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Searching at least 3 hours about the connection between beta distribution and dirichlet. My problem is:

I have a collection of random variables $X_i \sim Beta(a_i, b_i)$. The parameters $a_i$ and $b_i$ are known, $\forall i=1,2,...K$. From $\{X_i\}$, I want the dirichlet distribution $(X_1,X_2,...,X_k) \sim Dir(a)$. However, I cannot find a connection between the scale parameters and the concentration vector $a$. The individual distribution of $X_i$ does not provide useful information, but through the dirichlet, the marginals give me the answer.

Any suggestion?

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Beta distribution has $(0, 1)$ support, same as each of the variables jointly distributed as Ditichlet. Given $X_i \sim \mathsf{Beta}(a_i, b_i)$, if you wanted to have something like $(X_1, X_2, \dots, X_k) \sim \mathsf{Dir}(\alpha)$, you would need to force $X_1 + X_2 + \dots + X_k = 1$, because Dirichlet distribution has such constraint. For each draw from the distribution, you would need $x_2 < 1-x_1$, $x_3 < 1 - (x_1 + x_2)$ etc., finally with $x_k = 1 - (x_1 + x_2 + \dots + x_{k-1})$ being deterministic rather then random. This means, that the beta distributions would not be independent any more. What follows, independent beta variables do not jointly follow Dirichlet distribution.

There is however the reverse relation, if

$$ (Y_1, Y_2, \dots, Y_k) \sim \mathsf{Dir}(\alpha) $$

then given $\alpha_0 = \sum_{i=1}^k \alpha_i$, marginally $Y_i$'s follow beta distributions

$$ Y_i \sim \mathsf{Beta}(\alpha_i, \alpha_0 - \alpha_i) $$

So if your variables jointly follow Dirichlet distribution, their marginals are beta distributed. However if the variables are independent and follow beta distributions, then they do not jointly follow Dirichlet distribution, because you wouldn't be able to guarantee the constraint that they sum to unity.

Example

To illustrate this, let's simulate three independent $X_i \sim \mathsf{Beta}(1, 2)$ random variables, and $(Y_1,Y_2,Y_3) \sim \mathsf{Dir}(1, 1, 1)$ variables

library("extraDistr")
n <- 50000
X <- data.frame(V1=rbeta(n, 1, 2), V2=rbeta(n, 1, 2), V3=rbeta(n, 1, 2))
Y <- as.data.frame(rdirichlet(n, c(1, 1, 1)))

If you look at their marginal plots, they all follow the $\mathsf{Beta}(1, 2)$ distribution.

enter image description here

However if you look at the sums of the samples taken from the beta distributions, they clearly do not sum to unity:

range(rowSums(X))
## [1] 0.07868429 2.56876122

Also if you'll compare the scatter plot (with n = 1000 to make it more readable) of the joint distribution of the beta variables, and Dirichlet, you will see that the independent beta variables are less uniformly distributed then Dirichlet.

enter image description here

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  • $\begingroup$ Instead of having $X_i \sim Beta(a_i,b_i)$, which similar distribution would enable retrieving the dirichlet distribution? My problem is that $X_i$ can not be interpreted, but a joint distribution constrained to $\sum X_i=1$, would allow me to get the marginals $\endgroup$ – AlexandrosB May 30 at 8:59
  • $\begingroup$ @AlexandrosB what do you mean by retrieving? Why can't you model those variables as a Dirichlet distribution? The problem in here is that with independent distributions you cannot fit the constraint, as the values would need to depend on each other to meet it. So you need a distribution that models the dependence between the variables. $\endgroup$ – Tim May 30 at 9:03
  • $\begingroup$ Is it correct, given the fact that the marginals of the dirichlet follow the beta distribution, to estimate the concentration vector by mapping $X_i \sim beta(a_i,b_i) \to beta(m_i , m_0-m_i)$, where $m_0=\sum m_i $ and $m=[m_1 m_2 … m_k] $ the concentration vector. $\endgroup$ – AlexandrosB May 30 at 10:16
  • $\begingroup$ @AlexandrosB I'm not sure if I understand your comment, but if you take three independent beta distributions, then (no matter of the parameters) you have no guarantee that their sum would be one, and this is the constraint in Dirichlet distribution. $\endgroup$ – Tim May 30 at 11:33
  • $\begingroup$ OP never specified the Betas were independent to begin with, so it’s strange to say they cannot be independent “any more.” Given the information in the OP, it follows immediately that the components of $\alpha$ are the $a_i$’s, assuming they satisfy the restriction that $b_i = \sum_{j\ne i} a_j$, otherwise the necessary Dirichlet does not exist. $\endgroup$ – guy May 30 at 13:18

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