Computation of the estimator of the ELBO in Variational Auto-encoder I am reading the paper below by Kingma et.al.
https://arxiv.org/pdf/1906.02691.pdf
Section 2.4.4 which is titled as "Computation of $\log q_{\phi}(z|x)$":
In Eqn. (2.33) the authors explain a relation between the densities of $\epsilon$ and $z$ as follows:
\begin{equation}
\log q_{\phi}(z|x) = \log p(\epsilon) − \log d_{\phi}(x, \epsilon)
\end{equation}
How is the second term of the RHS derived?
Let me explain my question with more details. 
Applying Bayes rule and joint probability definition on $\log q_{\phi}(z|x)$ implies
$\log \frac{q_{\phi}(z, x)}{q_{\phi}(x)} = \log \frac{q_{\phi}(x|z) q_{\phi}(z)}{q_{\phi}(x)}$
applying log yields
$\log q_{\phi}(z|x) = \log q_{\phi}(x|z) + \log q_{\phi}(z) - \log q_{\phi}(x)$
Replacing $\log q_{\phi}(z)$ with $\log p(\epsilon)$ (because of reparametriztion) and rearranging leads to 
$\log q_{\phi}(z|x) = \log p(\epsilon) + \log q_{\phi}(x|z) - \log q_{\phi}(x)$
However, Eqn. 2.33 in the paper is:
$\log q_{\phi}(z|x) = \log p(\epsilon) - \log d_{\phi}(x, \epsilon)$.
I know this is a variable change but I have two questions:
1)  Am I right with what I explained above? 
2) I can't understand the sentence which is mentioned after it. That is, "where the second term is the log of the absolute value of the determinant
of the Jacobian matrix $(∂z/∂\epsilon)$"
 A: I was also confused about this, but as I read through the tutorial link which I put there in the comment, I think I have understood it. I want to share my notes here.
Your derivation is right, but it has no way leading to the desired form.
The above equation is a result from formula:
$$
\tag{1}
\newcommand{\d}{\mathrm{d}}
p(y) = p(f^{-1}(y)) \cdot \left| \frac{\d f^{-1}(y)}{\d y}  \right|
$$
more specifically, in its more general form (due to implicit function theorem from Manifolds, relevant post here found by Approach0):
$$
\tag{2}
p(y) = p(f^{-1}(y)) \cdot \left| \det \frac{\partial f^{-1}(y)}{\partial y} \right| = p(f^{-1}(y)) \cdot \left| \det \frac{\partial y}{\partial f^{-1}(y)} \right|^{-1}
$$
To see this, you can think of $q_\phi(z|x)$ as $f(\epsilon; \phi, x)$ where $x, \phi$ are given and the random variable $z$ only depends on $\epsilon$. 
Then the desired formula
$$
\log q_{\phi}(z|x) = \log p(\epsilon) − \log d_{\phi}(x, \epsilon) = \log p(\epsilon) + \log \left| \det \frac{\partial \epsilon}{ \partial z} \right|
$$
is essentially applying equation (2) 
$$
q_{\phi}(z|x) = p(\epsilon) \cdot \left| \det \frac{\partial z}{\partial \epsilon} \right|^{-1}
$$
(notice $f^{-1}(z|x) = \epsilon$)
Now, to see why equation (1) holds:
$$
F_Y(y) = p(Y \le y) = p(f(x) \le y)=p(x \le f^{-1}(y)) = F_X(f^{-1}(y))
$$
thus
$$
F_Y(y)=F_X(f^{-1}(y)) = \left\{ \begin{array}{ll}
\int_{-\infty}^{f^{-1}(y)} \; p(x) \; \d x && \text{if f(x) is increasing} \\
&\\
\int_{-\infty}^{-f^{-1}(y)} \; p(x) \; \d x && \text{if f(x) is decreasing}\\
\end{array}
\right.
$$
According to Fundamental Theorem of Calculus and the chain rule, you get equation (1).
