Laplace distribution

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

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

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

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

${\displaystyle \mu _{r}'={\bigg (}{\frac {1}{2}}{\bigg )}\sum _{k=0}^{r}{\bigg [}{\frac {r!}{(r-k)!}}b^{k}\mu ^{(r-k)}\{1+(-1)^{k}\}{\bigg ]}.}$

• If ${\displaystyle X\sim {\textrm {Laplace}}(\mu ,b)}$ then ${\displaystyle kX+c\sim {\textrm {Laplace}}(k\mu +c,|k|b)}$.
• If ${\displaystyle X\sim {\textrm {Laplace}}(0,1)}$ then ${\displaystyle bX\sim {\textrm {Laplace}}(0,b)}$.
• If ${\displaystyle X\sim {\textrm {Laplace}}(0,b)}$ then ${\displaystyle \left|X\right|\sim {\textrm {Exponential}}\left(b^{-1}\right)}$ (exponential distribution).
• If ${\displaystyle X,Y\sim {\textrm {Exponential}}(\lambda )}$ then ${\displaystyle X-Y\sim {\textrm {Laplace}}\left(0,\lambda ^{-1}\right)}$．
• If ${\displaystyle X\sim {\textrm {Laplace}}(\mu ,b)}$ then ${\displaystyle \left|X-\mu \right|\sim {\textrm {Exponential}}(b^{-1})}$.
• If ${\displaystyle X\sim {\textrm {Laplace}}(\mu ,b)}$ then ${\displaystyle X\sim {\textrm {EPD}}(\mu ,b,1)}$ (exponential power distribution).
• If ${\displaystyle X_{1},...,X_{4}\sim {\textrm {N}}(0,1)}$ (normal distribution) then ${\displaystyle X_{1}X_{2}-X_{3}X_{4}\sim {\textrm {Laplace}}(0,1)}$ and ${\displaystyle (X_{1}^{2}-X_{2}^{2}+X_{3}^{2}-X_{4}^{2})/2\sim {\textrm {Laplace}}(0,1)}$.
• If ${\displaystyle X_{i}\sim {\textrm {Laplace}}(\mu ,b)}$ then ${\displaystyle {\frac {\displaystyle 2}{b}}\sum _{i=1}^{n}|X_{i}-\mu |\sim \chi ^{2}(2n)}$ (chi-squared distribution).
• If ${\displaystyle X,Y\sim {\textrm {Laplace}}(\mu ,b)}$ then ${\displaystyle {\tfrac {|X-\mu |}{|Y-\mu |}}\sim \operatorname {F} (2,2)}$. (F-distribution)
• If ${\displaystyle X,Y\sim {\textrm {U}}(0,1)}$ (uniform distribution) then ${\displaystyle \log(X/Y)\sim {\textrm {Laplace}}(0,1)}$.
• If ${\displaystyle X\sim {\textrm {Exponential}}(\lambda )}$ and ${\displaystyle Y\sim {\textrm {Bernoulli}}(0.5)}$ (Bernoulli distribution) independent of ${\displaystyle X}$, then ${\displaystyle X(2Y-1)\sim {\textrm {Laplace}}\left(0,\lambda ^{-1}\right)}$.
• If ${\displaystyle X\sim {\textrm {Exponential}}(\lambda )}$ and ${\displaystyle Y\sim {\textrm {Exponential}}(\nu )}$ independent of ${\displaystyle X}$, then ${\displaystyle \lambda X-\nu Y\sim {\textrm {Laplace}}(0,1)}$．
• If ${\displaystyle X}$ has a Rademacher distribution and ${\displaystyle Y\sim {\textrm {Exponential}}(\lambda )}$ then ${\displaystyle XY\sim {\textrm {Laplace}}(0,1/\lambda )}$.
• If ${\displaystyle V\sim {\textrm {Exponential}}(1)}$ and ${\displaystyle Z\sim N(0,1)}$ independent of ${\displaystyle V}$, then ${\displaystyle X=\mu +b{\sqrt {2V}}Z\sim \mathrm {Laplace} (\mu ,b)}$.
• If ${\displaystyle X\sim {\textrm {GeometricStable}}(2,0,\lambda ,0)}$ (geometric stable distribution) then ${\displaystyle X\sim {\textrm {Laplace}}(0,\lambda )}$.
• The Laplace distribution is a limiting case of the hyperbolic distribution.
• If ${\displaystyle X|Y\sim {\textrm {N}}(\mu ,Y^{2})}$ with ${\displaystyle Y\sim {\textrm {Rayleigh}}(b)}$ (Rayleigh distribution) then ${\displaystyle X\sim {\textrm {Laplace}}(\mu ,b)}$. Note that if ${\displaystyle Y\sim {\textrm {Rayleigh}}(b)}$, then ${\displaystyle Y^{2}\sim {\textrm {Gamma}}(1,2b^{2})}$ with ${\displaystyle {\textrm {E}}(Y^{2})=2b^{2}}$, which in turn equals the exponential distribution ${\displaystyle {\textrm {Exp}}(1/(2b^{2}))}$.
• Given an integer ${\displaystyle n\geq 1}$, if ${\displaystyle X_{i},Y_{i}\sim \Gamma \left({\frac {1}{n}},b\right)}$ (gamma distribution, using ${\displaystyle k,\theta }$ characterization), then ${\displaystyle \sum _{i=1}^{n}\left({\frac {\mu }{n}}+X_{i}-Y_{i}\right)\sim {\textrm {Laplace}}(\mu ,b)}$ (infinite divisibility)[1]
• If X has a Laplace distribution, then Y = e^X has a log-Laplace distribution; conversely, if X has a log-Laplace distribution, then its logarithm has a Laplace distribution.

Let ${\displaystyle X,Y}$ be independent laplace random variables: ${\displaystyle X\sim {\textrm {Laplace}}(\mu _{X},b_{X})}$ and ${\displaystyle Y\sim {\textrm {Laplace}}(\mu _{Y},b_{Y})}$, and we want to compute ${\displaystyle P(X>Y)}$.

The probability of ${\displaystyle P(X>Y)}$ can be reduced (using the properties below) to ${\displaystyle P(\mu +bZ_{1}>Z_{2})}$, where ${\displaystyle Z_{1},Z_{2}\sim {\textrm {Laplace}}(0,1)}$. This probability is equal to

${\displaystyle P(\mu +bZ_{1}>Z_{2})={\begin{cases}{\frac {b^{2}e^{\mu /b}-e^{\mu }}{2(b^{2}-1)}},&{\text{when }}\mu <0\\1-{\frac {b^{2}e^{-\mu /b}-e^{-\mu }}{2(b^{2}-1)}},&{\text{when }}\mu >0\\\end{cases}}}$

When ${\displaystyle b=1}$, both expressions are replaced by their limit as ${\displaystyle b\to 1}$:

${\displaystyle P(\mu +Z_{1}>Z_{2})={\begin{cases}e^{\mu }{\frac {(2-\mu )}{4}},&{\text{when }}\mu <0\\1-e^{-\mu }{\frac {(2+\mu )}{4}},&{\text{when }}\mu >0\\\end{cases}}}$

To compute the case for ${\displaystyle \mu >0}$, note that ${\displaystyle P(\mu +Z_{1}>Z_{2})=1-P(\mu +Z_{1}<Z_{2})=1-P(-\mu -Z_{1}>-Z_{2})=1-P(-\mu +Z_{1}>Z_{2})}$

since ${\displaystyle Z\sim -Z}$ when ${\displaystyle Z\sim {\textrm {Laplace}}(0,1)}$

A Laplace random variable can be represented as the difference of two independent and identically distributed (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 ${\displaystyle p=0}$.