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2

The notation in the paper uses $\mathcal J_k$ instead of $\lambda_k$. I am going to use $\lambda_k$ as in the question. I am going to drop subscript $k$ for simplicity. The model is then \begin{align*} \beta \mid \lambda &\sim N(0, \lambda \tau^2) \\ \lambda &\sim (1-w) \delta_{\nu_0} + w \delta_1. \end{align*} The rest requires some measure theory ...


1

You don't need to have a "named" functional form for your conjugate prior, so @Fodor1234's comment points the way towards an answer. However, the conjugate prior $p$ is slightly different than in the comment: $$p(\alpha|k, C) \propto \left( {\ln \alpha \over \alpha-1}\right)^k \alpha^C$$ The posterior $p'$ for $\alpha$, assuming (as in the ...


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Given a closed-form prior $\pi(\theta_1,\theta_2,\ldots)$ and a closed-form likelihood $\ell(\theta_1,\theta_2,\ldots)$, the conditional densities remain available up to a constant: $$\pi(\theta_1|\theta_2,\ldots) \propto \pi(\theta_1,\theta_2,\ldots) \times \ell(\theta_1,\theta_2,\ldots)$$ Therefore any simulation method targeting this density can apply (if ...


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If the prior and likelihood aren't conjugate, you can't easily derive an expression for the conditional distribution, so you can't use Gibbs sampling (you can, as @Xi'an says obtain the density as a function, but that's more complicated!). The usual solution to this is to use a different sampling algorithm, like Metropolis-Hastings (which is simple enough to ...


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In most cases we don't have priors that would be conjugate to the likelihood function and result in a closed-form solution. In such a case, we are left with approximate solutions, such as grid approximation for simple problems, and for more realistic ones Laplace approximation, Markov Chain Monte Carlo (MCMC) (see mcmc) sampling, variational inference (see ...


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There are at least two ways to express my prior belief of mu2: I believe very strongly that mu2 can only be a single value, 1.27 say, in which case I can use the assignment operator as shown in chunk2. I believe that mu2 is most probably some value, 0 say, but with ~95% probability I think it could be between -0.0002 and 0.0002. In that case I can specify a ...


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