# Understanding transformation to Dirichlet distribution to inverted Dirichlet Distribution

This is essentially a question about the paper "On integrals of dirichlet distributions and their applications" by Hedayat Yassaee (1981).

Let $$K=\frac{\prod_{i=1}^{k+1}\Gamma(v_i)}{\Gamma(\sum_{i=1}^{k+1}v_i)}$$. Then we define the standard Dirichlet distribution $$Dir$$ via the following pdf:

$$f(x_1,\dots,x_k;v_1,\dots,v_k;v_{k+1} = K (\prod_{i=1}^k x_i^{v_i-1} )(1-\sum_{i=1}^kx_i)^{v_{k+1}-1}$$

for $$x_i\geq 0$$ and $$\sum_{i=1}^kx_i \leq 1$$, else $$f(\cdot)=0$$. Further, we define the inverted Dirichlet distribution (also called Dirichlet distribution of type II) $$Dir'$$ via the pdf:

$$g(x_1,\dots,x_k;v_1,\dots,v_k;v_{k+1} = K (\prod_{i=1}^k x_i^{v_i-1} )(1+\sum_{i=1}^kx_i)^{-\sum_{i=1}^{k+1}{v_i}}$$

for $$x_i\geq 0$$, else $$g(\cdot)=0$$.

Let $$X=(X_1,\dots,X_k) \sim Dir(v_1,\dots,v_k)$$ and define $$Y_i = \frac{X_i}{1-\sum_{j=1}^k X_j}$$. Then it can be shown that $$Y=(Y_1,\dots,Y_k) \sim Dir'(v_1,\dots,v_k)$$.

Let $$a_1,\dots,a_k$$ be real numbers between $$0$$ and $$1$$ and define $$R = \{(x_1,\dots,x_k)\mid x_i \leq a_i\}$$. Now Yassaee claims that:

$$Pr(X \in R) = Pr(Y \in R')$$

where $$R' = \{(x_1,\dots,x_k)\mid x_i \leq b_i\}$$ and $$b_i = \frac{a_i}{1-\sum_{j=1}^k a_j}$$.

I'm very much confused by the latter statement. Clearly this does not hold for all $$a_i$$: If $$\sum_{i=1}^k a_i > 1$$ then each $$b_i < 0$$ and since $$Y_i \geq 0$$ we clearly have $$Pr(Y \in R')=0$$, whereas $$Pr(X \in R) > 0$$. In particular, in case all $$a_i=1$$ we have $$Pr(X \in R) = 1$$.

But even if we add the restriction $$\sum_{i=1}^k a_i < 1$$, the transformation $$Y_i = \frac{X_i}{1-\sum_{j=1}^k X_j}$$ seems very much non-linear. So I don't see why we would even expect that a rectangular region like $$R$$ would still be rectangular after the transformation.

Am I misunderstanding something fundamental here or is the paper actually just flawed?

You’re right; this is wrong. $$X_i\le a_i$$ does imply $$Y_i\le b_i$$, since
$$Y_i=\frac{X_i}{1-\sum X_i}\le\frac{a_i}{1-\sum X_i}\le\frac{a_i}{1-\sum a_i}=b_i\;,$$
but for $$X_1=a_1+\epsilon$$ and all other $$X_i=0$$ we have
$$Y_1=\frac{X_1}{1-\sum X_i}=\frac{X_1}{1-X_1}=\frac{a_1+\epsilon}{1-X_1-\epsilon}\gt\frac{a_1}{1-X_1}=\frac{a_1}{1-\sum a_i}=b_i\;,$$
• Your first statement then is under the restriction that $\sum a_i < 1$, right (used in the second inequality)? In that case $Pr(Y \in R') \leq Pr(X \in R)$, i.e., we obtain a lower bound at least, correct? Mar 15, 2022 at 15:51
• @PattuX: First question: yes. Second question: I think it's the other way around? $X\in R$ implies $Y\in R'$, so $\mathsf P(X\in R)\le\mathsf P(Y\in R')$. Mar 15, 2022 at 16:24