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 is a continuous distribution on \( \R \) with probability density function \( g \) given by \[ g(u) = \frac{1}{2} e^{-\left|u\right|}, \quad u \in \R \]
It's easy to see that \( g \) is a valid PDF. By symmetry \[ \int_{-\infty}^\infty \frac{1}{2} e^{-\left|u\right|} du = \int_0^\infty e^{-u} du = 1 \]
The probability density function \( g \) satisfies the following properties:
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 \[ G(u) = \begin{cases} \frac{1}{2} e^u, & u \in (-\infty, 0] \\ 1 - \frac{1}{2} e^{-u}, & u \in [0, \infty) \end{cases} \]
Again this follows from basic calculus, since \( g(u) = \frac{1}{2} e^u \) for \( u \le 0 \) and \( g(u) = \frac{1}{2} e^{-u} \) for \( u \ge 0 \). Of course \( G(u) = \int_{-\infty} ^u g(t) \, dt \).
The quantile function \(G^{-1}\) given by \[ G^{-1}(p) = \begin{cases} \ln(2 p), & p \in \left[0, \frac{1}{2}\right] \\ -\ln[2(1 - p)], & p \in \left[\frac{1}{2}, 1\right] \end{cases} \]
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.
Suppose that \( U \) has the standard Laplace distribution.
\(U\) has moment generating function \(m\) given by \[ m(t) = \E\left(e^{t U}\right) = \frac{1}{1 - t^2}, \quad t \in (-1, 1) \]
For \( t \in (-1, 1) \), \[ m(t) = \int_{-\infty}^\infty e^{t u} g(u) \, du = \int_{-\infty}^0 \frac{1}{2} e^{(t + 1)u} du + \int_0^\infty \frac{1}{2} e^{(t - 1)u} du = \frac{1}{2(t + 1)} - \frac{1}{2(t - 1)} = \frac{1}{1 - t^2}\]
The moments of \( U \) are
This result can be obtained from the moment generating function in 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 \[ \E(U^n) = \frac{1}{2} \int_{-\infty}^0 u^n e^u du + \frac{1}{2} \int_0^\infty u^n e^{-u} du = \int_0^\infty u^n e^{-u} du = n! \]
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
It follows that the excess kurtosis is \( \kur(U) - 3 = 3 \).
Of course, the standard Laplace distribution has simple connections to the standard exponential distribution.
If \( U \) has the standard Laplace distribution then \( V = |U| \) has 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.
We give two proofs.
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.
If \( u \ge 0 \) then \[ \P(U \le u) = \P(I = 0) + \P(I = 1, V \le u) = \P(I = 0) + \P(I = 1) \P(V \le u) = \frac{1}{2} + \frac{1}{2}(1 - e^{-u}) = 1 - \frac{1}{2} e^{-u} \] If \( u \lt 0 \), \[ \P(U \le u) = \P(I = 0, V \gt -u) = \P(I = 0) \P(V \gt -u) = \frac{1}{2} e^{u} \]
The standard Laplace distribution has a curious connection to the standard normal distribution.
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.
\( Z_1 Z_2 \) and \( Z_3 Z_4 \) are independent, and each has a distribution known as the product normal distribution. The MGF of this distribution is \[ m_0(t) = \E\left(e^{t Z_1 Z_2}\right) = \int_{\R^2} e^{t x y} \frac{1}{2 \pi} e^{-(x^2 + y^2)/2} d(x, y) \] Changing to polar coordinates gives \[ m_0(t) = \frac{1}{2 \pi} \int_0^{2 \pi} \int_0^\infty e^{t r^2 \cos \theta \sin \theta} e^{-r^2/2} r \, dr \, d\theta = \frac{1}{2 \pi} \int_0^{2 \pi} \int_0^\infty \exp\left[r^2\left(t \cos \theta \sin\theta - \frac{1}{2}\right)\right] r \, dr \, d\theta \] The inside integral can be done with a simple substitution for \( \left|t\right| \lt 1 \), yielding \[ m_0(t) = \frac{1}{2 \pi} \int_0^{2 \pi} \frac{1}{1 - t \sin(2 \theta)} d\theta = \frac{1}{\sqrt{1 - t^2}} \] Hence \( U \) has MGF \( m_0^2(t) = \frac{1}{1 - t^2} \) for \( \left|t\right| \lt 1 \), which again is the standard Laplace MGF in .
The standard Laplace distribution has the usual connections to the standard uniform distribution by means of the distribution function and the quantile function .
Connections to 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 standard Laplace distribution is generalized by adding location and scale parameters.
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 \(b\).
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 \[ f(x) = \frac{1}{2 b} \exp\left(-\frac{\left|x - a\right|}{b}\right), \quad x \in \R \]
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 \[ F(x) = \begin{cases} \frac{1}{2} \exp\left(\frac{x - a}{b}\right), & x \in (-\infty, a] \\ 1 - \frac{1}{2} \exp\left(-\frac{x - a}{b}\right), & x \in [a, \infty) \end{cases} \]
\(X\) has quantile function \(F^{-1}\) given by \[ F^{-1}(p) = \begin{cases} a + b \ln(2 p), & 0 \le p \le \frac{1}{2} \\ a - b \ln[2(1 - p)], & \frac{1}{2} \le p \lt 1 \end{cases} \]
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.
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 , \( X = a + b U \) where \( U \) has the standard Lapalce distribution.
\(X\) has moment generating function \(M\) given by \[ M(t) = \E\left(e^{t X}\right) = \frac{e^{a t}}{1 - b^2 t^2}, \quad t \in (-1/b, 1/b) \]
The moments of \( X \) about the location parameter have a simple form.
The moments of \( X \) about \( a \) are
The mean and variance of \(X\) are
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
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 .
As before, the excess kurtosis is \( \kur(X) - 3 = 3 \).
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 \(b d\).
Once again, the Laplace distribution has the usual connections to the standard uniform distribution by means of the distribution function in and the quantile function in . The latter leads to the usual random quantile method of simulation.
Suppose that \( a \in \R \) and \( b \in (0, \infty) \).
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.
The Laplace distribution is also a member of the general exponential family of distributions.
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 \( \left|X - a\right| \).