I need to compute the Dirichlet CDF, but I can only find implementations of the PDF.
Do you guys know of any library (preferably in R) implementing it?
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2 Answers
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Remember that, if $Y_i$ are independent $\mathrm{Gamma}(a_i,b)$, for $i=1,\dots,k$, then $$ (X_1,\dots,X_k) = \left(\frac{Y_1}{\sum_{j=1}^k Y_j}, \dots, \frac{Y_k}{\sum_{j=1}^k Y_j} \right) \sim \mathrm{Dirichlet}(a_1,\dots,a_k) \, .$$
The proof can be found on page 594 of Luc Devroye's masterpiece.
Therefore, one possibility is to compute a Monte Carlo approximation of
$$
F_{X_1,\dots,X_k}(t_1,\dots,t_k)=P\left\{X_1\leq t_1,\dots, X_k\leq t_k\right\} \, ,
$$
starting with gammas. In R, try this:
pdirichlet <- function(a, t, N = 10^5) {
rdirichlet <- function(a) { y <- rgamma(length(a), a, 1); y / sum(y) }
x <- replicate(N, rdirichlet(a), simplify = FALSE)
mean(sapply(x, function(x) prod(x <= t)))
}
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3$\begingroup$ There's a vectorized rdirichlet function in R already - in fact several of them (in
gtools,MCMCpackanddirmultfor example). $\endgroup$– Glen_bCommented Apr 27, 2013 at 0:43 -
$\begingroup$ @Zen I have
a <- c(6, 20,2)how to obtain the Drichelt cdf? is t 2 by 2 matrix? $\endgroup$– score324Commented Oct 4, 2018 at 13:33 -
$\begingroup$ I used the above code, but its throwing an error. $\endgroup$– score324Commented Oct 4, 2018 at 19:08
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$\begingroup$ @score324 A Dirichlet vector usually has each element in $[0,1]$ and their sum being $1$, so $(6, 20,2)$ is outside the support $\endgroup$– HenryCommented Mar 10, 2020 at 15:28
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Any library? Mathematica has it. Here's the code for an example plot of a Dirichlet CDF from the documentation:
Plot3D[CDF[DirichletDistribution[{1, 3, 2}], {x, y}], {x, 0, 1}, {y, 0, 1}]
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1$\begingroup$ How to get expression for CDF[DirichletDistribution[{1, 3, 2}]] ? $\endgroup$– AIBCommented May 20, 2013 at 10:30
rdirichlet. If it's only 3-variate or possibly 4-variate (the last component, of course, being redundant) it may be worth trying numerical quadrature. $\endgroup$