I'm reading through a tutorial on the Dirichlet process (http://www.stats.ox.ac.uk/~teh/research/npbayes/Teh2010a.pdf) and have a small question. Given a draw from a Dirichlet process $G\sim\textsf{DP}(\alpha,H)$, and observations $\theta_1,\cdots,\theta_n\sim G$, we have that for a partition $A_1,\cdots,A_r$ of our probability space,

$$(G(A_1),\cdots,G(A_r))|\theta_1,\cdots,\theta_n\sim \textsf{Dir}(\alpha H(A_1)+\sum_{i=1}^{n}\delta_{\theta_i}(A_1),\cdots,\alpha H(A_r)+\sum_{i=1}^{n}\delta_{\theta_i}(A_r))$$

Since this holds for every partition, the posterior $G|\theta_1,\cdots,\theta_n$ is itself a Dirichlet process.

I can see that for this to hold, the product of the posterior parameters of the DP is $\alpha H+\sum_{i=1}^{n}\delta_{\theta_i}(\cdot)$, but I don't see why the posterior concentration parameter is $\alpha+n$ and the posterior base distribution is $\frac{\alpha}{\alpha+n} H+\frac{1}{\alpha+n}\sum_{i=1}^{n}\delta_{\theta_i}(\cdot)$. It seems kind of arbitrary to me that the concentration parameter is just pulled out of the product as $\alpha+n$. I'm sure this is something simple that I'm missing but I've been trying to figure it out for a bit and can't seem to get it (this is similarly skated over in the DP chapter of BDA as "it's straightforward to obtain" -_-).

Edit: Just thought of this. Given our partition, if we add together the parameters of the posterior Dirichlet

$$\alpha H(A_1)+\sum_{i=1}^{n}\delta_{\theta_i}(A_1)+\cdots+\alpha H(A_r)+\sum_{i=1}^{n}\delta_{\theta_i}(A_r)$$

we get $\alpha(\sum_{j=1}^rH(A_j))+n=\alpha+n$ since $\{A_j\}$ partition our probability space and $H$ is a probability distribution. Is this correct, and if so is there a reason this should equal the posterior concentration parameter?


1 Answer 1


The posterior base distribution must be a probability distribution. Let $M$ be the concentration parameter. For any $M$ $$\frac{\alpha}{M}H+\frac{1}{M}\sum_i \delta_{\theta_i}(\cdot)$$ is a finite measure, but it's only a probability measure if its total mass is 1, which will clearly be true for a unique $M$, and is true for $M=\alpha+n$.


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