diff --git a/slides/sections/0.md b/slides/sections/0.md new file mode 100644 index 0000000..e94a124 --- /dev/null +++ b/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} + } + + ``` +... diff --git a/slides/sections/1.md b/slides/sections/1.md index 66085a7..b6bb369 100644 --- a/slides/sections/1.md +++ b/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 @@ -46,16 +14,30 @@ How? - Applying some hypothesis testings -Why? +## Why? -- They are really similar. Historically, the Moyal distribution was utilized in - the approximation of the Landau Distribution. +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%} +::: {.column width=50%} \begin{center} Landau PDF $$ @@ -64,27 +46,27 @@ Why? $$ \end{center} ::: -::: {.column width=50%} +::: {.column width=50%} \begin{center} Moyal PDF $$ - M(x) = \frac{1}{\sqrt{2 \pi \sigma}} \exp \left( - \frac{x - \mu }{2 \sigma} - - \frac{1}{2} e^{- \frac{x -\mu}{\sigma}} \right) + M(x) = \frac{1}{\sqrt{2 \pi}} \exp \left[ - \frac{1}{2} + \left( x + e^{- x} \right) \right] $$ \end{center} ::: :::: :::: {.columns .c} -::: {.column width=50%} +::: {.column width=50%} ![](images/landau-pdf.pdf) ::: -::: {.column width=50%} +::: {.column width=50%} ![](images/moyal-pdf.pdf) ::: :::: ## Two similar distributions -grafici sovrapposti - +\centering +![](images/both-pdf.pdf) diff --git a/slides/sections/2.md b/slides/sections/2.md index d94c928..d7011bd 100644 --- a/slides/sections/2.md +++ b/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} -::: {.column width=30%} - \begin{center} - More generally: - \end{center} -::: -::: {.column width=70%} - \begin{center} - $$ - \begin{cases} - \text{location parameter} \mu \\ - \text{scale parameter} \sigma - \end{cases} - $$ - \end{center} -::: -:::: -\vspace{20pt} +\vspace{10pt} $$ - x \rightarrow \frac{x - \mu}{\sigma} + \text{More generally:} \hspace{15pt} + \begin{cases} + \text{location parameter:} \quad &\mu_M \\ + \text{scale parameter:} \quad &\sigma_M + \end{cases} +$$ +\vspace{10pt} +$$ + x \rightarrow \frac{x - \mu}{\sigma_M} \thus + M_{\mu \sigma_M}(x) = \frac{1}{\sqrt{2 \pi} \sigma_M} + e^{-\frac{1}{2} \left[ \frac{x - \mu}{\sigma_M} + e^{- \frac{x - \mu}{\sigma_M}} \right]} $$ ## 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: $$ - \mathscr{M}(x) = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, M(y) - = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{1}{2}} - e^{- \frac{1}{2} e^{-y}} + F_M(x) = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, M(y) + = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{y}{2}} + e^{- \frac{1}{2} e^{-y}} $$ + + +## 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)} - 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} +the CDF can be rewritten as: $$ -since the `erf` is defines as: + F_M(x) = + \frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)} dz \, e^{- z^2} + \with f(x) = \frac{e^{- \frac{x}{2}}}{\sqrt{2}} +$$ + + +## 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: $$ - 1 = \frac{2}{\sqrt{\pi}} \int_0^{+ \infty} dy \, e^{-y^2} - = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} + - \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2} - = \text{erf}(x) + \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2} + F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right) $$ -thus: + + +## Moyal QDF + +The quantile is defined as the inverse of the CDF: $$ - \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2} - 1 = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} + + F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right) +$$ +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... $$ diff --git a/slides/sections/3.md b/slides/sections/3.md new file mode 100644 index 0000000..ee9dcaa --- /dev/null +++ b/slides/sections/3.md @@ -0,0 +1,14 @@ +# Data sample + +## Data sample + +The $M(x)$ most similar to $L(x)$ is found by imposing: +\vspace{15pt} +$$ + \mu_M = \mu_L \sim = −0.22278298... +$$ +\vspace{15pt} +$$ + \text{FWHM}_M = \text{FWHM}_L = 4.0186457... + \thus \sigma_M = 1.1191486 +$$