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?


Your Answer

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

Browse other questions tagged or ask your own question.