slides: write about the Moyal PDF parameters
Also add the Landau, Moyal and 'both' plots.
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slides/sections/0.md
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slides/sections/0.md
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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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aspectratio: 169
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fontsize: 14pt
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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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% 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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}
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% "with" in formulas
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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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@ -1,35 +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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aspectratio: 169
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fontsize: 14pt
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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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% 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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}
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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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@ -46,16 +14,30 @@ How?
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- Applying some hypothesis testings
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Why?
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## Why?
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- They are really similar. Historically, the Moyal distribution was utilized in
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the approximation of the Landau Distribution.
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The Landau and Moyal PDFs are really similar. Historically, the latter distribution was utilized in
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the approximation of the Landau Distribution.
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:::: {.columns}
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::: {.column width=33%}
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\centering
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{height=130pt}
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:::
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::: {.column width=33%}
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\centering
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{height=130pt}
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:::
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::: {.column width=33%}
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\centering
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{height=130pt}
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:::
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::::
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# Two similar distributions
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## Two similar distributions
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:::: {.columns .c}
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::: {.column width=50%}
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::: {.column width=50%}
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\begin{center}
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Landau PDF
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$$
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@ -64,27 +46,27 @@ Why?
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$$
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\end{center}
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:::
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::: {.column width=50%}
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::: {.column width=50%}
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\begin{center}
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Moyal PDF
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$$
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M(x) = \frac{1}{\sqrt{2 \pi \sigma}} \exp \left( - \frac{x - \mu }{2 \sigma}
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- \frac{1}{2} e^{- \frac{x -\mu}{\sigma}} \right)
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M(x) = \frac{1}{\sqrt{2 \pi}} \exp \left[ - \frac{1}{2}
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\left( x + e^{- x} \right) \right]
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$$
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\end{center}
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:::
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::::
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:::: {.columns .c}
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::: {.column width=50%}
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::: {.column width=50%}
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:::
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::: {.column width=50%}
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::: {.column width=50%}
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:::
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::::
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## Two similar distributions
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grafici sovrapposti
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\centering
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@ -1,81 +1,97 @@
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# Moyal PDF
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# Moyal distribution
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# Moyal PDF
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## Moyal PDF
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$$
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M(x) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left[ x + e^{-x} \right]}
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$$
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:::: {.columns .c}
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::: {.column width=30%}
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\begin{center}
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More generally:
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\end{center}
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:::
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::: {.column width=70%}
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\begin{center}
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$$
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\begin{cases}
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\text{location parameter} \mu \\
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\text{scale parameter} \sigma
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\end{cases}
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$$
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\end{center}
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:::
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::::
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\vspace{20pt}
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\vspace{10pt}
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$$
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x \rightarrow \frac{x - \mu}{\sigma}
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\text{More generally:} \hspace{15pt}
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\begin{cases}
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\text{location parameter:} \quad &\mu_M \\
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\text{scale parameter:} \quad &\sigma_M
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\end{cases}
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$$
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\vspace{10pt}
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$$
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x \rightarrow \frac{x - \mu}{\sigma_M} \thus
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M_{\mu \sigma_M}(x) = \frac{1}{\sqrt{2 \pi} \sigma_M}
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e^{-\frac{1}{2} \left[ \frac{x - \mu}{\sigma_M} + e^{- \frac{x - \mu}{\sigma_M}} \right]}
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$$
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## Moyal CDF
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The cumulative distribution function $\mathscr{M}(x)$ can be derived from the
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The cumulative distribution function $F_M(x)$ can be derived from the
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pdf $M(x)$ integrating:
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$$
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\mathscr{M}(x) = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, M(y)
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= \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{1}{2}}
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e^{- \frac{1}{2} e^{-y}}
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F_M(x) = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, M(y)
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= \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{y}{2}}
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e^{- \frac{1}{2} e^{-y}}
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$$
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## Moyal CDF
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with the change of variable:
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\begin{align}
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$$
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z = \frac{1}{\sqrt{2}} e^{-\frac{y}{2}}
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&\thus \frac{dz}{dy} = \frac{-1}{2 \sqrt{2}} e^{-\frac{y}{2}} \\
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&\thus dy = -2 \sqrt{2} e^{\frac{y}{2}} dz
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\end{align}
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hence, the limits of the integral become:
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\begin{align}
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y \rightarrow - \infty &\thus z \rightarrow + \infty \\
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y = x &\thus z = \frac{1}{\sqrt{2}} e^{-\frac{x}{2}} = f(x)
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\end{align}
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and the CDF can be rewritten as:
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$$
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\mathscr{M}(x) = \frac{1}{2 \pi} \int\limits_{+ \infty}^{f(x)}
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dz \, (- 2 \sqrt{2}) e^{\frac{y}{2}} e^{- \frac{y}{2}} e^{- z^2}
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= \frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)}
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dz e^{- z^2}
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the CDF can be rewritten as:
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$$
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since the `erf` is defines as:
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F_M(x) =
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\frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)} dz \, e^{- z^2}
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\with f(x) = \frac{e^{- \frac{x}{2}}}{\sqrt{2}}
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$$
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## Moyal CDF
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given the definition of the error function `erf`:
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$$
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\text{erf} = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2}
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$$
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one finally gets:
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$$
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1 = \frac{2}{\sqrt{\pi}} \int_0^{+ \infty} dy \, e^{-y^2}
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= \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} +
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\frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
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= \text{erf}(x) + \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
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F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
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$$
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thus:
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## Moyal QDF
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The quantile is defined as the inverse of the CDF:
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$$
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\frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
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1 = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} +
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F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
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$$
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hence:
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$$
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Q(x) = -2 \ln \left[ \sqrt{2} \, \text{erf}^{(-1)} (1 - F_M(x)) \right]
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$$
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## Moyal median
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The median is defined as the point at which $F_M(x) = 1/2$:
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\vspace{15pt}
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$$
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M(x) \thus
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m_M = -2 \ln \left[ \sqrt{2} \,
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\text{erf}^{(-1)} \left( \frac{1}{2} \right) \right]
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$$
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\vspace{15pt}
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$$
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M_{\mu \sigma_M}(x) \thus
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m_M = \mu -2 \sigma_M \ln \left[ \sqrt{2} \,
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\text{erf}^{(-1)} \left( \frac{1}{2} \right) \right]
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$$
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## Moyal mode
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Peak of the PDF
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\vspace{10pt}
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$$
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\partial_x M(x) = \partial_x \left( \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
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\left( x + e^{-x} \right)} \right)
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@ -83,11 +99,43 @@ $$
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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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$$
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\vspace{15pt}
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\vspace{10pt}
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$$
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\partial_x M(x) = 0 \thus x = 0 \thus x = \mu
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\partial_x M(x) = 0 \thus \mu_M = 0
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$$
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\vspace{10pt}
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$$
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\partial_x M_{\mu \sigma_M}(x) = 0 \thus \mu_M = \mu
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$$
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## Moyal FWHM
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We need to compute the maximum value:
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$$
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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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$$
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which leads to:
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$$
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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_+ = a + W_0 \left( - \frac{1}{4 e} \right) \\
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x_- = a + W_{-1} \left( - \frac{1}{4 e} \right)
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\end{cases}
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$$
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with $a = 1 + 2 \ln(2)$
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## Moyal FWHM
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$$
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\text{FWHM}_L = x_+ - x_- = 3.590806098...
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$$
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\vspace{15pt}
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$$
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\thus \mu = m_L
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M(x) \thus \text{FWHM}_M = 3.590806098...
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$$
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\vspace{15pt}
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$$
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M_{\mu \sigma_M}(x) \thus \text{FWHM}_M = \sigma_M \cdot 3.590806098...
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$$
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14
slides/sections/3.md
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slides/sections/3.md
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# Data sample
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## Data sample
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The $M(x)$ most similar to $L(x)$ is found by imposing:
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\vspace{15pt}
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$$
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\mu_M = \mu_L \sim = −0.22278298...
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$$
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\vspace{15pt}
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$$
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\text{FWHM}_M = \text{FWHM}_L = 4.0186457...
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\thus \sigma_M = 1.1191486
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$$
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