Laplace distribution

A random variable has a ${\displaystyle {\textrm {Laplace}}(\mu ,b)}$ distribution if its probability density function is

${\displaystyle f(x\mid \mu ,b)={\frac {1}{2b}}\exp \left(-{\frac {|x-\mu |}{b}}\right)\,\!}$

${\displaystyle ={\frac {1}{2b}}\left\{{\begin{matrix}\exp \left(-{\frac {\mu -x}{b}}\right)&{\text{if }}x<\mu \\[8pt]\exp \left(-{\frac {x-\mu }{b}}\right)&{\text{if }}x\geq \mu \end{matrix}}\right.}$

Here, ${\displaystyle \mu }$ is a location parameter and ${\displaystyle b>0}$, which is sometimes referred to as the diversity, is a scale parameter. If ${\displaystyle \mu =0}$ and ${\displaystyle b=1}$, the positive half-line is exactly an exponential distribution scaled by 1/2.

The probability density function of the Laplace distribution is also reminiscent of the normal distribution; however, whereas the normal distribution is expressed in terms of the squared difference from the mean ${\displaystyle \mu }$, the Laplace density is expressed in terms of the absolute difference from the mean. Consequently, the Laplace distribution has fatter tails than the normal distribution.

The Laplace distribution is easy to integrate (if one distinguishes two symmetric cases) due to the use of the absolute value function. Its cumulative distribution function is as follows:

${\displaystyle {\begin{aligned}F(x)&=\int _{-\infty }^{x}\!\!f(u)\,\mathrm {d} u={\begin{cases}{\frac {1}{2}}\exp \left({\frac {x-\mu }{b}}\right)&{\mbox{if }}x<\mu \\1-{\frac {1}{2}}\exp \left(-{\frac {x-\mu }{b}}\right)&{\mbox{if }}x\geq \mu \end{cases}}\\&={\tfrac {1}{2}}+{\tfrac {1}{2}}\operatorname {sgn}(x-\mu )\left(1-\exp \left(-{\frac {|x-\mu |}{b}}\right)\right).\end{aligned}}}$

The inverse cumulative distribution function is given by

${\displaystyle F^{-1}(p)=\mu -b\,\operatorname {sgn}(p-0.5)\,\ln(1-2|p-0.5|).}$

${\displaystyle \mu _{r}'={\bigg (}{\frac {1}{2}}{\bigg )}\sum _{k=0}^{r}{\bigg [}{\frac {r!}{(r-k)!}}b^{k}\mu ^{(r-k)}\{1+(-1)^{k}\}{\bigg ]}={\frac {m^{n+1}}{2b}}\left(e^{m/b}E_{-n}(m/b)-e^{-m/b}E_{-n}(-m/b)\right)}$

where ${\displaystyle E_{n}()}$ is the generalized exponential integral function ${\displaystyle E_{n}(x)=x^{n-1}\Gamma (1-n,x)}$.

A Laplace random variable can be represented as the difference of two iid exponential random variables.[1] One way to show this is by using the characteristic function approach. For any set of independent continuous random variables, for any linear combination of those variables, its characteristic function (which uniquely determines the distribution) can be acquired by multiplying the corresponding characteristic functions.

Consider two i.i.d random variables ${\displaystyle X,Y\sim {\textrm {Exponential}}(\lambda )}$. The characteristic functions for ${\displaystyle X,-Y}$ are

${\displaystyle {\frac {\lambda }{-it+\lambda }},\quad {\frac {\lambda }{it+\lambda }}}$

respectively. On multiplying these characteristic functions (equivalent to the characteristic function of the sum of the random variables ${\displaystyle X+(-Y)}$), the result is

${\displaystyle {\frac {\lambda ^{2}}{(-it+\lambda )(it+\lambda )}}={\frac {\lambda ^{2}}{t^{2}+\lambda ^{2}}}.}$

This is the same as the characteristic function for ${\displaystyle Z\sim {\textrm {Laplace}}(0,1/\lambda )}$, which is

${\displaystyle {\frac {1}{1+{\frac {t^{2}}{\lambda ^{2}}}}}.}$

Sargan distributions are a system of distributions of which the Laplace distribution is a core member. A ${\displaystyle p}$th order Sargan distribution has density[2][3]

${\displaystyle f_{p}(x)={\tfrac {1}{2}}\exp(-\alpha |x|){\frac {\displaystyle 1+\sum _{j=1}^{p}\beta _{j}\alpha ^{j}|x|^{j}}{\displaystyle 1+\sum _{j=1}^{p}j!\beta _{j}}},}$

for parameters ${\displaystyle \alpha \geq 0,\beta _{j}\geq 0}$. The Laplace distribution results for ${\displaystyle p=0}$.

Given ${\displaystyle N}$ independent and identically distributed samples ${\displaystyle x_{1},x_{2},...,x_{N}}$, the maximum likelihood estimator ${\displaystyle {\hat {\mu }}}$ of ${\displaystyle \mu }$ is the sample median,[4] and the maximum likelihood estimator ${\displaystyle {\hat {b}}}$ of ${\displaystyle b}$ is the Mean Absolute Deviation from the Median

${\displaystyle {\hat {b}}={\frac {1}{N}}\sum _{i=1}^{N}|x_{i}-{\hat {\mu }}|}$

(revealing a link between the Laplace distribution and least absolute deviations).

Given a random variable ${\displaystyle U}$ drawn from the uniform distribution in the interval ${\displaystyle \left(-1/2,1/2\right)}$, the random variable

${\displaystyle X=\mu -b\,\operatorname {sgn}(U)\,\ln(1-2|U|)}$

has a Laplace distribution with parameters ${\displaystyle \mu }$ and ${\displaystyle b}$. This follows from the inverse cumulative distribution function given above.

A ${\displaystyle {\textrm {Laplace}}(0,b)}$ variate can also be generated as the difference of two i.i.d. ${\displaystyle {\textrm {Exponential}}(1/b)}$ random variables. Equivalently, ${\displaystyle {\textrm {Laplace}}(0,1)}$ can also be generated as the logarithm of the ratio of two i.i.d. uniform random variables.