Laplace Distribution
A continuous random variable $X$ is said to have a Laplace
distribution, if its p.d.f. is given by
$$ \begin{equation*} f(x;\mu, \lambda)=\left\{ \begin{array}{ll} \frac{1}{2\lambda}e^{-\frac{|x-\mu|}{\lambda}}, & \hbox{$-\infty < x< \infty$;} \\ & \hbox{$-\infty < \mu < \infty$, $\lambda >0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$
Distribution Function
Distribution function of $L(\mu,\lambda)$ distribution is
$$ \begin{equation*} F(x) = \left\{ \begin{array}{ll} \frac{1}{2}e^{\frac{(x-\mu)}{\lambda}}, & \hbox{$x< \mu$;} \\ &\\ 1-\frac{1}{2}e^{-\frac{(x-\mu)}{\lambda} }, & \hbox{$x\geq \mu$;} \end{array} \right. \end{equation*} $$
Mean of Laplace Distribution
The mean of Laplace distribution is $E(X) = \mu$.
Variance of Laplace Distribution
The variance of Laplace distribution is $V(X) = 2\lambda^2$.
Example
A random variable $X$ follows a Laplace distribution with parameter $\mu =5$ and $\lambda=2$.
Find the probability that
a. $X$ is less than 1,
b. $X$ is less than 6,
c. $X$ is between 6 and 10,
d. $X$ is greater than 3.5.
Solution
If $X\sim L(\mu,\lambda)$ then the distribution function of $X$ is
$F(x) =\dfrac{1}{2}e^{\dfrac{(x-\mu)}{\lambda}}$ for $x< \mu$
and
$F(x)=1-\dfrac{1}{2}e^{-\dfrac{(x-\mu)}{\lambda}}$ for $x\geq \mu$.
a. The probability that $X$ is less than $1$ is
$$ \begin{aligned} P(X \leq 1) &=F(1)\\ &=\frac{1}{2}e^{\dfrac{(1-5)}{2}}\\ &\qquad (\because 1< mu)\\ &=\frac{1}{2}e^{\dfrac{(1-5)}{2}}\\ &= 0.0677 \end{aligned} $$
b. The probability that $X$ is less than $6$ is
$$ \begin{aligned} P(X \leq 6) &=F(6)\\ &=1-\frac{1}{2}e^{\dfrac{-(6-5)}{2}}\\ &\qquad (\because 6> mu)\\ &=1-\frac{1}{2}e^{\dfrac{-(6-5)}{2}}\\ &= 1-0.3033\\ &= 0.6967 \end{aligned} $$
c. The probability that $X$ is between $6$ and $10$ is
$$ \begin{aligned} P(6 \leq X \leq 10)&=P(X\leq 10)-P(X\leq 6)\\ &=F(10) -F(6)\\ &=\bigg(1-\frac{1}{2}e^{\dfrac{-(10-5)}{2}}\bigg)-\bigg(1-\frac{1}{2}e^{\dfrac{-(6-5)}{2}}\bigg)\\ &=\frac{1}{2}e^{\dfrac{-(6-5)}{2}}-\frac{1}{2}e^{\dfrac{-(10-5)}{2}}\\ &= 0.3033-0.041\\ &=0.2623 \end{aligned} $$
d. The probability that $X$ is greater than $3.5$ is
$$ \begin{aligned} P(X > 3.5) &=1-P(X< 3.5)\\ &=1-F(3.5)\\ &=1-\frac{1}{2}e^{\dfrac{(3.5-5)}{2}}\\ &\qquad (\because 3.5< mu)\\ &= 1-0.2362\\ &= 0.7638 \end{aligned} $$
