Update.
authorFrançois Fleuret <francois@fleuret.org>
Sat, 24 Feb 2024 11:11:36 +0000 (12:11 +0100)
committerFrançois Fleuret <francois@fleuret.org>
Sat, 24 Feb 2024 11:11:36 +0000 (12:11 +0100)
elbo.tex

index 6875ddf..239a657 100644 (file)
--- a/elbo.tex
+++ b/elbo.tex
 
 \begin{document}
 
-\vspace*{0ex}
+\setlength{\abovedisplayskip}{2ex}
+\setlength{\belowdisplayskip}{2ex}
+\setlength{\abovedisplayshortskip}{2ex}
+\setlength{\belowdisplayshortskip}{2ex}
+
+\vspace*{-4ex}
 
 \begin{center}
 {\Large The Evidence Lower Bound}
 
+\vspace*{1ex}
+
 Fran\c cois Fleuret
 
 \today
 
-\vspace*{1ex}
+\vspace*{-1ex}
 
 \end{center}
 
@@ -102,12 +109,14 @@ p_\theta(x_n) & = \int_z p_\theta(x_n,z) dz                   \\
               & = \expect_{Z \sim q(z)} \left[\frac{p_\theta(x_n,Z)}{q(Z)}\right].
 \end{align*}
 %
-So if we wanted to maximize $p_\theta(x_n)$ alone, we could sample a
+So if we sample a
 $Z$ with $q$ and maximize
 %
 \begin{equation*}
-\frac{p_\theta(x_n,Z)}{q(Z)}.\label{eq:estimator}
+\frac{p_\theta(x_n,Z)}{q(Z)},
 \end{equation*}
+%
+we do maximize $p_\theta(x_n)$ on average.
 
 But we want to maximize $\sum_n \log \, p_\theta(x_n)$. If we use the
 $\log$ of the previous expression, we can decompose its average value