analistica/slides/sections/2.md

Moyal PDF

Moyal PDF

  M(x) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left[ x + e^{-x} \right]}

:::: {.columns .c} ::: {.column width=30%} \begin{center} More generally: \end{center} ::: ::: {.column width=70%} \begin{center} $$ \begin{cases} \text{location parameter} \mu \ \text{scale parameter} \sigma \end{cases} $$ \end{center} ::: :::: \vspace{20pt}

  x \rightarrow \frac{x - \mu}{\sigma}

Moyal CDF

The cumulative distribution function \mathscr{M}(x) can be derived from the pdf M(x) integrating:

  \mathscr{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{1}{2}}
    e^{- \frac{1}{2} e^{-y}}

with the change of variable: \begin{align} z = \frac{1}{\sqrt{2}} e^{-\frac{y}{2}} &\thus \frac{dz}{dy} = \frac{-1}{2 \sqrt{2}} e^{-\frac{y}{2}} \ &\thus dy = -2 \sqrt{2} e^{\frac{y}{2}} dz \end{align} hence, the limits of the integral become: \begin{align} y \rightarrow - \infty &\thus z \rightarrow + \infty \ y = x &\thus z = \frac{1}{\sqrt{2}} e^{-\frac{x}{2}} = f(x) \end{align} and the CDF can be rewritten as:

  \mathscr{M}(x) = \frac{1}{2 \pi} \int\limits_{+ \infty}^{f(x)}
  dz \, (- 2 \sqrt{2}) e^{\frac{y}{2}} e^{- \frac{y}{2}} e^{- z^2}
  = \frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)}
  dz e^{- z^2}

since the erf is defines as:

  \text{erf} = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2}

  1 = \frac{2}{\sqrt{\pi}} \int_0^{+ \infty} dy \, e^{-y^2}
    = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} + 
      \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
    = \text{erf}(x) + \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}

thus:

  \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
  1 = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} + 

Moyal mode

Peak of the PDF

  \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{15pt}

  \partial_x M(x) = 0 \thus x = 0 \thus x = \mu

\vspace{15pt}

  \thus \mu = m_L