 0\leqslant s_i\leqslant 1,\,s_1+\cdots+s_n\leqslant1\} $ and the volume of $D$ is $1/n!$ hence $f_s=n!\,\mathbf 1_D$, "so that it integrates to unity over its domain", not $f_s=\mathbf 1_D/n!$.

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