From acd4dc3d07992c93f08b0cdbd05a3c0205bb7ba5 Mon Sep 17 00:00:00 2001
From: Brian Ward <bward@flatironinstitute.org>
Date: Mon, 3 Jun 2024 11:53:56 -0400
Subject: [PATCH] Latex fixes

---
 src/header.tex                           | 1 +
 src/stan-users-guide/finite-mixtures.qmd | 6 +++---
 2 files changed, 4 insertions(+), 3 deletions(-)

diff --git a/src/header.tex b/src/header.tex
index 314c7362f..83127ee80 100644
--- a/src/header.tex
+++ b/src/header.tex
@@ -3,6 +3,7 @@
 \usepackage{mathpazo}
 \usepackage[scale=0.9]{sourcecodepro}
 \usepackage{amssymb}
+\usepackage{mathtools}
 \linespread{1.03}
 
 \usepackage{titlesec}
diff --git a/src/stan-users-guide/finite-mixtures.qmd b/src/stan-users-guide/finite-mixtures.qmd
index 226639075..e6b3bfdd5 100644
--- a/src/stan-users-guide/finite-mixtures.qmd
+++ b/src/stan-users-guide/finite-mixtures.qmd
@@ -435,7 +435,7 @@ $$
 As with other event probabilities, this can be calculated in the
 generated quantities block either by sampling $z_i$ and $z_j$ and
 using the indicator function on their equality, or by computing the
-term inside the integral as a generated quantity.  As with posterior 
+term inside the integral as a generated quantity.  As with posterior
 predictive distribute, working in expectation is more statistically
 efficient than sampling.
 
@@ -575,12 +575,12 @@ The hurdle model is similar to the zero-inflated model, but more
 flexible in that the zero outcomes can be deflated as well as
 inflated. Given the probability $\theta$ and the intensity $\lambda$,
 the distribution for $y_n$ can be written as
-$$
+\[
 \begin{align*}
 y_n & = 0 \quad\text{with probability } \theta, \text{ and}\\
 y_n & \sim \textsf{Poisson}_{x\neq 0}(y_n \mid \lambda) \quad\text{with probability } 1-\theta,
 \end{align*}
-$$
+\]
 Where $\textsf{Poisson}_{x\neq 0}$ is a truncated Poisson distribution, truncated at $0$.
 
 The corresponding likelihood function for the hurdle model is