Exponential distribution
A continuous random variable 𝑋𝑋 is said to have exponential distribution if it has
the following pdf:
𝑓𝑓(𝑥𝑥) = 𝜆𝜆 𝑒𝑒 −𝜆𝜆𝜆𝜆 ; 0 < 𝑥𝑥 < ∞
𝑓𝑓(𝑥𝑥)
0 𝑥𝑥
We write 𝑋𝑋~ exponential(𝜆𝜆).
Example
Lifetime of a machine, waiting time for an event, etc. sometimes follow exponential
distribution.
Mean and variance
1
𝐸𝐸(𝑋𝑋) =
𝜆𝜆
1
𝑉𝑉(𝑋𝑋) = 2
𝜆𝜆
That is, mean and SD are equal.
Note
∞
𝑃𝑃(𝑋𝑋 > 𝑘𝑘) = � 𝜆𝜆 𝑒𝑒 −𝜆𝜆𝜆𝜆 𝑑𝑑𝑑𝑑 = 𝑒𝑒 −𝜆𝜆𝜆𝜆
𝑘𝑘
Exercise
Survival time after a particular operation follows exponential distribution with an
average survival time of 3 years. What is the probability that a patient will survive
at least 2 years?
Solution
𝑋𝑋 = survival time of the patient
𝑋𝑋~ exponential(𝜆𝜆 = 1/3)
∞
1 −1𝑥𝑥 1
𝑃𝑃(𝑋𝑋 > 2) = � 𝑒𝑒 3 𝑑𝑑𝑑𝑑 = 𝑒𝑒 −3(2) = 0.5134
3
2
Memoryless property
𝑃𝑃(𝑋𝑋 > 𝑠𝑠 + 𝑡𝑡 | 𝑋𝑋 > 𝑠𝑠) = 𝑃𝑃(𝑋𝑋 > 𝑡𝑡)
Proof
𝑃𝑃(𝑋𝑋 > 𝑠𝑠 + 𝑡𝑡 | 𝑋𝑋 > 𝑠𝑠)
𝑃𝑃(𝑋𝑋 > 𝑠𝑠 + 𝑡𝑡)
=
𝑃𝑃(𝑋𝑋 > 𝑠𝑠)
𝑒𝑒 −𝜆𝜆(𝑠𝑠+𝑡𝑡)
=
𝑒𝑒 −𝜆𝜆𝜆𝜆
= 𝑒𝑒 −𝜆𝜆𝜆𝜆
= 𝑃𝑃(𝑋𝑋 > 𝑡𝑡)
Explanation
Suppose, lifetime (𝑋𝑋) of television of a particular brand follows exponential
distribution. You bought a 2-year-old TV while your friend bought a new one.
According to the memoryless property:
𝑃𝑃(𝑋𝑋 > 2 + 3 | 𝑋𝑋 > 2) = 𝑃𝑃(𝑋𝑋 > 3)
This means, the probability that your 2-year-old TV will survive at least 3 more years
is equal to the probability that your friend’s new TV will survive at least 3 years.
That is, the system has ‘forgotten’ that your TV has already worked for 2 years.
Exercise
Survival time after a particular operation follows exponential distribution with an
average survival time of 3 years. If a patient has already survived 1 year, what is the
probability that (s)he will survive at least 2 more years?
Solution
𝑋𝑋 = survival time of the patient
𝑋𝑋~ exponential(𝜆𝜆 = 1/3)
𝑃𝑃(𝑋𝑋 > 1 + 2 | 𝑋𝑋 > 1)
= 𝑃𝑃(𝑋𝑋 > 2) [memoryless property]
∞
1 −1𝑥𝑥 1
(2)
=� 𝑒𝑒 3 𝑑𝑑𝑑𝑑 = 𝑒𝑒 −3
3
2
= 0.5134
