158 lines
3.3 KiB
Markdown
158 lines
3.3 KiB
Markdown
# Moyal distribution
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## Moyal PDF {.b}
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:::: {.columns align=center}
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::: {.column width=50%}
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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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. . .
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More generally:
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- location parameter $\mu$
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- scale parameter $\sigma$
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$$
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M_{\mu \sigma}(x) = \frac{1}{\sqrt{2 \pi} \sigma} \exp
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\left[ - \frac{1}{2} \left(
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\frac{x - \mu}{\sigma}
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+ e^{-\frac{x - \mu}{\sigma}} \right) \right]
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$$
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:::
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::: {.column width=50%}
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:::
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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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after a bit of math, 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 for $x$:
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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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$$
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Q_M(y) = -2 \ln \left[ \sqrt{2} \, \text{erf}^{-1} (1 - y) \right]
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$$
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## Moyal median
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Defined by $F(m) = \frac{1}{2}$ or $m = Q \left( \frac{1}{2} \right)$:
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\begin{align*}
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M(z)
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&\thus m_M\ex = -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\ex = \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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\setbeamercovered{}
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\begin{center}
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\begin{tikzpicture}[overlay]
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\pause
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\node [opacity=0.5, xscale=0.55, yscale=0.4 ] at (1.85,1.1) {\includegraphics{images/high.png}};
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\end{tikzpicture}
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\end{center}
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\setbeamercovered{transparent}
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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\ex = 0 \\
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\partial_x M_{\mu \sigma}(x) = 0 &\thus \mu_M\ex = \mu \\
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\end{align*}
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\setbeamercovered{}
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\begin{center}
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\begin{tikzpicture}[overlay]
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\pause
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\node [opacity=0.5, xscale=0.18, yscale=0.25 ] at (2.4,1.8) {\includegraphics{images/high.png}};
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\end{tikzpicture}
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\end{center}
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\setbeamercovered{transparent}
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## Moyal FWHM
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We need to compute the maximum value:
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$$
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M(\mu_M\ex) = \frac{1}{\sqrt{2 \pi e}} \thus M(x_{\pm}) = \frac{1}{2 \, \sqrt{2 \pi e}}
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$$
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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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$$
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x_+ - x_- = 3.590806098... = a
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$$
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\begin{align*}
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M(z)
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&\thus w_M^{\text{exp}} = a \\
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M_{\mu \sigma}(x)
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&\thus w_M^{\text{exp}} = \sigma \cdot a \\
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\end{align*}
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\setbeamercovered{}
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\begin{center}
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\begin{tikzpicture}[overlay]
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\pause
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\node [opacity=0.5, xscale=0.2, yscale=0.25 ] at (1.9,1.9) {\includegraphics{images/high.png}};
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\end{tikzpicture}
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\end{center}
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\setbeamercovered{transparent}
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