Let random K-dimensional variable $\mathbf{v} \sim \mathbf{Dir}(\mathbf{\alpha})$. I want to find a one-to-one mapping $f(\cdot)$ such that $f(\mathbf{v})$ is a random variable on whole $R^{K-1}$.

  • $\begingroup$ Do you expect some specific properties of $f(v)$? Do you wish it to cover whole $R^{K-1}$? $\endgroup$ Jul 7, 2017 at 10:32
  • $\begingroup$ yes. I edited my question. $\endgroup$
    – messcode
    Jul 7, 2017 at 12:28
  • $\begingroup$ One standard solution is described, in detail, at stats.stackexchange.com/questions/259208/…. Once you have one such mapping, you may compose it with any bijective measurable map from $\mathbb{R}^{K-1}$ to itself to generate all possible solutions. Thus, there is tremendous ambiguity in your question. As a result, I take the link to be a valid answer, but if it doesn't suit you, I invite you to edit your question to clarify what properties you need $f$ to have. $\endgroup$
    – whuber
    Jul 7, 2017 at 13:41

1 Answer 1


Maybe something like this:

$f_1(v_1,\ldots,v_K) = (v_1,\ldots, v_{K-1})$ . Since in Dirichlet distribution $\sum_{i=1}^Kv_i=1$, this is one-to one mapping.

$f_1(v)$ is now random variable on a "hyper-triangle" in $\mathbb{R}^{K-1}$.

Let's now map it onto $[0,1]^{K-1}$. We can leave $v_1$ as it is, and compute what proportion of "what is left by $v_1$", $v_2$ took. This would be $v_2/(1-v_1)$ and this takes values from whole $[0,1]$, and this is one-to-one projection. Then, similarly, we can take $v_3/(1-v_1-v_2)$ and so on. We have then

$f_2(v_1,\ldots,v_{K-1}) = (v_1',\ldots,v_{K-1}')$ , where $v_1'=v_1$ and $v_i'=v_i/(1-v_1-\ldots-v_{i-1})$ for $i=2,3,,\ldots,K-1$.

Mapping from $[0,1]^{K-1}$ to whole space is easy now. We can rescale our $f_2(f_1(v))$ to $[-\pi/2,\pi/2]^{K-1}$ and use tangent.

So we have $f_3(v_1',\ldots,v_{K-1}') = (v_1'',\ldots,v_{K-1}'')$, where $v_{K-1}'' = \tan (\pi v_i'-\pi/2)$ for $i=1,2,3,,\ldots,K-1$.

Finally, $f(v)=f_3(f_2(f_1(v)))$ takes values from whole $\mathbb{R}^{K-1}$, and since $f_1$, $f_2$ and $f_3$ are one-to-one projections, $f$ is so.


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