# Can Bayesian Optimization solve this problem?

Suppose $${\bf{x}} = (x_1,\ldots,x_n)$$ and $$f({\bf{x}})\propto 1_A({\bf{x}}) \prod_{i=1}^n {x_i}^{\alpha_i-1} e^{-\beta_i x_i}$$ , i.e. $$f$$ is proportional to the product of independent gamma distributions (with not necessarily equal parameters) truncated on a set A. In other words, we are truncating the range of original product distribution to a nontrivial subset of $$\mathbb{R}^n$$ and then normalize to get a probability distribution.

I can use gibbs sampling to draw samples from $$f$$ and find the empirical mean. This truncation and normalization changes $$\mathbb{E}(X)$$ to some complicated function of all $$(\alpha_i,\beta_i)$$.

Suppose an arbitrary $$y \in A$$ is given. How to set the hyperparameters so that the empirical marginal mean converges to $$y$$? Basically we want to invert this map to set $$\alpha$$ and $$\beta$$.

Since the mean is a black-box function, can bayesian optimization solve this problem? (I've just come across it) or maybe hierarchical modeling? Random search is not computationally feasible. Thanks.

• If "$X$" refers to a vector of $n$ independent random variables distributed with the given parameters, then note that the means do not determine the parameters: you are free to specify another $n$ independent constraints. Thus "inversion" in the sense you seem to intend is not possible. – whuber Oct 11 '18 at 19:14
• @whuber "X " refers to a vector of dependent random variables, as we are forcing them to belong to A – ie86 Oct 11 '18 at 19:40
• That contradicts your description. If there's a dependence, then the likelihood will not factor (this is essentially the definition of statistical dependence), whence $f$ cannot be the likelihood, raising concerns about how to interpret it. Could you please clarify your question by explaining the notation? – whuber Oct 11 '18 at 20:00
• Thank you for the edits. They helped to focus my attention on the role played by $1_A$ and the fact that it does not necessarily factor. I think my initial remark about the underdetermined nature of this problem still applies, though: you are attempting to determine $2n$ parameters from just $n$ constraints. – whuber Oct 11 '18 at 20:25
• Thanks. Yes, it does not factorize. We start off with independent random variables but after forcing them to belong to set A, they become dependent. – ie86 Oct 11 '18 at 20:29