Let $X_i\sim\text{Gamma}(\alpha,p_i),i=1,2,...,n+1$ be independent random variables.

Define $Z_1=\sum_{i=1}^{n+1}X_i$ and $Z_i=\frac{X_i}{\sum_{j=1}^iX_j},\quad i=2,3,...,n+1$. Then show that $Z_1,Z_2,...,Z_{n+1}$ are independently distributed.

The joint density of $(X_1,...,X_{n+1})$ is given by

$$f_{\bf X}(x_1,...,x_{n+1})=\left[\frac{\alpha^{\sum_{i=1}^{n+1}p_i}}{\prod_{i=1}^{n+1}\Gamma(p_i)}\exp\left(-\alpha\sum_{i=1}^{n+1}x_i\right)\prod_{i=1}^{n+1}x_i^{p_i-1}\right]\mathbf I_{x_i>0}\quad,\alpha>0,p_i>0$$

We transform $\mathbf X=(X_1,\cdots,X_{n+1})\mapsto\mathbf Z=(Z_1,\cdots,Z_{n+1})$ such that

$Z_1=\sum_{i=1}^{n+1}X_i$ and $Z_i=\frac{X_i}{\sum_{j=1}^iX_j},\quad i=2,3,...,n+1$

$\implies x_{n+1}=z_1z_{n+1},$

$\qquad x_n=z_1z_n(1-z_{n+1}),$

$\qquad x_{n-1}=z_1z_{n-1}(1-z_n)(1-x_{n+1}),$


$\qquad x_3=z_1z_3\prod_{j=4}^{n+1}(1-z_j)$

$\qquad x_2=z_1z_2\prod_{j=3}^{n+1}(1-z_j)$

$\qquad x_1=z_1\prod_{j=2}^{n+1}(1-z_j)$, where $0<z_1<\infty$ and $0<z_i<1,\quad i=2,3,\cdots,n+1$

The Jacobian of the transformation is $J=\dfrac{\partial(x_1,...,x_{n+1})}{\partial(z_1,...,z_{n+1})}=\det\begin{pmatrix}\frac{\partial x_1}{\partial z_1} &\cdots& \frac{\partial x_1}{\partial z_{n+1}} \\ &\ddots\\\frac{\partial x_{n+1}}{\partial z_1} &\cdots& \frac{\partial x_{n+1}}{\partial z_{n+1}}\\\end{pmatrix}$

Performing the operation $R_1'=\sum_1^{n+1}R_i$, we get $J$ as the determinant of

$\begin{pmatrix}1&0&0&\cdots&0&0 \\z_2\prod_3^{n+1}(1-z_j)& z_1\prod_3^{n+1}(1-z_j) \\ z_3\prod_4^{n+1}(1-z_j)&0& z_1\prod_4^{n+1}(1-z_j)\\&&&\ddots\\z_n(1-z_{n+1})&0&0&\cdots&z_1(1-z_{n+1})& -z_1z_n \\z_{n+1}&0&0 &\cdots&0& z_1\\\end{pmatrix}$

which equals $z_1^n(1-z_3)(1-z_4)^2...(1-z_n)^{n-2}(1-z_{n+1})^{n-1}$.

After some simplification we get the joint density of $\bf Z$ as

$f_{\bf Z}(z_1,...,z_{n+1})=\prod_{i=1}^{n+1}f_{Z_i}(z_i)$

where $Z_1\sim \text{Gamma}(\alpha,\sum_1^{n+1}p_i),$

$\qquad Z_2\sim \text{Beta}_1(p_1,p_2),$

$\qquad Z_3\sim \text{Beta}_1(p_3,p_1+p_2),$

$\qquad \vdots$

$\qquad Z_{n+1}\sim \text{Beta}_1(p_{n+1},\sum_1^np_i)$,

with $0<z_1<\infty$ and $0<z_i<1,\quad i=2,3,\cdots,n+1$,

$\alpha>0$ and $p_i>0$ for $i=1,2,...,n+1$.

Needless to say, finding the inverse solutions $x_i$'s and evaluating the Jacobian was cumbersome and time consuming. Besides getting the job done, it also determines the distributions of the $Z_i$'s.

Is there any simpler way to show just the independence of the $Z_i$'s?

  • $\begingroup$ I meant $...(1-z_{n+1})$ in the rhs of $x_{n-1}$. $\endgroup$ Commented Dec 9, 2017 at 19:39
  • $\begingroup$ The only way I could think of simplifying this is using induction, i.e. starting with the n=1 case, and finding $Z_1$ and $Z_2$ are independent, then using the independence of $X_i$ with the $Z_j$ ($j<i$) to just add in the $X_i$ one by one as needed. $\endgroup$
    – aleshing
    Commented Dec 9, 2017 at 19:41
  • $\begingroup$ Related with the Dirichlet distribution. $\endgroup$ Commented Dec 24, 2017 at 19:48

1 Answer 1


I'll prove the equivalent statement.

Let $n\ge 1$, $X_k\sim \Gamma(\alpha,p_k)$, $k=1,\dots,n+1$, are independent. Denote $S_k = X_1+\dots+X_k$, $k=1,\dots,n+1$; $R_k = \frac{S_k}{S_{k+1}}$, $k=1,\dots,n$. Then $R_1$, $R_2$, $\dots$, $R_{n}$, and $S_{n+1}$ are independent and $S_{n+1}\sim \Gamma(\alpha,p_1+\dots + p_{n+1})$.

Remark In OP notation, $Z_1=S_{n+1}$, $Z_k = 1 - R_{k-1}$, $k=2,\dots,n+1$.

Proof. $n=1$ is easy (and well known).

$n-1\Rightarrow n$.

By induction hypothesis and independence, the vectors $\mathbf{R} = (R_1,\dots,R_{n-1})$ and $(S_{n},X_{n+1})$ are independent. Therefore, the vectors $\mathbf{R}$ and $\big(\frac{S_{n}}{S_{n}+X_{n+1}},S_{n}+X_{n+1}\big) = (R_n,S_{n+1})$ are independent. Both vectors have independent components: $\mathbf R$ by induction hypotheses, $(R_n,S_{n+1})$ by induction base. Therefore, their components are independent random variables.

It is no problem to include the distribution of $R_n$ to the statement, the proof won't change. The distribution of $S_n$ is already there: I need it for saying that the independence of $R_n$ and $S_{n+1}$ follows from the induction base.


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