# Moyal distribution


## Moyal PDF

$$
  M(x) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left[ x + e^{-x} \right]}
$$
\vspace{10pt}
$$
  \text{More generally:} \hspace{15pt}
  \begin{cases}
    \text{location parameter:} \quad &\mu_M \\
    \text{scale parameter:} \quad &\sigma_M
  \end{cases}
$$
\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 $F_M(x)$ can be derived from the
pdf $M(x)$ integrating:
$$
  F_M(x) = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, M(y)
  = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{y}{2}}
  e^{- \frac{1}{2} e^{-y}}
$$


## Moyal CDF

with the change of variable:
$$
  z = \frac{1}{\sqrt{2}} e^{-\frac{y}{2}}
$$
the CDF can be rewritten 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:
$$
  F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
$$


## Moyal QDF

The quantile is defined as the inverse of the CDF:
$$
  F_M(x) = 1 - \text{erf} \left( \frac{e^{- \frac{x}{2}}}{\sqrt{2}} \right)
$$
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}
$$
  \partial_x M(x) = \partial_x \left( \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
                    \left( x + e^{-x} \right)} \right)
                  = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}
                    \left( x + e^{-x} \right)} \left( -\frac{1}{2} \right)
                    \left( 1 - e^{-x} \right)
$$
\vspace{10pt}
$$
  \partial_x M(x) = 0 \thus \mu_M = 0
$$
\vspace{10pt}
$$
  \partial_x M_{\mu \sigma_M}(x) = 0 \thus \mu_M = \mu
$$


## 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}
$$
  M(x) \thus \text{FWHM}_M = 3.590806098...
$$
\vspace{15pt}
$$
  M_{\mu \sigma_M}(x) \thus \text{FWHM}_M = \sigma_M \cdot 3.590806098...
$$