Chapter 2
Discrete Random Variables
Random Variables
Informally: A rule for assigning a real number
to each outcome of a random experiment
– A fish is randomly selected from a lake. Let the
random variable
Z = the length of the fish in inches
– A voter is randomly selected and asked if he or she
supports a candidate running for office. Let the
random variable
1 if the person supports the candidate
W 
0 otherwise
Random Variables
Definition 2.1.1 Let S be the sample space of a
random experiment. A random variable is a
function
X :S 
The set of possible values of X is called the
range of X (also called the support of X).
Random Variables
1. A random variable is a function.
2. A random variable is not a probability.
3. Random variables can be defined in practically any
way. Their values do not have to be positive or
between 0 and 1 as with probabilities.
4. Random variables are typically named using capital
letters such as X, Y , or Z. Values of random variables
are denoted with their respective lower-case letters.
Thus, the expression
X=x
means that the random variable X has the value x.
Discrete Random Variables
Definition 2.1.2 A random variable is discrete if
its range is either finite or countable.
Chapter 3 – Continuous random variables
2.2 – Probability Mass Functions
Definition 2.2.1 Let X be a discrete random
variable and let R be the range of X. The
probability mass function (abbreviated p.m.f.) of
X is a function f : R → ℝ that satisfies the
following three properties:
1. f  x   0 for all x  R
2.  f ( x)  1
xR
3. If A  R, then P ( X  A)   f ( x )
xA
Distribution
• Definition 2.2.2 The distribution of a random
variable is a description of the probabilities of
the values of variable.
Example 2.2.1
A bag contains one red cube and one blue cube.
Consider the random experiment of selecting two
cubes with replacement. The two cubes we select
are called the sample. The same space is
S = {RR, RB, BR, BB},
and we assume that each outcome is equally
likely. Define the random variable
X = the number of red cubes in the sample
Example 2.2.1
Values of X:
X(RR) = 2, X(RB) = 1, X(BR) = 1, X(BB) = 0
Values of the p.m.f.:
1
f (0)  P( X  0)  P( BB) 
4
2 1
f (1)  P( X  1)  P({RB}  {BR})  
4 2
1
f (2)  P( X  2)  P ( RR ) 
4
Example 2.2.1
The Distribution
Table
Probability Histogram
2  x 1
, for x  0,1, 2
Formula f ( x) 
4
Uniform Distribution
Definition 2.2.3 Let X be a discrete random
variable with k elements in its range R. X has a
uniform distribution (or be uniformally
distributed) if its p.m.f. is
1
f ( x)  for each x  R
k
Example 2.2.2
Consider the random experiment of rolling a fair
six-sided die. The sample space is
S ={1, 2, 3, 4, 5, 6}
Let the random variable X be the number of dots
on the side that lands up. Then
1
f ( x)  P( X  x)  for each x  {1,
6
(i.e. X has a uniform distribution)
, 6}
Cumulative Distribution Function
Definition 2.2.4 The cumulative distribution
function (abbreviated c.d.f.) of a discrete random
variable X is
F (b)  P( X  b)   f ( x) for all real numbers b
x b
Example 2.2.4
Consider the random experiment of rolling two
fair six-sided dice and calculating the sum. Let
the random variable X be this sum.
F (4)  P( X  4)
 f (2)  f (3)  f (4) by property 3 of the p.m.f.
1

6
Example 2.2.4
Mode and Median
• Definition 2.2.5 The mode of a discrete
random variable X is a value of X at which the
p.m.f. f is maximized. The median of X is the
smallest number m such that
P( X  m)  0.5
and
P( X  m)  0.5
Example 2.2.1
• Mode = 1
• Median = 1 since
P( X  1)  1/ 4  1/ 2  3 / 4
and
P( X  1)  1/ 2  1/ 4  3 / 4
2.3 – Hypergeometric and Binomial
Distributions
Definition 2.3.1 Consider a random experiment that
meets the following requirements:
1. We select n objects from a population of N without
replacement (n ≤ N).
2. The N objects are of two types, call them type I and
type II, where N1 are of type I and N2 are of type II
(N1 + N2 = N).
Define the random variable
X = the number of type I objects in the sample of n.
Hypergeometric Distribution
Definition 2.3.1 (continued) Then X has a
hypergeometric distribution and its p.m.f. is
 N1  N 2 
 

x
n

x
,
f ( x)  P ( X  x)   
N
 
n
x  0,1, , n
– The numbers N, n, N1, and N2 are called the
parameters of the random variable.
Example 2.3.2
A manufacturer of radios receives a shipment of
200 transistors, four of which are defective. To
determine if they will accept the shipment, they
randomly select 10 transistors and test each. If
there is more than one defective transistor in the
sample, they reject the entire shipment. Find the
probability that the shipment is not rejected.
Example 2.3.2
• X = number of defectives in the sample of 10
• X has a hypergeometric distribution with
N = 200, n = 10, N1 = 4, and N2 = 196.
P(shipment is not rejected)  P( X  0  X  1)
 P  X  0   P  X  1
 f  0   f 1
 4  196   4  196 
 
  

0
10

0
1
10

1
   

  
 200 
 200 




 10 
 10 
 0.813  0.173  0.986
Bernoulli Experiment
Definition 2.3.2 A random experiment is called a
Bernoulli experiment if each outcome is classified
into exactly one of two distinct categories. These
two categories are often called success and failure.
If a Bernoulli experiment is repeated several times
in such a way that the probability of a success does
not change from one iteration to the next, the
experiments are said to be independent. A sequence
of n Bernoulli trials is a sequence of n independent
Bernoulli experiments.
Binomial Distribution
Definition 2.3.3 Consider a random experiment
that meets the following requirements:
1. A sequence of n Bernoulli trials is performed
2. The probability of a success in any one trial is p
Define the random variable
X = the number of successes in the n trials
Binomial Distribution
Definition 2.3.3 (continued) Then X has a
binomial distribution and its p.m.f. is
n x
f ( x)  P( X  x)    p (1  p ) n  x , x  0,1, , n
 x
– The numbers n and p are the parameters of the
distribution
– The phrase “X is b(n, p)” means the variable X has
a binomial distribution with parameters n and p
Example 2.3.4
If a family with nine children is randomly
chosen, find the probability of selecting a family
with exactly four boys.
– Let X = the number of boys in the family
– X is b(9, 0.5)
9 4
94
P( X  4)  f (4)    0.5 (1  0.5)  0.246
 4
Unusual Events
Unusual Event Principle: If we make an
assumption (or a claim) about a random
experiment, and then observe an event with a
very small probability based on that assumption
(called an “unusual event”), then we conclude
that the assumption is most likely incorrect.
Unusual Events
Unusual Event Guideline: If X is b(n, p) and
represents the number of successes in a random
experiment, then an observed event consisting of
exactly x successes is considered “unusual” if
P( X  x)  0.05
or
P( X  x)  0.05
Example 2.3.7
Suppose a university student body consists of
70% females. To discuss ways of improving
dormitory policies, the President selects a panel
of 15 students. He claims to have randomly
selected the students, however, one administrator
questions this claim because there are only five
females on the panel. Use the rare event
principle to test the claim of randomness.
Example 2.3.7
• Let X = number of females on the panel of 15
• If randomly selected, then X would be
approximately b(15, 0.7)
P( X  5)  P( X  0)   P( X  5)
 0  0  0  0  0.0006  0.003
 0.0036  0.05
• Conclusion: Reject the claim of randomness
2.4 – The Poisson Distribution
Definition 2.4.1 A random variable X has a
Poisson distribution if its p.m.f. is
f ( x)  P( X  x) 

x
x!
e   , x  0,1, 
where λ > 0 is a constant.
– Describes the number of “occurrences” over a
randomly selected “interval”
– The parameter λ is the “average” number of
occurrences per interval
Example 2.4.1
Suppose a 1000 ft2 lawn contains 3000
dandelions. Find the probability that a randomly
chosen 1 ft2 section of lawn contains exactly five
dandelions.
– “Occurrence” = dandelion
– “Interval” = 1 ft2 section of lawn
– X = number of dandelions in a 1 ft2 section of
lawn
– Assume X is Poisson
Example 2.4.1
3000 dandelions
2


3
dandelions/ft
1000 ft 2
5
3 3
P( X  5)  e  0.101
5!
2.5 – Mean and Variance
Definition 2.5.1 The mean (or expected value) of
a discrete random variable X with range R and
p.m.f. f (x) is
  E ( X )   x  f ( x)
xR
provided this series converges absolutely
Variance
Definition 2.5.2 The variance of a discrete
random variable X with range R and p.m.f. f (x)
is
 2  Var ( X )   ( x   ) 2 f ( x)
xR
provided this series converges absolutely. The
standard deviation of X, denoted σ, is the squareroot of the variance
Example 2.5.2
The p.m.f. for a random variable X is
Mean
  0(0.19)  1(0.35)  2(0.33)  3(0.13)  1.4
Example 2.5.2
Variance
Example 2.5.3
Consider a random variable X with range
R = {1, 2,…, k} and p.m.f. f (x) = 1/k
(X has a uniform distribution)
Mean:
1 1 k
1 k (k  1) k  1
  E ( X )   x  f ( x)   x     x  

k k x 1
k
2
2
xR
x 1
k
Example 2.5.3
Variance:
Alternate Formula :  2   x 2 f ( x)   2
xR
1 1 k 2 1 k (k  1)(2k  1) ( k  1)(2k  1)
x f ( x)   x   x  


k k x 1
k
6
6
xR
x 1
k
2
2
k
 2   x 2 f ( x)   2
x 1
(k  1)(2k  1)  k  1 
2(k  1)(2k  1)  3(k  1) 2


 
6
12
 2 
2

(k  1)  2(2k  1)  3(k  1) 
12
(k  1)(k  1) k 2  1


12
12
2.6 – Functions of a Random
Variable
Example 2.6.1: Suppose a man is injured in a
car wreck. The doctor tells him that he needs to
spend between one and four days in the hospital
and that if X is the number of days, its
distribution is given below
Example 2.6.1
The man’s supplemental insurance policy will
give him $500 plus $100 per day in the hospital
to cover expenses. How much should the man
expect to get from the insurance company?
– Y = amount received from the insurance company
Y  500  100 X
Example 2.6.1
• Note
P Y  500  100 x   P  X  x   f  x 
  E (Y )  600(0.3)  700(0.4)  800(0.2)  900(0.1)
 [500  100(1)] f (1) 
 710
 [500  100(4)] f (4)
Mathematical Expectation
Definition 2.6.1 Let X be a discrete random
variable with range R and p.m.f. f(x). Also let
u(X) be a function of X. The mathematical
expectation or the expected value of u(X) is
E u ( X )    u ( x) f ( x)
xR
provided this series converges absolutely
Example 2.6.2
Let X be a random variable with distribution
a. E  X 2    x 2 f ( x)
xR
 (2) 2 (0.25)  (1) 2 (0.5)  (5) 2 (0.25)  7.75
b. E ( X  1) 2    ( x  1) 2 f ( x)
xR
 (2  1) 2 (0.25)  (1  1) 2 (0.5)  (5  1) 2 (0.25)  6.25
Properties
Theorem 2.6.1 Let X be a discrete random.
Whenever the expectations exist, the following
three properties hold:
1. If a is a constant, then E  a   a
2. If a is a constant and u ( X ) is a function of X , then
E  a u  X    aE u  X  
3. If a1 and a2 are constants, and u1 ( X ) and u2 ( X ) are
functions of X , then
E  a1u1 ( X )  a2u2 ( X )   a1E u1 ( X )   a2 E u2 ( X ) 
Example 2.6.3
Let X be a random variable with distribution
a. E  X    2  0.25   1 0.5    5  0.25   1.25
b. E 3 X  4  E 3 X   E  4  3E  X   4  3 1.25   4  7.75
c. E[ X ( X  2)]  E  X 2  2 X   E  X 2   2 E[ X ]  7.75  2.5  10.25
2.7 – Moment-Generating Function
Definition 2.7.1 Let X be a discrete random
variable with p.m.f f(x) and range R. The
moment-generating function (m.g.f.) of X is
M (t )  E etX    etx f ( x)
xR
for all values of t for which this mathematical
expectation exists.
Example 2.7.1
Consider a random variable X with range
R = {2, 5} and p.m.f. f (2) = 0.25, f (5) = 0.75
Its m.g.f. is
M (t )  e (0.25)  e (0.75)
2t
5t
Theorem 2.7.1
If X is a random variable and its m.g.f. M(t)
exists for all t in an open interval containing 0,
then
1. M (t ) uniquely determines the distribution of X and
2. M '  0   E  X  and M ''(0)  E  X 2 
Uses of m.g.f.
1. If we know the m.g.f. of a random variable,
then we can use the first and second
derivatives to find the mean and variance of
the variable.
2. If we can show that two random variables
have the same m.g.f., then we can conclude
that they have the same distribution.
Example 2.7.3
Consider a random variable X with a binomial
distribution. Its p.m.f. is
 n  x n x
f ( x)    p q , x  0,1, , n
 x
where 𝑞 = 1 − 𝑝
Example 2.7.3
1. Find the m.g.f.
 n  k nk
Binomial Theorem : (a  b)     a b
k 0  k 
n
n
 n  x n x
M (t )  E e    e   p q
x 0
 x
n
n
t x n x
     pe  q
x 0  x 
n
tX
tx
  q  pe 
t
n
Example 2.7.3
2. Find M’ and M’’
M (t )  n  q  pe 
t
n 1
 pe 
t
M ''(t )  n(n  1)  q  pe 
t
n2
 pe

t 2
 n  q  pe 
t
n 1
t
pe
 
3. Find mean
  E ( X )  M (0)  n  q  pe 
0
 n q  p 
n 1
 p   np
n 1
0
pe
 
Example 2.7.3
4. Find variance
2
2
2



  E  X     M (0)  
n2 2
n 1
2
 n(n  1)(q  p ) p  n(q  p ) p  (np )
2
 n(n  1) p 2  np  (np ) 2
 np   n  1 p  1  np 
 np 1  p   npq