Transcript Chapter 4

Chapter 4.
Continuous Probability
Distributions
4.1 The Uniform Distribution
4.2 The Exponential Distribution
4.3 The Gamma Distribution
4.4 The Weibull Distribution
4.5 The Beta Distribution
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4.1 The Uniform Distribution
4.1.1 Definition of the Uniform Distribution(1/2)
• It has a flat pdf over a region.
U (a, b), X takes on values between a and b,
– if X
and
1
f ( x) 
ba
, for a  x  b
d c
P (c  X  d ) 
,
ba
fo r a  c < d  b
– Mean and Variance
a b
2
(b  a ) 2
V (X ) 
12
E( X ) 
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Figure 4.1 Probability density function of a
U(a, b) distribution
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4.2 The Exponential Distribution
4.2.1 Definition of the Exponential Distribution
•
Pdf:
f ( x)   e
•
•
 x
,
for x  0
F ( x)  1  e   x ,
0,
for x  0
o therw ise
Cdf:
Mean and variance:
E( X ) 
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1

and
V (X ) 
1
2
Figure 4.3
Probability density
function of an
exponential
distribution with
parameter  = 1
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• The exponential distribution often arises, in practice, as being of
the amount of time until some specific event occurs.
For example,
the amount of time until an earthquake occurs,
the amount of time until a new war breaks out, or
the amount of time until a telephone call you receive turns out to
be a wrong number, etc.
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4.2.2 The memoryless property of the Exponential Distribution
• For any non-negative x and y
P( X  x  y | X  x)  P( X  y )
• The exponential distribution is the only continuous distribution
that has the memoryless property
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• Memoryless property of exponential random variable:
P{ X  x  y | X  x}  P{ X  y}
This is equivalent to
P{ X  x  y, X  x}  P{ X  x  y}  P{ X  x}P{ X  y}
When X is an exponential random variable,
P{ X  x}  e x
The memoryless condition is satisfied since
e
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 ( x y )
e
 x  y
e
• Example: Suppose that a number of miles that a car run before
its battery wears out is exponentially distributed with an average
value of 10,000 miles. If a person desires to take a 5,000 mile
trip, what is the probability that he will be able to complete his
trip without to replace the battery?
(Sol)
Let X be a random variable including the remaining lifetime (in
thousand miles) of the battery. Then,
E[ X ]  1/   10    1/10
P{ X  5}  1  F (5)  e5  e1/ 2  0.604
What if X is not exponential random variable?
1  F (t  5)
P{ X  t  5 | X  t} 
1  F (t )
That is, an additional information of t should be known.
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• Proposition: If X1 , X 2 , , X n are independent exponential
random variables having respective parameters 1 , 2 , , n ,
then min( X1 , X 2 , , X n ) is the exponential random variable with
n
parameter

i 1
i
(proof)
P{min( X 1 , X 2 ,
n
  P{ X i  x}
i 1
n
  exp(i x)
i 1
n
 exp( i x)
i 1
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, X n )}  P{ X 1  x, X 2  x,
, X n  x}
• Example: A series system is one that all of its components to
function in order for the system itself to be functional. For an n
component series system in which the component lifetimes are
independent exponential random variables with respective
parameters 1 , 2 , , n .
What is the probability the system serves for a time t?
(Sol)
Let X be a random variable indicating the system lifetime. n
Then, X is an exponential random variable with parameter
i
Hence,
i 1
n

P{ X  t}  exp( i t )
i 1
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4.2.3 The Poisson process
• A stochastic process is a sequence of random events
• A Poisson process with parameter  is a stochastic process
where the time (or space) intervals between event-occurrences
follow the Exponential distribution with parameter  .
• If X is the number of events occurring within a fixed time (or
space) interval of length t, then
X
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P ( t )
Figure 4.7 A Poisson process. The number of events
occurring in a time interval of length t has a Poisson
distribution with mean t
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• The Poisson Process
Suppose that events are occurring at random time points and
let N(t) denote the number of points that occur in time interval
[0,t].
Then, A Poisson process having rate  (  0) is defined if
(a) N(0)=0,
(b) the number of events that occur in disjoint time intervals are
independent,
(c) the distribution of N(t) depends only on the length of interval,
(d)
P{N (h)  1}
lim
  , and
h 
(e)
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h
P{N (h)  2}
lim
 0.
h 
h
• Let us break the interval [0,t] into n non-overlapping
subintervals each of length t/n. Now, there will be k events in
[0,t] if either
(1) N(t) equals to k and there is at most one event in each
subinterval, or
(2) N(t) equals to k and at least one of subintervals contain 2 or
more events. Then,
P{exactly \1\ event \ in \ a \ subin} 
P{0 \ event \ in \ a \ subin}  1 
t
n
t
n
\ and
.
Since the number of events in the different subintervals are
independent, it follows that
P{k \ of \ subin \ contain \ exactly \1\ event \ and \
 n   t 
the \ other \ n  k \ contain \ 0 \ event}    
 k  n 
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k
t 

1  
n

nk
.
Hence, as n approach infinity,
P{N (t )  k}  e t
( t ) k
, \ k  0,1,
k!
That is, the number of events in any interval of length t has a
Poisson distribution with mean t.
For a Poisson process, let X 1 denote the time of the first event
and for n>1, X n denote the elapsed time between the (n-1)st
and nth event. Then, the sequence { X n , n  1, 2, }
is called the sequence of inter-arrival times.
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The distribution of X n :
The event { X 1  t} takes place if and only if no events of
the Poisson process occur in [0,t] and thus,
P{ X 1  t}  P{N (t )  0}  e  t
Likewise,
P{ X 2  t | X 1  s}  P{0 \ event \ in \ ( s, s  t ] | X 1  s}
 P{0 \ event \ in \ ( s, s  t ]}
 e  t .
Repeating the same argument yields X 1 , X 2 ,
are
independent exponential random variables with mean
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1/ .
• Example 32 (Steel Girder Fractures, p.209)
– 42 fractures on average on a 10m long girder
between-fracture length:
10/43=0.23m on average
– If the between-fracture length (X) follows an exponential
distribution, how would you define the gap?
X
E ( ) w ith  = 43/10
– How would you define the number of fractures (Y) per 1m
steel girder?
Y
P ( )
– How are the Exponential distribution and the Poisson
distribution related?
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Figure 4.9 Poisson process modeling fracture
locations on a steel girder
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• P(the length of a gap is less than 10cm)=?
P{ X  0.1}  1  e4.3 0.1  0.35
• P(a 25-cm segment of a girder contains at least two
fractures)=?
 0.25  4.3 0.25  1.075,
P{Y  2}  1  P{Y  0}  P{Y  1}
e 1.075 1.0750 e 1.075 1.0751
 1

0!
1!
 1  0.341  0.367  0.292
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Figure 4.10 The number of fractures in a 25-cm
segment of the steel girder has a Poisson distribution
with mean 1.075
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4.3 The Gamma Distribution
4.3.1 Definition of the Gamma distribution
• Useful for reliability theory and life-testing and has several
important sister distributions
• The Gamma function:

(k )   x e dx for k  0
k 1  x
0
• The Gamma pdf with parameters k>0 and
f ( x) 
 ( x)k 1 e  x
( k )
 >0:
fo r
x0
• Mean and variance:
E( X )  k / 
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and
V (X )  k / 
2
Curves of Gamma pdf
f ( x) 
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 k x k 1e  x
( k )
fo r x  0
• Calculation of Gamma function


0

0
(k )    e   x ( x) k 1 dx   e  y y k 1dy \ \(by \ letting \ y   x).

(k )   e  y y k 1dy  e  y y k 1 |0   e  y (k  1) y k  2 dy  (k  1)(k  1)
0
0
When k is an integer,

(k )  (k  1)!(1) // and // (1)   e  y dy  1
0
That \ is, \ \ (k )  (k  1)!
When k=1, the gamma distribution is reduced to the exponential
with mean 1/ .
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• Properties of Gamma random variables
If X i , i  1, , n are independent gamma random variables with
respective parameters (ki ,  ), then
n
X
i 1
i
n
is a gamma random variable with parameters (
 k ,  ).
i 1
i
• The gamma random variable with parameters (1,  ) is equivalent
to the exponential random variable with parameter.
• If X i , i  1, , n are independent exponential random variables,
each having rate  , then
n
X
i 1
i
is a gamma random variable with parameters
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( n,  ).
4.3.2 Examples of the Gamma distribution
• Example 32 (Steel Girder Fractures) :
, X k iid with E ( )
X1,
then
X 
k
X
i
G (k ,  )
( why ?)
i 1
• Y= the number of fractures within 1m of the girder
Then
Y Poisson( )
and
P ( X  1)  P (Y  k ) ( why ?)
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Figure 4.15 Distance to fifth fracture has a
gamma distribution with parameters k = 5 and  = 4.3
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Xi
E ( ), \   4.3,
5
Then, \ X   X i \ is
i 1
Gamma \ r.v. \ with \ k  5 \ and \   4.3.
k
5
E[ X ]  
 1.16[m]
 4.3
P{ X  1}  F (1)  0.4296 \ ( from \ numerical \ computation)
Let \ Y
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P( ). \ Then, \ P{Y  5}  0.4296
4.4 Weibull distribution
4.4.1 Definition of the Weibull distribution
• Useful for modeling failure and waiting times
(see Examples 33 & 34)
• The pdf:
f ( x)  a ( x) e
a 1  (  x )a
for a  0,   0
• Mean and variance:
1
1
E ( X )  (1  )

a
2
1 
2 
1  
V ( X )  2 (1  )  (1  )  
 
a 
a  
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Curves of the Weibull distribution
f ( x)  a ( x)
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a 1  (  x )a
e
The c.d.f. of Weibull distribution:
F ( x)  1  e
 (  x )a
The pth quantile of Weibull distribution:
Find x such that F ( x)  p.
1 e
e
 (  x )a
 (  x )a
p
 1 p
( x) a  ln(1  p )
x
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( ln(1  p ))1/ a

4.5 The Beta distribution
• Useful for modeling proportions and personal probability
(See Examples 35 & 36.)
• Pdf:
(a  b) a 1
f ( x) 
x (1  x)( b 1)
(a )(b)
• Mean and variance:
a
E( X ) 
ab
1
a
b
V (X ) 
a  b 1 a  b a  b
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Figure 4.21 Probability density functions of
the beta distribution
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Figure 4.22 Probability density functions of
the beta distribution
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• Exponential and Weibull random variables have (0, ) as their
set of possible values.
• In engineering applications of probability theory, it is
occasionally helpful to have available family of distributions
whose set of possible values is finite interval.
• One of such family is the beta family of distributions.
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• Example: the beta distribution and rainstorms.
Data gathered by the U.S. Weather Service in Alberquerque,
concern the fraction of the total rainfall falling during the first 5
minutes of storms occurring during both summer and
nonsummer seasons. The data for 14 nonsummer storms can
be described reasonably well by a standard beta distribution
with a=2.0 and b=8.8.
• Let X be the fraction of the storm’s rainfall falling during the first
5 minutes. Then, the probability that more than 20% of the
storm’s rainfall during the first 5 minutes is determined by
(2.0  8.8)
1.0
7.8
P{ X  0.2} 
u
(1

u
)
du  (86.24)(0.0045)  0.39.

(2.0)(8.8) 0.2
1.0
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