sections: fix a lot of things

slides/sections/0.md

@@ -1,5 +1,5 @@
 ---
-title: Title
+title: Testing for a Landau distribution
 date: \today
 author:
   - Giulia Marcer
@@ -16,9 +16,19 @@ fontsize: 12pt
 mainfont: Fira Sans
 mainfontoptions:
   - BoldFont=Fira Sans
-
 mathfont: FiraMath-Regular
 
+references:
+  - type: article-journal
+    id: trapani15
+    author:
+      family: Trapani
+      given: Lorenzo
+    title: testing for (in)finite moments
+    container-title: Journal of Econometrix
+    issued:
+      year: 2015
+
 header-includes: |
   ```{=latex}
   %% Colors
@@ -32,6 +42,11 @@ header-includes: |
   \definecolor{yellow}{HTML}{CFB017}
 
   \setbeamercolor{frametitle}{bg=mDarkRed}
+  \definecolor{cyclamen}{RGB}{146,24,43}
+
+  \usepackage{ulem}
+  \newcommand\strike{\bgroup\markoverwith{%
+    \textcolor{mDarkRed}{\rule[0.5ex]{2pt}{1pt}}}\ULon}
 
   % center images
   \LetLtxMacro{\oldIncludegraphics}{\includegraphics}
@@ -40,6 +55,7 @@ header-includes: |
     \oldIncludegraphics[#1]{#2}
   }
 
+
   % "thus" in formulas
   \DeclareMathOperator{\thus}{%
     \hspace{30pt} \Longrightarrow \hspace{30pt}
@@ -69,5 +85,9 @@ header-includes: |
   \DeclareMathOperator{\ob}{%
     ^{\text{obs}}
   }
+
+  \setbeamercovered{transparent}
   ```
+
+csl: ../notes/docs/bibliography.csl
 ...

slides/sections/1.md

@@ -3,24 +3,26 @@
 
 ## Goal
 
-- Generate a sample $L$ of points from a Landau PDF
+Construct six statistical tests to assert whether a sample comes from a Landau
-- Generate a sample $M$ of points from a Moyal PDF
+distribution
 
 . . .
 
-- Implement a bunch of statistical tests
+- Generate a sample $L$ from a Landau PDF
+- Generate a sample $M$ from a Moyal PDF
 
 . . .
 
-- Check if they work:
+$H_0$: sample following Landau PDF
-  - the sample $L$ truly comes from a Landau PDF
+
-  - the sample $M$ does not come from a Landau PDF
+ - can we accept $H_0$ for $L$?
+ - can we reject $H_0$ for $M$?
 
 
-## Why?
+## Why Moyal?
 
 The Landau and Moyal PDFs are really similar. Historically, the latter was
-utilized in the approximation of the former.
+utilized as an approximation of the former.
 
 :::: {.columns}
 :::  {.column width=33%}
@@ -79,15 +81,18 @@ utilized in the approximation of the former.
 
 . . .
 
-- Parameters comparison:
+- **Properties test**:
-  - compatibility between expected and observed PDF parameters
+
+  compatibility between expected and observed PDF properties
 
 . . .
 
-- Kolmogorov - Smirnov test:
+- **Kolmogorov - Smirnov test**:
-  - compatibility between expected and observed CDF
+
+  compatibility between expected and empirical CDF
 
 . . .
 
-- Trapani test:
+- **Trapani test**:
-  - compatibility between expected and observed moments
+
+  test for finite or infinite moments

slides/sections/2.md

@@ -1,18 +1,24 @@
 # Landau PDF
 
 
-## A pathological distribution
+## Landau PDF
 
-Because of its fat tail:
+:::: {.columns}
+:::  {.column width=50% .c}
+  $$
+    L(x) = \frac{1}{\pi} \int \limits_{0}^{+ \infty}
+           dt \, e^{-t \ln(t) -xt} \sin (\pi t)
+  $$
+:::
 
-\begin{align*}
+:::  {.column width=50%}
-  E[x] &\longrightarrow + \infty  \\
+  ![](images/landau-pdf.pdf)
-  V[x] &\longrightarrow + \infty
+:::
-\end{align*}
+::::
 
 . . .
 
-No closed form for parameters $\thus$ numerical estimations
+No closed form for \textcolor{cyclamen}{ANYTHING}
 
 
 ## Landau median
@@ -28,9 +34,17 @@ $$
 - CDF computed by numerical integration
 - QDF computed by numerical root-finding (Brent)
 
-$$
+\setbeamercovered{}
-  m_L\ex = 1.3557804...
+
-$$
+\begin{center}
+  \begin{tikzpicture}[remember picture]
+    \node at (0,0) (here) {$m_L\ex = 1.3557804...$};
+    \pause
+    \node [opacity=0.5, xscale=0.35, yscale=0.25 ] at (here) {\includegraphics{images/high.png}};
+  \end{tikzpicture}
+\end{center}
+
+\setbeamercovered{transparent}
 
 
 ## Landau mode
@@ -41,9 +55,17 @@ $$
 
 - Computed by numerical minimization (Brent)
 
-$$
+\setbeamercovered{}
-  \mu_L\ex = − 0.22278...
+
-$$
+\begin{center}
+  \begin{tikzpicture}[remember picture]
+    \node at (0,0) (here) {$\mu_L\ex = − 0.22278...$};
+    \pause
+    \node [opacity=0.5, xscale=0.32, yscale=0.25 ] at (here) {\includegraphics{images/high.png}};
+  \end{tikzpicture}
+\end{center}
+
+\setbeamercovered{transparent}
 
 
 ## Landau FWHM
@@ -62,6 +84,14 @@ $$
 
 - Computed by numerical root finding (Brent)
 
-$$
+\setbeamercovered{}
-  w_L\ex = 4.018645...
+
-$$
+\begin{center}
+  \begin{tikzpicture}[remember picture]
+    \node at (0,0) (here) {$w_L\ex = 4.018645...$};
+    \pause
+    \node [opacity=0.5, xscale=0.32, yscale=0.25 ] at (here) {\includegraphics{images/high.png}};
+  \end{tikzpicture}
+\end{center}
+
+\setbeamercovered{transparent}

slides/sections/3.md

@@ -121,12 +121,8 @@ $$
 ## Moyal FWHM
 
 $$
-  x_+ - x_- = W_0 \left( - \frac{1}{4 e} \right)
+  x_+ - x_- = 3.590806098... = a
-          - W_{-1} \left( - \frac{1}{4 e} \right)
-          = 3.590806098...
-          = a
 $$
-
 \begin{align*}
   M(z)
     &\thus w_M^{\text{exp}} = a              \\

slides/sections/4.md

@@ -1,26 +1,31 @@
-# Sample parameters estimation
+# Sample statistics
 
 
-## Sample parameters estimation
+## Sample statistics
 
-Once the points are sampled,  
+How to estimate sample median, mode and FWHM?
-how to estimate their median, mode and FWHM?
 
 . . .
 
-- Binning data $\hence$ result depending on bin-width
+- \only<3>\strike{Binning data $\hence$ depends wildly on bin-width}
 
 . . .
 
 - Alternative solutions
+  - Robust estimators
+  - Kernel density estimation
 
 
 ## Sample median
 
+:::: {.columns}
+:::  {.column width=50% .c}
   $$
-  m = Q \left( \frac{1}{2} \right)
+    F(m) = \frac{1}{2}
   $$
 
+  \vspace{20pt}
+
   . . .
 
   - Sort points in ascending order
@@ -28,7 +33,14 @@ $$
   . . .
 
   - Middle element if odd
-- Average of the two central elements if even
+
+    Average of the two central elements if even
+:::
+
+:::  {.column width=50%}
+  ![](images/median.pdf)
+:::
+::::
 
 
 ## Sample mode
@@ -37,11 +49,78 @@ Most probable value
 
 . . .
 
-HSM
+Half Sample Mode
 
 - Iteratively identify the smallest interval containing half points
 - Once the sample is reduced to less than three points, take average
 
+. . .
+
+\setbeamercovered{}
+
+\begin{center}
+\begin{tikzpicture}[remember picture]
+  % line
+  \draw [line width=3, ->, cyclamen] (-5,0) -- (5,0);
+  \node [right] at (5,0) {$x$};
+  % points
+  \draw [blue, fill=blue] (-4.6,-0.1) rectangle (-4.8,0.1);
+  \draw [blue, fill=blue] (-4,-0.1)   rectangle (-4.2,0.1);
+  \draw [blue, fill=blue] (-3.3,-0.1) rectangle (-3.5,0.1);
+  \draw [blue, fill=blue] (-2.3,-0.1) rectangle (-2.5,0.1);
+  \draw [blue, fill=blue] (-0.6,-0.1) rectangle (-0.8,0.1);
+  \draw [blue, fill=blue] (-0.1,-0.1) rectangle (0.1,0.1);
+  \draw [blue, fill=blue] (1.1,-0.1)  rectangle (1.3,0.1);
+  \draw [blue, fill=blue] (2  ,-0.1)  rectangle (2.2,0.1);
+  \draw [blue, fill=blue] (2.7,-0.1)  rectangle (2.9,0.1);
+  \draw [blue, fill=blue] (4,-0.1)    rectangle (4.2,0.1);
+  % future nodes
+  \node at (-1,-0.3)  (1a) {};
+  \node at (3.1,0.3)  (1b) {};
+  \node at (0.9,-0.3) (2a) {};
+  \node at (1.8,-0.3) (3a) {};
+  % result nodes
+  \node at (2.45,-0.7) (f1) {};
+  \node at (2.45,0.7)  (f2) {};
+\end{tikzpicture}
+\end{center}
+
+. . .
+
+\begin{center}
+\begin{tikzpicture}[remember picture, overlay]
+  % region
+  \draw [orange, fill=orange, opacity=0.5] (1a) rectangle (1b);
+\end{tikzpicture}
+\end{center}
+
+. . .
+
+\begin{center}
+\begin{tikzpicture}[remember picture, overlay]
+  % region
+  \draw [orange, fill=orange, opacity=0.5] (2a) rectangle (1b);
+\end{tikzpicture}
+\end{center}
+
+. . .
+
+\begin{center}
+\begin{tikzpicture}[remember picture, overlay]
+  % region
+  \draw [orange, fill=orange, opacity=0.5] (3a) rectangle (1b);
+\end{tikzpicture}
+\end{center}
+
+. . .
+
+\begin{center}
+\begin{tikzpicture}[remember picture, overlay]
+  % region
+  \draw [cyclamen, ultra thick] (f1) -- (f2);
+\end{tikzpicture}
+\end{center}
+
 
 ## Sample FWHM
 
@@ -49,9 +128,10 @@ $$
   \text{FWHM} = x_+ - x_- \with L(x_{\pm}) = \frac{L_{\text{max}}}{2}
 $$
 
+\setbeamercovered{transparent}
 . . .
 
-KDE
+Kernel Density Estimation
 
 - empirical PDF construction:
 
@@ -82,9 +162,9 @@ with:
 
 . . .
 
-\vspace{10pt}
+Numerical minimization (Brent) for $\quad f_{\varepsilon_{\text{max}}}$  
-
+Numerical root finding (Brent) for $\quad f_{\varepsilon}(x_{\pm}) =
-Numerical root finding (Brent)
+\frac{f_{\varepsilon_{\text{max}}}}{2}$
 
 
 ## Sample FWHM

slides/sections/5.md

@@ -1,81 +1,43 @@
-# MC simulations
+# Kolmogorov - Smirnov test
 
 
-## In summary
+## KS
 
------------------------------------------------------
+Quantify distance between expected and observed CDF
-                  Landau            Moyal
------------------ ----------------- -----------------
-median             $m_L\ex$          $m_M\ex (μ, σ)$
 
-mode               $\mu_L\ex$        $\mu_M\ex (μ)$
+. . .
 
-FWHM               $w_L\ex$          $w_M\ex (σ)$
+KS statistic:
------------------------------------------------------
 
-
-## Moyal parameters
-
-A $M(x)$ similar to $L(x)$ can be found by imposing:
-
-\vspace{15pt}
-
-- equal mode
 $$
-  \mu_M\ex = \mu_L\ex \approx −0.22278298...
+  D_N = \text{sup}_x |F_N(x) - F(x)|
+$$
+
+- $F(x)$ is the expected CDF
+- $F_N(x)$ is the empirical CDF of $N$ sampled points
+  - sort points in ascending order
+  - number of points preceding the point normalized by $N$
+
+
+## KS
+
+$H_0$: points sampled according to $F(x)$
+
+. . .
+
+If $H_0$ is true:
+
+- $\sqrt{N}D_N \xrightarrow{N \rightarrow + \infty} K$
+
+Kolmogorov distribution with CDF:
+
+$$
+  P(K \leqslant K_0) = 1 - p = \frac{\sqrt{2 \pi}}{K_0}
+  \sum_{j = 1}^{+ \infty} e^{-(2j - 1)^2 \pi^2 / 8 K_0^2}
 $$
 
 . . .
 
-- equal width
+a $p$-value can be computed
-$$
-  w_M\ex = w_L\ex = \sigma \cdot a
-$$
 
-$$
+- At 95% confidence level, $H_0$ cannot be disproved if $p > 0.05$
-  \implies \sigma_M \approx 1.1191486...
-$$
-
-
-## Moyal parameters
-
-:::: {.columns}
-:::  {.column width=50%}
-  ![](images/both-pdf-bef.pdf)
-:::
-
-:::  {.column width=50%}
-  ![](images/both-pdf-aft.pdf)
-:::
-::::
-
-
-## Moyal parameters
-
-This leads to more different medians:
-
-\begin{align*}
-  m_M = 0.787... \thus &m_M = 0.658... \\
-                       &m_L = 1.355...
-\end{align*}
-
-
-## Compatibility test
-
-Comparing results:
-
-$$
-  p = 1 - \text{erf} \left( \frac{t}{\sqrt{2}} \right)\ \with
-  t = \frac{|x\ex - x\ob|}{\sqrt{\sigma\ex^2 + \sigma\ob^2}}
-$$
-
-- $x\ex$ and $x\ob$ are the expected and observed values
-- $\sigma\ex$ and $\sigma\ob$ are their absolute errors
-
-. . .
-
-At 95% confidence level, the values are compatible if:
-
-$$
-  p > 0.05
-$$

slides/sections/6.md

@@ -1,148 +1,181 @@
-# Landau sample
+# Trapani test
 
 
-## Sample
+## A pathological distribution
 
-Sample N = 50'000 random points following $L(x)$
+Because of its fat tail:
+\begin{align*}
+  \mu_1 &= \text{E}\left[|x|\right] \longrightarrow + \infty  \\
+  \mu_2 &= \text{E}\left[|x|^2\right] \longrightarrow + \infty
+\end{align*}
 
+. . .
+
+No closed form for parameters $\thus$ numerical estimations
+
+. . .
+
+For a Moyal PDF:
+\begin{align*}
+  E_M[x] &= \mu + \sigma [ \gamma + \ln(2) ]  \\
+  V_M[x] &= \frac{\pi^2 \sigma^2}{2}
+\end{align*}
+
+
+## Infinite moments
+
+- Check whether a moment is finite or infinite
+\begin{align*}
+  \text{infinite} &\thus Landau \\
+  \text{finite}   &\thus Moyal
+\end{align*}
+
+. . .
+
+
+# Trapani test
+
+
+## Trapani test
+
+::: incremental
+
+  - Random variable $\left\{ x_i \right\}$ sampled from a distribution $f$
+  - Sample moments according to $f$ moments
+  - $H_0$: $\mu_k \longrightarrow + \infty$
+  - Statistic with 1 dof chi-squared distribution
+
+:::
+
+
+## Trapani test
+
+- Start with $\left\{ x_i \right\}^N$ and compute $\mu_k$ as:
   $$
-  L(x) = \frac{1}{\pi} \int \limits_{0}^{+ \infty}
+    \mu_k = \frac{1}{N} \sum_{i = 1}^N |x_i|^k
-         dt \, e^{-t \ln(t) -xt} \sin (\pi t)
   $$
 
 . . .
 
-gsl_ran_Landau(gsl_rng)
+- Generate $r$ points $\left\{ \xi_j\right\}^r$ according to $G(0, 1)$ and define
-
+  $\left\{ a_j \right\}^r$ as:
-
-## Compatibility results:
-
-Median:
-
-:::: {.columns}
-:::  {.column width=50%}
-  - $t = 0.761$
-  - $p = 0.446$
-:::
-
-:::  {.column width=50%}
   $$
-  \hence \text{Compatible!}
+    a_j = \sqrt{e^{\mu_k}} \cdot \xi_j
-  $$
+    \thus G'\left( 0, \sqrt{e^{\mu_k}} \right)
-:::
-::::
-
-\vspace{10pt}
-
-. . .
-
-Mode:
-
-:::: {.columns}
-:::  {.column width=50%}
-  - $t = 1.012$
-  - $p = 0.311$
-:::
-
-:::  {.column width=50%}
-  $$
-  \hence \text{Compatible!}
-  $$
-:::
-::::
-
-\vspace{10pt}
-
-. . .
-
-FWHM:
-
-:::: {.columns}
-:::  {.column width=50%}
-  - $t=1.338$
-  - $p=0.181$
-:::
-
-:::  {.column width=50%}
-  $$
-  \hence \text{Compatible!}
-  $$
-:::
-::::
-
-
-# Moyal sample
-
-
-## Sample
-
-Sample N = 50'000 random points following $M_{\mu \sigma}(x)$
-
-$$
-  M_{\mu \sigma}(x) = \frac{1}{\sqrt{2 \pi} \sigma} \exp
-         \left[ - \frac{1}{2} \left(
-           \frac{x - \mu}{\sigma}
-           + e^{-\frac{x - \mu}{\sigma}} \right) \right]
   $$
 
 . . .
 
-reverse sampling
+The greater $\mu^k$, the 'larger' $G'$
 
-- sampling $y$ uniformly in [0, 1] $\hence x = Q_M(y)$
+- if $\mu_k \longrightarrow + \infty \thus a_j$ distributed uniformly
 
 
-## Compatibility results:
+## Trapani test
 
-Median:
+- Define the sequence: $\left\{ \zeta_j (u) \right\}^r$ as:
-
-:::: {.columns}
-:::  {.column width=50%}
-  - $t = 669.940$
-  - $p = 0.000$
-:::
-
-:::  {.column width=50%}
   $$
-  \hence \text{Not compatible!}
+    \zeta_j (u) = \theta( u -  a_j) \with \theta - \text{Heaviside}
   $$
-:::
-::::
-
-\vspace{10pt}
 
 . . .
 
-Mode:
+\begin{center}
-
+\begin{tikzpicture}
-:::: {.columns}
+  % line
-:::  {.column width=50%}
+  \draw [line width=3, ->, cyclamen] (0,0) -- (10,0);
-  - $t = 0.732$
+  \node [right] at (10,0) {$u$};
-  - $p = 0.464$
+  % tic
-:::
+  \draw [thick] (5,-0.3) -- (5,0.3);
-
+  \node [above] at (5,0.3) {$u_0$};
-:::  {.column width=50%}
+  % aj tics
-  $$
+  \draw [thick, cyclamen]          (1,-0.2) -- (1,0.2);
-  \hence \text{Compatible!}
+  \node [below right, cyclamen] at (1,-0.2) {$a_{j+2}$};
-  $$
+  \draw [thick, cyclamen]          (2,-0.2) -- (2,0.2);
-:::
+  \node [below right, cyclamen] at (2,-0.2) {$a_j$};
-::::
+  \draw [thick, cyclamen]          (5.2,-0.2) -- (5.2,0.2);
-
+  \node [below right, cyclamen] at (5.2,-0.2) {$a_{j+2}$};
-\vspace{10pt}
+  \draw [thick, cyclamen]          (6,-0.2) -- (6,0.2);
+  \node [below right, cyclamen] at (6,-0.2) {$a_{j+3}$};
+  \draw [thick, cyclamen]          (8.5,-0.2) -- (8.5,0.2);
+  \node [below right, cyclamen] at (8.5,-0.2) {$a_{j+4}$};
+  % notes
+  \node [below] at (1,-1)   {0};
+  \node [below] at (2,-1)   {0};
+  \node [below] at (5.2,-1) {1};
+  \node [below] at (6,-1)   {1};
+  \node [below] at (8.5,-1) {1};
+  \draw [thick, ->] (1,-0.5)   -- (1,-1);
+  \draw [thick, ->] (2,-0.5)   -- (2,-1);
+  \draw [thick, ->] (5.2,-0.5) -- (5.2,-1);
+  \draw [thick, ->] (6,-0.5)   -- (6,-1);
+  \draw [thick, ->] (8.5,-0.5) -- (8.5,-1);
+\end{tikzpicture}
+\end{center}
 
 . . .
 
-FWHM:
+If $a_j$ uniformly distributed and $N \rightarrow + \infty$:
 
-:::: {.columns}
+- $\zeta_j (u)$ Bernoulli PDF with $P(\zeta_j (u) = 1) = \frac{1}{2}$
-:::  {.column width=50%}
-  - $t = 1.329$
-  - $p = 0.184$
-:::
 
-:::  {.column width=50%}
+
+## Trapani test
+
+- Define the function $\vartheta (u)$ as:
 $$
-  \hence \text{Compatible!}
+  \vartheta (u) = \frac{2}{\sqrt{r}}
+                  \left[ \sum_{j} \zeta_j (u) - \frac{r}{2} \right]
 $$
-:::
+
-::::
+. . .
+
+If $a_j$ uniformly distributed and $N \rightarrow + \infty$, for the CLT:
+$$
+  \sum_j \zeta_j (u) \hence
+    G \left( \frac{r}{2}, \frac{r}{4} \right)
+  \thus \vartheta (u) \hence
+    G \left( 0, 1 \right)
+$$
+
+. . .
+
+- Test statistic:
+$$
+  \Theta = \int_{\underbar{u}}^{\bar{u}} du \, \vartheta^2 (u) \psi(u)
+$$
+
+
+## Trapani test
+
+According to L. Trapani [@trapani15]:
+
+- $r = o(N) \hence r = N^{0.75}$
+- $\underbar{u} = 1 \quad \wedge \quad \bar{u} = 1$
+- $\psi(u) = \chi_{[\underbar{u}, \bar{u}]}$
+
+. . .
+
+$\mu_k$ must be scale invariant for $k > 1$:
+
+$$
+  \tilde{\mu_k} = \frac{\mu_k}{ \left( \mu_{\phi} \right)^{k/\phi} }
+  \with \phi \in (0, k)
+$$
+
+
+## Trapani test
+
+If $\mu_k \ne + \infty \hence \left\{ a_j \right\}$ are not uniformly distributed
+\vspace{20pt}
+Rewriting:
+$$
+  \vartheta (u) = \frac{2}{\sqrt{r}}
+                  \left[ \sum_{j} \zeta_j (u) - \frac{r}{2} \right]
+                = \frac{2}{\sqrt{r}}
+                  \sum_{j} \left[ \zeta_j (u) - \frac{1}{2} \right]
+$$
+
+\vspace{20pt}
+
+Residues become very large $\hence$ $p$-values decreases.

slides/sections/7.md

@@ -1,87 +1,81 @@
-# Kolmogorov - Smirnov test
+# MC simulations
 
 
-## KS
+## In summary
 
-Quantify distance between expected and observed CDF
+-----------------------------------------------------
+                  Landau            Moyal
+----------------- ----------------- -----------------
+median             $m_L\ex$          $m_M\ex (μ, σ)$
 
-. . .
+mode               $\mu_L\ex$        $\mu_M\ex (μ)$
 
-KS statistic:
+FWHM               $w_L\ex$          $w_M\ex (σ)$
+-----------------------------------------------------
 
+
+## Moyal parameters
+
+A $M(x)$ similar to $L(x)$ can be found by imposing:
+
+\vspace{15pt}
+
+- equal mode
 $$
-  D_N = \text{sup}_x |F_N(x) - F(x)|
+  \mu_M\ex = \mu_L\ex \approx −0.22278298...
-$$
-
-- $F(x)$ is the expected CDF
-- $F_N(x)$ is the empirical CDF of $N$ sampled points
-  - sort points in ascending order
-  - number of points preceding the point normalized by $N$
-
-
-## KS
-
-$H_0$: points sampled according to $F(x)$
-
-. . .
-
-If $H_0$ is true:
-
-- $\sqrt{N}D_N \xrightarrow{N \rightarrow + \infty} K$
-
-Kolmogorov distribution with CDF:
-
-$$
-  P(K \leqslant K_0) = 1 - p = \frac{\sqrt{2 \pi}}{K_0}
-  \sum_{j = 1}^{+ \infty} e^{-(2j - 1)^2 \pi^2 / 8 K_0^2}
 $$
 
 . . .
 
-a $p$-value can be computed
+- equal width
+$$
+  w_M\ex = w_L\ex = \sigma \cdot a
+$$
 
-- At 95% confidence level, $H_0$ cannot be disproved if $p > 0.05$
+$$
+  \implies \sigma_M \approx 1.1191486...
+$$
 
 
-# Samples results
+## Moyal parameters
-
-
-## Samples results
-
-$N = 50000$ sampled points
-
-. . .
-
-Landau sample:
 
 :::: {.columns}
 :::  {.column width=50%}
-  - $D = 0.004$
+  ![](images/both-pdf-bef.pdf)
-  - $p = 0.379$
 :::
 
 :::  {.column width=50%}
-  $$
+  ![](images/both-pdf-aft.pdf)
-  \hence \text{Compatible!}
-  $$
 :::
 ::::
 
-\vspace{10pt}
+
+## Moyal parameters
+
+This leads to more different medians:
+
+\begin{align*}
+  m_M = 0.787... \thus &m_M = 0.658... \\
+                       &m_L = 1.355...
+\end{align*}
+
+
+## Compatibility test
+
+Comparing results:
+
+$$
+  p = 1 - \text{erf} \left( \frac{t}{\sqrt{2}} \right)\ \with
+  t = \frac{|x\ex - x\ob|}{\sqrt{\sigma\ex^2 + \sigma\ob^2}}
+$$
+
+- $x\ex$ and $x\ob$ are the expected and observed values
+- $\sigma\ex$ and $\sigma\ob$ are their absolute errors
 
 . . .
 
-Moyal sample:
+At 95% confidence level, the values are compatible if:
 
-:::: {.columns}
-:::  {.column width=50%}
-  - $D = 0.153$
-  - $p = 0.000$
-:::
-
-:::  {.column width=50%}
 $$
-  \hence \text{Not compatible!}
+  p > 0.05
 $$
-:::
-::::

slides/sections/8.md

@@ -1,184 +1,198 @@
-# Trapani test
+# Landau sample
 
 
-## Infinite moments
+## Sample
 
-For a Landau PDF:
+Sample N = 50'000 random points following $L(x)$
-\begin{align*}
+
-  E_L[x] &\longrightarrow + \infty  \\
+$$
-  V_L[x] \text{undefined}
+  L(x) = \frac{1}{\pi} \int \limits_{0}^{+ \infty}
-\end{align*}
+         dt \, e^{-t \ln(t) -xt} \sin (\pi t)
+$$
 
 . . .
 
-For a Moyal PDF:
+gsl_ran_Landau(gsl_rng)
-\begin{align*}
-  E_M[x] &= \mu + \sigma [ \gamma + \ln(2) ]  \\
-  V_M[x] &= \frac{\pi^2 \sigma^2}{2}
-\end{align*}
 
 
-## Infinite moments
+## Compatibility results:
 
-- Check whether a moment is finite or infinite
+Median:
-\begin{align*}
-  \text{infinite} &\thus Landau \\
-  \text{finite}   &\thus Moyal
-\end{align*}
-
-. . .
-
-
-# Trapani test
-
-
-## Trapani test
-
-::: incremental
-
-  - Random variable $\left\{ x_i \right\}$ sampled from a distribution $f$
-  - Sample moments according to $f$ moments
-  - $H_0$: $\mu_k \longrightarrow + \infty$
-  - Statistic with 1 dof chi-squared distribution
 
+:::: {.columns}
+:::  {.column width=50%}
+  - $t = 0.761$
+  - $p = 0.446$
 :::
 
-
+:::  {.column width=50%}
-## Trapani test
-
-- Start with $\left\{ x_i \right\}^N$ and compute $\mu_k$ as:
   $$
-    \mu_k = \frac{1}{N} \sum_{i = 1}^N |x_i|^k
+  \hence \text{Compatible!}
+  $$
+:::
+::::
+
+\vspace{10pt}
+
+. . .
+
+Mode:
+
+:::: {.columns}
+:::  {.column width=50%}
+  - $t = 1.012$
+  - $p = 0.311$
+:::
+
+:::  {.column width=50%}
+  $$
+  \hence \text{Compatible!}
+  $$
+:::
+::::
+
+\vspace{10pt}
+
+. . .
+
+FWHM:
+
+:::: {.columns}
+:::  {.column width=50%}
+  - $t=1.338$
+  - $p=0.181$
+:::
+
+:::  {.column width=50%}
+  $$
+  \hence \text{Compatible!}
+  $$
+:::
+::::
+
+
+# Moyal sample
+
+
+## Sample
+
+Sample N = 50'000 random points following $M_{\mu \sigma}(x)$
+
+$$
+  M_{\mu \sigma}(x) = \frac{1}{\sqrt{2 \pi} \sigma} \exp
+         \left[ - \frac{1}{2} \left(
+           \frac{x - \mu}{\sigma}
+           + e^{-\frac{x - \mu}{\sigma}} \right) \right]
 $$
 
 . . .
 
-- Generate $r$ points $\left\{ \xi_j\right\}^r$ according to $G(0, 1)$ and define
+reverse sampling
-  $\left\{ a_j \right\}^r$ as:
+
+- sampling $y$ uniformly in [0, 1] $\hence x = Q_M(y)$
+
+
+## Compatibility results:
+
+Median:
+
+:::: {.columns}
+:::  {.column width=50%}
+  - $t = 669.940$
+  - $p = 0.000$
+:::
+
+:::  {.column width=50%}
   $$
-    a_j = \sqrt{e^{\mu_k}} \cdot \xi_j
+  \hence \text{Not compatible!}
-    \thus G'\left( 0, \sqrt{e^{\mu_k}} \right)
   $$
+:::
+::::
+
+\vspace{10pt}
 
 . . .
 
-The greater $\mu^k$, the 'larger' $G'$
+Mode:
 
-- if $\mu_k \longrightarrow + \infty \thus a_j$ distributed uniformly
+:::: {.columns}
+:::  {.column width=50%}
+  - $t = 0.732$
+  - $p = 0.464$
+:::
 
-
+:::  {.column width=50%}
-## Trapani test
-
-- Define the sequence: $\left\{ \zeta_j (u) \right\}^r$ as:
   $$
-    \zeta_j (u) = \theta( u -  a_j) \with \theta - \text{Heaviside}
+  \hence \text{Compatible!}
   $$
+:::
+::::
+
+\vspace{10pt}
 
 . . .
 
-\begin{center}
+FWHM:
-\begin{tikzpicture}
+
-  \definecolor{cyclamen}{RGB}{146,24,43}
+:::: {.columns}
-  % line
+:::  {.column width=50%}
-  \draw [line width=3, ->, cyclamen] (0,0) -- (10,0);
+  - $t = 1.329$
-  \node [right] at (10,0) {$u$};
+  - $p = 0.184$
-  % tic
+:::
-  \draw [thick] (5,-0.3) -- (5,0.3);
+
-  \node [above] at (5,0.3) {$u_0$};
+:::  {.column width=50%}
-  % aj tics
+  $$
-  \draw [thick, cyclamen]          (1,-0.2) -- (1,0.2);
+  \hence \text{Compatible!}
-  \node [below right, cyclamen] at (1,-0.2) {$a_{j+2}$};
+  $$
-  \draw [thick, cyclamen]          (2,-0.2) -- (2,0.2);
+:::
-  \node [below right, cyclamen] at (2,-0.2) {$a_j$};
+::::
-  \draw [thick, cyclamen]          (5.2,-0.2) -- (5.2,0.2);
+
-  \node [below right, cyclamen] at (5.2,-0.2) {$a_{j+2}$};
+
-  \draw [thick, cyclamen]          (6,-0.2) -- (6,0.2);
+# KS results
-  \node [below right, cyclamen] at (6,-0.2) {$a_{j+3}$};
+
-  \draw [thick, cyclamen]          (8.5,-0.2) -- (8.5,0.2);
+
-  \node [below right, cyclamen] at (8.5,-0.2) {$a_{j+4}$};
+## Samples results
-  % notes
+
-  \node [below] at (1,-1)   {0};
+$N = 50000$ sampled points
-  \node [below] at (2,-1)   {0};
-  \node [below] at (5.2,-1) {1};
-  \node [below] at (6,-1)   {1};
-  \node [below] at (8.5,-1) {1};
-  \draw [thick, ->] (1,-0.5)   -- (1,-1);
-  \draw [thick, ->] (2,-0.5)   -- (2,-1);
-  \draw [thick, ->] (5.2,-0.5) -- (5.2,-1);
-  \draw [thick, ->] (6,-0.5)   -- (6,-1);
-  \draw [thick, ->] (8.5,-0.5) -- (8.5,-1);
-\end{tikzpicture}
-\end{center}
 
 . . .
 
-If $a_j$ uniformly distributed and $N \rightarrow + \infty$:
+Landau sample:
 
-- $\zeta_j (u)$ Bernoulli PDF with $P(\zeta_j (u) = 1) = \frac{1}{2}$
+:::: {.columns}
+:::  {.column width=50%}
+  - $D = 0.004$
+  - $p = 0.379$
+:::
 
-
+:::  {.column width=50%}
-## Trapani test
-
-- Define the function $\vartheta (u)$ as:
   $$
-  \vartheta (u) = \frac{2}{\sqrt{r}}
+  \hence \text{Compatible!}
-                  \left[ \sum_{j} \zeta_j (u) - \frac{r}{2} \right]
   $$
+:::
+::::
+
+\vspace{10pt}
 
 . . .
 
-If $a_j$ uniformly distributed and $N \rightarrow + \infty$, for the CLT:
+Moyal sample:
+
+:::: {.columns}
+:::  {.column width=50%}
+  - $D = 0.153$
+  - $p = 0.000$
+:::
+
+:::  {.column width=50%}
   $$
-  \sum_j \zeta_j (u) \hence
+  \hence \text{Not compatible!}
-    G \left( \frac{r}{2}, \frac{r}{4} \right)
-  \thus \vartheta (u) \hence
-    G \left( 0, 1 \right)
-$$
-
-. . .
-
-- Test statistic:
-$$
-  \Theta = \int_{\underbar{u}}^{\bar{u}} du \, \vartheta^2 (u) \psi(u)
   $$
+:::
+::::
 
 
-## Trapani test
+# Trapani results
-
-According to L. Trapani (10.1016/j.jeconom.2015.08.006):
-
-- $r = o(N) \hence r = N^{0.75}$
-- $\underbar{u} = 1 \quad \wedge \quad \bar{u} = 1$
-- $\psi(u) = \chi_{[\underbar{u}, \bar{u}]}$
-
-. . .
-
-$\mu_k$ must be scale invariant for $k > 1$:
-
-$$
-  \tilde{\mu_k} = \frac{\mu_k}{ \left( \mu_{\phi} \right)^{k/\phi} }
-  \with \phi \in (0, k)
-$$
-
-
-## Trapani test
-
-If $\mu_k \ne + \infty \hence \left\{ a_j \right\}$ are not uniformly distributed
-\vspace{20pt}
-Rewriting:
-$$
-  \vartheta (u) = \frac{2}{\sqrt{r}}
-                  \left[ \sum_{j} \zeta_j (u) - \frac{r}{2} \right]
-                = \frac{2}{\sqrt{r}}
-                  \sum_{j} \left[ \zeta_j (u) - \frac{1}{2} \right]
-$$
-
-\vspace{20pt}
-
-Residues become very large $\hence$ $p$-values decreases.
-
-
-# Samples results
 
 
 ## Samples results