From 43b0cb04eae4537d95775038d9e700e642087d6d Mon Sep 17 00:00:00 2001
From: =?utf8?q?Fran=C3=A7ois=20Fleuret?= <francois@fleuret.org>
Date: Sat, 24 Feb 2024 12:11:36 +0100
Subject: [PATCH] Update.

---
 elbo.tex | 17 +++++++++++++----
 1 file changed, 13 insertions(+), 4 deletions(-)

diff --git a/elbo.tex b/elbo.tex
index 6875ddf..239a657 100644
--- a/elbo.tex
+++ b/elbo.tex
@@ -71,16 +71,23 @@
 
 \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
-- 
2.20.1