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