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Suppose you sample $M$ vectors from $Dirichlet_K(\alpha)$. You then show a histogram summarizing the distribution of the $M$ values that were sampled for dimension $k = 1$ (i.e. the first dimension, which is the first index of each vector). This marginal distribution should be a Beta distribution since the Dirichlet is a multivariate Beta distribution.

  1. I am not sure how to describe the distribution depicted in the histogram. Would it be accurate to describe the distribution depicted in the histogram as the "marginal probability distribution for $K = 1$ implied by the $Dirichlet(\alpha)$"? Alternatively, is this showing the sampling distribution for the marginal probability $K = 1$?

  2. How does the value in each vector sampled from a Dirichlet relate to the parameters of the underlying Beta distributions? The Beta has two parameters, but the Dirichlet only draws one value and so it is unclear whether this is e.g. the expected value of the Beta or some other quantity affecting the shape of the Beta distributions.

socialscientist
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1 Answers1

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  1. Both descriptions seem fine, though you might want to use something other than K as your index variable since K is the (fixed) dimensionality of the Dirichlet.

  2. You are correct that the marginal distributions of a Dirichlet distribution are Beta distributions. Specifically, if $$\mathbf X = (X_1, X_2... X_K) \sim Dirichlet (\alpha)$$ and $$ \alpha_0 = \sum_{i=1}^K \alpha_i $$ then the marginal Beta distributions are (per Wikipedia): $$X_i = Beta(\alpha_i, \alpha_0 - \alpha_i)$$

In other words, the beta parameter of each marginal Beta distribution equals the sum of all the Dirichlet's parameters except for the parameter corresponding to that particular index i. As Wikipedia notes, this fact is due to the Dirichlet's aggregation property: by aggregating all of the Dirichlet's parameters except one, we reduce the K-dimensional Dirichlet to a 2-dimensional Dirichlet or, in other words, a Beta. Since the expected value of a Beta is $\frac{\alpha}{\alpha + \beta}$, the expected value of each Beta marginal matches the expected value of each $X_i$ (and those expected values necessarily sum to 1): $$E(X_i) = \frac{\alpha_i}{\alpha_i + (\alpha_0 - \alpha_i)} = \frac{\alpha_i}{\alpha_0} $$

Bill Vander Lugt
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