94 lines
2.3 KiB
Markdown
94 lines
2.3 KiB
Markdown
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# Moyal PDF
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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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:::: {.columns .c}
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::: {.column width=30%}
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\begin{center}
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More generally:
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\end{center}
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:::
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::: {.column width=70%}
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\begin{center}
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$$
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\begin{cases}
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\text{location parameter} \mu \\
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\text{scale parameter} \sigma
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\end{cases}
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$$
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\end{center}
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:::
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::::
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\vspace{20pt}
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$$
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x \rightarrow \frac{x - \mu}{\sigma}
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$$
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## Moyal CDF
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The cumulative distribution function $\mathscr{M}(x)$ can be derived from the
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pdf $M(x)$ integrating:
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$$
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\mathscr{M}(x) = \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, M(y)
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= \frac{1}{\sqrt{2 \pi}} \int\limits_{- \infty}^x dy \, e^{- \frac{1}{2}}
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e^{- \frac{1}{2} e^{-y}}
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$$
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with the change of variable:
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\begin{align}
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z = \frac{1}{\sqrt{2}} e^{-\frac{y}{2}}
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&\thus \frac{dz}{dy} = \frac{-1}{2 \sqrt{2}} e^{-\frac{y}{2}} \\
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&\thus dy = -2 \sqrt{2} e^{\frac{y}{2}} dz
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\end{align}
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hence, the limits of the integral become:
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\begin{align}
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y \rightarrow - \infty &\thus z \rightarrow + \infty \\
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y = x &\thus z = \frac{1}{\sqrt{2}} e^{-\frac{x}{2}} = f(x)
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\end{align}
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and the CDF can be rewritten as:
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$$
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\mathscr{M}(x) = \frac{1}{2 \pi} \int\limits_{+ \infty}^{f(x)}
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dz \, (- 2 \sqrt{2}) e^{\frac{y}{2}} e^{- \frac{y}{2}} e^{- z^2}
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= \frac{-2 \sqrt{2}}{\sqrt{2 \pi}} \int\limits_{+ \infty}^{f(x)}
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dz e^{- z^2}
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$$
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since the `erf` is defines as:
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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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$$
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1 = \frac{2}{\sqrt{\pi}} \int_0^{+ \infty} dy \, e^{-y^2}
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= \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} +
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\frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
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= \text{erf}(x) + \frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
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$$
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thus:
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$$
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\frac{2}{\sqrt{\pi}} \int_x^{+ \infty} dy \, e^{-y^2}
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1 = \frac{2}{\sqrt{\pi}} \int_0^x dy \, e^{-y^2} +
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$$
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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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\vspace{15pt}
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$$
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\partial_x M(x) = 0 \thus x = 0 \thus x = \mu
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$$
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\vspace{15pt}
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$$
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\thus \mu = m_L
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$$
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