sections: reorder the sections in order to add the Landau paragraph

slides/sections/0.md

@@ -8,16 +8,38 @@ institute:
   - Università di Milano-Bicocca
 
 theme: metropolis
+themeoptions:
+  - titleformat=allcaps
 aspectratio: 169
 
-fontsize: 14pt
+fontsize: 12pt
+mainfont: Fira Sans
+mainfontoptions:
+  - BoldFont=Fira Sans
+
 mathfont: FiraMath-Regular
-sansfont: Fira Sans
 
 header-includes: |
   ```{=latex}
+  %% Colors
+  \definecolor{mDarkTeal}  {HTML}{020202}
+  \definecolor{mLightBrown}{HTML}{C49D4A}
+  \definecolor{mDarkRed}   {HTML}{92182B}
+
+  \definecolor{green} {HTML}{60AC39}
+  \definecolor{red}   {HTML}{D73737}
+  \definecolor{blue}  {HTML}{6684E1}
+  \definecolor{yellow}{HTML}{CFB017}
+
+  \setbeamercolor{frametitle}{bg=mDarkRed}
+
+  % center images
+  \LetLtxMacro{\oldIncludegraphics}{\includegraphics}
+  \renewcommand{\includegraphics}[2][]{
+    \centering
+    \oldIncludegraphics[#1]{#2}
+  }
 
-  % Misc
   % "thus" in formulas
   \DeclareMathOperator{\thus}{%
     \hspace{30pt} \Longrightarrow \hspace{30pt}
@@ -27,6 +49,5 @@ header-includes: |
   \DeclareMathOperator{\with}{%
   \hspace{30pt} \text{with} \hspace{30pt}
   }
-
   ```
 ...

slides/sections/1.md

@@ -1,63 +1,3 @@
----
-title: Randomness tests of a non-uniform distribution
-date: \today
-author:
-  - Giulia Marcer
-  - Michele Guerini Rocco
-institute:
-  - Università di Milano-Bicocca
-
-theme: metropolis
-themeoptions:
-  - titleformat=allcaps
-aspectratio: 169
-
-fontsize: 12pt
-mainfont: Fira Sans
-mainfontoptions:
-  - BoldFont=Fira Sans
-
-mathfont: FiraMath-Regular
-
-header-includes: |
-  ```{=latex}
-  %% Colors
-  \definecolor{mDarkTeal}  {HTML}{020202}
-  \definecolor{mLightBrown}{HTML}{C49D4A}
-  \definecolor{mDarkRed}   {HTML}{92182B}
-
-  \definecolor{green} {HTML}{60AC39}
-  \definecolor{red}   {HTML}{D73737}
-  \definecolor{blue}  {HTML}{6684E1}
-  \definecolor{yellow}{HTML}{CFB017}
-
-  \setbeamercolor{frametitle}{bg=mDarkRed}
-
-  % center images
-  \LetLtxMacro{\oldIncludegraphics}{\includegraphics}
-  \renewcommand{\includegraphics}[2][]{
-    \centering
-    \oldIncludegraphics[#1]{#2}
-  }
-
-  %% customer macros
-  \DeclareMathOperator{\with}{%
-    \hspace{30pt} \text{with} \hspace{30pt}
-  }
-
-  % "thus" in formulas
-  \DeclareMathOperator{\thus}{%
-    \hspace{30pt} \Longrightarrow \hspace{30pt}
-  }
-
-  % "et" in formulas
-  \DeclareMathOperator{\et}{%
-  \hspace{30pt} \wedge \hspace{30pt}
-  }
-  ```
-...
-
-
 # Goal
  
 

slides/sections/2.md

@@ -1,135 +1,32 @@
-# Moyal distribution
+# Landau PDF
 
 
-## Moyal PDF
+## Pathological probability distribution
 
+Because of its fat tail:
 
-Standard form:
+\begin{align*}
-$$
+  E[x] &\longrightarrow + \infty  \\
-  M(z) = \frac{1}{\sqrt{2 \pi}} \exp
+  V[x] &\longrightarrow + \infty
-         \left[ - \frac{1}{2} \left( z + e^{-z} \right) \right]
+\end{align*}
-$$
 
-. . .
+No closed form for parameters.
 
-More generally:
+## Landau median
 
-   - location parameter $\mu$
+The median of a PDF is defined as:
-   - scale parameter $\sigma$
 
 $$
-  z = \frac{x - \mu}{\sigma}
+  Q_L(x) = \frac{1}{2}
-  \thus
-  M(x) = \frac{1}{\sqrt{2 \pi} \sigma} \exp
-         \left[ - \frac{1}{2} \left(
-           \frac{x - \mu}{\sigma}
-           + e^{-\frac{x - \mu}{\sigma}} \right) \right]
 $$
 
+- CDF computed by numerical integration,
+- QDF computed by numerical root-finding (Brent)
 
-## Moyal CDF
-
-The CDF $F_M(x)$ can be derived by direct integration:
-$$
-  F_M(x) = \int\limits_{- \infty}^x dy \, M(y)
-  = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{y}{2}}
-  e^{- \frac{1}{2} e^{-y}}
-$$
-
-. . .
-
-With the change of variable $z = e^{-\frac{y}{2}}/\sqrt{2}$:
-$$
-  F_M(x) =
-  \frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)} dz \, e^{- z^2}
-  \with f(x) = \frac{e^{- \frac{x}{2}}}{\sqrt{2}}
-$$
-
-
-## Moyal CDF
-
-Remembering the error function
-$$
-  \text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2},
-$$
-one finally gets:
-$$
-  F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
-$$
-
-
-## Moyal QDF
-
-The quantile (CDF\textsuperscript{-1}) is found solving:
-$$
-  y = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
-$$
 hence:
-$$
-  Q_M(x) = -2 \ln \left[ \sqrt{2} \, \text{erf}^{-1} (1 - F_M(x)) \right]
-$$
-
-
-## Moyal median
-
-Defined by $\text{CDF}(m) = 1/2$, or $m=\text{QDF}(1/2)$.
-
-\begin{align*}
-  M(z)
-    &\thus m_M = -2 \ln \left[ \sqrt{2} \,
-    \text{erf}^{-1} \left( \frac{1}{2} \right) \right] \\
-  M_{\mu \sigma}(x)
-    &\thus m_M = \mu -2 \sigma \ln \left[ \sqrt{2} \,
-    \text{erf}^{-1} \left( \frac{1}{2} \right) \right]
-\end{align*}
-
-## Moyal mode
-
-Peak of the PDF:
-$$
-  \partial_x M(x) = \partial_x \left( \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
-                    \left( x + e^{-x} \right)} \right)
-                  = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
-                    \left( x + e^{-x} \right)} \left( -\frac{1}{2} \right)
-                    \left( 1 - e^{-x} \right)
-$$
-
-\begin{align*}
-  \partial_x M(z) = 0 &\thus \mu_M = 0                \\
-  \partial_x M_{\mu \sigma}(x) = 0 &\thus \mu_M = \mu \\
-\end{align*}
-
-
-## Moyal FWHM
-
-We need to compute the maximum value:
-$$
-  M(\mu) = \frac{1}{\sqrt{2 \pi e}} \thus M(x_{\pm}) = \frac{1}{\sqrt{8 \pi e}}
-$$
-
-. . .
-
-which leads to:
-$$
-  x_{\pm} + e^{-x_{\pm}} = 1 + 2 \ln(2) \thus
-  \begin{cases}
-    x_+ = 1 + 2 \ln(2) + W_0 \left( - \frac{1}{4 e} \right)    \\
-    x_- = 1 + 2 \ln(2) + W_{-1} \left( - \frac{1}{4 e} \right)
-  \end{cases}
-$$
-
-## Moyal FWHM
 
 $$
-  x_+ - x_- = W_0 \left( - \frac{1}{4 e} \right)
+  m_L = 1.3557804...
-          - W_{-1} \left( - \frac{1}{4 e} \right)
-          = 3.590806098...
-          = a
 $$
 
-\begin{align*}
+o
-  M(z)
-    &\thus \text{FWHM}_M = a              \\
-  M_{\mu \sigma}(x)
-    &\thus \text{FWHM}_M = \sigma \cdot a \\
-\end{align*}

slides/sections/3.md

@@ -1,21 +1,135 @@
-# Data sample
+# Moyal distribution
 
-## Data sample
 
-The $M(x)$ most similar to $L(x)$ is found by imposing:
+## Moyal PDF
 
-- equal mode
+
+Standard form:
 $$
-  \mu_M = M_L \approx −0.22278298...
+  M(z) = \frac{1}{\sqrt{2 \pi}} \exp
-$$
+         \left[ - \frac{1}{2} \left( z + e^{-z} \right) \right]
-
-- equal width
-$$
-  \text{FWHM}_M = \text{FWHM}_L = \sigma \cdot a
 $$
 
 . . .
 
+More generally:
+
+   - location parameter $\mu$
+   - scale parameter $\sigma$
+
 $$
-  \implies \sigma_M \approx 1.1191486
+  z = \frac{x - \mu}{\sigma}
+  \thus
+  M(x) = \frac{1}{\sqrt{2 \pi} \sigma} \exp
+         \left[ - \frac{1}{2} \left(
+           \frac{x - \mu}{\sigma}
+           + e^{-\frac{x - \mu}{\sigma}} \right) \right]
 $$
+
+
+## Moyal CDF
+
+The CDF $F_M(x)$ can be derived by direct integration:
+$$
+  F_M(x) = \int\limits_{- \infty}^x dy \, M(y)
+  = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{y}{2}}
+  e^{- \frac{1}{2} e^{-y}}
+$$
+
+. . .
+
+With the change of variable $z = e^{-\frac{y}{2}}/\sqrt{2}$:
+$$
+  F_M(x) =
+  \frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)} dz \, e^{- z^2}
+  \with f(x) = \frac{e^{- \frac{x}{2}}}{\sqrt{2}}
+$$
+
+
+## Moyal CDF
+
+Remembering the error function
+$$
+  \text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2},
+$$
+one finally gets:
+$$
+  F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
+$$
+
+
+## Moyal QDF
+
+The quantile (CDF\textsuperscript{-1}) is found solving:
+$$
+  y = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
+$$
+hence:
+$$
+  Q_M(x) = -2 \ln \left[ \sqrt{2} \, \text{erf}^{-1} (1 - F_M(x)) \right]
+$$
+
+
+## Moyal median
+
+Defined by $\text{CDF}(m) = 1/2$, or $m=\text{QDF}(1/2)$.
+
+\begin{align*}
+  M(z)
+    &\thus m_M = -2 \ln \left[ \sqrt{2} \,
+    \text{erf}^{-1} \left( \frac{1}{2} \right) \right] \\
+  M_{\mu \sigma}(x)
+    &\thus m_M = \mu -2 \sigma \ln \left[ \sqrt{2} \,
+    \text{erf}^{-1} \left( \frac{1}{2} \right) \right]
+\end{align*}
+
+## Moyal mode
+
+Peak of the PDF:
+$$
+  \partial_x M(x) = \partial_x \left( \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
+                    \left( x + e^{-x} \right)} \right)
+                  = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
+                    \left( x + e^{-x} \right)} \left( -\frac{1}{2} \right)
+                    \left( 1 - e^{-x} \right)
+$$
+
+\begin{align*}
+  \partial_x M(z) = 0 &\thus \mu_M = 0                \\
+  \partial_x M_{\mu \sigma}(x) = 0 &\thus \mu_M = \mu \\
+\end{align*}
+
+
+## Moyal FWHM
+
+We need to compute the maximum value:
+$$
+  M(\mu) = \frac{1}{\sqrt{2 \pi e}} \thus M(x_{\pm}) = \frac{1}{\sqrt{8 \pi e}}
+$$
+
+. . .
+
+which leads to:
+$$
+  x_{\pm} + e^{-x_{\pm}} = 1 + 2 \ln(2) \thus
+  \begin{cases}
+    x_+ = 1 + 2 \ln(2) + W_0 \left( - \frac{1}{4 e} \right)    \\
+    x_- = 1 + 2 \ln(2) + W_{-1} \left( - \frac{1}{4 e} \right)
+  \end{cases}
+$$
+
+## Moyal FWHM
+
+$$
+  x_+ - x_- = W_0 \left( - \frac{1}{4 e} \right)
+          - W_{-1} \left( - \frac{1}{4 e} \right)
+          = 3.590806098...
+          = a
+$$
+
+\begin{align*}
+  M(z)
+    &\thus \text{FWHM}_M = a              \\
+  M_{\mu \sigma}(x)
+    &\thus \text{FWHM}_M = \sigma \cdot a \\
+\end{align*}

slides/sections/4.md

@@ -0,0 +1,21 @@
+# Data sample
+
+## Data sample
+
+The $M(x)$ most similar to $L(x)$ is found by imposing:
+
+- equal mode
+$$
+  \mu_M = \mu_L \approx −0.22278298...
+$$
+
+- equal width
+$$
+  \text{FWHM}_M = \text{FWHM}_L = \sigma \cdot a
+$$
+
+. . .
+
+$$
+  \implies \sigma_M \approx 1.1191486
+$$