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. $\endgroup$ Apr 11 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.