I have to count Jacobians on my fingers.

Overall, what needs to happen is the K free unconstrained parameters map to the scaled simplex \beta \alpha, where \alpha is a simplex and \beta is positive. I just meant that the Jacobian for the K - 1 unconstrained parameters to simplex \alpha and 1 unconstrained variable to positive \beta are already included.

But now that I write it out, I see you took this into account in the mapping from the K - 1 outputs of the simplex and \beta to \beta \alpha.

Now if I lay out another transform as

(\alpha_1, \ldots, \alpha_{K - 1}, \beta)
\mapsto (\beta \alpha_1, \ldots, \beta \alpha_K)

then we need to add the log Jacobian determinant of this map to the implicit log Jacobian applied to the simplex and positive transforms. The Jacobian is lower triangular [edit: not, see below!] with the first K - 1 elements of the diagonal being for k \in 1:(K - 1),

J_{k, k} = \frac{\partial}{\partial \alpha_k} \beta \alpha_k = \beta

and the K-th element being

J_{K, K} = \frac{\partial}{\partial \beta} \, \beta \cdot \alpha_K
\ = \ \frac{\partial}{\partial \beta} \, \beta \cdot \left( 1 - \sum_{k=1}^{K-1} \alpha_k \right)
\ = \ \alpha_K

[edit: this is right as far as it goes, but it misses the non-zero elements in the last row and column.]

The log Jacobian determinant of a lower-triangular matrix is just the sum of the logs of the elements [edit: correct], so it looks like it should be [edit: but is not]:

\log \mathrm{det}(J) = (K - 1) \log \beta + \log \alpha_K

So it looks to me like the log Jacobian should is missing a \log \alpha_K term in the final log Jacobian determinant [edit: itās not].

But as I said, Iām really bad at this. These problems are like homework to me and I keep hoping to get better at it. So please correct my homework if itās wrong. [edit: Thanks!]