142 lines
2.9 KiB
Markdown
142 lines
2.9 KiB
Markdown
# Moyal distribution
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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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\vspace{10pt}
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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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$$
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x \rightarrow \frac{x - \mu}{\sigma_M} \thus
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M_{\mu \sigma_M}(x) = \frac{1}{\sqrt{2 \pi} \sigma_M}
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e^{-\frac{1}{2} \left[ \frac{x - \mu}{\sigma_M} + e^{- \frac{x - \mu}{\sigma_M}} \right]}
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$$
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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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$$
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F_M(x) = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, M(y)
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= \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{y}{2}}
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e^{- \frac{1}{2} e^{-y}}
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$$
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## Moyal CDF
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with the change of variable:
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$$
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z = \frac{1}{\sqrt{2}} e^{-\frac{y}{2}}
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$$
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the CDF can be rewritten as:
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$$
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F_M(x) =
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\frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)} dz \, e^{- z^2}
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\with f(x) = \frac{e^{- \frac{x}{2}}}{\sqrt{2}}
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$$
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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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$$
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F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
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$$
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## Moyal QDF
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The quantile is defined as the inverse of the CDF:
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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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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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$$
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\partial_x M(x) = \partial_x \left( \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
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\left( x + e^{-x} \right)} \right)
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= \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
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\left( x + e^{-x} \right)} \left( -\frac{1}{2} \right)
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\left( 1 - e^{-x} \right)
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$$
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\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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## 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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$$
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M(x) \thus \text{FWHM}_M = 3.590806098...
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$$
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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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