slides: write about the Moyal PDF parameters

Also add the Landau, Moyal and 'both' plots.

slides/sections/0.md

@@ -0,0 +1,32 @@
+---
+title: Randomness tests of a non-uniform distribution
+date: \today
+author:
+  - Giulia Marcer
+  - Michele Guerini Rocco
+institute:
+  - Università di Milano-Bicocca
+
+theme: metropolis
+aspectratio: 169
+
+fontsize: 14pt
+mathfont: FiraMath-Regular
+sansfont: Fira Sans
+
+header-includes: |
+  ```{=latex}
+
+  % Misc
+  % "thus" in formulas
+  \DeclareMathOperator{\thus}{%
+    \hspace{30pt} \Longrightarrow \hspace{30pt}
+  }
+
+  % "with" in formulas
+  \DeclareMathOperator{\with}{%
+  \hspace{30pt} \text{with} \hspace{30pt}
+  }
+
+  ```
+...

slides/sections/1.md

@@ -1,35 +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
-aspectratio: 169
-
-fontsize: 14pt
-mathfont: FiraMath-Regular
-sansfont: Fira Sans
-
-header-includes: |
-  ```{=latex}
-
-  % Misc
-  % "thus" in formulas
-  \DeclareMathOperator{\thus}{%
-    \hspace{30pt} \Longrightarrow \hspace{30pt}
-  }
-
-  % "et" in formulas
-  \DeclareMathOperator{\et}{%
-  \hspace{30pt} \wedge \hspace{30pt}
-  }
-
-  ```
-...
 
 
 # Goal
@@ -46,13 +14,27 @@ How?
 
 - Applying some hypothesis testings  
 
-Why?
+## Why?
 
-- They are really similar. Historically, the Moyal distribution was utilized in
+The Landau and Moyal PDFs are really similar. Historically, the latter distribution was utilized in
 the approximation of the Landau Distribution.
 
+:::: {.columns}
+:::  {.column width=33%}
+  \centering
+  ![](images/moyal-photo.jpg){height=130pt}
+:::
+:::  {.column width=33%}
+  \centering
+  ![](images/mondau-photo.jpg){height=130pt}
+:::
+:::  {.column width=33%}
+  \centering
+  ![](images/landau-photo.jpg){height=130pt}
+:::
+::::
 
-# Two similar distributions
+## Two similar distributions
 
 :::: {.columns .c}
 :::  {.column width=50%}
@@ -68,8 +50,8 @@ Why?
   \begin{center}
   Moyal PDF
   $$
-  M(x) = \frac{1}{\sqrt{2 \pi \sigma}} \exp \left( - \frac{x - \mu }{2 \sigma}
+  M(x) = \frac{1}{\sqrt{2 \pi}} \exp \left[ - \frac{1}{2}
-         - \frac{1}{2} e^{- \frac{x -\mu}{\sigma}} \right)
+         \left( x + e^{- x} \right) \right]
   $$
   \end{center}
 :::
@@ -86,5 +68,5 @@ Why?
 
 ## Two similar distributions
 
-grafici sovrapposti
+\centering
-
+![](images/both-pdf.pdf)

slides/sections/2.md

@@ -1,81 +1,97 @@
-# Moyal PDF
+# Moyal distribution
 
 
-# Moyal PDF
+## Moyal PDF
 
 $$
   M(x) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left[ x + e^{-x} \right]}
 $$
-:::: {.columns .c}
+\vspace{10pt}
-::: {.column width=30%}
-  \begin{center}
-    More generally:
-  \end{center}
-:::
-::: {.column width=70%}
-  \begin{center}
 $$
+  \text{More generally:} \hspace{15pt}
   \begin{cases}
-      \text{location parameter} \mu \\
+    \text{location parameter:} \quad &\mu_M \\
-      \text{scale parameter} \sigma
+    \text{scale parameter:} \quad &\sigma_M
   \end{cases}
 $$
-  \end{center}
+\vspace{10pt}
-:::
-::::
-\vspace{20pt}
 $$
-  x \rightarrow \frac{x - \mu}{\sigma}
+  x \rightarrow \frac{x - \mu}{\sigma_M} \thus
+  M_{\mu \sigma_M}(x) = \frac{1}{\sqrt{2 \pi} \sigma_M}
+  e^{-\frac{1}{2} \left[ \frac{x - \mu}{\sigma_M} + e^{- \frac{x - \mu}{\sigma_M}} \right]}
 $$
 
 
 ## Moyal CDF
 
-The cumulative distribution function $\mathscr{M}(x)$ can be derived from the
+The cumulative distribution function $F_M(x)$ can be derived from the
 pdf $M(x)$ integrating:
 $$
-  \mathscr{M}(x) = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, M(y)
+  F_M(x) = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, M(y)
-  = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{1}{2}}
+  = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{y}{2}}
   e^{- \frac{1}{2} e^{-y}}
 $$
+
+
+## Moyal CDF
+
 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)}
+the CDF can be rewritten as:
-  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}
 $$
-since the `erf` is defines as:
+  F_M(x) =
+  \frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)} dz \, e^{- z^2}
+  \with f(x) = \frac{e^{- \frac{x}{2}}}{\sqrt{2}}
+$$
+
+
+## Moyal CDF
+
+given the definition of the error function `erf`:
 $$
   \text{erf} = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2}
 $$
+one finally gets:
 $$
-  1 = \frac{2}{\sqrt{\pi}} \int_0^{+ \infty} dy \, e^{-y^2}
+  F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
-    = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} + 
-      \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
-    = \text{erf}(x) + \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
 $$
-thus:
+
+
+## Moyal QDF
+
+The quantile is defined as the inverse of the CDF:
 $$
-  \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
+  F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
-  1 = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} + 
+$$
+hence:
+$$
+  Q(x) = -2 \ln \left[ \sqrt{2} \, \text{erf}^{(-1)} (1 - F_M(x)) \right]
+$$
+
+
+## Moyal median
+
+The median is defined as the point at which $F_M(x) = 1/2$:
+\vspace{15pt}
+$$
+  M(x) \thus
+  m_M = -2 \ln \left[ \sqrt{2} \,
+  \text{erf}^{(-1)} \left( \frac{1}{2} \right) \right]
+$$
+\vspace{15pt}
+$$
+  M_{\mu \sigma_M}(x) \thus
+  m_M = \mu -2 \sigma_M \ln \left[ \sqrt{2} \,
+  \text{erf}^{(-1)} \left( \frac{1}{2} \right) \right]
 $$
 
 
 ## Moyal mode
 
 Peak of the PDF
+\vspace{10pt}
 $$
   \partial_x M(x) = \partial_x \left( \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
                     \left( x + e^{-x} \right)} \right)
@@ -83,11 +99,43 @@ $$
                     \left( x + e^{-x} \right)} \left( -\frac{1}{2} \right)
                     \left( 1 - e^{-x} \right)
 $$
-\vspace{15pt}
+\vspace{10pt}
 $$
-  \partial_x M(x) = 0 \thus x = 0 \thus x = \mu
+  \partial_x M(x) = 0 \thus \mu_M = 0
+$$
+\vspace{10pt}
+$$
+  \partial_x M_{\mu \sigma_M}(x) = 0 \thus \mu_M = \mu
+$$
+
+
+## Moyal FWHM
+
+We need to compute the maximum value:
+$$
+  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_+ = a + W_0 \left( - \frac{1}{4 e} \right)    \\
+    x_- = a + W_{-1} \left( - \frac{1}{4 e} \right)
+  \end{cases}
+$$
+with $a =  1 + 2 \ln(2)$
+
+
+## Moyal FWHM
+
+$$
+  \text{FWHM}_L = x_+ - x_- = 3.590806098...
 $$
 \vspace{15pt}
 $$
-  \thus \mu = m_L
+  M(x) \thus \text{FWHM}_M = 3.590806098...
+$$
+\vspace{15pt}
+$$
+  M_{\mu \sigma_M}(x) \thus \text{FWHM}_M = \sigma_M \cdot 3.590806098...
 $$

slides/sections/3.md

@@ -0,0 +1,14 @@
+# Data sample
+
+## Data sample
+
+The $M(x)$ most similar to $L(x)$ is found by imposing:
+\vspace{15pt}
+$$
+  \mu_M = \mu_L \sim = −0.22278298...
+$$
+\vspace{15pt}
+$$
+  \text{FWHM}_M = \text{FWHM}_L = 4.0186457...
+  \thus \sigma_M = 1.1191486
+$$