Download The Binomial and Poisson Distribution

ST2334: SOME NOTES ON THE BINOMIAL AND POISSON
DISTRIBUTIONS
Binomial Distribution
We first note the binomial coefficient: the number of ways, disregarding order,
that x objects can be chosen from amongst n objects:
n
n!
=
.
x
(n − x)!x!
Let n ≥ 1 be fixed and finite integer and let X = {0, 1, . . . , n}. Then a random
variable X has a binomial distribution, if its PMF is:
n x
p (1 − p)n−x x ∈ X, p ∈ (0, 1).
f (x) =
x
Such a PMF is suitable for a random variable that counts the ‘successes’ in a
collection of n independent and identical Bernoulli trials (an experiment where
there are only two possible outcomes), where the probability of success is p ∈ (0, 1).
An example of which, can be found in the notes; flipping a coin n times which
lands heads with probability p and counting the number of heads which lands. In
this example, the Bernoulli trial is that of flipping the coin; there are only two
possible outcomes, head or tail. The success is a ‘heads’ landing and one would
expect that these consecutive Bernoulli trials are independent.
We remark that f (x) is indeed a valid PMF. First
f (x) ≥ 0
∀x ∈ X.
This is because 1 > p > 0 and so for any x ∈ X, px (1 − p)n−x > 0. The binomial
coefficient is always positive in our scenario. Second,
n
n X
X
n x
f (x) =
p (1 − p)n−x = 1.
x
x=0
x=0
Here we use the fact that for any fixed a, b ∈ (−∞, ∞) and poitive integer n,
n X
n x n−x
a b
.
(a + b)n =
x
x=0
In our setting a = p and b = 1 − p and for any n ≥ 1, (1)n = 1. Thus, our PMF is
such that 1 ≥ f (x) ≥ 0 for each x ∈ X.
The distribution function is for x ∈ X
x X
n y
P(X ≤ x) = F (x) =
p (1 − p)n−y .
y
y=0
This latter summation does not always have an analytic expression. One can use
the distribution function to answer various questions of interest. For example; if
one flips a coin 10 times and the probability of heads is p, what is the probability
of:
1
2
ST2334
• 1 head
• at least 9 heads
For the first question we have
P(X = 1) = f (1) = F (1) − F (0) =
10
p(1 − p)9 .
1
For the second question, this is
10 9
10 10
P(X ≥ 9) = f (9) + f (10) = F (10) − F (8) =
p (1 − p) +
p .
9
10
Poisson Distribution
Let X be the set of non-negative integers: X = {0, 1, . . . }. Then a random
variable X ∈ X is said to have a Poisson distribution of parameter ∞ > λ > 0 if
the PMF is
λx e−λ
f (x) =
x ∈ X.
x!
In general, the Poisson distribution is often used for random variables associated
to counts (for example, counts which can grow indefinitely). This is a PMF as
f (x) ≥ 0 ∀x ∈ X: λx /x! > 0 and e−λ > 0. In addition
∞
X
X
λx e−λ
f (x) =
x!
x=0
x∈X
= e−λ
= e
=
∞
X
λx
x=0
−λ λ
x!
e
1.
Here we have used the Taylor series for the exponential function.
The distribution function is for x ∈ X
x
X
λy e−λ
P(X ≤ x) = F (x) =
y!
y=0
This latter summation does not always have an analytic expression. One can use
the distribution function or the PMF to answer questions of interest:
• what is the probability that X is 1?
• what is the probability that X is positive?
For the first question:
P(X = 1) = f (1) = F (1) − F (0) = λe−λ .
For the second question:
P(X ≥ 1) = 1 − P(X = 0) = 1 − f (0) = 1 − e−λ .