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?


1 Answer 1


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

$$ Y_1=\frac{X_1}{1-\sum X_i}=\frac{X_1}{1-X_1}=\frac{a_1+\epsilon}{1-X_1-\epsilon}\gt\frac{a_1}{1-X_1}=\frac{a_1}{1-\sum a_i}=b_i\;, $$

so one of the sets is contained in the other but not vice versa.

  • $\begingroup$ 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? $\endgroup$
    – PattuX
    Mar 15, 2022 at 15:51
  • 1
    $\begingroup$ @PattuX: First question: yes. Second question: I think it's the other way around? $X\in R$ implies $Y\in R'$, so $\mathsf P(X\in R)\le\mathsf P(Y\in R')$. $\endgroup$
    – joriki
    Mar 15, 2022 at 16:24

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