Suppose $\mathbf{X} = (X_1, \cdots, X_K)$ follows a Dirichlet distribution with parameters $(\alpha_1, \cdots, \alpha_K)$. Partition $\mathbf{X} = (\mathbf{X}_{(1)}, \mathbf{X}_{(2)})$ for subvectors $\mathbf{X}_{(1)}, \mathbf{X}_{(2)}$. What is the distribution of $\mathbf{X}_{(1)}|\mathbf{X}_{(2)}$?


1 Answer 1


WLOG just relabel those random variables such that $X_1|X_2$ is the random variable of interest, and remaining random variables are labeled from $X_3, X_4, \ldots, X_k$


By the aggregation property, we know that $$\left(X_1, X_2, X_0 = 1 - X_1 - X_2 = \sum_{i=3}^k X_i\right) \sim\text{Dir}\left(\alpha_1, \alpha_2, \alpha_0 = \sum_{i=3}^k \alpha_i\right)$$

And therefore their joint pdf is $$ f_{X_1, X_2}(x_1, x_2) = \frac {\Gamma(\alpha_1 + \alpha_2 + \alpha_0)} {\Gamma(\alpha_1)\Gamma(\alpha_2)\Gamma(\alpha_0)} x_1^{\alpha_1-1}x_2^{\alpha_2-1}(1-x_1-x_2)^{\alpha_0-1} $$

where $x_1, x_2 \in (0, 1), x_1 + x_2 \leq 1$. Similarly, the marginal pdf of $X_2$ is $$ f_{X_2}(x_2) = \frac {\Gamma(\alpha_2 + \alpha_1 + \alpha_0)} {\Gamma(\alpha_2)\Gamma(\alpha_1+\alpha_0)} x_2^{\alpha_2-1}(1-x_2)^{\alpha_1+\alpha_0-1}, ~~ x_2 \in (0, 1) $$

i.e. $\text{Beta}(\alpha_2, \alpha_1 + \alpha_0) $. Therefore the conditional pdf of $X_1|X_2 = x_2$ is $$ f_{X_1|X_2=x_2}(x_1|x_2) = \frac {f_{X_1,X_2}(x_1,x_2)} {f_{X_2}(x_2)} = \frac {\Gamma(\alpha_1+\alpha_0)} {\Gamma(\alpha_1) \Gamma(\alpha_0)} \left(\frac {x_1} {1-x_2}\right)^{\alpha_1-1} \left(1 - \frac {x_1} {1-x_2}\right)^{\alpha_0-1}\frac {1} {1 - x_2}$$

where $x_1 \in (0, 1-x_2)$. So $X_1|X_2 = x_2$ has a scaled Beta distribution, i.e. $$ \frac {1} {1 - x_2} X_1|X_2 = x_2 \sim \text{Beta}(\alpha_1, \alpha_0)$$

Edit: It seems that I have only finished a univariate version of the problem.

Let $m \in \{1, \ldots, k-1\}$ be the dimension of $\mathbf{X}_{(1)} = (X_1, \ldots X_m)$, such that $\mathbf{X}_{(2)} = (X_{m+1}, \ldots X_k)$ (WLOG)

By definition the joint pdf of $(X_1, \ldots, X_k)$ is $$ f_{X_1, \ldots, X_k}(x_1, \ldots, x_k) = \frac {\Gamma(\sum_{i=1}^k \alpha_i)} {\prod_{i=1}^k \Gamma(\alpha_i)} \prod_{i=1}^k x_i^{\alpha_i-1}, x_i \in (0, 1), \sum_{i=1}^k x_i = 1 $$

Similarly, the joint pdf of $(X_{m+1}, \ldots, X_k)$ is $$ f_{X_{m+1}, \ldots, X_k}(x_{m+1}, \ldots, x_k) = \frac {\Gamma(\sum_{i=1}^k \alpha_i)} {\Gamma(\alpha_0)\prod_{i=m+1}^k \Gamma(\alpha_i)} \prod_{i=m+1}^k x_i^{\alpha_i-1} \left(1 - \sum_{j=m+1}^k x_j\right)^{\alpha_0-1} $$

where $x_i \in (0, 1), \sum_{i=m+1}^k x_i < 1$ and $\alpha_0 = \sum_{i=1}^m \alpha_i$. Again the conditional pdf is the ratio of these two:

$$ \begin{align} &~ f_{X_1, \ldots, X_m|X_{m+1}, \ldots, X_k}(x_1, \ldots, x_m|x_{m+1},\ldots, x_k) \\ =&~ \frac {\Gamma(\sum_{i=1}^m \alpha_i)} {\prod_{i=1}^m\Gamma(\alpha_i)} \prod_{i=1}^m x_i^{\alpha_i-1} \left(1 - \sum_{j=m+1}^k x_j\right)^{-(\alpha_0-1)} \\ =&~ \frac {\Gamma(\sum_{i=1}^m \alpha_i)} {\prod_{i=1}^m\Gamma(\alpha_i)} \prod_{i=1}^m \left[x_i \left(1 - \sum_{j=m+1}^k x_j\right)^{-1}\right]^{\alpha_i-1} \left(1 - \sum_{j=m+1}^k x_j\right)^{-(m-1)} \end{align}$$

So $\mathbf{X}_{(1)}|\mathbf{X}_{(2)} = \mathbf{x}_2$ has a scaled Dirichlet distribution, or equivalently, $$ \frac {1} {1 - \mathbf{1}^T\mathbf{x}_2}\mathbf{X}_{(1)}|\mathbf{X}_{(2)} = \mathbf{x}_2 \sim \text{Dir}(\alpha_1, \ldots, \alpha_m)$$

where $\mathbf{1}$ is the vector with length $k-m$ and all entries equal to 1

  • $\begingroup$ Thanks for a great answer--very helpful. $\endgroup$ Commented Apr 11, 2022 at 21:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .