While reading through Yee Whye Teh's tutorial on Dirichlet Process, I came across the Posterior distribution for DP. After observing some draws $\theta_1,\theta_2, ... ,$, the posterior is defined as $$G | \theta_1,...,\theta_n \sim DP(\alpha + n, \frac{\alpha}{\alpha + n}H +\frac{n}{\alpha + n}\frac{\sum_{i=1}^n \delta_{\theta_i}}{n})$$ I understand the posterior distribution of the DP is thus generated in the following manner: Draw an infinity of $\rho's$: $\rho_1,\rho_2, ...$ following a $GEM(\alpha + n)$ distribution. And then draw an infinity of $\theta$'s from the mixture of base distribution $H$ and the empirical distribution.

Can someone explain what $\delta_{\theta_i}$ means in this case ?


1 Answer 1


$\delta_{\theta_{i}}$ is the dirac measure on $\theta_{i}$:

$$\delta_{\theta_{i}}(A) = 1_{A}(\theta_{i})$$

It is present in the posterior of the DP because, for a measurable partition $B_1,...,B_k$ of $\Theta$, the marginal distribution becomes:

$$(G(B_1),...,G(B_k)) | \theta_{1},...,\theta_{n} \sim DP(\alpha H + \sum_{i=1}^n \delta_{\theta_i}(B_1), ..., \alpha H + \sum_{i=1}^n \delta_{\theta_i}(B_k)) \sim DP(\alpha H + n_1, ..., \alpha H + n_k)$$

since $n_j = \sum_{i=1}^n 1_{B_j}(\theta_i) = \sum_{i=1}^n \delta_{\theta_i}(B_j)$


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