Lectures on Elementary Probability

6.3. POISSON RANDOM VARIABLES 39
6.3 Poisson random variables
The first equation is homogeneous and has the solution
P [N(t) = 0] = e −λt . (6.14)
The second equation is inhomogeneous with source term λP [N(t) = k − 1] and
has the solution
P [N(t) = k] =
∫ t
These equations may be solved to find P [N(t) = k].
0
e −λ(t−s) λP [N(s) = k − 1] ds. (6.15)
Theorem 6.1 The solution of the differential equations is
P [N(t) = k] = (λt)k e −λt . (6.16)
k!
The result is that N(t) is a Poisson random variable with parameter λt.
Proof: This is proved by noting that it is true for k = 0. Furthermore, we
shall see that for each k ≥ 1, if it is true for k − 1 it is also true for k. Therefore
it must be true for all k.
To show that if it is true for k − 1 it is also true for k, suppose that
P [N(s) = k − 1] = (λs)k−1
(k − 1)! e−λs (6.17)
for all s ≥ 0. Then from the solution of the differential equation
This can be written
P [N(t) = k] =
∫ t
0
e −λ(t−s) λ (λs)k−1
(k − 1)! e−λs ds. (6.18)
∫ t
P [N(t) = k] =
λk
(k − 1)! e−λt s k−1 ds =
λk
(k − 1)! e−λt tk k . (6.19)
This immediately gives the result that
for all t ≥ 0.
0
P [N(t) = k] = (λt)k e −λt (6.20)
k!
Theorem 6.2 The expectation of the Poisson random variable N(t) is
E[N(t)] = λt. (6.21)