Poisson Arrivals ↔ Exponential Waiting Times
• Theorem: (Exponential Wating Times → Poisson Arrivals)
– Let Y1 , Y2 , . . . be iid Exponential(λ), and let T > 0. Define
W = largest j such that Y1 + . . . + Yj ≤ T.
Then,
W ∼ Poisson(λT ).
• Theorem: (Poisson Arrivals → Exponential Waiting Times)
(Hofmann, Thrm. 2.4.3)
– Let X = number of arrivals in one unit. Assume X ∼ Poisson(λ).
Let Ti be the time between arrival i − 1 and arrival i, for i = 1, 2, . . ..
Then,
T1 , T2 , . . . ∼ iid Exponential(λ).
Erlang Distribution
• Definition: Let Y1 , . . . , Yk be iid Exponential(λ) random variables. Define
X = Y1 + . . . + Yk .
Then,
X ∼ Erlang(k, λ),
where k is the stage parameter, and λ is the rate parameter.
• PDF:
fX (x; k, λ) =

k−1
−λx (λx)

λe


(k−1)! ,
if x > 0


0
if x < 0
• R.V.: If X has pdf fX (x; λ), then X ∼ Erlang(k, λ)
• E[X] = λk , V ar(X) =
k
λ2
– Because X ∼ Erlang(k, λ) means X = Y1 + . . . + Yk , where Yi ∼
iid Expo(λ),
E[X] = E[Y1 + . . . + Yk ]
= E[Y1 ] + . . . + E[Yk ]
1
k
1
+ ... + = ,
=
λ
λ λ
and
V ar(X) = V ar(Y1 + . . . + Yk )
= V ar(Y1 ) + . . . + V ar(Yk ) by INDEPENDENCE
1
1
k
= 2 + . . . + 2 = 2.
λ
λ
λ
• CDF:
Erlangk,λ (x) =


1 − Poλx (k − 1) ifx > 0


0
otherwise.
Why?
P (X > x) = P (Y1 + . . . + Yk > x),
where Yi ∼ iid Expo(λ)
= P (W ≤ k − 1),
where W ∼ Poisson(λx)
= P oλx (k − 1).
So, P (X ≤ x) = 1 − P oλx (k − 1).
• Example: A safety system works if component A works or if the
backup component B works. We expect each component to function for 5 years. The lifetimes of the two components are independent exponential random variables.
(a.) For how long do we expect the safety system to continue to
function?
(b.) What is the probability that the safety system will fork for
more than 5 years?
Soln:
– Let X =lifetime of component A.
– Let Y =lifetime of component B.
– Because E[X] = E[Y ] = 5years, X, Y ∼ iid Exponential(1/5).
(a.) Let Z =time safety system functions. Then,
Z = X + Y, so Z ∼ Erlang(2, 1/5), and E[Z] = 2 ∗ 5 = 10years.
(b.)
P (Z > 5) = 1 − P (Z ≤ 5)
= 1 − P (W ≥ 2),
where W ∼ Poisson(5/5)
11e−1 10e−1
= 1 − (1 − P o1(1)) =
+
= 2e−1
1!
0!