An exponential random variable is a continuous random variable, described by a single parameter λ. The pdf that describes an exponential random variable is
f(x) = $\lambda e^{-\lambda x}$ for x > 0.
Correspondingly the cdf of exponential random variable is
F(x) = 1 - $e^{-\lambda x}$.
An exponential random variable can be viewed as the time between events described by a Poisson random variable and the geometric random variable can be considered as its discrete counterpart.
The Mean and the variance of an exponential distribution is
Mean = E(X) = $\frac{1}{\lambda }$ and Var(X) = $\frac{1}{\lambda ^{2}}$

## Sum of Exponential Random Variables

Suppose X1, X2.....Xn be independent and identically distributed random variables with parameter λ >0.
Then the sum = X1 + X2 + .....Xn is a gamma distribution with parameters n, λ.
The pdf of the sum is given by
f(x) = $\lambda e^{-\lambda x}$$\frac{(\lambda x)^{n-1}}{(n-1)!} ## Exponential Random Variable Examples ### Solved Examples Question 1: The length of a shower in a hypothetical state during rainy season form an exponential random variable with parameter λ = \frac{1}{2}. (a) What is the probability that a shower will last for more than 2 minutes. (b) Find the probability that a shower will last between 2 and 3 minutes. Solution: The pdf for the exponential random variable is f(x) = \frac{1}{2} e^{\frac{-x}{2}} Probability the shower will last for more than 2 minutes P (x > 2) = \int_{2}^{\infty }\frac{1}{2}e^{\frac{-2}{2}}$$dx$

= $\frac{1}{e}$ ≈ 0.3679.

Probability the shower will last between 2 and 3 minutes P(2 < X < 3)

= $\int_{2}^{3}\frac{1}{2}e^{\frac{-x}{2}}dx$

= $\frac{1}{e}$ - $\frac{1}{e^{\frac{3}{2}}}$

≈ 0.14475.

Using the cdf function F(x) = 1 - $e^{-\lambda x}$ for an exponential random variable, it can be proved

P(x > a) = $e^{-\lambda x}$

Note: for the above example P(X > 2) = $e^{-(\frac{2}{2})}$ = e-1 = $\frac{1}{e}$ ≈ 0.3679.

P(2 < X < 3) = F(3) - F(2) = (1 - $e^{-(\frac{3}{2})}$) - (1 - $e^{-(\frac{2}{2})}$)

= e-1 - e-1.5  = 1/e - $\frac{1}{e^{\frac{3}{2}}}$  ≈ 0.14475.

Question 2: Suppose the wait time X for filling gas in a gas filling station is an exponential random variable with average wait time = 2 minutes. What is the probability a customer has to wait over 4 minutes before he got attended? Suppose he has already waited for 2  minutes, what is the probability that he has to wait 4 more minutes before getting attended?
Solution:

Here the mean wait time = E( x) = $\frac{1}{\lambda }$ = 2.    Hence λ = $\frac{1}{2}$.

P(x > a) = $e^{-\lambda x}$

P( X > 4) = $e^{-(\frac{1}{2})4}$ = e-2 ≈ 0.1353.

An important property of exponential distribution is if the random variable represents a waiting time, the probability of  waiting for a certain length of time is not altered if you have already waited for some time.

Thus the probability of a customer needs to wait for more than 4 minutes when he has already waited for 4 minutes
= P( X > 4) = 0.1353.