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```Chapter 4-1
Continuous Random Variables

Content


Random Variables and Distribution Functions
Probability Density Functions of Continuous Random
Variables

The Exponential Distributions

The Reliability and Failure Rate

The Erlang Distributions

The Gamma Distributions

The Gaussian or Normal Distributions

The Uniform Distributions
Chapter 4-1
Continuous Random Variables
Random Variables and
Distribution Functions
The Temperature in Taipei

25C之機率為何?

Renewed Definition of Random Variables
A random variable X on a probability space (, A, P) is
a function
X : R
that assigns a real number X() to each sample point
, such that for every real number x, the set
{|X()  x} is an event, i.e., a member of A.
The (Cumulative) Distribution Functions
The (cumulative) distribution function FX of a
random variable X is defined to be the function
FX(x) = P(X  x), − < x < .
Example 1
Example 1
x
p X ( x)
0 1 2 3
1
8
3
8
3
8
1
8
F X (x )
1
0
1
 8
FX ( x)   84
7
8
1
x0
0  x 1
1 x  2
2 x3
3 x
x
-3 -2 -1
0
1
2
3
4
5
6
7
Example 1
FY ( y)  P(Y  y)
y

 R2
2
R
y
2
y
 2
R
0 yR
Example 1
FY ( y)  P(Y  y)
0
 2
y
 2
R
 1
y0
0 yR
1 y
R
y
Example 1
FY ( y)  P(Y  y)
0
 2
y
 2
R
 1
y0
F Y (y )
1
0 yR
0.5
1 y
y
0
R /2
R
Example 1
FZ ( z )  P( Z  z )
0
 ( R  z )2
1  R2
5

  98
9
1  ( R  z )2
R2

1
RY
z0
R
0 z
R
3
R
3
z
R
2
R
2
z
2R
3
2R
3
zR
Rz
R/2
Example 1
FZ ( z )  P( Z  z )
0
 ( R  z )2
1  R2
5

  98
9
1  ( R  z )2
R2

1
z0
F Z (y )
1
0 z
R
3
R
3
z
R
2
R
2
z
2R
3
2R
3
zR
Rz
0.5
z
0
R /3 R /2 2R /3
R
Example 1
FX ( x)
11
FY ( y )
FFXX(x(x))
FZ ( z )
FFZ (y(y) )
Z
FFY (y(y) )
Y
11
11
0.5
0.5
0.5
0.5
xx
-3
-3 -2
-2 -1
-1 00 11 22 33 44 55 66 77
00
yy
RR/2/2
RR
00
zz
R /3 R /2 2R /3
R /3 R /2 2R /3
R
R
Properties of Distribution Functions
0  F(x)  1 for all x;
F is monotonically nondecreasing;
F() = 0 and F() =1;
F(x+) = F(x) for all x.
1.
2.
3.
4.
11
FFXX(x(x))
FFZ (y(y) )
Z
FFY (y(y) )
Y
11
11
0.5
0.5
0.5
0.5
xx
-3
-3 -2
-2 -1
-1 00 11 22 33 44 55 66 77
00
yy
RR/2/2
RR
00
zz
R /3 R /2 2R /3
R /3 R /2 2R /3
R
R
Definition 
Continuous Random Variables
A random variable X is called a continuous
random variable if
P ( X  x )  F ( x  )  F ( x )  0
11
FFXX(x(x))
FFZ (y(y) )
Z
FFY (y(y) )
Y
11
11
0.5
0.5
0.5
0.5
xx
-3
-3 -2
-2 -1
-1 00 11 22 33 44 55 66 77
00
yy
RR/2/2
RR
00
zz
R /3 R /2 2R /3
R /3 R /2 2R /3
R
R
Example 2
11
FFXX(x(x))
FFZ (y(y) )
Z
FFY (y(y) )
Y
11
11
0.5
0.5
0.5
0.5
xx
-3
-3 -2
-2 -1
-1 00 11 22 33 44 55 66 77
00
yy
RR/2/2
RR
00
zz
R /3 R /2 2R /3
R /3 R /2 2R /3
R
R
Chapter 4-1
Continuous Random Variables
Probability Density
Functions of Continuous
Random Variables
Probability Density Functions of
Continuous Random Variables
A probability density function (pdf) fX(x) of a
continuous random variable X is a nonnegative
function f such that
x
FX ( x)   f X (u)du

Probability Density Functions of
Continuous Random Variables
A probability density function (pdf) fX(x) of a
continuous random variable X is a nonnegative
function f such that
x
FX ( x)   f X (u)du

x
FX ( x)   f X (u)du

Properties of Pdf's
1. f ( x)  0;
Remark: f(x) can be larger than 1.

2.  f ( x)dx  1;

3. P(a  X  b)  P(a  X  b)  P(a  X  b)  P(a  X  b)
 P( X  b)  P( X  a)  F (b)  F (a)
b
a
b


a
  f ( x)dx   f ( x)dx   f ( x)dx
dF ( x)
4. f ( x) 
 F ( x).
dx
Example 3
k 6
Example 3
kx(1  x) 0  x  1
f ( x)  
otherwise
0
k 6
1
x
x  k
1   f ( x)dx   kx(1  x)dx  k    

0
 2 3 0 6

1
2
3
k 6
Example 3
6 x(1  x) 0  x  1
f ( x)  
otherwise
0
x0
0
x0
0
 x
 2
3
F ( x)    6u (1  u )du 0  x  1  3x  2 x 0  x  1
0
1

1 x

1 x
1
k 6
Example 3
6 x(1  x) 0  x  1
f ( x)  
otherwise
0
2
f (x )
1.5
1
x0
0

F ( x)  3x 2  2 x3 0  x  1
1
1 x

0.5
x
0
-1
0
1.2
1
2
F (x )
1
0.8
0.6
0.4
0.2
x
0
-1
0
1
2
k 6
Example 3
6 x(1  x) 0  x  1
f ( x)  
otherwise
0
2
f (x )
1.5
1
x0
0

F ( x)  3x 2  2 x3 0  x  1
1
1 x

0.5
x
0
-1
0
1.2
1
2
F (x )
1
0.8
P( X  13 )  3( 13 ) 2  2( 13 )3
7

 0.25926
27
0.6
0.4
0.25926
0.2
x
0
-1
0
1/3
1
2
Chapter 4-1
Continuous Random Variables
The Exponential Distributions
The Exponential Distributions

The following r.v.’s are often modelled as
exponential:
1.
Interarrival time between two successive
job arrivals.
2.
Service time at a server in a queuing
network.
3.
Life time of a component.
The Exponential Distributions
A r.v. X is said to possess an exponential distribution
and to be exponentially distributed, denoted by
X ~ Exp(), if it possesses the density
f ( x)   e
 x
, x0

: arriving rate
: failure rate
The Exponential Distributions
X ~ Exp( )
pdf
f ( x)   e
 x
1  e
cdf F ( x)  
0
, x0
 x
x0
x0

3.5
: arriving rate
: failure rate
f (x )
3
2.5
2
X ~ Exp (2)
1.5
1
The Exponential Distributions
0.5
x
0
-2
0
1.2
2
4
6
F (x )
1
X ~ Exp( )
0.8
0.6
0.4
0.2
x
0
-2
0
2
4
-0.2
pdf
f ( x)   e
 x
1  e
cdf F ( x)  
0
, x0
 x
x0
x0
6
X ~ Exp( )
a0
P ( X  a  b | X  a )  P ( X  b)
b0
Memoryless or Markov Property
X ~ Exp( )
a0
P ( X  a  b | X  a )  P ( X  b)
b0
Memoryless or Markov Property
P( X  a  b | X  a ) 
P( X  a  b and X  a)
P( X  a )
P ( X  a  b ) 1  P ( X  a  b ) 1  F ( a  b)



P( X  a)
1  P( X  a )
1  F (a)
e   ( a b )
   a  e  b
e
 P ( X  b)
X ~ Exp( )
pdf
f ( x)   e   x , x  0
1  e  x
cdf F ( x)  
0
x0
x0
X ~ Exp( )
a0
P ( X  a  b | X  a )  P ( X  b)
b0
Memoryless or Markov Property
Exercise:

The Relation Between Poisson and
Exponential Distributions
: arriving rate

: failure rate
Let r.v. Nt denote #jobs arriving to a computer system
in the interval (0, t].
Nt ~ ?P(t )
Nt
0
t
The Relation Between Poisson and
Exponential Distributions
: arriving rate

: failure rate
Let r.v. Nt denote #jobs arriving to a computer system
in the interval (0, t].
Nt ~ ?P(t )
Nt
The next
arrival t
0
X
Let X denote the time of the next arrival.
X ~?
f (t )  ? or F (t )  ?
P( X  t )  P( Nt  0)
The Relation Between Poisson and
Exponential Distributions
: arriving rate

: failure rate
Let r.v. Nt denote #jobs arriving to a computer system
in the interval (0, t].
Nt ~ ?P(t )
Nt
0

t
The next
arrival
X
Let X denote the time of the next arrival.
X ~?
f (t )  ? or F (t )  ?
P( X  t )  P( Nt  0)
X ~ Exp( )
pdf
cdf
f ( x)   e   x , x  0
The Relation
Between
Poisson
and
 x

1

e
x  Distributions
0
Exponential
: arriving rate
F ( x)  
0

x0
: failure rate
PLet
( X r.v.
 t )Nt P
( Nt  0)
denote
#jobs arriving to a computer system
t
in the interval
(t )0(0,
e t].

 e  t t  0
t
0! N
N ~ ?P(t )
 t
t
F (t )  1  e
t 0
0

t 0
f (t )  e t
t
The next
arrival
X
Let X denote the time of the next arrival.
X ~?
f (t )  ? or F (t )  ?
The Relation Between Poisson and
Exponential Distributions
: arriving rate

: failure rate
Let r.v. Nt denote #jobs arriving to a computer system
in the interval (0, t].
Nt
0
Nt ~ ?P(t )
The next
arrival t
X
Let X denote the time of the next arrival.
X ~ Exp( )
The Relation Between Poisson and
Exponential Distributions
: arriving rate

t1
t2
t3
t4
: failure rate
t5
The interarrival times of a Poisson process are
exponentially distributed.
P(“No job”) = ?
0
Example 5
10 secs
 = 0.1 job/sec
P(“No job”) = ?
0
Example 5
10 secs
 = 0.1 job/sec
Method 1: Let N10 represent #jobs arriving in the 10 secs.
P("No job")  P( N10  0)
10 e 1
 e 1

0!
N10 ~ P(1)
Method 2: Let X represent the time of the next arriving job.
P("No job")  P( X  10)
 1  P( X  10)
 1  (1  e10 )
 e 1
X ~ Exp(0.1)
Chapter 4-1
Continuous Random Variables
The Reliability
and
Failure Rate
Definition  Reliability
Let r.v. X be the lifetime or time to failure of a component. The
probability that the component survives until some time t is
called the reliability R(t) of the component, i.e.,
R(t) = P(X > t) = 1  F(t)
Remarks:
1.
F(t) is, hence, called unreliability.
2.
R’(t) = F’(t) = f(t) is called the failure density function.
The Instantaneous Failure Rate
The Instantaneous Failure Rate

P(t  X  t  t | X  t )
t
0
t
t+t
F (t  t )  F (t )
P(t  X  t  t | X  t ) 
R(t )
The Instantaneous Failure Rate

P(t  X  t  t | X  t )
P(t  X  t  t and X  t )

P( X  t )
P(t  X  t  t ) F (t  t )  F (t )


P( X  t )
R(t )
F (t  t )  F (t )
P(t  X  t  t | X  t ) 
R(t )
The Instantaneous Failure Rate

P(t  X  t  t | X  t )
h(t )  lim
t 0
t
F (t  t )  F (t )
F (t )
 lim

t 0
tR(t )
R(t )
f (t )

R(t )
The Instantaneous Failure Rate

f (t )
h(t ) 
R(t )
X ~ Exp( )
pdf
Example 6
f ( x)   e   x , x  0
1  e  x
cdf F ( x)  
0
x0
x0
Show that the failure rate of exponential distribution
is characterized by a constant failure rate.
f (t )
 e  t
f (t )

  t  
h(t ) 
e
R(t ) 1  F (t )
t 0

More on Failure Rates
h(t)
CFR
h(t )  0

t
More on Failure Rates
h(t)
IFR
DFR
h(t )  0
Useful Life
CFR
h(t )  0
h(t )  0

t
More on Failure Rates
h(t)
DFR
h(t )  0

?
Exponential
Distribution
Useful Life
CFR
h(t )  0
IFR
h(t )  0
?
t
Relationships among F(t), f(t), R(t), h(t)
1  F (t )
F (t )
dF (t )
dt
f (t )
R (t )
f (t )
R (t )
h(t )
Relationships among F(t), f(t), R(t), h(t)
F (t )

f (t )
f (t )
R (t )
1  F (t )
R (t )
f (t )
h(t )
Relationships among F(t), f(t), R(t), h(t)
F (t )
1  R (t )
d
F (t )
dt
f (t )
R (t )
f (t )
R (t )
h(t )
Relationships among F(t), f(t), R(t), h(t)
F (t )
R (t )
?
f (t )
?
?
h(t )
Cumulative Hazard
t
H (t )   h( x)dx  0
t
0
t R( x)
f ( x)
dx   
dx
0
R( x)
R( x)
t
1
 
dR( x)   ln R( x) 0
0 R( x)
t
  ln R(t )  ln R(0)  ln1  ln R(t )   ln R (t )
R(t )  e
 H (t )
t

0
e

h ( x ) dx
Relationships among F(t), f(t), R(t), h(t)
F (t )
R (t )
1  R (t )
f (t )
d
F (t )
dt
t
h ( x ) dx

e 0

h(t )
Example 7
2 t
x
0t 2
H (t )   h( x)dx  0  xdx  0

, t0
0
0
2 0
2
t
 0t 2 
R(t )  exp 
,
 2 
t
t 0
Chapter 4-1
Continuous Random Variables
The Erlang
Distributions

The Erlang Distributions
time
The lifetime of my flash (X)
[0, )
I(X)=?
fX(t)=?
Nt ~ P(t)
The Erlang Distributions




Consider a component subjected to an environment so that
Nt, the number of peak stresses in the interval (0, t], is
Poisson distributed with parameter t.
Suppose that the rth peak will cause a failure.
Let X denote the lifetime of the component.
Then,
( t ) k e   t
P( X  t )  ?P( Nt  r )  
k!
k 0
r 1
(t )k e t
,
cdf F (t )  1  
k!
k 0
r 1
t 0
Nt ~ P(t)
The Erlang Distributions




Exercise of
Consider a component subjected to an environment so that
Chapter
Nt, the number of peak stresses in the interval
(0, t],2is
Poisson distributed with parameter t.
Suppose that the rth peak will cause a failure.
Let X denote the lifetime of the component.
Then,
pdf f (t ) 
 r t r 1et
(r  1)!
,
t 0
(t )k e t
,
cdf F (t )  1  
k!
k 0
r 1
t 0
The r-Stage Erlang Distributions




Consider a component subjected to an environment so that
Nt, the number of peak stresses in the interval (0, t], is
Poisson distributed with parameter t.
Suppose that the rth peak will cause a failure.
Let X denote the lifetime of the component.
Then,
pdf f (t ) 
 r t r 1et
(r  1)!
,
t 0
(t )k e t
,
cdf F (t )  1  
k!
k 0
r 1
t 0
Exp( )  Erlang (1,  )
The r-Stage Erlang Distributions
X
Erlang (r ,  )
pdf f (t ) 
 r t r 1et
(r  1)!
,
t 0
(t )k e t
,
cdf F (t )  1  
k!
k 0
r 1
t 0
Exp( )  Erlang (1,  )
The r-Stage Erlang Distributions
X
Erlang (r ,  )

t
e
f (t ) 
, t 0
r r 1  t
pdf
 r t r 1et
(r  1)!
(r  1)!
,
t 0
X
Example 8
Erlang (r ,  )
f (t ) 
 r x r 1e x
(r  1)!
,
x0
In a batch processing environment, the number of jobs arriving for
service is 9 per hour. If the arrival process satisfies the requirement of
a Poisson experiment. Find the probability that the elapse time between
a given arrival and the fifth subsequent arrival is less than 10 minutes.
 = 9 jobs/hr.
Let X represent the time of the 5th arrival.
P( X  16 hr)  ?
1/ 6
0
5
4 9 x
9 xe
4!
X ~ Erlang (5,9)
k
(9
/
6)
9/ 6
 0.0285
dx  1  e 
k!
k 0
4
Chapter 4-1
Continuous Random Variables
The Gamma
Distributions
Review
X
r為一正整數

Erlang (r ,  )
pdf f (t ) 
 x e
r
r 1   x
(r  1)!
,
x0
 0
Review
X
 
Erlang (
r,  )
pdf f (t ) 
 
r 1   x
r 
 x e
,
(r  1)!( )
x0
The Gamma Distributions
X
( ,  ),   0
 x e
pdf f ( x) 
( )
  1   x
,
x0
X
( ,  )
 x e
f (t ) 
( )
  1   x
Review

( )   x
 1  x
0
e dx
1. (1)  1
2. ( )  (  1)( )
3. ( )  (  1)!,   N
1
4.     
2
 (n  1)!
n
5.     n1 n1
 2  2  2 !
,
x0
X
Chi-Square
Distributions
( ,  )
 x e
f (t ) 
( )
  1   x
X
1 1

   , 
2 2
X
v 1

   , 
2 2
2
2
v
,
x0
Chapter 4-1
Continuous Random Variables
The Gaussian or
Normal Distributions
The Gaussian or Normal Distributions

The Gaussian or Normal Distributions
X
N ( ,  )
2

1
pdf f ( x)  n( x;  ,  ) 
e
2
2
( x   )2
2 2
  x  
 : mean
 : standard deviation
2: variance
The Gaussian or Normal Distributions
X
N ( ,  )
Inflection
point
2
f ( x)  n( x;  ,  )
2
1
e

2

( x )
2 2
  x  
2
Inflection
point
 : mean
 : standard deviation
2: variance
The Gaussian or Normal Distributions
X
N ( ,  )
2
  15
varying 
f ( x)  n( x;  ,  )
2

1
e

2
( x   )2
2 2
  x  
  100
varying 
 : mean
 : standard deviation
2: variance
The Gaussian or Normal Distributions
X
N ( ,  )
2
Facts:



f ( x)  n( x;  ,  )
e
 x2 / 2
dx  2
2

1
e

2
( x   )2
2
  x  
2
1
2

1
2



e
 x2 / 2


e
dx  1
 ( x   )2 / 2 2
dx  1
 : mean
 : standard deviation
2: variance
The Gaussian or Normal Distributions
X
N ( ,  )  e


2
f ( x)  n( x;  ,  )
2

1
e

2
( x   )2
2
  x  
2


0
e
 ( x 9)2 /8
 x2 /18
1
2

dx  2? 2
3 2
dx  ?
2


e
 ( x   )2 / 2 2
dx  1
Standard Normal Distribution
X
N ( ,  )
2
f ( x)  n( x;  ,  )
2

1
e

2
( x   )2
2 2
  x  
Z
N (0,1)
f Z ( z )  n( z;0,1)
1  z2 / 2

e
2
  z  
Table of N(0, 1)
Z
N (0,1)
f Z ( z )  n( z;0,1)
1  z2 / 2
FZ ( z )  ( z )

e
2

1 z  x2 / 2

e
dx
  z  


2
z
Table of N(0, 1)
 (1.625)  ?0.0521
(1.625)  ?0.9479
z
Fact:  (  z )  1   ( z )
FZ ( z )  ( z )
1 z  x2 / 2

e
dx

2 
Probability Evaluation for N(, 2)
f ( x) 
x
1
e
2

( x   )2
  x  
2 2

1
FX ( x) 
2
(t   )2
y2


2
2
2

x


( t   )2
e
2 2
t

1
dt  ?
2
y

t  y  
tx
t 

e
( t   )2
2 2
dt
Probability Evaluation for N(, 2)
f ( x) 
x
1
e
2

( x   )2
2 2
  x  
t  y  

( t   )2
( t   )2


x
tx
1
1
2 2
2 2

?
FX ( x) 
e
dt
e
dt



t 
2
2
x
 y x
2
1
1
 y2 / 2
y /2

e
dy

e
  dy 


2 
2  y   
 x 
 




Fact:
X 

N (0,1)
Probability Evaluation for N(, 2)
X
x
N ( ,  )
2

 x 
FX ( x)   

  
Z-Score:表距離中心若干個標準差
Example 9
X ~ N(12.00, 0.202)
1. P(11.92  X  12.27)  ?
2. P( X  12.45)  ?
3. P( X  11.70)  ?
X ~ N(12.00, 0.202)
Example 9
P(11.92  X  12.27)  P( X  12.27)  P( X  11.92)
 12.27  12 
 11.92  12 
 
 

0.2
0.2




  1.35    0.40  0.9115  0.3446  0.5669
1. P(11.92  X  12.27)  ?0.5669
2. P( X  12.45)  ?
3. P( X  11.70)  ?
X ~ N(12.00, 0.202)
Example 9
P( X  12.45)  1  P( X  12.45)
 12.45  12 
 1  
  1    2.25
0.2


 1  0.9878  0.0122
1. P(11.92  X  12.27)  ?0.5669
2. P( X  12.45)  ?0.0122
3. P( X  11.70)  ?
X ~ N(12.00, 0.202)
Example 9
 11.70  12    1.50
P( X  11.70)   



0.2


 0.0668
1. P(11.92  X  12.27)  ?0.5669
2. P( X  12.45)  ?0.0122
3. P( X  11.70)  ?0.0668
Example 10
N ( ,  2 )
X




|X  | < 
|X  | < 2
|X  | < 3


N ( ,  2 )
X


Example 10

|X  | < 
|X  | < 2
|X  | < 3
P(| X   | k )  P(k  X    k )
X 


 P  k 
k



  (k )   ( k )
 1  (k )  (k )
 1  2 (  k )



N ( ,  2 )
X

Example 10





|X  | < 
|X  | < 2
|X  | < 3
P(| X   | k )  1  2 ( k )
P(| X   |  )  1  2 (1)  1  2  0.1587  0.6826
P(| X   | 2 )  1  2(2)  1  2  0.0228  0.9544
P (| X   | 3 )  1  2(3)  1  2  0.0013  0.9974
Example 10
P(| X   | k )  1  2 ( k )
P(| X   |  )  1  2 (1)  1  2  0.1587  0.6826
P(| X   | 2 )  1  2(2)  1  2  0.0228  0.9544
P (| X   | 3 )  1  2(3)  1  2  0.0013  0.9974
Chapter 4-1
Continuous Random Variables
The Uniform
Distributions
The Uniform Distributions
X ~ U (a, b), a  b
f(x)
1
, a xb
pdf f ( x) 
ba
0
xa

F
(
x
)

cdf

b  a
1
1
ba
x
a
ax
1
b
F(x)
a xb
bx
x
a
b
Summary

The Exponential Distributions

The Erlang Distributions

The Gamma Distributions

The Gaussian or Normal Distributions

The Uniform Distributions
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