Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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 $f_s=\mathbf 1_D/n!$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.