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Suppose that we have have some form of compositional data $x_i, i\in[1, n]$ which we propose comes from a Dirichlet distribution such that $$ x_i \sim \mbox{Dir}(\lambda \alpha),$$ where $x_i=(x_1^{(i)}, x_2^{(i)}, x_3^{(i)})$, $\lambda>0$, and that $\alpha=(\alpha_1, \alpha_2, \alpha_3)$ such that $\sum_{i=1}^3\alpha_i=1$. Due to the constraint on $\alpha$, we do not have an identifiability between $\alpha$ and $\lambda$. Therefore using MCMC we can get posterior distribution for all parameters involved. My question is, given the posterior distributions that we have for $\alpha_1, \alpha_2, \alpha_3$, how can we determine the probability that say $\alpha_1$ is greater than $\alpha_2$ and $\alpha_3$?

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Given that you would be using MCMC sampling, you can simply calculate the probabilities from the samples. If you draw $n$ MCMC samples from the posterior distribution of $\alpha$, than the alpha matrix would be $n \times 3$, so you can calculate the probability by counting such cases and dividing by total (i.e. taking mean of booleans). In R, this translates to:

mean(alpha[1,] > alpha[2,] & alpha[1,] > alpha[3,])
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