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I learned from this blog: https://vasishth.github.io/bayescogsci/ by Prof. Dr. Shravan Vasishth that In recent years, Bayesian methods have come to be widely adopted in all areas of science. This is in large part due to the development of sophisticated software for probabilisic programming; a recent example is the astonishing computing capability afforded ...


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The expectation you wrote can't be computed efficiently and with an acceptable degree of accuracy -- about 3 paragraphs above where you quote, the authors describe this: Intractability: the case where the integral of the marginal likelihood $p_\theta(x) = \int p_\theta(z) p_\theta(x|z) dz$ is intractable (so we cannot evaluate or differentiate the marginal ...


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The objective would look exactly the same as above -- the KL divergence is still closed form (albeit the form now includes $m$ and $s$), and $p_\theta(z)$ doesn't really show up elsewhere. A really easy way to think about this is to imagine you add a "scaling layer" at the very end of your encoder which adds $m$ to your mean and $\log s^2$ to your ...


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Start with your initial expression for the log marginal likelihood: $$\log p(y|X) = \log \int_{f,u} p(y|f) p(f|u,X,Z) p(u|Z)$$ Multiply by $\frac{q(f,u)}{q(f,u)} = 1$: $$= \log \int \frac{q(f,u)}{q(f,u)}p(y|f) p(f|u,X,Z) p(u|Z)$$ The integral can be seen as an expectation w.r.t. $q(f,u)$: $$= \log E_{q(f,u)} \left[ \frac{p(y|f) p(f|u,X,Z) p(u|Z)}{q(f,u)} \...


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I'll go through some of your questions: Why can we use this reconstruction loss to judge the pixel-wise differences between the input x and the decoded latent sample z? Implicitly, we assume that pixel values are Normally distributed with uniform diagonal covariance. I mean, x gets encoded and specifies some latent distribution over z's. Now we sample ...


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Joint distribution. Using the graphical model you provided, we get the following joint distribution over all variables of interest, conditioning on model parameters. $$p(\Theta, \mathbf{v} | a_0, b_0, c_0, d_0, \left\{e_0^s, f_0^s \right\}_{s = 0,1}, \left \{ e_0^{s0}, f_0^{s0}, e_0^{s1}, f_0^{s1} \right\}_{s=2:L})$$ In more detail, this gives the ...


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