diff --git a/slides/sections/1.md b/slides/sections/1.md index 2d85cc5..38d172a 100644 --- a/slides/sections/1.md +++ b/slides/sections/1.md @@ -72,6 +72,7 @@ How? - Applying some hypothesis testings + ## Why? The Landau and Moyal PDFs are really similar. Historically, the latter distribution was utilized in @@ -79,32 +80,31 @@ 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 -:::: {.columns .c} +:::: {.columns} ::: {.column width=50%} - \begin{center} Landau PDF $$ L(x) = \frac{1}{\pi} \int \limits_{0}^{+ \infty} dt \, e^{-t \ln(t) -xt} \sin (\pi t) $$ ::: + ::: {.column width=50%} - \begin{center} Moyal PDF $$ M(x) = \frac{1}{\sqrt{2 \pi}} \exp \left[ - \frac{1}{2} @@ -113,10 +113,11 @@ the approximation of the Landau Distribution. ::: :::: -:::: {.columns .c} +:::: {.columns} ::: {.column width=50%} ![](images/landau-pdf.pdf) ::: + ::: {.column width=50%} ![](images/moyal-pdf.pdf) ::: @@ -124,5 +125,4 @@ the approximation of the Landau Distribution. ## Two similar distributions -\centering ![](images/both-pdf.pdf) diff --git a/slides/sections/2.md b/slides/sections/2.md index d7011bd..f9eb128 100644 --- a/slides/sections/2.md +++ b/slides/sections/2.md @@ -3,43 +3,42 @@ ## Moyal PDF + +Standard form: $$ - M(x) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left[ x + e^{-x} \right]} + M(z) = \frac{1}{\sqrt{2 \pi}} \exp + \left[ - \frac{1}{2} \left( z + e^{-z} \right) \right] $$ -\vspace{10pt} + +. . . + +More generally: + + - location parameter $\mu$ + - scale parameter $\sigma$ + $$ - \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]} + z = \frac{x - \mu}{\sigma} + \thus + M(x) = \frac{1}{\sqrt{2 \pi} \sigma} \exp + \left[ - \frac{1}{2} \left( + \frac{x - \mu}{\sigma} + + e^{-\frac{x - \mu}{\sigma}} \right) \right] $$ ## Moyal CDF -The cumulative distribution function $F_M(x)$ can be derived from the -pdf $M(x)$ integrating: +The CDF $F_M(x)$ can be derived by direct integration: $$ - F_M(x) = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, M(y) + F_M(x) = \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}} $$ +. . . -## Moyal CDF - -with the change of variable: -$$ - z = \frac{1}{\sqrt{2}} e^{-\frac{y}{2}} -$$ -the CDF can be rewritten as: +With the change of variable $z = e^{-\frac{y}{2}}/\sqrt{2}$: $$ F_M(x) = \frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)} dz \, e^{- z^2} @@ -49,9 +48,9 @@ $$ ## Moyal CDF -given the definition of the error function `erf`: +Remembering the error function $$ - \text{erf} = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} + \text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2}, $$ one finally gets: $$ @@ -61,37 +60,32 @@ $$ ## Moyal QDF -The quantile is defined as the inverse of the CDF: +The quantile (CDF\textsuperscript{-1}) is found solving: $$ - F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right) + y = 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] + Q_M(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] -$$ +Defined by $\text{CDF}(m) = 1/2$, or $m=\text{QDF}(1/2)$. +\begin{align*} + M(z) + &\thus m_M = -2 \ln \left[ \sqrt{2} \, + \text{erf}^{-1} \left( \frac{1}{2} \right) \right] \\ + M_{\mu \sigma}(x) + &\thus m_M = \mu -2 \sigma \ln \left[ \sqrt{2} \, + \text{erf}^{-1} \left( \frac{1}{2} \right) \right] +\end{align*} ## Moyal mode -Peak of the PDF -\vspace{10pt} +Peak of the PDF: $$ \partial_x M(x) = \partial_x \left( \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left( x + e^{-x} \right)} \right) @@ -99,14 +93,11 @@ $$ \left( x + e^{-x} \right)} \left( -\frac{1}{2} \right) \left( 1 - e^{-x} \right) $$ -\vspace{10pt} -$$ - \partial_x M(x) = 0 \thus \mu_M = 0 -$$ -\vspace{10pt} -$$ - \partial_x M_{\mu \sigma_M}(x) = 0 \thus \mu_M = \mu -$$ + +\begin{align*} + \partial_x M(z) = 0 &\thus \mu_M = 0 \\ + \partial_x M_{\mu \sigma}(x) = 0 &\thus \mu_M = \mu \\ +\end{align*} ## Moyal FWHM @@ -115,27 +106,30 @@ 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: $$ x_{\pm} + e^{-x_{\pm}} = 1 + 2 \ln(2) \thus \begin{cases} - x_+ = a + W_0 \left( - \frac{1}{4 e} \right) \\ - x_- = a + W_{-1} \left( - \frac{1}{4 e} \right) + x_+ = 1 + 2 \ln(2) + W_0 \left( - \frac{1}{4 e} \right) \\ + x_- = 1 + 2 \ln(2) + W_{-1} \left( - \frac{1}{4 e} \right) \end{cases} $$ -with $a = 1 + 2 \ln(2)$ - ## Moyal FWHM $$ - \text{FWHM}_L = x_+ - x_- = 3.590806098... -$$ -\vspace{15pt} -$$ - M(x) \thus \text{FWHM}_M = 3.590806098... -$$ -\vspace{15pt} -$$ - M_{\mu \sigma_M}(x) \thus \text{FWHM}_M = \sigma_M \cdot 3.590806098... + x_+ - x_- = W_0 \left( - \frac{1}{4 e} \right) + - W_{-1} \left( - \frac{1}{4 e} \right) + = 3.590806098... + = a $$ + +\begin{align*} + M(z) + &\thus \text{FWHM}_M = a \\ + M_{\mu \sigma}(x) + &\thus \text{FWHM}_M = \sigma \cdot a \\ +\end{align*} diff --git a/slides/sections/3.md b/slides/sections/3.md index ee9dcaa..ace417b 100644 --- a/slides/sections/3.md +++ b/slides/sections/3.md @@ -3,12 +3,19 @@ ## Data sample The $M(x)$ most similar to $L(x)$ is found by imposing: -\vspace{15pt} + +- equal mode $$ - \mu_M = \mu_L \sim = −0.22278298... + \mu_M = M_L \approx −0.22278298... $$ -\vspace{15pt} + +- equal width $$ - \text{FWHM}_M = \text{FWHM}_L = 4.0186457... - \thus \sigma_M = 1.1191486 + \text{FWHM}_M = \text{FWHM}_L = \sigma \cdot a +$$ + +. . . + +$$ + \implies \sigma_M \approx 1.1191486 $$