# Exponential Distribution

 English Português Français ‎Español Italiano Nederlands

A continuous random variable X has and exponential distribution with parameter ${\displaystyle \lambda >0}$ if its probability density function can be defined as: ${\displaystyle f_{EX}(x;\lambda )=\left\{{\begin{array}{ll}\lambda e^{-\lambda x}\quad &{\text{for}}\ x=\geq 0;\lambda >0\\&\\0\quad &{\text{for}}\ x<0\end{array}}\right.}$ This is denoted as ${\displaystyle X\sim EX(\lambda )}$. The distribution function is given as: ${\displaystyle F_{EX}(x;\lambda )=\left\{{\begin{array}{ll}1-e^{-\lambda x}\quad &{\text{for}}\ x=\geq 0;\lambda >0\\&\\0\quad &{\text{for}}\ x<0\end{array}}\right.}$ The expected value and variance of an exponentially distributed random variable are: ${\displaystyle E(X)={\frac {1}{\lambda }}\quad Var(X)={\frac {1}{\lambda ^{2}}}}$ As ${\displaystyle \lambda }$ ${\displaystyle \rightarrow \infty }$, the faster the density function approaches 0 and the distribution function approaches 1. The exponential distribution depends on the parameter ${\displaystyle \lambda }$. The following example allows you to alter ${\displaystyle \lambda }$ and see its effect on the density function. In a manufacturing plant, 48 equipment failures are expected per day (=24 hours). These failures are purely randomly and independent. On average, ${\displaystyle \lambda }$=48/24 =2 failures are expected per hour. Define the random variable T for the time between 2 failures, which will have an exponential distribution: ${\displaystyle T\sim EX(2)}$.

The probability density function for EX(2) is displayed below;

The probability the next equipment failure will occur in two 2 hours is: ${\displaystyle P(t>2)=1-F_{EX}(2)=1-(1-e^{-2\cdot 2})=e^{-4}=0.01832}$ Suppose that a plant uses two of these systems. The plant comes to a halt as soon as one of the systems stops to function. Let T1 = ”Time between 2 failures for the first component”. T2 = ”Time between 2 failures for the second component”. ${\displaystyle T_{1}\sim EX(2)}$ and ${\displaystyle T_{2}\sim EX(2)}$ Since the plant can only function while both components are operating, both will need more than 2 hours to operate; {\displaystyle {\begin{aligned}P({\text{The system operates for more than 2 hours}})&=&\\&=&P[({\text{first component operates for more than 2 hours}})\cap P({\text{second component operates for more than 2 hours}})]\\&=&P({\text{first component operates for more than 2 hours}})\cdot P({\text{second component operates for more than 2 hours}})\\&=&P(T_{1}\geq 2)\cdot P(T_{2}\geq 2)=(0.01832)^{2}=0.000336\\&&\end{aligned}}} We use the multiplicative property for independent outcomes here since both components operate independently. On the basis of the relationship between the exponential and the Poisson distribution; the Poisson distribution defines the probability of the number of outcomes Y of a specific phenomenon, in a fixed and continuous length or interval with the intensity ${\displaystyle \lambda }$. The following example illustrate a Poisson distribution. Suppose there is a machine in which 2 defects, on the average, are recorded per week. Let t = number of intervals with fixed length (in weeks). a). The probability that no defects are recorded in a week is: ${\displaystyle Y_{1}}$: “number of defects per week” with t=1. ${\displaystyle E(Y_{1})=\lambda =2\,,\qquad Y_{1}\sim PO(2)}$ ${\displaystyle f_{PO}(y_{1};\lambda )={\frac {(\lambda t)^{y_{1}}}{y_{1}!}}e^{-\lambda x}={\frac {(2\cdot 1)^{0}}{0!}}e^{-2\cdot 1}=e^{-2}=0.1353}$ b). The probability of recording no defects in 2 weeks: ${\displaystyle Y_{2}}$: “number of defects in 2 weeks” with t=2 ${\displaystyle E(Y_{2})=\lambda t=2\cdot 2\,,\qquad Y_{2}\sim PO(4)}$ ${\displaystyle P(Y_{2}=0)={\frac {(2\cdot 2)^{0}}{0!}}e^{-2\cdot 2}={\frac {4^{0}}{0!}}e^{-4}=e^{-4}=0.0183}$ In general, the probability that no defect is recorded in t weeks is given as: ${\displaystyle Y}$ : “number of defects in t weeks”. E(Y) = ${\displaystyle \lambda }$ t ${\displaystyle \qquad }$ Y ${\displaystyle \sim }$ PO(${\displaystyle \lambda }$ t) ${\displaystyle P(Y=0)=frac{(\lambda t)^{0}}{0!}e^{-\lambda t}=e^{-\lambda t}}$ If we are interested in finding the probability associated with the next defect occurs, for example, that it will be at least 2 weeks before the next defect; X: ”Waiting time till next defect”. To calculate P(${\displaystyle X>2}$) we use the exponential distribution. ${\displaystyle P(X>2)=1-P(X\leq 2)=1-F_{EX}(x;\lambda )=1-(1-e^{-\lambda x})=e^{-\lambda x}=e^{-2\cdot 2}=0.0183}$ This value is the same as the probability P(${\displaystyle Y_{2}}$ = 0) from the Poisson distribution, for the random variable Y ”in 2 weeks no defects is recorded”. The probability density function of EX(2).

The distribution function of EX(2).

Note!! f(x) = F(X) The Poisson distribution is to be used to compute the probability associated with a random variable Y that defines the number of occurrences of a given outcome within a defined continuous length with an intensity ${\displaystyle \lambda }$. But if we are interested in the time between these outcomes then an exponential distribution can be used to make probability statements. The exponential distribution provides the probability that the ”distance” between two subsequent Poisson random variables. We denote this new continuous random variable as X ”the interval between 2 subsequent outcomes”. The probability that X takes on a maximum value of x is P(X${\displaystyle \leq }$x) = 1 - P (no outcome within the interval of length x). But P(no outcome within the interval of length x) simply represents the probability that a Poisson distributed random variable Y takes on a value 0 with the interval of length x ; P(Y=0) so that ${\displaystyle f_{PO}(y;\lambda x)={\frac {(\lambda x)^{y}}{y!}}e^{-\lambda x}}$ ${\displaystyle P(Y=0)=f_{PO}(0;\lambda x)={\frac {(\lambda x)^{0}}{0!}}e^{-\lambda x}=e^{-\lambda x}}$ This is the distribution function of the exponential distribution, i.e. X is exponentially distributed. ${\displaystyle P(X\leq x)=1-e^{-\lambda x}}$ Therefore, there exists a relationship between the exponential and Poisson distributions. The exponential distribution is often used to model the length of time for continuous processes as well as the waiting times. For example:

• The waiting time before service in a restaurant, bank or filling station.
• The time taken before a component within a technical system fails.
• Service time (time to load a truck, time to carry out a repair).
• Half life (life span) of a component (person).
• Time taken for a telephone conversation.
• Time taken before the next report on damages at property insurance firm.

The following condition is often associated with an exponential distribution. P(X ${\displaystyle \leq }$ t + s ${\displaystyle |}$ X ${\displaystyle \geq }$ t) = P( X ${\displaystyle \leq }$ s). This condition means that the time associated with an outcome does not depend on previous times. This means that the exponential distribution is memoryless. The graphical presentation of an exponentially distributed random variable will be given in the form of a density function, since it refers to the case of a random variable.