show/hide this revision's text 2 added 1008 characters in body

One cannot conclude $X\sim BP(\alpha H_0)$ just by knowing the marginal distribution of each $X(B_k)$, separately. Your calculation is not wrong as the univariate marginal distribution and conditional distribution of a Dirichlet distribution are Beta distributed.

In particular, in a Dirichlet process, samples correspond to the density function $$f(\theta)=\sum \beta_i \delta_{\theta_i}$$ (Here the $\beta_i$ are constructed as $\beta_i=\beta_i'\prod_{j < i}(1-\beta_j')$, and $\beta_j'\sim \text{Beta}(1,\alpha)=Y$)

While in a Beta process, given an infinitesimal partition $(B_1,...,B_K)$ with $K\to \infty$ and $H(B_k)\to 0$ the samples correspond to the density function $$H(B)=\sum \pi_i\delta_{B_i}$$ where $\pi_i\sim \text{Beta}(\alpha H_0(B_i),\alpha (1-H_0(B_i)))$

I hope you can see the difference. One more thing, the reason why the Dirichlet process is defined in terms of finitely dimensional distributions is because Kolmogorov extension theorem guarantees that it defines a stochastic process. Unfortunately the Beta process, does not verify the conditions of this theorem, and as a continuous time Levy process must be defined directly in the infinitesimal limit.

show/hide this revision's text 1

One cannot conclude $X\sim BP(\alpha H_0)$ just by knowing the marginal distribution of each $X(B_k)$, separately. Your calculation is not wrong as the univariate marginal distribution and conditional distribution of a Dirichlet distribution are Beta distributed.