sethaxen
commented
2 years ago Continuing further, if we let $s_i = \sin^2(\phi_i) \in (0, 1]$, then the transform from $s$ is
$$ xi = \left(\prod{j=1}^{i-1} s_j \right) \begin{cases} 1 & \text{ if } i = N\ 1 - s_i & \text{otherwise} \end{cases}, $$
and its absolute Jacobian determinant is
$$ |J| = \prod_{i=1}^{N-2}s_i^{N-i - 1}. $$
Finally, letting $s_i = \operatorname{logistic}(y_i)$, we have the transform
$$ xi = \left(\prod{j=1}^{i-1} \operatorname{logistic}(y_j) \right) \begin{cases} 1 & \text{ if } i = N\ 1 - \operatorname{logistic}(y_i) & \text{otherwise} \end{cases}, $$
$$ |J| = \prod_{i=1}^{N-1}\operatorname{logistic}(y_i)^{N-i} (1-\operatorname{logistic}(y_i)) $$
mjhajharia
As noted in an earlier e-mail, if $z$ is a unit vector of length $N$, then squaring each element so $x_i = z_i^2$ gives us a point $x$ on the simplex. If all elements of $z$ are positive (i.e. $z$ is on the positive orthant of the hypersphere), then this map is bijective. So if we can generate unit vectors with positive elements, we can use that transform to also sample on the simplex.
As mentioned in #1, using
expto make each element positive in the end gives us the augmented softmax parameterization.We can do this as well with the hyperspherical transform, which is written
$$ zi = \left(\prod{j=1}^{i-1} \sin(\phi_j)\right) \begin{cases} 1 & \text{ if } i = N\ \cos(\phi_i) & \text{otherwise} \end{cases}, $$
where $\phi_j$ is in $(0, 2\pi]$ when $j=N-1$ and $(0, \pi]$ otherwise.
The requirement that $z_i$ be positive for $i < N-1$ is satisfied by requiring $\phi_i \in (0, \pi/2]$ for all $i$.
It turns out that the Jacobian determinant here is
$$ \begin{aligned} |J| &= 2^{N-1} \left(\prod_{i=1}^{N-2} \sin(\phii)^{N-i-1}\right) \left(\prod{i=1}^N zi \right)\ &= 2^{N-1} \prod{i=1}^{N-1}\tan(\phi_i) xi\ &= 2^{N-1} \prod{i=1}^{N-1}\sin^{2(N-i)-1}(\phi_i) \cos(\phi_i). \end{aligned} $$
I suspect due to the trigonometric functions, it may not be extremely efficient. However, all of these trigonometric functions are bijective over this range, so we can replace sines or cosines with numbers constrained to the interval $(0, 1)$ to get a new transform.
I'll shortly add Stan functions for both of these.