sections: reorder the sections in order to add the Landau paragraph
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@ -8,16 +8,38 @@ institute:
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- Università di Milano-Bicocca
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theme: metropolis
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themeoptions:
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- titleformat=allcaps
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aspectratio: 169
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fontsize: 14pt
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fontsize: 12pt
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mainfont: Fira Sans
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mainfontoptions:
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- BoldFont=Fira Sans
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mathfont: FiraMath-Regular
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sansfont: Fira Sans
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header-includes: |
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```{=latex}
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%% Colors
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\definecolor{mDarkTeal} {HTML}{020202}
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\definecolor{mLightBrown}{HTML}{C49D4A}
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\definecolor{mDarkRed} {HTML}{92182B}
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\definecolor{green} {HTML}{60AC39}
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\definecolor{red} {HTML}{D73737}
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\definecolor{blue} {HTML}{6684E1}
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\definecolor{yellow}{HTML}{CFB017}
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\setbeamercolor{frametitle}{bg=mDarkRed}
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% center images
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\LetLtxMacro{\oldIncludegraphics}{\includegraphics}
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\renewcommand{\includegraphics}[2][]{
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\centering
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\oldIncludegraphics[#1]{#2}
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}
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% Misc
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% "thus" in formulas
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\DeclareMathOperator{\thus}{%
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\hspace{30pt} \Longrightarrow \hspace{30pt}
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@ -27,6 +49,5 @@ header-includes: |
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\DeclareMathOperator{\with}{%
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\hspace{30pt} \text{with} \hspace{30pt}
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}
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```
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...
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@ -1,63 +1,3 @@
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---
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title: Randomness tests of a non-uniform distribution
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date: \today
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author:
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- Giulia Marcer
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- Michele Guerini Rocco
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institute:
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- Università di Milano-Bicocca
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theme: metropolis
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themeoptions:
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- titleformat=allcaps
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aspectratio: 169
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fontsize: 12pt
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mainfont: Fira Sans
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mainfontoptions:
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- BoldFont=Fira Sans
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mathfont: FiraMath-Regular
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header-includes: |
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```{=latex}
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%% Colors
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\definecolor{mDarkTeal} {HTML}{020202}
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\definecolor{mLightBrown}{HTML}{C49D4A}
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\definecolor{mDarkRed} {HTML}{92182B}
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\definecolor{green} {HTML}{60AC39}
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\definecolor{red} {HTML}{D73737}
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\definecolor{blue} {HTML}{6684E1}
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\definecolor{yellow}{HTML}{CFB017}
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\setbeamercolor{frametitle}{bg=mDarkRed}
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% center images
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\LetLtxMacro{\oldIncludegraphics}{\includegraphics}
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\renewcommand{\includegraphics}[2][]{
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\centering
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\oldIncludegraphics[#1]{#2}
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}
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%% customer macros
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\DeclareMathOperator{\with}{%
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\hspace{30pt} \text{with} \hspace{30pt}
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}
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% "thus" in formulas
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\DeclareMathOperator{\thus}{%
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\hspace{30pt} \Longrightarrow \hspace{30pt}
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}
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% "et" in formulas
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\DeclareMathOperator{\et}{%
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\hspace{30pt} \wedge \hspace{30pt}
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}
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```
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...
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# Goal
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@ -1,135 +1,32 @@
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# Moyal distribution
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# Landau PDF
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## Moyal PDF
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## Pathological probability distribution
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Because of its fat tail:
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Standard form:
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$$
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M(z) = \frac{1}{\sqrt{2 \pi}} \exp
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\left[ - \frac{1}{2} \left( z + e^{-z} \right) \right]
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$$
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\begin{align*}
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E[x] &\longrightarrow + \infty \\
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V[x] &\longrightarrow + \infty
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\end{align*}
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. . .
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No closed form for parameters.
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More generally:
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## Landau median
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- location parameter $\mu$
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- scale parameter $\sigma$
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The median of a PDF is defined as:
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$$
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z = \frac{x - \mu}{\sigma}
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\thus
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M(x) = \frac{1}{\sqrt{2 \pi} \sigma} \exp
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\left[ - \frac{1}{2} \left(
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\frac{x - \mu}{\sigma}
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+ e^{-\frac{x - \mu}{\sigma}} \right) \right]
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Q_L(x) = \frac{1}{2}
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$$
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- CDF computed by numerical integration,
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- QDF computed by numerical root-finding (Brent)
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## Moyal CDF
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The CDF $F_M(x)$ can be derived by direct integration:
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$$
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F_M(x) = \int\limits_{- \infty}^x dy \, M(y)
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= \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{y}{2}}
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e^{- \frac{1}{2} e^{-y}}
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$$
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. . .
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With the change of variable $z = e^{-\frac{y}{2}}/\sqrt{2}$:
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$$
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F_M(x) =
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\frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)} dz \, e^{- z^2}
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\with f(x) = \frac{e^{- \frac{x}{2}}}{\sqrt{2}}
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$$
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## Moyal CDF
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Remembering the error function
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$$
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\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2},
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$$
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one finally gets:
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$$
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F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
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$$
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## Moyal QDF
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The quantile (CDF\textsuperscript{-1}) is found solving:
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$$
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y = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
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$$
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hence:
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$$
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Q_M(x) = -2 \ln \left[ \sqrt{2} \, \text{erf}^{-1} (1 - F_M(x)) \right]
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$$
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## Moyal median
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Defined by $\text{CDF}(m) = 1/2$, or $m=\text{QDF}(1/2)$.
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\begin{align*}
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M(z)
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&\thus m_M = -2 \ln \left[ \sqrt{2} \,
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\text{erf}^{-1} \left( \frac{1}{2} \right) \right] \\
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M_{\mu \sigma}(x)
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&\thus m_M = \mu -2 \sigma \ln \left[ \sqrt{2} \,
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\text{erf}^{-1} \left( \frac{1}{2} \right) \right]
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\end{align*}
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## Moyal mode
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Peak of the PDF:
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$$
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\partial_x M(x) = \partial_x \left( \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
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\left( x + e^{-x} \right)} \right)
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= \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
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\left( x + e^{-x} \right)} \left( -\frac{1}{2} \right)
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\left( 1 - e^{-x} \right)
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$$
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\begin{align*}
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\partial_x M(z) = 0 &\thus \mu_M = 0 \\
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\partial_x M_{\mu \sigma}(x) = 0 &\thus \mu_M = \mu \\
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\end{align*}
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## Moyal FWHM
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We need to compute the maximum value:
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$$
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M(\mu) = \frac{1}{\sqrt{2 \pi e}} \thus M(x_{\pm}) = \frac{1}{\sqrt{8 \pi e}}
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$$
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. . .
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which leads to:
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$$
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x_{\pm} + e^{-x_{\pm}} = 1 + 2 \ln(2) \thus
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\begin{cases}
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x_+ = 1 + 2 \ln(2) + W_0 \left( - \frac{1}{4 e} \right) \\
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x_- = 1 + 2 \ln(2) + W_{-1} \left( - \frac{1}{4 e} \right)
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\end{cases}
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$$
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## Moyal FWHM
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$$
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x_+ - x_- = W_0 \left( - \frac{1}{4 e} \right)
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- W_{-1} \left( - \frac{1}{4 e} \right)
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= 3.590806098...
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= a
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m_L = 1.3557804...
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$$
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\begin{align*}
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M(z)
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&\thus \text{FWHM}_M = a \\
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M_{\mu \sigma}(x)
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&\thus \text{FWHM}_M = \sigma \cdot a \\
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\end{align*}
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o
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@ -1,21 +1,135 @@
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# Data sample
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# Moyal distribution
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## Data sample
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The $M(x)$ most similar to $L(x)$ is found by imposing:
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## Moyal PDF
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- equal mode
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Standard form:
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$$
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\mu_M = M_L \approx −0.22278298...
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$$
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- equal width
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$$
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\text{FWHM}_M = \text{FWHM}_L = \sigma \cdot a
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M(z) = \frac{1}{\sqrt{2 \pi}} \exp
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\left[ - \frac{1}{2} \left( z + e^{-z} \right) \right]
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$$
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. . .
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More generally:
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- location parameter $\mu$
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- scale parameter $\sigma$
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$$
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\implies \sigma_M \approx 1.1191486
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z = \frac{x - \mu}{\sigma}
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\thus
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M(x) = \frac{1}{\sqrt{2 \pi} \sigma} \exp
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\left[ - \frac{1}{2} \left(
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\frac{x - \mu}{\sigma}
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+ e^{-\frac{x - \mu}{\sigma}} \right) \right]
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$$
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## Moyal CDF
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The CDF $F_M(x)$ can be derived by direct integration:
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$$
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F_M(x) = \int\limits_{- \infty}^x dy \, M(y)
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= \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{y}{2}}
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e^{- \frac{1}{2} e^{-y}}
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$$
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. . .
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With the change of variable $z = e^{-\frac{y}{2}}/\sqrt{2}$:
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$$
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F_M(x) =
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\frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)} dz \, e^{- z^2}
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\with f(x) = \frac{e^{- \frac{x}{2}}}{\sqrt{2}}
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$$
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## Moyal CDF
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Remembering the error function
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$$
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\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2},
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$$
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one finally gets:
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$$
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F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
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$$
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## Moyal QDF
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The quantile (CDF\textsuperscript{-1}) is found solving:
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$$
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y = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
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$$
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hence:
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$$
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Q_M(x) = -2 \ln \left[ \sqrt{2} \, \text{erf}^{-1} (1 - F_M(x)) \right]
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$$
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## Moyal median
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Defined by $\text{CDF}(m) = 1/2$, or $m=\text{QDF}(1/2)$.
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\begin{align*}
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M(z)
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&\thus m_M = -2 \ln \left[ \sqrt{2} \,
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\text{erf}^{-1} \left( \frac{1}{2} \right) \right] \\
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M_{\mu \sigma}(x)
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&\thus m_M = \mu -2 \sigma \ln \left[ \sqrt{2} \,
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\text{erf}^{-1} \left( \frac{1}{2} \right) \right]
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\end{align*}
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## Moyal mode
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Peak of the PDF:
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$$
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\partial_x M(x) = \partial_x \left( \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
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\left( x + e^{-x} \right)} \right)
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= \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
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\left( x + e^{-x} \right)} \left( -\frac{1}{2} \right)
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\left( 1 - e^{-x} \right)
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$$
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\begin{align*}
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\partial_x M(z) = 0 &\thus \mu_M = 0 \\
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\partial_x M_{\mu \sigma}(x) = 0 &\thus \mu_M = \mu \\
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\end{align*}
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## Moyal FWHM
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We need to compute the maximum value:
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$$
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M(\mu) = \frac{1}{\sqrt{2 \pi e}} \thus M(x_{\pm}) = \frac{1}{\sqrt{8 \pi e}}
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$$
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. . .
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which leads to:
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$$
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x_{\pm} + e^{-x_{\pm}} = 1 + 2 \ln(2) \thus
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\begin{cases}
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x_+ = 1 + 2 \ln(2) + W_0 \left( - \frac{1}{4 e} \right) \\
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x_- = 1 + 2 \ln(2) + W_{-1} \left( - \frac{1}{4 e} \right)
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\end{cases}
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$$
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## Moyal FWHM
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$$
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x_+ - x_- = W_0 \left( - \frac{1}{4 e} \right)
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- W_{-1} \left( - \frac{1}{4 e} \right)
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= 3.590806098...
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= a
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$$
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\begin{align*}
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M(z)
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&\thus \text{FWHM}_M = a \\
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M_{\mu \sigma}(x)
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&\thus \text{FWHM}_M = \sigma \cdot a \\
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\end{align*}
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21
slides/sections/4.md
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21
slides/sections/4.md
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@ -0,0 +1,21 @@
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# Data sample
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## Data sample
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The $M(x)$ most similar to $L(x)$ is found by imposing:
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- equal mode
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$$
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\mu_M = \mu_L \approx −0.22278298...
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$$
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- equal width
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$$
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\text{FWHM}_M = \text{FWHM}_L = \sigma \cdot a
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$$
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. . .
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$$
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\implies \sigma_M \approx 1.1191486
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$$
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