Index: The Book of Statistical ProofsProbability Distributions ▷ Multivariate continuous distributions ▷ Dirichlet distribution ▷ Exceedance probabilities

Theorem: Let $r = [r_1, \ldots, r_k]$ be a random vector following a Dirichlet distribution with concentration parameters $\alpha = [\alpha_1, \ldots, \alpha_k]$:

$\label{eq:r-Dir} r \sim \mathrm{Dir}(\alpha) \; .$

1) If $k = 2$, then the exceedance probability for $r_1$ is

$\label{eq:Dir2-EP} \varphi_1 = 1 - \frac{\mathrm{B}\left( \frac{1}{2};\alpha_1,\alpha_2 \right)}{\mathrm{B}(\alpha_1,\alpha_2)}$

where $\mathrm{B}(x,y)$ is the beta function and $\mathrm{B}(x;a,b)$ is the incomplete beta function.

2) If $k > 2$, then the exceedance probability for $r_i$ is

$\label{eq:Dir-EP} \varphi_i = \int_0^\infty \prod_{j \neq i} \left( \frac{\gamma(\alpha_j,q_i)}{\Gamma(\alpha_j)} \right) \, \frac{q_i^{\alpha_i-1} \exp[-q_i]}{\Gamma(\alpha_i)} \, \mathrm{d}q_i \; .$

where $\Gamma(x)$ is the gamma function and $\gamma(s,x)$ is the lowerr incomplete gamma function.

Proof: In the context of the Dirichlet distribution, the exceedance probability for a particular $r_i$ is defined as:

$\label{eq:Dir-EP-def} \begin{split} \varphi_i &= p \Bigl( \forall j \in \left\lbrace 1, \ldots, k \Bigm| j \neq i \right\rbrace: \, r_i > r_j |\alpha \bigr) \\ &= p \Bigl( \bigwedge_{j \neq i} r_i > r_j \Bigm| \alpha \Bigr) \; . \end{split}$

The probability density function of the Dirichlet distribution is given by:

$\label{eq:Dir-pdf} \mathrm{Dir}(r; \alpha) = \frac{\Gamma\left( \sum_{i=1}^k \alpha_i \right)}{\prod_{i=1}^k \Gamma(\alpha_i)} \, \prod_{i=1}^k {r_i}^{\alpha_i-1} \; .$

Note that the probability density function is only calculated, if

$\label{eq:Dir-req} r_i \in [0,1] \quad \text{for} \quad i = 1,\ldots,k \quad \text{and} \quad \sum_{i=1}^k r_i = 1 \; ,$

1) If $k = 2$, the probability density function of the Dirichlet distribution reduces to

$\label{eq:Dir2-pdf} p(r) = \frac{\Gamma(\alpha_1 + \alpha_2)}{\Gamma(\alpha_1) \, \Gamma(\alpha_2)} \, r_1^{\alpha_1-1} \, r_2^{\alpha_2-1}$

which is equivalent to the probability density function of the beta distribution

$\label{eq:Beta-pdf} p(r_1) = \frac{r_1^{\alpha_1-1} \, (1-r_1)^{\alpha_2-1}}{\mathrm{B}(\alpha_1,\alpha_2)}$

with the beta function given by

$\label{eq:beta-fct} \mathrm{B}(x,y) = \frac{\Gamma(x) \, \Gamma(y)}{\Gamma(x + y)} \; .$

With \eqref{eq:Dir-req}, the exceedance probability for this bivariate case simplifies to

$\label{eq:Dir2-EP-def} \varphi_1 = p(r_1 > r_2) = p(r_1 > 1 - r_1) = p(r_1 > 1/2) = \int_{\frac{1}{2}}^1 p(r_1) \, \mathrm{d}r_1 \; .$

Using the cumulative distribution function of the beta distribution, it evaluates to

$\label{eq:Dir2-EP-qed} \varphi_1 = 1 - \int_0^{\frac{1}{2}} p(r_1) \, \mathrm{d}r_1 = 1 - \frac{\mathrm{B}\left( \frac{1}{2};\alpha_1,\alpha_2 \right)}{\mathrm{B}(\alpha_1,\alpha_2)}$

with the incomplete beta function

$\label{eq:inc-beta-fct} \mathrm{B}(x; a, b) = \int_0^x x^{a-1} \, (1-x)^{b-1} \, \mathrm{d}x \; .$

2) If $k > 2$, there is no similarly simple expression, because in general

$\label{eq:Dir-EP-ineq} \varphi_i = p(r_i = \mathrm{max}(r)) > p(r_i > 1/2) \quad \text{for} \quad i = 1, \ldots, k \; ,$

i.e. exceedance probabilities cannot be evaluated using a simple threshold on $r_i$, because $r_i$ might be the maximal element in $r$ without being larger than $1/2$. Instead, we make use of the relationship between the Dirichlet and the gamma distribution which states that

$\label{eq:Gam-Dir} \begin{split} & Y_1 \sim \mathrm{Gam}(\alpha_1,\beta), \, \ldots, \, Y_k \sim \mathrm{Gam}(\alpha_k,\beta), \, Y_s = \sum_{i=1}^k Y_j \\ \Rightarrow \; & X = (X_1, \ldots, X_k) = \left( \frac{Y_1}{Y_s}, \ldots, \frac{Y_k}{Y_s} \right) \sim \mathrm{Dir}(\alpha_1, \ldots, \alpha_k) \; . \end{split}$

The probability density function of the gamma distribution is given by

$\label{eq:Gam-pdf} \mathrm{Gam}(x; a, b) = \frac{ {b}^{a} }{\Gamma(a)} \, x^{a-1} \, \exp[-b x] \quad \text{for} \quad x > 0 \; .$

Consider the gamma random variables

$\label{eq:Gam-Dir-A} q_1 \sim \mathrm{Gam}(\alpha_1,1), \, \ldots, \, q_k \sim \mathrm{Gam}(\alpha_k,1), \, q_s = \sum_{j=1}^k q_j$

and the Dirichlet random vector

$\label{eq:Gam-Dir-B} r = (r_1, \ldots, r_k) = \left( \frac{q_1}{q_s}, \ldots, \frac{q_k}{q_s} \right) \sim \mathrm{Dir}(\alpha_1, \ldots, \alpha_k) \; .$

Obviously, it holds that

$\label{eq:Gam-Dir-eq} r_i > r_j \; \Leftrightarrow \; q_i > q_j \quad \text{for} \quad i,j = 1, \ldots, k \quad \text{with} \quad j \neq i \; .$

Therefore, consider the probability that $q_i$ is larger than $q_j$, given $q_i$ is known. This probability is equal to the probability that $q_j$ is smaller than $q_i$, given $q_i$ is known

$\label{eq:Gam-EP0} p(q_i > q_j|q_i) = p(q_j < q_i|q_i)$

which can be expressed in terms of the cumulative distribution function of the gamma distribution as

$\label{eq:Gam-EP1} p(q_j < q_i|q_i) = \int_0^{q_i} \mathrm{Gam}(q_j;\alpha_j,1) \, \mathrm{d}q_j = \frac{\gamma(\alpha_j,q_i)}{\Gamma(\alpha_j)}$

where $\Gamma(x)$ is the gamma function and $\gamma(s,x)$ is the lower incomplete gamma function. Since the gamma variates are independent of each other, these probabilties factorize:

$\label{eq:Gam-EP2} p(\forall_{j \neq i} \left[ q_i > q_j \right]|q_i) = \prod_{j \neq i} p(q_i > q_j|q_i) = \prod_{j \neq i} \frac{\gamma(\alpha_j,q_i)}{\Gamma(\alpha_j)} \; .$

In order to obtain the exceedance probability $\varphi_i$, the dependency on $q_i$ in this probability still has to be removed. From equations (\ref{eq:Dir-EP-def}) and (\ref{eq:Gam-Dir-eq}), it follows that

$\label{eq:Dir-EP2a} \varphi_i = p(\forall_{j \neq i} \left[ r_i > r_j \right]) = p(\forall_{j \neq i} \left[ q_i > q_j \right]) \; .$

Using the law of marginal probability, we have

$\label{eq:Dir-EP2b} \varphi_i = \int_0^\infty p(\forall_{j \neq i} \left[ q_i > q_j \right]|q_i) \, p(q_i) \, \mathrm{d}q_i \; .$

With (\ref{eq:Gam-EP2}) and (\ref{eq:Gam-Dir-A}), this becomes

$\label{eq:Dir-EP2c} \varphi_i = \int_0^\infty \prod_{j \neq i} \left( p(q_i > q_j|q_i) \right) \cdot \mathrm{Gam}(q_i;\alpha_i,1) \, \mathrm{d}q_i \; .$

And with (\ref{eq:Gam-EP1}) and (\ref{eq:Gam-pdf}), it becomes

$\label{eq:Dir-EP-qed} \varphi_i = \int_0^\infty \prod_{j \neq i} \left( \frac{\gamma(\alpha_j,q_i)}{\Gamma(\alpha_j)} \right) \cdot \frac{q_i^{\alpha_i-1} \exp[-q_i]}{\Gamma(\alpha_i)} \, \mathrm{d}q_i \; .$

In other words, the exceedance probability for one element from a Dirichlet-distributed random vector is an integral from zero to infinity where the first term in the integrand conforms to a product of gamma cumulative distribution functions and the second term is a gamma probability density function.

Sources:
• Soch J, Allefeld C (2016): "Exceedance Probabilities for the Dirichlet Distribution" ; in: arXiv stat.AP , 1611.01439 ; URL: https://arxiv.org/abs/1611.01439 .

Metadata: ID: P181 | shortcut: dir-ep | author: JoramSoch | date: 2020-10-22, 08:04.