Chapter 3
Continuous Random Variables
3.2 – Definitions
Definition 3.2.1 A random variable X is said to
be continuous if there exists a function f called
the probability density function (p.d.f ) which is
continuous at all but a finite number of points
and satisfies the following three properties:
1. f ( x)  0 for all x
2. 


f ( x) dx  1
b
3. P(a  X  b)   f ( x) dx
a
The c.d.f.
• The cumulative distribution function (c.d.f.) of
a continuous random variable X is
F ( x)  P( X  x)  
x

f (t ) dt
• An important relationship:
d x
F ( x)   f (t ) dt  f ( x)
dx 
Example 3.2.1
Suppose that X is a continuous random variable
with range [1, 2] and p.d.f
2 x  2 1  x  2
f ( x)  
elsewhere
0
Verify Property 2:



2
f ( x) dx   2 x  2 dx  x  2 x  (4  4)  (1  2)  1
1
2
2
1
Example 3.2.1
• Calculate 𝑃(1 < 𝑋 ≤ 1.25):
P(1  X  1.25)  
1.25
1
2 x  2 dx  x  2 x
2
1.25
1
 (1.252  2(1.25))  (1  2)
 0.0625
• Find the c.d.f.:
F ( x)  P( X  x)  
x

f (t ) dt
x 1
0
 x
   2t  2 dt  x 2  2 x  1 1  x  2
1

x2
1
Definition 3.2.2
Let X be a continuous random variable with
p.d.f. f (x)
Mean :

  E ( X )   x f ( x) dx


Variance :   Var ( X )  E ( X   )    ( x   ) 2 f ( x) dx

2
2
Standard Deviation :   Var ( X )
Moment - generating function : M (t )  E e     etx  f ( x ) dx

tX


Expected value of u ( X ) : E[u ( X )]   u ( x) f ( x) dx

Example 3.2.2
2 x  2 1  x  2
f ( x)  
elsewhere
0
Mean :   
2
1
2
2 3
5
2
x(2 x  2) dx  x  x 
3
3
1
Variance :   
2
2
1
1
( x  5 / 3) (2 x  2) dx 
18
2
x2
m.g.f.: M (t )  
2
1
 x 1 1 
e (2 x  2) dx  2e 
 2
t  x 1
 t
tx
tx
1 1 
t 1
 2e   2   2e 2 for t  0
t
t t 
2t
Percentiles
Definition 3.2.3 Let X be continuous random
variable with p.d.f. f and c.d.f. F, and let p be a
number between 0 and 1. The (100p)th percentile is
a value of X, denoted πp, such that
p
p

f ( x) dx  F  p 
• The 50th percentile is called the median and is
denoted m.
• The mode of X is the value x for which f is
maximum.
Special Percentiles
• 25th – First Quartile – p1 = π0.25
• 50th – Second Quartile – p2 = π0.50 = m
• 75th – Third Quartile – p3 = π0.75
Example 3.2.4
2 x  2 1  x  2
f ( x)  
 F ( x)  x 2  2 x  1 for 1  x  2
elsewhere
0
Find p1   0.25 : Solve the quadratic
0.25  F  0.25    0.25   2  0.25   1   0.25  1.5
2
3.3 – Uniform and Exponential
Definition 3.3.1 A continuous random variable X
has a uniform distribution if its p.d.f. is
 1

f ( x)   b  a
0
a xb
otherwise
where a < b are any real numbers. The phrases
“X is U(a, b)” and “X is uniformly distributed
over [a, b]” mean that X has this type of
distribution.
Uniform Distribution
Mean :

b
a
b
1
1 x 
x
dx 
 
ba
ba 2 a
1  b2 a 2 

  
ba 2 2 
 (b  a )(b  a )  1


2

ba
ab

2
(b  a ) 2
2
Variance :  
12
2
etb  eta
m.g.f.: M (t ) 
,t 0
t (b  a )
Example 3.3.2
The owner of a local gas station observes that
customers typically purchase between five and
20 gallons of gasoline at each fill-up and that no
amount is more frequent than any other. If gas
costs $2.50 per gallon, find the probability that a
randomly selected customer spends more than
$30 on gas.
Example 3.3.2
Let X = gallons of gas purchased by a customer
– Assume X is U(5, 20)
– 𝑓 𝑥 = 1/15 for 5 < 𝑥 ≤ 20
Let Y = amount of money spent by a customer
– Y = 2.5X
P(Y  30)  P(2.5 X  30)  P( X  12)
20
1
8

dx   0.533
12 20  5
15
Exponential Distribution
Definition 3.3.2 A random variable X has an
exponential distribution if its p.d.f. is
 e   x
f  x  
0
x0
otherwise
where 𝜆 > 0
– Describes waiting time between events that are
described by a Poisson distribution
Exponential Distribution
m.g.f.: M (t ) 
Mean :  

 t
1

Variance :  2 
1
2
c.d.f.: F ( x)  P( X  x)  F  x  
x


x
f  t  dt    e   t dt  1  e   x for x  0
0
P( X  x)  1  P ( X  x)  1  1  e   x   e   x for x  0
Example 3.3.6
Suppose a manufacturer of electrical wire
observes a mean of 0.5 defects per 100 ft of wire.
Find the mean distance between defects (in ft)
and find the probability that the distance between
one defect and the next is less than 125 ft.
– Assume the number of defects per ft has a Poisson
distribution
0.5 defects

 0.005 defects/ft
100 ft
Example 3.3.6
Let X = the distance in ft between defects
– “Wait time” between defects
– X is exponential with 𝜆 = 0.005
1
Mean :  
 200
0.005
P( X  125)  F (125)  1  e 0.005(125)  0.465
3.4 – The Normal Distribution
Definition 3.4.1 A continuous random variable Z
has a standard normal distribution if its p.d.f. is
1  z 2 /2
f ( z) 
e
for    z  
2
m.g.f.: M (t )  e
Mean :   0
t 2 /2
Variance :  2  1
The c.d.f.
( z )  P( Z  z )  
z

1  t 2 /2
e
dt
2
See Table C.1 – Two important properties
1. P( Z  z )  1  P( Z  z )  1   ( z )
2. P(a  Z  b)   (b)   (a )
Example 3.4.1
P( Z  2.05)  0.9798
P( Z  1.23)  1   (1.23)  1  0.1093  0.8907
P(2.65  Z  0.23)   (0.23)   (2.65)  0.5910  0.0040  0.5870
The Normal Distribution
Definition 3.4.2 A continuous random variable X
has a normal distribution if its p.d.f. is
 ( x   )2 /  2 2 
1
f ( x) 
e
for    x  
 2
where μ is any real number and σ > 0
– The phrase “X is 𝑁 𝜇, 𝜎 2 ” means that X has a
normal distribution with parameters μ and σ
Mean : 
Variance :  2

 2t 2 
m.g.f. : M (t )  exp   t 

2 

The Normal Distribution
Empirical Rule
P(     X     )  0.68, P(   2  X    2 )  0.95
and P(   3  X    3 )  0.997
Relationship
Theorem 3.4.1 If X is 𝑁 𝜇, 𝜎 2 then the random
𝑋−𝜇
variable 𝑍 =
is 𝑁 0, 1
𝜎
Proof: We find the c.d.f. of Z:
 X 

F ( z )  P( Z  z )  P 
 z   P ( X  z   )
 

z  
 ( x   )2 /  2 2 
1

e
dx

 2
Let 𝑦 = (𝑥 − 𝜇)/𝜎 so that 𝑑𝑦 = 1/𝜎 𝑑𝑥,
x    y  , and
( z   )  
x  z    y 
z

Thus
1  y 2 /2
F ( z)  
e
dy   ( z )

2
so Z has the same c.d.f. as that of a standard normal
random variable. Therefore, it must have the same
p.d.f., and we conclude that Z is N (0, 1).
z
Application
 X  x 
P( X  x)  P 






The “z-score” of x is
z
x

 P( X  x)  P  Z  z    ( z )
Example 3.4.2
Consider the random experiment of choosing a
woman at random and measuring her height. Let
the random variable X be the height in inches.
Assuming that X is 𝑁 63.6, 2.52 , find the
probability of selecting a woman who is between
60 and 62 inches tall.
Example 3.4.2
We want 𝑃(60 < 𝑋 ≤ 62)
– Calculate z-scores:
60  63.6
62  63.6
z1 
 1.44 and z2 
 0.64
2.5
2.5
 P(60  X  62)  P(1.44  Z  0.64)
  (0.64)   (1.44)
 0.2611  0.0749  0.1862
3.5 – Functions of Continuous
Random Variables
Goal: To find the p.d.f. of a variable defined in
terms of another variable
– An important relationship
d x
F ( x) 
f (t ) dt  f ( x)

dx 
– Second Fundamental Theorem of Calculus
d u ( x)
du
f (t ) dt  f (u ( x))

dx a
dx
Example 3.5.2
Consider the random variable X with range 0 <
𝑥 < 1 and p.d.f. 𝑓𝑥 𝑥 = 5𝑥 4 . Define the
variable 𝑌 = 𝑋 2 . Find the p.d.f. of Y.
– Find the c.d.f.


FY ( y )  P(Y  y )  P  X  y   P X  y   5 x 4 dx
2
y
0
– Find the p.d.f.
d y 4
fY ( y ) 
5 x dx  5

0
dy
 y
4
1
5 3/2
 y , 0  y 1
2 y 2
Simulations
Theorem 3.5.1 Let Y be a random variable that
is 𝑈(0, 1) and let 𝐹(𝑥) satisfy the requirements
of a c.d.f. of a continuous random variable with
certain properties.
– Let 𝐹 −1 denote the inverse of the function 𝐹
– Define the random variable 𝑋 = 𝐹 −1 (𝑌)
– Then 𝑋 has the c.d.f. 𝐹(𝑥)
Example 3.5.4
Generate values of a random variable X with an
exponential distribution
1. Find the c.d.f.: 𝐹 𝑥 = 1 − 𝑒 −𝜆𝑥
2. Find 𝐹 −1 (𝑦)
y  1 e
 x
 x
1

ln 1  y   F
1
 y  
1

ln 1  y 
Example 3.5.4
3. Use software to generate values of a random
variable Y that is 𝑈(0, 1)
– Suppose 𝜆 = 2
– 𝑦𝑖 - Values of Y
– 𝑥𝑖 - Values of X
3.6 – Joint Distributions
Definition 3.6.2 Let X and Y be two continuous
random variables. The joint probability density
function, or joint p.d.f., is an integrable function
f(x, y) that satisfies the following three
properties:
1. 0  f ( x, y ) for all    x   and    y  
2. 



 
f ( x, y ) dy dx  1
3. if S is a subset of the two-dimensional plane, then
P  ( X ,Y )  S    f ( x, y ) dy dx
S
Marginal Distributions
The marginal probability density functions, or
marginal p.d.f.’s, of X and Y are
f X ( x)  


f ( x, y ) dy
and
fY ( y )  


f ( x, y ) dx
Definition 3.6.3 Two discrete or continuous
random variables X and Y are said to be
independent if
f ( x, y )  f X ( x ) fY ( y )
Example 3.6.3
Suppose the joint p.d.f. of X and Y is
1/ 2 0  x  1, 0  y  2
f  x, y   
otherwise
0
Note

 

 
f ( x, y ) dy dx  
1

2
0 0
1
1 1
1
dy dx   2 dx  (2)  1
2
2 0
2
1
f X ( x)   dy  1 for 0  x  1 and
0 2
11
1
fY ( y )   dx  for 0  y  2
0 2
2
2
Marginal p.d.f.’s:
Example 3.6.4
Suppose a man and a woman agree to meet at a
restaurant sometime between 6:00 and 6:15 PM.
Find the probability that the man has to wait
longer than five minutes for the woman to arrive.
– Let X and Y denote the number of minutes past
6:00 PM that the man and the woman arrive,
respectively.
Example 3.6.4
Assume
1. Both X and Y are U  0, 15  so that f X  x   f Y  y   1/15
for 0  x  15, 0  x  15
2. X and Y are independent so that f ( x, y )  f X ( x) fY ( y )  (1/15) 2
Then
P(Y  X  5)  
10
0
2
1
1
x5  15  dy dx   15 
15
1
 
 15 
2
10

x 
2
10 x    .
2 0 9

2
2
 15  ( x  5) dx
10
0
3.7 – Functions of Independent
Random Variables
Definition 3.7.1 The mean or expected value of a
function of two random variables 𝑋 and 𝑌,
𝑢(𝑋, 𝑌) is
E u ( X ,Y )   



 
u ( x, y ) f ( x, y ) dy dx
where 𝑓(𝑥, 𝑦) is the joint p.d.f. of 𝑋 and 𝑌
Example 3.7.1
Suppose the joint p.d.f. of X and Y is
1/ 2 0  x  1, 0  y  2
f  x, y   
otherwise
0
E  XY  3 X  2Y   
2



 
2
xy
  3x  2 y  f ( x, y) dy dx
1
    xy 2  3x  2 y  dy dx
0 0 2
1 4x
25

 3 x  2 dx 
0 3
6
1
2
Properties
Theorem 3.7.1 For any discrete or continuous
random variables,
E  X  Y   E  X   E Y 
If X and Y are independent, then
E  XY   E  X  E Y 
Theorem 3.7.2 If X and Y are independent, then
Var ( X  Y )  Var ( X )  Var (Y )
Properties
Theorem 3.7.3 If 𝑋 and 𝑌 are independent
random variables with respective m.g.f.’s 𝑀𝑋 (𝑡)
and 𝑀𝑌 𝑡 , and a and b are constants, then the
m.g.f. of 𝑊 = 𝑎𝑋 + 𝑏𝑌 is
M W  t   M X  at  M Y  bt 
Properties
Theorem 3.7.4 If X is N   X ,  X2  , Y is N  Y ,  Y2  , X and Y
are independent, and a and b are constants, then the variable
W  aX  bY
is N  a  X  bY , a 2 X2  b 2 Y2  .
Theorem 3.7.5 Let X and Y be independent discrete or
continuous random variables and let g ( x) and h( x) be any
functions. Then the random variables Z  g ( X ) and W  h( y )
are independent.
Example 3.7.5
Let X and Y be independent and both 𝑁 220, 292
Define 𝑊 = 𝑋 + 𝑌. Then 𝑊 is
N  220  220, 29  29
2
2
  N  440,1682 
So that
500  440 

P(W  500)  P  Z 
  P( Z  1.46)  0.0721
1682 

3.8 – The Central Limit Theorem
Definition 3.8.1 When a set of objects is selected
from a larger set of objects, the larger set is
called the population and the smaller set is called
the sample. The number of objects in the sample
is called the sample size.
The Central Limit Theorem
Theorem 3.8.2 If X1, X2,…, Xn are mutually
independent random variables with a common
distribution, mean μ, and variance σ2, then as
𝑛 → ∞ the distribution of the sample mean
Xn 
X1 
 Xn
n
approaches
 2 
N  , 
n 

Example 3.8.5
Suppose people at a certain movie theater wait a
mean of 30 sec with a variance of 52 sec2 to buy
tickets. Find the probability that the mean wait
time of a group of 50 movie-goers is less than 28
sec.
– We want to find P  X 50  28 
Example 3.8.5
By CLT:
1. X 50 is approximately normally distributed
2. X 50 has mean 30
3. X 50
Then
52
has variance
 0.5
50
28  30 

P  X 50  28   P  Z 
  P ( Z  2.83)  0.0023
0.5 

3.9 – The Gamma and Related
Distributions

The Gamma Function: (t )   y t 1 e  y dy for 0  t
0
 If n is an integer: (n)  (n  1)!
Definition 3.9.2 A random variable X has a
gamma distribution with parameters r and λ
where 𝑟 > 0 and 𝜆 > 0 if its p.d.f. is
f ( x) 
r
(r )
x r 1 e   x
for x  0
Chi-Square Distribution
Theorem 3.9.1 Let 𝑌1 , 𝑌2 , … , 𝑌𝑛 be independent
random variables each with a standard normal
distribution. Define the random variable
X  Y12  Y22 
 Yn2
1
n /2 1  x /2
The p.d.f. of X is f ( x) 
x
e
n /2
(n / 2)2
for x  0
Definition 3.9.3 A random variable with this
p.d.f. is said to have a chi-square distribution
with n degrees of freedom.
Chi-Square Distribution
Definition 3.9.4 Let p be a number between 0 and 1 and let
X be  2 (n). Define the critical value  p2 (n) to be a positive
number such that
P  X   p2 (n)   1  p
 See Table C.3
Student-t Distribution
Definition 3.9.5 Let 𝑍 and 𝐶 be independent
random variables where 𝑍 is 𝑁(0, 1) and 𝐶 is
𝜒 2 (𝑛). The random variable
Z
T
C/n
is called the Student-t ratio with n degrees of
freedom.
Student-t Distribution
Theorem 3.9.2 The p.d.f. for the Student-t ratio
with n degrees of freedom is
 n 1 


2


f (t ) 
,   t  
(
n

1)/2
2
 n  t 
n    1  
n
 2 
Definition A random variable with this p.d.f. is
said to have a Student-t distribution with n
degrees of freedom.
Student-t Distribution
Definition 3.9.6 Let 𝑝 be a number between 0 and 0.5
and let 𝑇 have a Student-t distribution with n degrees of
freedom. A critical t-value is a number 𝑡𝑝 (𝑛) such that
– See Table C.2
P T  t p ( n)   1  p
F-Distribution
Theorem 3.9.3 Let 𝑈 and V be two independent
chi-square random variables with n and d
degrees of freedom, respectively. Define the
random variable
U /n
F
V /d
 n  d  n /2 d /2 ( n /2) 1

n d x
2 
f ( x)  
, x0
n d 
      (d  nx)( n  d )/2
2  2
The p.d.f. of 𝐹 is
F-Distribution
Definition 3.9.7 A random variable with this
p.d.f. is said to have an F-distribution with n and
d degrees of freedom. The number n is called the
numerator degrees of freedom while d is called
the denominator degrees of freedom.
F-Distribution
Definition 3.9.8 Let p be a number between 0
and 1 and let 𝐹 have an F-distribution with n and
d degrees of freedom. A critical F-value is a
number 𝑓𝑝 (𝑛, 𝑑) such that
P  F  f p ( n, d )   1  p
– See Table C.4
3.10 – Approximating the Binomial
Distribution
Theorem 3.10.1 (Limit Theorem of De Moivre
and Laplace) Let 𝑋 be 𝑏(𝑛, 𝑝). Then as 𝑛 → ∞
the distribution of 𝑋 approaches
N  np, np (1  p ) 
Application: Let 𝑌 be 𝑁 𝑛𝑝, 𝑛𝑝 1 − 𝑝 . Then
P(a  X  b)  P(a  0.5  Y  b  0.5)
Example 3.10.1
An airline has a policy of selling 375 tickets for a
plane that seats only 365 people (called “overbooking”). Records indicate that about 95% of
people who buy tickets for this flight actually
show up. Find the probability that there are
enough seats.
Example 3.10.1
Let X = number of ticket-holders who show up
– X is 𝑏(375, 0.95)
– Let Y be
N (375  0.95,375  0.95  0.05)  N (356.25,17.81)
Then
365.5  356.25 

P( X  365)  P(Y  365.5)  P  Z 

17.81


 P( Z  2.19)  0.9857
Bernoulli’s Law of Large Numbers
Theorem 3.10.2 Let the random variable 𝑋 be
the number of times a specified event 𝐴 is
observed in n trials of a random experiment. Let
𝑝 = 𝑃(𝐴). Then for any small number 𝜀 > 0,
 X

P   p  ò   1 as n  
 n
