Poisson Distribution
Consider the following random variables:
1. The number of people arriving for treatment at an emergency room in
each hour.
2. The number of drivers who travel between Salt Lake City and Sandy
during each day.
3. The number of trees in each square mile in a forest.
None of them are binomial, hypergeometric or negative binomial random
variables.
In fact, the experiments associated with above random variables DO NOT
involve trials. We use Poisson distribution to model the experiment for
occurence of events of some type over time or area.
Liang Zhang (UofU)
Applied Statistics I
June 23, 2008
1 / 10
Poisson Distribution
Definition
A random variable X is said to have a Posiion distribution with
parameter λ (λ > 0) if the pmf of X is
p(x; λ) =
e −λ λx
x!
x = 0, 1, 2, . . .
1. The value λ is frequently a rate per unit time or per unit area.
2. e is the base of the natural
P∞ logarithm system.
3. It is guaranteed that x=0 p(x; λ) = 1.
∞
eλ = 1 + λ +
X λx
λ2 λ3
+
+ ··· =
2!
3!
x!
x=0
Liang Zhang (UofU)
Applied Statistics I
June 23, 2008
2 / 10
Poisson Distribution
Example:
The red blood cell (RBC) density in blood is estimated by means of a
hematometer. A blood sample is thoroughly mixed with a saline solution,
and then pipetted onto a slide. The RBC’s are counted under a
microscope through a square grid. Because the solution is throughly
mixed, the RBC’s have an equal chance of being in a particular square in
the grid. It is known that the number of cells counted in a given square
follows a Poisson distribution and the parameter λ for certain blood
sample is believed to be 1.5.
Then what is the probability that there is no RBC in a given square?
What is the probability for a square containing exactly 2 RBC’s?
What is the probability for a square containing at most 2 RBC’s?
Liang Zhang (UofU)
Applied Statistics I
June 23, 2008
3 / 10
Poisson Distribution
Proposition
If X has a Poisson distribution with parameter λ, then E (X ) = V (X ) = λ.
We see that the parameter λ equals to the mean and variance of the
Poisson random variable X .
e.g. for the previous example, the expected number of RBC’s per square is
thus 1.5 and the variance is also 1.5.
In practice, the parameter usually is unknown to us. However, we can use
the sample mean to estimate it. For example, if we observed 15 RBC’s
over 10 squares, then we can use x̄ = 15
10 = 1.5 to estimate λ.
Liang Zhang (UofU)
Applied Statistics I
June 23, 2008
4 / 10
Poisson Distribution
Poisson Process: the occurrence of events over time.
1. There exists a parameter α > 0 such that for any short time interval of
length ∆t, the probability that exactly one event is received is
α · ∆t + o(∆t).
2. The probability of more than one event being received during ∆t is
o(∆t) [which, along with Assumption 1, implies that the probability of no
events during ∆t] is 1 − α · ∆t − o(∆t)].
3. The number of events received during the time interval ∆t is
independent of the number received prior to this time interval.
Liang Zhang (UofU)
Applied Statistics I
June 23, 2008
5 / 10
Poisson Distribution
Proposition
Let Pk (t) denote the probability that k events will be observed during any
particular time interval of length t. Then
Pk (t) = e −αt ·
(αt)k
.
k!
In words, the number of events during a time interval of length t is a
Poisson rv with parameter λ = αt. The expected number of events during
any such time interval is then αt, so the expected number during a unit
interval of time is α.
Liang Zhang (UofU)
Applied Statistics I
June 23, 2008
6 / 10
Poisson Distribution
Example: (Problem 92)
Automobiles arrive at a vehicle equipment inspection station according to
a Poisson process with rate α = 10 per hour. Suppose that with
probability 0.5 an arriving vehicle will have no equipemnt violations.
a. What is the probability that exactly ten arrive during the hour and all
ten have no violations?
b. For any fixed y ≥ 10, what is the probability that y arrive during the
hour, of which ten have no violations?
c. What is the probability that ten “no-violation” cars arrive during the
next 45 minutes?
Liang Zhang (UofU)
Applied Statistics I
June 23, 2008
7 / 10
Poisson Distribution
In some sense, the Poisson distribution can be recognized as the limit of a
binomial experiment.
Proposition
Suppose that in the binomial pmf b(x; n, p), we let n → ∞ and p → 0 in
such a way that np approaches a value λ > 0. Then b(x; n, p) → p(x; λ).
This tells us in any binomial experiment in which n is large
and p is small, b(x; n, p) ≈ p(x; λ), where λ = np.
As a rule of thumb, this approximation can safely be applied if n > 50 and
np < 5.
Liang Zhang (UofU)
Applied Statistics I
June 23, 2008
8 / 10
Poisson Distribution
Example 3.40:
If a publisher of nontechnical books takes great pains to ensure that its
books are free of typographical errors, so that the probability of any given
page containing at least one such error is 0.005 and errors are independent
from page to page, what is the probability that one of its 400-page novels
will contain exactly one page with errors?
Let S denote a page containing at least one error, F denote an error-free
page and X denote the number of pages containing at least one error.
Then X is a binomial rv, and
P(X = 1) = b(1; 400, 0.005) ≈ p(1; 400 · 0.005) = p(1; 2) =
e −2 (2)
1!
= 0.270671
Liang Zhang (UofU)
Applied Statistics I
June 23, 2008
9 / 10
Poisson Distribution
A proof for b(x; n, p) → p(x; λ) as n → ∞ and p → 0 with np → λ.
n!
p x (1 − p)n−x
x!(n − x)!
n(n − 1) · · · (n − x + 1) x
n!
p x = lim
p
lim
n→∞
n→∞ x!(n − x)!
x!
(np)[(n − 1)p] · · · [(n − x + 1)p]
= lim
n→∞
x!
λx
=
x!
np n−x
n−x
lim (1 − p)
= lim {1 −
}
n→∞
n→∞
n
λ
= lim {1 − }n−x
n→∞
n
= e −λ
b(x; n, p) =
Liang Zhang (UofU)
Applied Statistics I
June 23, 2008
10 / 10