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.

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    $\begingroup$ 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. $\endgroup$ – whuber Oct 11 '18 at 19:14
  • $\begingroup$ @whuber "X " refers to a vector of dependent random variables, as we are forcing them to belong to A $\endgroup$ – ie86 Oct 11 '18 at 19:40
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    $\begingroup$ 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? $\endgroup$ – whuber Oct 11 '18 at 20:00
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    $\begingroup$ 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. $\endgroup$ – whuber Oct 11 '18 at 20:25
  • $\begingroup$ 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. $\endgroup$ – ie86 Oct 11 '18 at 20:29

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