X = 0, with probability 27
8 3
2 3
1 3
3
X = 1, with probability 27
12 3
2 3
1 1
3
2 1
X = 2, with probability 27
6 3
2 3
1 2
3
1 2
X = 3, with probability 27
1 3
2 3
1 3
3
3
We can calculate that EX =
1 27
1 3
27 6
2 27
12 1
. Why can we leave at the term when X = 0? Also, why is this value in tune with our intuitive
idea of what we should expect? We can formally prove this intuitive notion, namely that for a binomial distribution X, EX = np.
H. The Poisson Random Variable
1. Definition of Poisson random variables
Definition 4.11
A random variable X taking on one of the values 0, 1, 2, ..., is said to be a Poisson random variable with parameter if for some
0, its pmf is of the following form:
p i = P{X = i} =
λ
λ
i
e i
i = 0, 1, 2, ...
4.3 Note: Poisson is pronounced as pwason.
A comment: The Poisson random variable has a lot of applications because it may be used as an
approximation of the binomial random variable with parameters n, p when n is large and p is small enough so that np is a moderate value. See the following fact.
2. Approximation of binomial random variables with Poisson random variables
Fact 4.8
When = np is moderate, we have P
{X = i}
λ
λ
i
e i
i =1, 2, ..., n
where X is a binomial random variable with parameters n, p.
Noteμ “” means “approximately equals.” Proof:
see the reference book.
The meaning of approximation indicated by Fact 4.8 above ---
If n independent trials are performed with each resulting in a success with probability p and a failure with probability 1
p, then when n is large and p small enough to make np moderate, the number of successes occurring is approximately a
Poisson random variable with parameter = np.
Applications of the Poisson random variable ---
There are a lot of the Poisson random variables: No. of misprints on a page of a book.
No. of people in a community living to the age of 100. No. of wrong telephone numbers that are dialed in a day.
....
Noteμ the abbreviation “No.” means “the number of,” and is equivalent to “” which we have used before.
Why ? Because the above numbers of various objects or peoples are all binomial
random variables which may be approximated by the Poisson random variable.
Example 4.14
Suppose that the probability that an item produced by a certain machine will be defective is 0.1. Find the probability that a sample of 10 items will contain at most 1
defective item. Solution:
According to the binomial random variable, the desired probability for 0 or 1 defective item is
P {X
1} = P{X = 0} + P{X = 1} = C10, 00.1
0.9
10
+ C10, 10.1
1
0.9
9
= 0.7361.
Poisson approximation using P{X = i}
λ
λ
i
e i
with = np = 10
0.1 = 1 is
P {X
1} = P{X = 0} + P{X = 1} =
1
1 e
+
1 1
1 1
e
= 2e = 0.7358 which is close to 0.7361 computed above
3. The mean and variance of a Poisson random variable
