## The simplex's missing coordinate

Consider the following two ways to write the $d$-dimensional simplex:

First as subset of $\mathbb{R}^{d+1}$

$$ \Delta^d = \bigg\{x\in\mathbb{R}^{d+1} \,\bigg|\, x_i>0,\sum_{i=1}^{d+1}x_i=1\bigg\},$$

and second as subset of $\mathbb{R}^d$

$$\Delta_\text{I}^d = \bigg\{x\in\mathbb{R}^d \,\bigg|\, x_i>0,\sum_{i=1}^d x_i\le 1\bigg\}.$$

Note that the first form is a subset of the $d$-dimensional affine subspace

$$ H = \bigg\{x\in\mathbb{R}^{d+1} \,\bigg|\, \sum_{i=1}^{d+1}x_i=1\bigg\},$$

Consider the projection $P:\mathbb{R}^{d+1}\rightarrow\mathbb{R}^d$ that simply sets the $(d+1)$st coordinate to zero and leaves the others unchanged. Note that $P$ is a bijection from $H$ to $\mathbb{R}^d$ and also from $\Delta^d$ to $\Delta^d_\text{I}$.

Consider a subset $S\subset H$ with $d$-dimensional Hausdorff volume $v$ (as induced by the Euclidean metric of $\mathbb{R}^{d+1}$) and note that $P(S)\subset\mathbb{R}^d$ has Lebesgue volume $v/\alpha$. (To get the value of $\alpha$, [this](https://chatgpt.com/share/67dbf3d4-6c7c-8005-a2b1-a876df0b2175) should help. I think, $\alpha=\sqrt{\frac{d+1}{d}}$.)

The Lebesgue volume of $\Delta_\text{I}^d$ is $1/d!$. (Proof left as an exercise. `;->`) Therefore, the Hausdorff volume of $\Delta^d$ is $\alpha/d!$ (and not $1/d!$ as claimed by sloppy textbooks).

Now let's consider the stick-breaking process, which we denote as the bijective transformation $\Phi:[0,1]^d\rightarrow\Delta^d$. This transformation takes a vector $(v_1,...,v_d)\in[0,1]^d$ and interprets its coordinates as fractions that should be successively broken of from a stick that initially has unit length, yielding $d+1$ pieces of lengths $(x_1,...,x_{d+1})=\Phi(v_1,...,v_d)$. We have $x_i=v_i\prod_{j=1}^{i-1}(1-v_j)$ for $i\le d$ and $x_{d+1}=\prod_{i=1}^d(1-v_i)$. (Consider replacing $v_i$ by $1-v_i$ throughout to simplify notation. Then, $v_i$ is not the fraction that is broken off but the fraction that remains.)

Now, consider the composite map $P\circ \Phi$, which is a bijection from $[0,1]^d\rightarrow\Delta^d_I$. Find its Jacobian's determinant.