LAPLACE DISTRIBUTION

Phitter implementation

Distribution Definition

import phitter

distribution = phitter.continuous.Laplace({"mu": *, "b": *})

💡 The distribution's parameters are defined equation section below

Distribution Methods and Attributes

## CDF, PDF, PPF receive float or numpy.ndarray.
distribution.cdf(float | numpy.ndarray) # -> float | numpy.ndarray
distribution.pdf(float | numpy.ndarray) # -> float | numpy.ndarray
distribution.ppf(float | numpy.ndarray) # -> float | numpy.ndarray
distribution.sample(int) # -> numpy.ndarray

## STATS
distribution.mean # -> float
distribution.variance # -> float
distribution.standard_deviation # -> float
distribution.skewness # -> float
distribution.kurtosis # -> float
distribution.median # -> float
distribution.mode # -> float

Equations

Distribution Definition

$$X\sim \mathrm{Laplace}(\mu ,b)$$

Distribution Domain

$$x\in (-\mathrm{\infty },\mathrm{\infty })$$

Parameters Domain and Constraints

$$\mu \in {\mathbb{R}}^{+},b\in {\mathbb{R}}^{+}$$

Cumulative Distribution Function

$$F_X\left(x\right)=\frac{1}{2}+\frac{1}{2}\mathrm{sign}(x-\mu )(1-\exp (-\frac{|x-\mu |}{b}))$$

Probability Density Function

$$f_X\left(x\right)=\frac{1}{2b}\exp (-\frac{|x-\mu |}{b})$$

Percent Point Function / Sample

$$F_X^{-1}\left(u\right)=\mu -b\times \mathrm{sign}(u-\frac{1}{2})\ln (1-2|p-\frac{1}{2}|)$$

Parametric Centered Moments

$${\mu }_k^{\prime }=E[X^k]={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^k{f}_X\left(x\right)dx=(\frac{1}{2})\sum _{k=0}^r[\frac{r!}{(r-k)!}{b}^k{\mu }^{(r-k)}\{1+(-1{)}^k\}]$$

Parametric Mean

$$\mathrm{Mean}(X)={\mu }_1^{\prime }=\mu$$

Parametric Variance

$$\mathrm{Variance}(X)={\mu }_2^{\prime }-{\mu }_1^{\prime 2}=2b^2$$

Parametric Skewness

$$\mathrm{Skewness}(X)=\frac{{\mu }_3^{\prime }-3{\mu }_2^{\prime }{\mu }_1^{\prime }+2{\mu }_1^{\prime 3}}{({\mu }_2^{\prime }-{\mu }_1^{\prime 2}{)}^{1.5}}=0$$

Parametric Kurtosis

$$\mathrm{Kurtosis}(X)=\frac{{\mu }_4^{\prime }-4{\mu }_1^{\prime }{\mu }_3^{\prime }+6{\mu }_1^{\prime 2}{\mu }_2^{\prime }-3{\mu }_1^{\prime 4}}{({\mu }_2^{\prime }-{\mu }_1^{\prime 2}{)}^2}=6$$

Parametric Median

$$\mathrm{Median}(X)=\mu$$

Parametric Mode

$$\mathrm{Mode}(X)=\mu$$

Additional Information and Definitions

• $\mu :\text{Location parameter}$
• $b:\text{Scale parameter}$
• $u:\text{Uniform[0,1] random varible}$

Spreadsheet Documents

• Phitter playground
• Download Excel Spreadsheet
• Excel file from GitHub repository
• Google spreadsheet document