analistica/slides/sections/2.md

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...