diff --git a/slides/sections/1.md b/slides/sections/1.md
index ccc621e..66085a7 100644
--- a/slides/sections/1.md
+++ b/slides/sections/1.md
@@ -16,6 +16,8 @@ sansfont: Fira Sans
 
 header-includes: |
   ```{=latex}
+
+  % Misc
   % "thus" in formulas
   \DeclareMathOperator{\thus}{%
     \hspace{30pt} \Longrightarrow \hspace{30pt}
@@ -25,7 +27,64 @@ header-includes: |
   \DeclareMathOperator{\et}{%
   \hspace{30pt} \wedge \hspace{30pt}
   }
+
   ```
 ...
 
+
 # Goal
+ 
+
+## Goal
+ 
+What?
+
+- Generate a sample of points from a Moyal PDF
+- Prove it truly comes from it and not from a Landau PDF
+
+How?
+
+- Applying some hypothesis testings  
+
+Why?
+
+- They are really similar. Historically, the Moyal distribution was utilized in
+  the approximation of the Landau Distribution.
+
+
+# Two similar distributions
+
+:::: {.columns .c}
+::: {.column width=50%}
+  \begin{center}
+  Landau PDF
+  $$
+    L(x) = \frac{1}{\pi} \int \limits_{0}^{+ \infty}
+           dt \, e^{-t \ln(t) -xt} \sin (\pi t)
+  $$
+  \end{center}
+:::
+::: {.column width=50%}
+  \begin{center}
+  Moyal PDF
+  $$
+  M(x) = \frac{1}{\sqrt{2 \pi \sigma}} \exp \left( - \frac{x - \mu }{2 \sigma}
+         - \frac{1}{2} e^{- \frac{x -\mu}{\sigma}} \right)
+  $$
+  \end{center}
+:::
+::::
+
+:::: {.columns .c}
+::: {.column width=50%}
+  ![](images/landau-pdf.pdf)
+:::
+::: {.column width=50%}
+  ![](images/moyal-pdf.pdf)
+:::
+::::
+
+## Two similar distributions
+
+grafici sovrapposti
+
diff --git a/slides/sections/2.md b/slides/sections/2.md
new file mode 100644
index 0000000..d94c928
--- /dev/null
+++ b/slides/sections/2.md
@@ -0,0 +1,93 @@
+# Moyal PDF
+
+
+# Moyal PDF
+
+$$
+  M(x) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left[ x + e^{-x} \right]}
+$$
+:::: {.columns .c}
+::: {.column width=30%}
+  \begin{center}
+    More generally:
+  \end{center}
+:::
+::: {.column width=70%}
+  \begin{center}
+    $$
+    \begin{cases}
+      \text{location parameter} \mu \\
+      \text{scale parameter} \sigma
+    \end{cases}
+    $$
+  \end{center}
+:::
+::::
+\vspace{20pt}
+$$
+  x \rightarrow \frac{x - \mu}{\sigma}
+$$
+
+
+## Moyal CDF
+
+The cumulative distribution function $\mathscr{M}(x)$ can be derived from the
+pdf $M(x)$ integrating:
+$$
+  \mathscr{M}(x) = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, M(y)
+  = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{1}{2}}
+    e^{- \frac{1}{2} e^{-y}}
+$$
+with the change of variable:
+\begin{align}
+  z = \frac{1}{\sqrt{2}} e^{-\frac{y}{2}}
+  &\thus \frac{dz}{dy} = \frac{-1}{2 \sqrt{2}} e^{-\frac{y}{2}} \\
+  &\thus dy = -2 \sqrt{2} e^{\frac{y}{2}} dz
+\end{align}
+hence, the limits of the integral become:
+\begin{align}
+  y \rightarrow - \infty &\thus z \rightarrow + \infty \\
+  y = x &\thus z =  \frac{1}{\sqrt{2}} e^{-\frac{x}{2}} = f(x)
+\end{align}
+and the CDF can be rewritten as:
+$$
+  \mathscr{M}(x) = \frac{1}{2 \pi} \int\limits_{+ \infty}^{f(x)}
+  dz \, (- 2 \sqrt{2}) e^{\frac{y}{2}} e^{- \frac{y}{2}} e^{- z^2}
+  = \frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)}
+  dz e^{- z^2}
+$$
+since the `erf` is defines as:
+$$
+  \text{erf} = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2}
+$$
+$$
+  1 = \frac{2}{\sqrt{\pi}} \int_0^{+ \infty} dy \, e^{-y^2}
+    = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} + 
+      \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
+    = \text{erf}(x) + \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
+$$
+thus:
+$$
+  \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
+  1 = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} + 
+$$
+
+
+## Moyal mode
+
+Peak of the PDF
+$$
+  \partial_x M(x) = \partial_x \left( \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
+                    \left( x + e^{-x} \right)} \right)
+                  = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
+                    \left( x + e^{-x} \right)} \left( -\frac{1}{2} \right)
+                    \left( 1 - e^{-x} \right)
+$$
+\vspace{15pt}
+$$
+  \partial_x M(x) = 0 \thus x = 0 \thus x = \mu
+$$
+\vspace{15pt}
+$$
+  \thus \mu = m_L
+$$