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Chapter 4-1
Continuous Random Variables
主講人:虞台文
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
今天中午台北市氣溫為
25C之機率為何?
今天中午台北市氣溫小於
或等於25C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
x0
0 x 1
1 x 2
2 x3
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 yR
Example 1
FY ( y) P(Y y)
0
2
y
2
R
1
y0
0 yR
1 y
R
y
Example 1
FY ( y) P(Y y)
0
2
y
2
R
1
y0
F Y (y )
1
0 yR
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
RY
z0
R
0 z
R
3
R
3
z
R
2
R
2
z
2R
3
2R
3
zR
Rz
R/2
Example 1
FZ ( z ) P( Z z )
0
( R z )2
1 R2
5
98
9
1 ( R z )2
R2
1
z0
F Z (y )
1
0 z
R
3
R
3
z
R
2
R
2
z
2R
3
2R
3
zR
Rz
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
x0
0
x0
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
x0
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
x0
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
, x0
: arriving rate
: failure rate
The Exponential Distributions
X ~ Exp( )
pdf
f ( x) e
x
1 e
cdf F ( x)
0
, x0
x
x0
x0
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
, x0
x
x0
x0
6
X ~ Exp( )
a0
P ( X a b | X a ) P ( X b)
b0
Memoryless or Markov Property
X ~ Exp( )
a0
P ( X a b | X a ) P ( X b)
b0
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
x0
x0
X ~ Exp( )
a0
P ( X a b | X a ) P ( X b)
b0
Memoryless or Markov Property
Exercise:
連續型隨機變數中,唯有指數分佈具備無記憶性。
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
能求出P(X > t)嗎?
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
x0
: failure rate
PLet
( X r.v.
t )Nt 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
能求出P(X > t)嗎?
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 e10 )
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
生命將在時間t後瞬間結束的機率
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
生命將在時間t後瞬間結束的機率
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
瞬間暴斃率h(t)
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
瞬間暴斃率h(t)
f (t )
h(t )
R(t )
X ~ Exp( )
pdf
Example 6
f ( x) e x , x 0
1 e x
cdf F ( x)
0
x0
x0
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
以指數分配來model物件壽命之機率分配合理嗎?
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
, t0
0
0
2 0
2
t
0t 2
R(t ) exp
,
2
t
t 0
Chapter 4-1
Continuous Random Variables
The Erlang
Distributions
我的老照相機與閃光燈
它只能使用四次
每使用一次後轉動九十度
使用四次後壽終正寢
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 1et
(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 1et
(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 1et
(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 1et
(r 1)!
(r 1)!
,
t 0
X
Example 8
Erlang (r , )
f (t )
r x r 1e x
(r 1)!
,
x0
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
欲將之推廣為正實數
Erlang (r , )
pdf f (t )
x e
r
r 1 x
(r 1)!
,
x0
0
Review
X
Erlang (
r, )
pdf f (t )
r 1 x
r
x e
,
(r 1)!( )
x0
The Gamma Distributions
X
( , ), 0
x e
pdf f ( x)
( )
1 x
,
x0
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. n1 n1
2 2 2 !
,
x0
X
Chi-Square
Distributions
( , )
x e
f (t )
( )
1 x
X
1 1
,
2 2
X
v 1
,
2 2
2
2
v
,
x0
Chapter 4-1
Continuous Random Variables
The Gaussian or
Normal Distributions
The Gaussian or Normal Distributions
德國的10馬克紙幣, 以高斯(Gauss, 1777-1855)為
人像, 人像左側有一常態分佈之p.d.f.及其圖形。
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
tx
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
tx
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 xb
pdf f ( x)
ba
0
xa
F
(
x
)
cdf
b a
1
1
ba
x
a
ax
1
b
F(x)
a xb
bx
x
a
b
Summary
The Exponential Distributions
The Erlang Distributions
The Gamma Distributions
The Gaussian or Normal Distributions
The Uniform Distributions