Laplace Distribution
A continuous random variable $X$ is said to have a Laplace distribution (Double exponential distribution or bilateral exponential distribution), if its p.d.f. is given by
$$ \begin{align*} f(x;\mu, \lambda)&= \begin{cases} \frac{\lambda}{2}e^{-\lambda|x-\mu|}, & -\infty < x< \infty; \\ & -\infty < \mu < \infty, \lambda >0; \\ 0, & Otherwise. \end{cases} \end{align*} $$
In Laplace distribution $\mu$ is called location parameter, since it locates the curve of the distribution, and $\lambda$ is called scale parameter, since the shape of the curve depends on the value of $\lambda$.
$$ \begin{eqnarray*} \int_{-\infty}^\infty f(x; \mu, \lambda) \; dx &=& \int_{-\infty}^\mu f(x; \mu, \lambda) \; dx + \int_\mu^\infty f(x; \mu, \lambda)\; dx \\ &=& \int_{-\infty}^\mu \frac{\lambda}{2}e^{-\lambda(\mu-x)} \; dx + \int_\mu^\infty \frac{\lambda}{2}e^{-\lambda(x-\mu)}\; dx \\ \\ &=& \frac{\lambda}{2}\bigg[\frac{e^{-\lambda(x-\mu)}}{\lambda}\bigg]_{-\infty}^\mu+ \frac{\lambda}{2}\bigg[\frac{e^{-\lambda(x-\mu)}}{-\lambda}\bigg]_\mu^{-\infty}\\ &=& \frac{1}{2} + \frac{1}{2}=1. \end{eqnarray*} $$
The function $f(x)$ defined above is a legitimate p.d.f.
In notation, $X\sim L(\mu, \lambda)$. The standard form of Laplace distribution is obtained by taking $\mu=0$ and $\lambda =1$. The p.d.f of standard Laplace distribution is
$$ \begin{align*} f(x)&= \begin{cases} \frac{1}{2}e^{-|x|}, & -\infty < x< \infty; \\ & \\ 0, & Otherwise. \end{cases} \end{align*} $$
Graph of Laplace Distribution
Distribution Function of Laplace Distribution
Distribution function is given by
$$ \begin{align*} F(x) &= \begin{cases} \frac{1}{2}e^{\lambda(x-\mu)}, & x< \mu; \\ &\\ 1-\frac{1}{2}e^{-\lambda(x-\mu)}, & x\geq \mu; \end{cases} \end{align*} $$
Proof
The distribution function of Laplace distribution is given by
$$ \begin{eqnarray*} F(x) &=& P(X\leq x) \\ &=& \int_{-\infty}^x f(t) \; dt\\ &=& \int_{-\infty}^x \frac{\lambda}{2}e^{-\lambda|t-\mu|} \; dt \end{eqnarray*} $$
There are two cases (I) when $x< \mu$ and (II) when $x\geq \mu$.
Case I:
When $x< \mu$, $\Rightarrow x-\mu < 0 \Rightarrow |x-\mu| = -(x-\mu)$,
$$ \begin{eqnarray*} \therefore F(x) &=& \int_{-\infty}^x \frac{\lambda}{2} e^{\lambda(t-\mu)}\; dt \\ &=& \frac{\lambda}{2} \bigg[\frac{1}{\lambda}e^{\lambda(t-\mu)}\bigg]_{-\infty}^x\\ &=& \frac{1}{2} e^{\lambda(x-\mu)},\qquad x<\mu. \end{eqnarray*} $$
Case II:
When $x\geq \mu$, $\Rightarrow x-\mu > 0 \Rightarrow |x-\mu| = (x-\mu)$,
$$ \begin{eqnarray*} \therefore F(x) &=& \int_{-\infty}^x \frac{\lambda}{2} e^{-\lambda(t-\mu)}\; dt \\ &=& \frac{\lambda}{2}\int_{-\infty}^\mu \frac{\lambda}{2} e^{\lambda(t-\mu)}\; dt+ \frac{\lambda}{2}\int_{\mu}^x \frac{\lambda}{2} e^{-\lambda(t-\mu)}\; dt \\ &=& \frac{\lambda}{2} \bigg[\frac{1}{\lambda}e^{\lambda(t-\mu)}\bigg]_{-\infty}^\mu+ \frac{\lambda}{2} \bigg[\frac{(-1)}{\lambda}e^{-\lambda(t-\mu)}\bigg]_{\mu}^x\\ &=& \frac{1}{2}[1-0]+\frac{1}{2} \big(-e^{-\lambda(x-\mu)}+1\big)\\ & =& 1- \frac{1}{2}e^{-\lambda(x-\mu)},\qquad x \geq \mu. \end{eqnarray*} $$
Therefore, the distribution function of Laplace distribution is
$$ \begin{align*} F(x) &= \begin{cases} \frac{1}{2}e^{\lambda(x-\mu)}, & x< \mu; \\ &\\ 1-\frac{1}{2}e^{-\lambda(x-\mu)}, & x\geq \mu; \end{cases} \end{align*} $$
Mean and Variance of Laplace Distribution
The mean and variance of Laplace distribution are $\mu$ and $\frac{2}{\lambda^2}$ respectively.
Proof
The mean of Laplace distribution is
$$ \begin{eqnarray*} \text{Mean = } E(X) &=& \int_{-\infty}^\infty xf(x)\; dx \\ &=& \int_{-\infty}^\infty (x-\mu+\mu)f(x)\; dx \\ &=& \int_{-\infty}^\infty (x-\mu)f(x)\; dx+\int_{-\infty}^\infty \mu f(x)\; dx \\ &=& I_1+\mu I_2, \end{eqnarray*} $$
In $I_1$, put $(x-\mu)=y$, therefore $dx=dy$,
$$ \begin{eqnarray*} \therefore I_1 & = & \frac{\lambda}{2}\int_{-\infty}^\infty y e^{-\lambda|y|}\; dy = 0,\qquad \text{ ($\because$ integrand is an odd function)}\\ \text{ and } I_2 & = & \int_{-\infty}^\infty f(x) dx =1. \end{eqnarray*} $$
Therefore, $E(X) = 0 + \mu \cdot 1 = \mu$. Hence, mean of Laplace
distribution is $\mu$.
Variance :
To find variance of $X$ we need to find $E(X^2)$.
$$ \begin{eqnarray*} E(X^2) &=& \int_{-\infty}^\infty x^2f(x)\; dx \\ &=& \int_{-\infty}^\infty x^2\frac{\lambda}{2} e^{-\lambda|x-\mu|}\; dx. \end{eqnarray*} $$
Let $z=\lambda(x-\mu)$, $\Rightarrow x = \mu+\dfrac{z}{\lambda}$. So, $dz = \lambda dx$. Hence,
$$ \begin{eqnarray*} E(X^2) & = & \frac{1}{2}\int_{-\infty}^\infty \bigg(\mu +\frac{z}{\lambda}\bigg)^2 e^{-|z|}\; dz.\\ & = &\frac{1}{2} \int_{-\infty}^\infty \bigg(\mu^2 +\frac{2\mu z}{\lambda} + \frac{z^2}{\lambda^2}\bigg) e^{-|z|}\; dz.\\ & = &\mu^2 \int_{-\infty}^\infty \frac{1}{2}e^{-|z|}\; dz +\frac{\mu}{\lambda} \int_{-\infty}^\infty z e^{-|z|}\; dz+\frac{1}{2\lambda^2}\int_{-\infty}^\infty z^2 e^{-|z|}\; dz.\\ & = & \mu^2 (1) + 0 + \frac{1}{2\lambda^2}\int_{-\infty}^\infty z^2 e^{-|z|}\; dz.\\ &=& \mu^2 +\frac{2}{2\lambda^2}\int_0^\infty z^2 e^{-z}\;dz\\ &=& \mu^2 +\frac{1}{\lambda^2}\frac{\Gamma 3}{1^3}\\ &=& \mu^2 +\frac{2}{\lambda^2}. \end{eqnarray*} $$
Thus,
$$ \begin{align*} V(X) &=E[X^2]-[E(X)]^2\\ &= \mu^2 -\mu^2+\frac{2}{\lambda^2}\\ &=\frac{2}{\lambda^2}. \end{align*} $$
Quartiles of Laplace Distribution
The quartiles of Laplace distribution are $Q_1 = \mu + \dfrac{1}{\lambda}\log_e(0.5)$, $Q_2 = \mu$ and $Q_3 = \mu - \dfrac{1}{\lambda}\log_e(0.5)$.
Proof
The distribution function of Laplace distribution is
$$ \begin{align*} F(x) &= \begin{cases} \frac{1}{2}e^{\lambda(x-\mu)}, & x< \mu; \\ &\\ 1-\frac{1}{2}e^{-\lambda(x-\mu)}, & x\geq \mu; \end{cases} \end{align*} $$
We have,
$$ \begin{equation*} F(Q_i) = P(X\leq Q_i) = \frac{i}{4},\quad i=1,2,3. \end{equation*} $$
For first quartile, $Q_1<\mu$,
$$ \begin{eqnarray*} F(Q_1) & = & \frac{1}{2}e^{\lambda(Q_1-\mu)}=\frac{1}{4}\\ \Rightarrow& & e^{\lambda(Q_1-\mu)}=\frac{1}{2} \\ \Rightarrow & &\lambda(Q_1-\mu)=\log_e(0.5)\\ \Rightarrow & & Q_1 = \mu + \frac{1}{\lambda}\log_e(0.5). \end{eqnarray*} $$
For second quartile, $Q_2\geq\mu$,
$$ \begin{eqnarray*} F(Q_2) & = & 1-\frac{1}{2}e^{-\lambda(Q_2-\mu)}=\frac{1}{2}\\ \Rightarrow& & e^{-\lambda(Q_2-\mu)}=1\\ \Rightarrow & & -\lambda(Q_2-\mu)=0\\ \Rightarrow & & Q_2 = \mu. \end{eqnarray*} $$
For third quartile, $Q_3\geq\mu$,
$$ \begin{eqnarray*} F(Q_3) & = & 1-\frac{1}{2}e^{-\lambda(Q_3-\mu)}=\frac{3}{4}\\ \Rightarrow& & \frac{1}{2}e^{-\lambda(Q_3-\mu)}=\frac{1}{4}\\ \Rightarrow & &-\lambda(Q_3-\mu)=\log_e(0.5)\\ \Rightarrow & & Q_3 = \mu-\frac{1}{\lambda}\log_e(0.5). \end{eqnarray*} $$
M.G.F. of Laplace Distribution
The moment generating function of Laplace distribution is
$$M_X(t)=e^{t\mu}\sum_{r=0}^\infty\bigg(\frac{t^2}{\lambda^2}\bigg)^r.$$
Proof
The moment generating function of Laplace distribution is
$$ \begin{eqnarray*} M_X(t) &=& E(e^{tX}) \\ &=& \int_{-\infty}^\infty e^{tx}f(x) \; dx\\ &=& \int_{-\infty}^\infty e^{t(x-\mu+\mu)}f(x) \; dx\\ &=& \frac{\lambda}{2}e^{t\mu}\int_{-\infty}^\infty e^{t(x-\mu)}e^{-\lambda|x-\mu|} \; dx \end{eqnarray*} $$
Let $(x-\mu) =y$, therefore $dx=dy$.
$$ \begin{eqnarray*} M_X(t) &=& \frac{\lambda}{2}e^{t\mu}\int_{-\infty}^\infty e^{ty}e^{-\lambda|y|} \; dy\\ &=& \frac{\lambda}{2}e^{t\mu}\bigg[\int_{-\infty}^0 e^{ty}e^{\lambda y} \;dy +\int_0^{\infty} e^{ty}e^{-\lambda y}\; dy\bigg]\\ &=& \frac{\lambda}{2}e^{t\mu}\bigg[\int_{-\infty}^0 e^{(t+\lambda)y}\;dy +\int_0^{\infty} e^{-(\lambda -t)y}\; dy \bigg]\\ &=& \frac{\lambda}{2}e^{t\mu}\bigg\{\bigg[\frac{e^{(t+\lambda)y}}{t+\lambda}\bigg]_{-\infty}^0+\bigg[\frac{e^{-(\lambda-t)y}}{-(\lambda-t)}\bigg]_0^{\infty}\bigg\}\\ &=& \frac{\lambda}{2}e^{t\mu}\bigg\{\bigg[\frac{1}{t+\lambda}-0\bigg]+\bigg[0-\frac{1}{t-\lambda}\bigg]\bigg\}\\ &=& \frac{\lambda}{2}e^{t\mu}\bigg[\frac{-2\lambda}{t^2-\lambda^2}\bigg]\\ &=& e^{t\mu}\bigg[\frac{\lambda^2}{\lambda^2-t^2}\bigg]\\ &=& e^{t\mu}\bigg[1-\frac{t^2}{\lambda^2}\bigg]^{-1}\\ &=& e^{t\mu}\sum_{r=0}^\infty \bigg(\frac{t^2}{\lambda^2}\bigg)^r. \end{eqnarray*} $$
C.G.F. of Laplace Distribution
The cumulant generating function of Laplace distribution $L(\mu,\lambda)$ is $K_X(t)=\mu t - \log_e \bigg(1-\frac{t^2}{r^2}\bigg)$.
Proof
The cumulant generating function of Laplace distribution is
$$ \begin{eqnarray*} K_X(t) &=& \log_e M_X(t) \\ &=& \log_e \bigg[e^{t\mu} \bigg(1-\frac{t^2}{r^2}\bigg)^{-1} \bigg]\\ &=& \mu t-\log_e \bigg(1-\frac{t^2}{r^2}\bigg)\\ &=& \mu t +\bigg(\frac{t^2}{\lambda^2} + \frac{t^4}{2\lambda^4}+\cdots\bigg)\\ & & \qquad (\because \log (1-x) = -(x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots). \end{eqnarray*} $$
Cumulants
The first cumulant of Laplace distribution is
$$ \begin{eqnarray*} \kappa_1 =\mu_1^\prime &=& \text{coefficient of $t$ in the expansion of $K_X(t)$}= \mu. \end{eqnarray*} $$
The second cumulant of Laplace distribution is
$$ \begin{eqnarray*} \kappa_2 =\mu_2 &=& \text{coefficient of $\frac{t^2}{2!}$ in the expansion of $K_X(t)$} \\ &=& \frac{2}{\lambda^2}=\text{ variance }. \end{eqnarray*} $$
The third cumulant of Laplace distribution is
$$ \begin{eqnarray*} \kappa_3 =\mu_3 &=& \text{coefficient of $\frac{t^3}{3!}$ in the expansion of $K_X(t)$} \\ &=& 0. \end{eqnarray*} $$
The fourth cumulant of Laplace distribution is
$$ \begin{eqnarray*} \kappa_4 =\mu_4-3\mu_2^2&=& \text{coefficient of $\frac{t^4}{4!}$ in the expansion of $K_X(t)$} \\ &=& \frac{12}{\lambda^4}. \end{eqnarray*} $$
Hence
$$ \begin{eqnarray*} \mu_4 & = & \kappa_4 + 3\kappa_2^2 =\frac{12}{\lambda^4}+\frac{12}{\lambda^4}=\frac{24}{\lambda^4}. \end{eqnarray*} $$
In general, the $r^{th}$ cumulant is given by
$$ \begin{align*} \kappa_r&= \begin{cases} 0, & \text{if } r \text{ is odd except 1;} \\ &\\ \dfrac{2(r-1)!}{\lambda^r}, & \text{ if } r \text{ is even} \end{cases} \end{align*} $$
Characteristics Function
The characteristics function of Laplace distribution $L(\mu,\lambda)$ is
$$ \begin{equation*} \phi_X(t) = e^{it\mu}\bigg(1+\frac{t^2}{\lambda^2}\bigg)^{-1}. \end{equation*} $$
Proof
Let $X\sim L(\mu,\lambda)$ distribution. Then the M.G.F. of Laplace distribution is
$$ \begin{equation*} M_X(t) = e^{t\mu} \bigg(1- \frac{t^2}{\lambda^2}\bigg)^{-1}. \end{equation*} $$
The characteristics function of $X$ is
$$ \begin{eqnarray*} \phi_{X}(t) &=& E(e^{itX}) \\ &=& M_X(it)\\ &=& e^{it\mu} \bigg(1- \frac{(it)^2}{\lambda^2}\bigg)^{-1}\\ &=& e^{it\mu} \bigg(1- \frac{i^2t^2}{\lambda^2}\bigg)^{-1}\\ &=& e^{it\mu} \bigg(1+ \frac{t^2}{\lambda^2}\bigg)^{-1}. \end{eqnarray*} $$
If $X\sim L(0,1)$, then the characteristics function of Standard Laplace distribution is
$$ \begin{equation*} \phi_X(t)=(1+t^2)^{-1} =\frac{1}{1+t^2}. \end{equation*} $$
$r^{th}$ moment of Laplace Distribution
The $r^{th}$ moment of Laplace Distribution is
$$ \begin{equation*} \mu_r^\prime =\frac{1}{2} \sum_{k=0}^r\binom{r}{k} \mu^{r-k} \frac{k!}{\lambda^k} ((-1)^k +1). \end{equation*} $$
Proof
Let $X\sim L(\mu,\lambda)$. Then the pdf of $X$ is
$$ \begin{equation*} f(x) =\frac{\lambda}{2}e^{-\lambda |x-\mu|}, -\infty < x < \infty, -\infty < \mu < \infty, \lambda>0. \end{equation*} $$
The $r^{th}$ moment of Laplace distribution is
$$ \begin{eqnarray*} \mu_r^\prime &=& E(X^r) \\ &=& \frac{\lambda}{2}\int_{-\infty}^\infty x^r e^{-\lambda|x-\mu|}\; dx. \end{eqnarray*} $$
Let $Z =\lambda (X-\mu)$, $x = \mu +\frac{z}{\lambda}$. Thus $dx = \frac{1}{\lambda} dz$.
$$ \begin{eqnarray*} \mu_r^\prime &=& \frac{\lambda}{2} \int_{-\infty}^\infty (\mu+\frac{z}{\lambda})^r e^{-|z|} \frac{1}{\lambda}\; dz\\ &=& \frac{1}{2}\int_{-\infty}^\infty \sum_{k=0}^r\binom{r}{k} \bigg(\frac{z}{\lambda}\bigg)^k \mu^{r-k} e^{-|z|}\; dz\\ &=& \frac{1}{2} \sum_{k=0}^r\binom{r}{k} \mu^{r-k} \frac{1}{\lambda^k} \int_{-\infty}^\infty z^k e^{-|z|}\;dz\\ &=& \frac{1}{2} \sum_{k=0}^r\binom{r}{k} \mu^{r-k} \frac{1}{\lambda^k} \bigg[(-1)^k \int_{0}^\infty z^k e^{-z}\;dz+\int_{0}^\infty z^k e^{-z}\;dz\bigg]\\ &=&\frac{1}{2} \sum_{k=0}^r\binom{r}{k} \mu^{r-k} \frac{1}{\lambda^k} \bigg[(-1)^k \Gamma(k+1) +\Gamma (k+1)\bigg]\\ &=&\frac{1}{2} \sum_{k=0}^r\binom{r}{k} \mu^{r-k} \frac{k!}{\lambda^k} ((-1)^k +1). \end{eqnarray*} $$
Result
Let $X$ and $Y$ be two i.i.d. exponential variate with parameter $\theta$. Obtain the distribution of $(X-Y)$.
Solution
$X$ and $Y$ be two i.i.d. exponential variate with parameter $\theta$. Then p.d.f. of $x$ is
$$ \begin{equation*} f_X(x) = \theta e^{-\theta x}; \quad x> 0, \theta>0 \end{equation*} $$
and p.d.f. of $Y$ is
$$ \begin{equation*} f_Y(y) = \theta e^{-\theta y}; \quad y> 0, \theta>0 \end{equation*} $$
The m.g. f. of $X$ is $M_X(t) = \bigg(1-\dfrac{t}{\theta}\bigg)^{-1}$. Let $U=X-Y$. Then the m.g.f. of $U$ is
$$ \begin{eqnarray*} M_U(t) = E(e^{tU}) &=& E(e^{t(X_Y)}\\ &=& E(e^{tX})\cdot E(e^{-tY})\\ &=& M_X(t)\cdot M_Y(-t)\\ &=& \bigg(1-\frac{t}{\theta}\bigg)^{-1}\cdot \bigg(1+\frac{t}{\theta}\bigg)^{-1}\\ &=&\bigg(1-\frac{t^2}{\theta^2}\bigg)^{-1}. \end{eqnarray*} $$
which is the m.g.f. of Laplace distribution with $\mu= 0$ and $\lambda = \theta$. Hence, by uniqueness property of M.G.F., $U=X-Y \sim L(0,\theta)$ distribution.
Conclusion
In this tutorial, you learned about theory of Laplace distribution like the probability density function, mean, median, median, quartiles, moment generating function, characteristics function and other properties of Laplace distribution.
To read more about the step by step examples and calculator for Laplace distribution refer the link Laplace Distribution Calculator with Examples. This tutorial will help you to understand how to calculate mean, variance of Laplace distribution and you will learn how to calculate probabilities and cumulative probabilities for Laplace distribution with the help of step by step examples.
To learn more about other probability distributions, please refer to the following tutorial:
Let me know in the comments if you have any questions on Laplace Distribution and your thought on this article.
