The Laplace Distribution

The Laplace distribution, named for Pierre Simon Laplace arises naturally as the distribution of the difference of two independent, identically distributed exponential variables. For this reason, it is also called the double exponential distribution.

The Standard Laplace Distribution

Distribution Functions

The standard Laplace distribution is a continuous distribution on $R$ with probability density function $g$ given by $g (u) = \frac{1}{2} e^{- | u |}, u \in R$

Details:

It's easy to see that $g$ is a valid PDF. By symmetry $\int_{- \infty}^{\infty} \frac{1}{2} e^{- | u |} d u = \int_{0}^{\infty} e^{- u} d u = 1$

The probability density function $g$ satisfies the following properties:

$g$ is symmetric about 0.
$g$ increases on $(- \infty, 0]$ and decreases on $[0, \infty)$ , with mode $u = 0$ .
$g$ is concave upward on $(- \infty, 0]$ and on $[0, \infty)$ with a cusp at $u = 0$

Details:

These results follow from standard calculus, since $g (u) = \frac{1}{2} e^{- u}$ for $u \in [0, \infty)$ and $g (u) = \frac{1}{2} e^{u}$ for $u \in (- \infty, 0]$ .

Open the special distribution simulator and select the Laplace distribution. Keep the default parameter value and note the shape of the probability density function. Run the simulation 1000 times and compare the emprical density function and the probability density function.

The standard Laplace distribution function $G$ is given by $G (u) = {\begin{cases} \frac{1}{2} e^{u}, & u \in (- \infty, 0] \\ 1 - \frac{1}{2} e^{- u}, & u \in [0, \infty) \end{cases}$

Details:

Again this follows from basic calculus, since $g (u) = \frac{1}{2} e^{u}$ for $u \leq 0$ and $g (u) = \frac{1}{2} e^{- u}$ for $u \geq 0$ . Of course $G (u) = \int_{- \infty}^{u} g (t) d t$ .

The quantile function $G^{- 1}$ given by $G^{- 1} (p) = {\begin{cases} \ln (2 p), & p \in [0, \frac{1}{2}] \\ - \ln [2 (1 - p)], & p \in [\frac{1}{2}, 1] \end{cases}$

$G^{- 1} (1 - p) = - G^{- 1} (p)$ for $p \in (0, 1)$
The first quartile is $q_{1} = - \ln 2 \approx - 0.6931$ .
The median is $q_{2} = 0$
The third quartile is $q_{3} = \ln 2 \approx 0.6931$ .

Details:

The formula for the quantile function follows immediately from [4] by solving $p = G (u)$ for $u$ in terms of $p \in (0, 1)$ . Part (a) is due to the symmetry of $g$ about 0.

Open the quantile app and select the Laplace distribution. Keep the default parameter value for the standard Laplace distribution and note the shape of the distribution function. Compute the quantiles of order 0.1 and 0.9.

Moments

$U$ has moment generating function $m$ given by $m (t) = E (e^{t U}) = \frac{1}{1 - t^{2}}, t \in (- 1, 1)$

Details:

For $t \in (- 1, 1)$ , $m (t) = \int_{- \infty}^{\infty} e^{t u} g (u) d u = \int_{- \infty}^{0} \frac{1}{2} e^{(t + 1) u} d u + \int_{0}^{\infty} \frac{1}{2} e^{(t - 1) u} d u = \frac{1}{2 (t + 1)} - \frac{1}{2 (t - 1)} = \frac{1}{1 - t^{2}}$

The moments of $U$ are

$E (U^{n}) = 0$ if $n \in N$ is odd.
$E (U^{n}) = n!$ if $n \in N$ is even.

Details:

This result can be obtained from the moment generating function in [7] or directly. That the odd order moments are 0 follows from the symmetry of the distribution. For the even order moments, symmetry and an integration by parts (or using the gamma function) gives $E (U^{n}) = \frac{1}{2} \int_{- \infty}^{0} u^{n} e^{u} d u + \frac{1}{2} \int_{0}^{\infty} u^{n} e^{- u} d u = \int_{0}^{\infty} u^{n} e^{- u} d u = n!$

The mean and variance of $U$ are

$E (U) = 0$
$var (U) = 2$

Open the special distribution simulator and select the Laplace distribution. Keep the default parameter value for the standard Laplace disstribution. Run the simulation 1000 times and compare the empirical mean and standard deviation to the distribution mean and standard deviation.

The skewness and kurtosis of $U$ are

$skew (U) = 0$
$kurt (U) = 6$

Details:

This follows from the symmetry of the distribution.
Since $E (U) = 0$ , we have $kurt (U) = \frac{E (U^{4})}{[E (U^{2})]^{2}} = \frac{4!}{(2!)^{2}} = 6$

Related Distributions

If $U$ has the standard Laplace distribution then $V = | U |$ has the standard exponential distribution.

Details:

From [4] we have $P (V \leq v) = P (- v \leq U \leq v) = G (v) - G (- v) = 1 - e^{- v}$ for $v \in [0, \infty)$ . This function is the CDF of the standard exponential distribution.

If $V$ and $W$ are independent and each has the standard exponential distribution, then $U = V - W$ has the standard Laplace distribution.

Details:

We give two proofs.

Our first proof uses PDFs. Let $h$ denote the standard exponential PDF, extended to all of $R$ , so that $h (v) = e^{- v}$ if $v \geq 0$ and $h (v) = 0$ if $v < 0$ . Using convolution, the PDF of $V - W$ is $g (u) = \int_{R} h (v) h (v - u) d v$ . If $v \geq 0$ , $g (u) = \int_{u}^{\infty} e^{- v} e^{- (v - u)} d v = e^{u} \int_{u}^{\infty} e^{- 2 v} d v = \frac{1}{2} e^{- u}$ If $u < 0$ then $g (u) = \int_{0}^{\infty} e^{- v} e^{- (v - u)} = e^{u} \int_{0}^{\infty} e^{- 2 v} d v = \frac{1}{2} e^{u}$
Our second proofs uses MGFs. The MGF of $V$ is $t \mapsto 1 / (1 - t)$ for $t < 1$ . The MGF of $- W$ is $t \mapsto 1 / (1 + t)$ for $t > - 1$ . Hence the MGF of $U$ is $t \mapsto 1 / (1 - t) (1 + t) = 1 / (1 - t^{2})$ for $- 1 < t < 1$ , which is the standard Laplace MGF in [7].

If $V$ has the standard exponential distribution, $I$ has the standard Bernoulli distribution, and $V$ and $I$ are independent, then $U = (2 I - 1) V$ has the standard Laplace distribution.

Details:

If $u \geq 0$ then $P (U \leq u) = P (I = 0) + P (I = 1, V \leq u) = P (I = 0) + P (I = 1) P (V \leq u) = \frac{1}{2} + \frac{1}{2} (1 - e^{- u}) = 1 - \frac{1}{2} e^{- u}$ If $u < 0$ , $P (U \leq u) = P (I = 0, V > - u) = P (I = 0) P (V > - u) = \frac{1}{2} e^{u}$

Suppose that $(Z_{1}, Z_{2}, Z_{3}, Z_{4})$ is a random sample of size 4 from the standard normal distribution. Then $U = Z_{1} Z_{2} + Z_{3} Z_{4}$ has the standard Laplace distribution.

Details:

$Z_{1} Z_{2}$ and $Z_{3} Z_{4}$ are independent, and each has a distribution known as the product normal distribution. The MGF of this distribution is $m_{0} (t) = E (e^{t Z_{1} Z_{2}}) = \int_{R^{2}} e^{t x y} \frac{1}{2 π} e^{- (x^{2} + y^{2}) / 2} d (x, y)$ Changing to polar coordinates gives $m_{0} (t) = \frac{1}{2 π} \int_{0}^{2 π} \int_{0}^{\infty} e^{t r^{2} \cos θ \sin θ} e^{- r^{2} / 2} r d r d θ = \frac{1}{2 π} \int_{0}^{2 π} \int_{0}^{\infty} \exp [r^{2} (t \cos θ \sin θ - \frac{1}{2})] r d r d θ$ The inside integral can be done with a simple substitution for $| t | < 1$ , yielding $m_{0} (t) = \frac{1}{2 π} \int_{0}^{2 π} \frac{1}{1 - t \sin (2 θ)} d θ = \frac{1}{\sqrt{1 - t^{2}}}$ Hence $U$ has MGF $m_{0}^{2} (t) = \frac{1}{1 - t^{2}}$ for $| t | < 1$ , which again is the standard Laplace MGF in [7].

The standard Laplace distribution has the usual connections to the standard uniform distribution by means of the distribution function [4] and the quantile function [5].

Connections to the standard uniform distribution.

If $V$ has the standard uniform distribution then $U = \ln (2 V) 1 (V < \frac{1}{2}) - \ln [2 (1 - V)] 1 (V \geq \frac{1}{2})$ has the standard Laplace distribution.
If $U$ has the standard Laplace distribution then $V = \frac{1}{2} e^{U} 1 (U < 0) + (1 - \frac{1}{2} e^{- U}) 1 (U \geq 0)$ has the standard uniform distribution.

From part (a), the standard Laplace distribution can be simulated with the usual random quantile method.

Open the random quantile app and select the Laplace distribution. Keep the default parameter values and note the shape of the probability density and distribution functions. Run the simulation 1000 times and compare the empirical density function, mean, and standard deviation to their distributional counterparts.

The General Laplace Distribution

Suppose that $U$ has the standard Laplace distribution. If $a \in R$ and $b \in (0, \infty)$ , then $X = a + b U$ has the Laplace distribution with location parameter $a$ and scale parameter $b$ .

Distribution Functions

Suppos that

X

has the Laplace distribution with location parameter

a \in R

and scale parameter

b \in (0, \infty)

$X$ has probability density function $f$ given by $f (x) = \frac{1}{2 b} \exp (- \frac{| x - a |}{b}), x \in R$

$f$ is symmetric about $a$ .
$f$ increases on $[0, a]$ and decreases on $[a, \infty)$ with mode $x = a$ .
$f$ is concave upward on $[0, a]$ and on $[a, \infty)$ with a cusp at $x = a$ .

Details:

Recall that $f (x) = \frac{1}{b} g (\frac{x - a}{b})$ where $g$ is the standard Laplace PDF in [1].

Open the special distribution simulator and select the Laplace distribution. Vary the parameters and note the shape and location of the probability density function. For various values of the parameters, run the simulation 1000 times and compare the emprical density function to the probability density function.

$X$ has distribution function $F$ given by $F (x) = {\begin{cases} \frac{1}{2} \exp (\frac{x - a}{b}), & x \in (- \infty, a] \\ 1 - \frac{1}{2} \exp (- \frac{x - a}{b}), & x \in [a, \infty) \end{cases}$

Details:

Recall that $F (x) = G (\frac{x - a}{b})$ where $G$ is the standard Laplace CDF in [4].

$X$ has quantile function $F^{- 1}$ given by $F^{- 1} (p) = {\begin{cases} a + b \ln (2 p), & 0 \leq p \leq \frac{1}{2} \\ a - b \ln [2 (1 - p)], & \frac{1}{2} \leq p < 1 \end{cases}$

$F^{- 1} (1 - p) = a - b F^{- 1} (p)$ for $p \in (0, 1)$
The first quartile is $q_{1} = a - b \ln 2$ .
The median is $q_{2} = a$
The third quartile is $q_{3} = a + b \ln 2$ .

Details:

Recall that $F^{- 1} (p) = a + b G^{- 1} (p)$ where $G^{- 1}$ is the standard Laplace quantile function in [5].

Open the quantile app and select the Laplace distribution. Vary the parameters and note the shape of the distribution function. For selected values of the parameters, compute the quantiles of order 0.1 and 0.9.

Moments

Again, we assume that

X

has the Laplace distribution with location parameter

a \in R

and scale parameter

b \in (0, \infty)

, so that by definition [18],

X = a + b U

where

U

has the standard Lapalce distribution.

$X$ has moment generating function $M$ given by $M (t) = E (e^{t X}) = \frac{e^{a t}}{1 - b^{2} t^{2}}, t \in (- 1 / b, 1 / b)$

Details:

Recall that $M (t) = e^{a t} m (b t)$ where $m$ is the standard Laplce MGF in [7].

The moments of $X$ about $a$ are

$E [(X - a)^{n}] = 0$ if $n \in N$ is odd.
$E [(X - a)^{n}] = b^{n} n!$ if $n \in N$ is even.

Details:

Note that $E [(X - a)^{n}] = b^{n} E (U^{n})$ so the results follow from [8].

The mean and variance of $X$ are

$E (X) = a$
$var (X) = 2 b^{2}$

Details:

Recall that $E (X) = a + b E (U)$ and $var (X) = b^{2} var (U)$ , so the results follow from [9].

Open the special distribution simulator and select the Laplace distribution. Vary the parameters and note the size and location of the mean $\pm$ standard deviation bar. For various values of the parameters, run the simulation 1000 times and compare the empirical mean and standard deviation to the distribution mean and standard deviation.

The skewness and kurtosis of $X$ are

$skew (X) = 0$
$kurt (X) = 6$

Details:

Recall that skewness and kurtosis are defined in terms of the standard score, and hence are unchanged by a location-scale transformation. Thus the results from .

Related Distributions

By construction, the Laplace distribution is a location-scale family, and so is closed under location-scale transformations.

Suppose that $X$ has the Laplace distribution with location parameter $a \in R$ and scale parameter $b \in (0, \infty)$ , and that $c \in R$ and $d \in (0, \infty)$ . Then $Y = c + d X$ has the Laplace distribution with location parameter $c + a d$ scale parameter $b d$ .

Details:

Again by definition [18], we can take $X = a + b U$ where $U$ has the standard Laplace distribution. Hence $Y = c + d X = (c + a d) + (b d) U$ .

Once again, the Laplace distribution has the usual connections to the standard uniform distribution by means of the distribution function in [21] and the quantile function in [22]. The latter leads to the usual random quantile method of simulation.

Suppose that $a \in R$ and $b \in (0, \infty)$ .

If $V$ has the standard uniform distribution then $U = [a + b \ln (2 V)] 1 (V < \frac{1}{2}) + (a - b \ln [2 (1 - V)]) 1 (V \geq \frac{1}{2})$ has the Laplace distribution with location parameter $a$ and scale parameter $b$ .
If $X$ has the Laplace distribution with location parameter $a$ and scale parameter $b$ , then $V = \frac{1}{2} \exp (\frac{X - a}{b}) 1 (X < a) + [1 - \frac{1}{2} \exp (- \frac{X - a}{b})] 1 (X \geq a)$ has the standard uniform distribution.

Open the random quantile experiment and select the Laplace distribution. Vary the parameters and note the shape of the probability density and distribution functions. For selected values of the parameters, run the simulation 1000 times and compare the empirical density function, mean, and standard deviation to their distributional counterparts.

Suppose that $X$ has the Laplace distribution with known location parameter $a \in R$ and unspecified scale parameter $b \in (0, \infty)$ . Then $X$ has a general exponential distribution in the scale parameter $b$ , with natural parameter $- 1 / b$ and natural statistics $| X - a |$ .

Details:

This follows from the definition of the general exponential family and the form of the probability density function in [19] $f$ .