Download Some important discrete random variables

Some important discrete random variables
We now describe some very important discrete random variables. The first type is very simple,
but it is a very important building block.
Suppose that a trial, or an experiment, whose outcome can be classified as either a success or a
failure is performed. If we let X = 1 when the outcome is a success and X = 0 when the outcome
is a failure, then the probability mass function of X is given by



1 − p, x = 0
p(x) =
p,


0,
x=1
(0.1)
otherwise
where p ∈ [0, 1] is the probability that the trial is a success.
Suppose that p ∈ (0, 1). A random variable X is called a Bernoulli random variable with
parameter p if its mass function is given by (0.1).
If X is a Bernoulli random variable with parameter p, then
E[X] = p,
Var(X) = p(1 − p).
Suppose now that n independent trials, each results in a success with probability p and in a
failure with probability 1 − p, are performed. If we let X denote the the number of successes in
these n trials, then the probability mass function of X is given by
Ã
!

n


px (1 − p)n−x , x = 0, 1, . . . , n,
p(x) =
(0.2)
x


0,
otherwise
where p ∈ [0, 1] is the probability that the trial is a success.
Suppose that n is a positive integer and p ∈ (0, 1). A random variable X is called a binomial
random variable with parameters n and p if its mass function is given by (0.2).
Obviously, a Bernoulli random variable with parameter p is simply a binomial random variable
with parameters 1 and p. From the definition of a binomial random variable, we can see that
any binomial random variable X with parameters n and p can always be written as the sum of n
Bernoulli random variables with parameter p:
X = X1 + · · · + Xn ,
where, for i = 1, . . . , n, Xi = 1 if the i-th trial is a success and Xi = 0 if the i-th trial is a failure.
If X is a binomial random variable with parameters n and p, then
E[X] = np,
Var(X) = np(1 − p).
1
In fact
!
Ã
!
n
X
n
n
i
n−i
E[X] =
i
p (1 − p)
=
i
pi (1 − p)n−i
i
k
i=0
i=1
Ã
!
Ã
!
n
n
X
X
n−1
n−1
i−1
n−i
= np
p (1 − p)
= np
pj (1 − p)n−1−j
i−1
j
i=1
j=0
n
X
Ã
= np.
Using a similar argument, we can find that E[X 2 ] = np[(n − 1)p + 1], and so Var(X) = np(1 − p).
Suppose that λ > 0 is a constant. A nonnegative integer valued random variable X is called a
Poisson random random variable with parameter λ if its probability mass function is given by
(
x
e−λ λx! , x = 0, 1, . . .
p(x) =
(0.3)
0,
otherwise.
(0.3) defines a probability mass function, since
∞
X
−λ
p(x) = e
x=0
∞
X
λx
x=0
i!
= e−λ eλ = 1.
Poisson random variables are very important in applications because it can be used as an
approximation of a binomial random variable with parameters n and p when n is large and p is
small so that np is of moderate size. To see this, suppose that X is binomial random variable with
parameters n and p, and let λ = np Then
P(X = i) =
=
=
Now, for n large and λ moderate,
µ
¶
λ n
1−
≈ e−λ ,
n
n!
pi (1 − p)i
(n − i)!i!
µ ¶i µ
¶
n!
λ
λ n−i
1−
(n − i)!i! n
n
n(n − 1) · · · (n − i + 1) λi (1 − λ/n)n
.
ni
i! (1 − λ/n)i
µ
λ
1−
n
¶i
≈ 1,
n(n − 1) · · · (n − i + 1)
≈ 1.
ni
Hence for n large and λ moderate,
λi
.
i!
Some examples of random variables that can be approximately modeled by Poisson random
variables are as follows:
P(X = i) ≈ e−λ
1. the number of misprints on a page of a book;
2. the number of people in a community over the age of 100;
3. the number of people entering a post office in a given hour.
2
Example 1 A machine produces screws, 1% of which are defective. Find the probability that in a
box of 200 screws there are no defective ones.
Solution. Assuming independence, the number of defective ones in the box is a binomial
random variable with with parameters 200 and .01. The probability that there are no defective
ones is
(1 − .01)200 = (.99)200 = .1340.
The Poisson approximation to this is given by
e−200(.01) = e−2 = .1353.
If X is a Poisson random variable with parameter λ. Then
E[X] = λ,
Var(X) = λ.
In fact,
E[X] =
∞
X
∞
ie−λ
i=0
= λe−λ
X λi−1
λi
= λe−λ
i!
(i − 1)!
i=1
∞
X
λj
j=0
j!
= λ.
Using a similar argument we can get that E[X 2 ] = λ(λ + 1), and consequently we get Var(X) = λ.
Another use of the Poisson distribution arises in situations where “events” occur at certain
points in time. One example of this is that an event is the occurrence of an earthquake, another
possibility would be for events to correspond to people entering a particular establishment (bank,
post office, etc). Let us suppose that events are indeed occurring at certain (random) points of
time, and let us assume that for some positive constant λ the following assumptions hold:
1. The probability that exactly one event occurs in a given interval of length h is equal to λh+o(h).
2. The probability that 2 or more events occur in an interval of length h is equal to o(h).
3. For any integers n > 1, j1 ≥ 0, . . . , jn ≥ 0, and any set of n non-overlapping intervals, if we
define Ei to be the event that exactly ji events occur in the i-th interval, then the events
E1 , . . . , En are independent.
Under assumptions 1, 2 and 3, we can show that the number of events occurring in any interval
of length t is a Poisson random variable with parameter λt.
Suppose that independent trials, each having probability p ∈ (0, 1) of being a success, are
performed until a success occurs. If we let X be the number of trials needed, then its probability
mass function is given by
(
(1 − p)x−1 p, x = 1, 2, . . .
p(x) =
(0.4)
0.
otherwise.
Any random variable whose probability mass function is given by (0.4) is called a geometric random
variable with parameter p.
3
Example 2 A box contains N white and M black balls. Balls are randomly selected, one at a time,
until a black is obtained. If we assume that each selected ball is returned to the box before the next
one is drawn, what is the probability that (a) exactly k draws are needed; (b) at least k draws are
needed?
Solution. If we let X be the numbers of draws needed, then X is a geometric random variable
with parameter p = M/(N + M ). Hence
µ
P(X = k) =
N
N +M
and
µ
P(X ≥ k) =
¶k−1
N
N +M
M
N +M
¶k−1
.
If X is a geometric random variable with parameter p, then we have already found that
E[X] =
1
.
p
Using a similar argument we can find that E[X 2 ] =
2
p2
Var(X) =
Remark on Notations
4
− p1 . Thus
1−p
.
p2