diff --git a/slides/misc/notes_moyal.md b/slides/misc/notes_moyal.md
index aa16f13..3c64a59 100644
--- a/slides/misc/notes_moyal.md
+++ b/slides/misc/notes_moyal.md
@@ -33,3 +33,59 @@ Moyal. The distribution models the energy lost by a fast charged particle
 the Moyal distribution has been utilized in the approximation of the Landau
 Distribution and has since found use in modeling a wide array of phenomena.
 
+
+
+# PDF
+
+The Moyal distribution is defined as:
+$$
+  M(x) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left[ x + e^{-x} \right]}
+$$
+More generally, it is defined with the location and scale parameters $\mu$ and
+$\sigma$ such as:
+$$
+  x \rightarrow \frac{x - \mu}{\sigma}
+$$
+
+# 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:
+$$
+  \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
+  1 = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} + 
+$$