Download 1 review of probability

Stochastic Processes and Queueing
Chapter 1: Review of Probability
1
REVIEW OF PROBABILITY ......................................................................................... 2
1.1
Some Mathematical Machinery ......................................................................................... 2
1.1.1
1.1.2
1.1.3
Set Notation ....................................................................................................................................2
Venn Diagrams ...............................................................................................................................2
Operations of Sets ...........................................................................................................................3
1.2
Random Experiment ........................................................................................................... 4
1.3
Outcomes ............................................................................................................................. 4
1.4
Sample Space ....................................................................................................................... 4
1.4.1
Examples ........................................................................................................................................5
1.5
Random Event ..................................................................................................................... 5
1.6
Concept of Probability ........................................................................................................ 5
1.7
Axioms of Probability ......................................................................................................... 5
1.8
Conditional Probability ...................................................................................................... 6
1.9
Law of Total Probability .................................................................................................... 7
1.10
Bayes Rule............................................................................................................................ 8
1.11
Random Variable .............................................................................................................. 10
1.12
Probability Function ......................................................................................................... 10
1.13
Probability Mass or Density Function............................................................................. 12
1.14
Cumulative Probability Distribution Function (Cumulative Distribution Function). 12
1.15
Discrete Random Variables.............................................................................................. 13
1.16
Continuous Random Variables ........................................................................................ 13
1.17
Expectation ........................................................................................................................ 14
1.18
Jointly Distributed Random Variables ........................................................................... 17
1.19
Some Important Discrete Distributions .......................................................................... 18
1.19.1
1.19.2
1.19.3
1.19.4
1.20
Some Important Continuous Distributions .................................................................... 19
1.20.1
1.20.2
1.20.2.1
1.20.2.2
1.20.2.3
1.20.2.4
1.20.2.5
1.20.3
1.20.4
1.21
Bernoulli Distribution .............................................................................................................. 18
Binomial Distribution .............................................................................................................. 18
Geometric Distribution ............................................................................................................ 18
Poisson Distribution ................................................................................................................. 19
Uniform Distribution ............................................................................................................... 20
Exponential Distribution .......................................................................................................... 20
Property 1: Lack of Memory .............................................................................................. 20
Property 2: Constant Failure Rate: ..................................................................................... 22
Property 3: Combined failure rate from more than one component ................................... 22
Property 4: Relation to Poisson Distribution ...................................................................... 23
Property 5: Detected and undetected failures. .................................................................... 25
Erlang Distribution .................................................................................................................. 26
Normal Distribution ................................................................................................................. 26
Convolution Integral ......................................................................................................... 26
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Chapter 1: Review of Probability
1
1.1
REVIEW OF PROBABILITY
Some Mathematical Machinery
Below is some useful notation and pictorial representations used in this review.
1.1.1 Set Notation
Consider the following set representation.
S={s: s is an outcome of the experiment}
Breaking down the notation:
1. { - means “the set of all elements”
2. : - means “such that”
3. text following the : - condition for inclusion of element s in the set S
Others
1.
2.
3.
4.
sA – “s is an element of the set A”
sA - “s is not an element of the set A”
AB – “A is a subset of B”
A - complement of the set A, i.e., A  {s : s  , s  A}
1.1.2 Venn Diagrams
Venn diagrams are pictorial representations of a set and its elements.
Ω
s
s
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Ω
s
sA and A
A
Ω
_
A
A
A
1.1.3 Operations of Sets
Unions
A  B  {s : s  A or
s  B}
Ω
A
B
If A1, A2, …, An are events in , then
n
 Ai  A1  A2    An
i 1
 {s : s  A1 or s  A2 or  or s  An }
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Intersections
A  B  {s : s  A and
s  B}
Ω
A
B
If A1, A2, …, An are events in , then
n
 Ai  A1  A2    An
i 1
 {s : s  A1 and s  A2 and  and s  An }
1.2
Random Experiment
A Random Experiment is an experiment or observation which can be performed (at least in
theory) any number of times under the same relevant conditions. Some examples include:
1.
2.
3.
4.
5.
1.3
Toss of a fair coin
Toss of a fair coin twice
Roll of a die
Measured times between arrivals in an intersection
Count the number of rainy days in a year
Outcomes
An Outcome is a possible result,  of a random experiment. For the examples listed above
respectively, some example outcomes are:
1.
2.
3.
4.
5.
1.4
Heads
Heads twice
Three dots on top
5.3 seconds
127 days
Sample Space
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The Sample Space is the collection, , of all possible outcomes of a random experiment. For
each of the examples above, respectively, the sample space is:
1. {Heads, Tails}
2. {Heads twice, Heads then Tails, Tails then Heads, Tails Twice}
3. {One dot on top, Two dots on top, Three dots on top, Four dots on top, Five dots on top,
Six dots on top}
4. {[0,]}
5. {0, 1, 2,…365 days}
1.4.1 Examples
1. Tossing a Fair Coin - ={: = H or =T} i.e., ={H,T}
2. Tossing a Fair Coin Twice - ={(1, 2): i= H or i =T for i=1,2} =
{(T,T),(T,H),(H,T),(H,H)}
3. Time to Failure of a Transistor - ={: 0<}
1.5
Random Event
An Event, E, is a collection of outcomes, , in , i.e. some subset of .
Ε  {  Ω}
 {ω1, ,ωn }
 { : (X ()  Re }
Define E to be the family of events.
1.6
Concept of Probability
1.7
Axioms of Probability
Let  be a sample space of a random experiment. A probability measure (function) P is an
assignment of a real value P(A) to each event A that satisfies the following:
1. P[Ø ] = 0 and P[  ] = 1
2. 0 P[ A ]  1
3. If A  B = Ø (mutually exclusive) then P[A  B ] = P[A ] + P[ B ]
Other basic properties:
1. If A  B then P[B - A] = P[B] - P[A], i.e., the probability of an outcome occurring which
is  B but not A is P[B] - P[A].
2. P[ A ]  1  P[ A]
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3. P[A  B] = P[A ] + P[B ] - P[A  B ] and, in general for n events ( A1 , A2 ,, An ),
n
n  n
n 1
P   Ai    P[ Ai ]   P[ Ai  A j ]   P[Ai  A j  Ak ]    (1)
P[  Ai ]
i j
i jk
i 1
i 1  i 1
4. P[A  B]  P[A ] + P[B ] This is known as Boole's Inequality.
n  n
5. If A1 , A2 ,, An are mutually exclusive events, then P  Ai    P[ Ai ]
 i1  i1
6. For events A1 and A2 , A1  A2 implies P[ A1 ]  P[ A2 ]
1.8
Conditional Probability
Let A and B be events in  then
P[ A  B]
P[ B]
or alternativ ely
P[ A B] 
P[ A B]P[ B]  P[ A  B] and
P[ B A] 
P[ B A] 
P[ A  B]
thus
P[ A]
P[ A B]P[ B]
P[ A]

B
A|B
A
Consider the following. I am going to choose, at random, one member of the class to come
to the board to answer a question next class period. While I will make the choice today, I
will not let the individual selected know until next class period. Assume that the probability
of a given individual being chosen depends upon the row they sit in, i.e., the further from the
front of the class, the more likely an individual will be chosen. The process is as follows. I
first randomly choose a row where the probability of selecting row i is as follows.
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P[Selecting row i out of n total rows] 
i
n
j

i
. Once a row is selected, the
n(n  1) / 2
j 1
individuals in that row are equally likely to be chosen.
Question? Does knowing today which row I selected change how you study before the next
class period? Explain how...
1.9
Law of Total Probability
n
Let A1 , A2 ,, An be mutually exclusive events such that
A
i
i 1
  (i.e. A1 , A2 ,, An
form an exhaustive partition of  as shown below).
A1
A4
A7
A9
A3
B
A6
A10
A2
A5
A8

We know due to mutual exclusivity that Ai  A j    i  j. Let B be some event  
then:
n
P[ B]   P[ B Ai ]P[ Ai ]
i 1
Proof:
n
B  B    B  ( Ai )
i 1
n
  ( B  Ai )
i 1
n
 P[ B]  P[ ( B  Ai )]
i 1
Now
( B  Ai )  Ai and since Ai  Aj  
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Then
( B  Ai ) ( B  Aj )  
From Axiom (3)
P[( B  Ai )  ( B  Aj )]  P[( B  Ai )]  P[( B  Aj )]
Thus
n
P[ B]   P[ B  Ai ]
i 1
From the definition of conditional probability:
n
P[ B]   P[ B / Ai ] P[ Ai ] QED
i 1
Example: A climber wants to summit Mount Rainier. Data is available on the percentage of
all climbers that summitted led by each of the Rainier Mountaineering, Inc. guides. Data is
also available on the percentage of all climbers led by each of the n possible guides. What is
the probability a climber will summit?
Let
A = event that a climber summits
Bi = event that a climber is led by guide i
The historical data gives us P[A|Bi] and P[Bi]. Now the Bi‘s form a mutually exclusive and
exhaustive set of events therefore using the Law of Total Probability
n
P[ A]   P[ A | Bi ]P[ Bi ] .
i 1
We can define a conditional Probability Distribution Function (PDF):
FX (t | B)  P[ X  t | B] 
P[ X  t , B]
P[ B]
1.10 Bayes Rule
Let A1An form a partition of  and let B be any event in . Then
P[ Ai | B] 
P[ Ai  B]
P[ B]
(1)
Using the law of total probability
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n
P[ B]   P[ B| Aj ] P[ A j ]
(2)
j 1
substituting (2) into (1) yields
P[ Ai | B] 
P[ Ai  B]
n
 P[ B| A ] P[ A ]
j 1
j
j
using P[ Ai  B]  P[ B  Ai ]  P[ B| Ai ] P[ Ai ] leads to
P[ Ai | B] 
P[ B| Ai ] P[ Ai ]
n
 P[ B| A ] P[ A ]
j 1
j
j
Example: A climber knows that he has been assigned to be led by guide Win Whittaker.
The climber is interested in the probability he will reach level i given Win is his guide. Data
is available on the percentage of time Win was the guide for those climbers who reached
level i and the overall percentage of climbers who reached level i but no higher. Let Ai be the
event that a climber climbs only as high as level i and B be the event Win is the guide. Let m
be the number of levels, then
P[ Ai | B] 
P[ B| Ai ]P[ Ai ]
m
 P[ B| A ]P[ A ]
j
j
j 1
Let A and B be events in . A and B are said to be independent if
P[ A  B]  P[ A]P[ B] i.e., P[ A | B]  P[ A] and P[ B | A]  P[ B].
Now if A and B are mutually exclusive then P[ A  B]  P[ A]  P[ B] . Does mutual
exclusion imply independence or vice-versa?
NO! If A, B are mutually exclusive then P[ A | B]  0!
A family of events A1 , A2 ,, An is mutually independent if for every suset of r different
events Ai1 , Ai2 ,, Air and for r = 2, 3, …, n we have
P[ Ai1  Ai2    Air ]  P[ Ai1 ]P[ Ai2 ]P[ Air ] . Typically we arrange for mutual
independence by performing replications of some basic random experiment. When people
talk about independent trials they typically mean mutually independent.
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1.11 Random Variable
A Random Variable is a numerical function that maps outcomes onto the real line, i.e., it
assigns a numerical value to the outcomes of some experiment as shown in the figure below.
Typically a random variable is denoted by an upper case letter (X, Y, Z, etc.) while a
particular outcome is denoted by a lower case letter (x, y, z, etc.). One way to think about it is
that X denotes the random function before the associated outcome of the experiment is
observed while x is the value associated with the observed outcome. Technically, since the
random variable is a function of the outcome, we should write it X(). However, typically
the parenthetic argument is dropped.
X()

Re
There are three different types of random variables.
1) Discrete, i.e., the value of the random variable takes on a discrete number of values. For
example, the number of arrivals to a bank in some time interval.
2) Continuous, i.e., there are an infinite number of values of the random variable. For
example, the service time at the bank.
3) Mixed, i.e., a combination of discrete and continuous. This will be made more clear
when we discuss probability. An example is the time spent waiting before being served
in the bank.
1.12 Probability Function
A Probability Function is a function that uniquely maps the events into the set of real values
in [0,1].
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P()

[0,1]
The Probability System is defined by (, P, E)
Example :
Coin Tossing (Single Toss)
Sample Space
 = { H,T }
Events Set
E  {H}, {T}, {H,T}, Ø
Probability Function P[{H}] = 1/2
P[{H,T}] = 1
P[{T}] = 1/2
P[Ø] = Ø
Random Variable
X(H) = 1 X(T) = 0
Notation
     is an element of the set 
     is not an element of the set 
Let A and B be events in  (subsets of ), then:
1.
2.
3.
4.
A  B = {    |   A OR   B }
A  B = {    |   A AND   B }
A c= {    AND   A }
A  B  all elements in the set A are also in B
Consider some event A in . Let fn(A) = the number of replications out of n total of which
the event A is said to occur. The “relative frequency” interpretation of probability would say
f ( A)
P[ A]  lim n
n  n
This is a bit more complicated than it may first appear. The overall random experiment has
 A sample space, *, consisting of infinite sequences (s1, s2, s3, …)
*  {s*  ( s1 , s 2 ,...) : si   for i  1,2...}

Random variables X1(s*), X2(s*), X3(s*), …as follows
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0, s  A
For s  , I A ( s)  
1, s  A
For s*  * ,
X n (s* )  X n ( s1 , s 2 ,...) 
f ( A)
1 n
I A ( si )  n

n i 1
n
The limit of the sequence X1(s*), X2(s*), …, Xn(s*), Xn+1(s*), is (we hope) a random variable
X(s*) where


 f ( A) 
X (s* )  lim X n (s* )  lim  n

n 
n   n 
This is not an ordinary limit in the calculus sense.
Let’s let the experiment be the toss of a fair coin and the event A be that a “heads” results. In
this case we hope that in the overall random experiment with sample space *, we observe
that X(s*) = 1/2. Define the event B in * as
1
B  {s* : X (s* )  }
2
For the relative frequency interpretation of probability to be meaningful, we must have
P[B]=1, i.e., B must be certain to occur.
Circular chain of reasoning inherent in the relative frequency approach leads to an axiomatic
approach to the assignment of probability.
1.13 Probability Mass or Density Function
Probability mass function, p X (t ) - discrete random variabiables.
p X (a)  P[ X  a]
Probability density function, f X (t ) - continuous random variables.
a
P[ X  a]   f (t )dt  0
a
b
P[a  X  b]   f (t )dt
a
1.14 Cumulative Probability Distribution Function (Cumulative Distribution
Function)
define:
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EtX  { : X ( )  t}
 { X  t}
 P[ EtX ]  P[ X  t ]  FX (t )
This is known as the Cumulative Probability Distribution Function, FX (t ) .
We will refer to it as the Distribution Function. Note that the (Cumulative Probability)
Distribution Function is an increasing function of the random variable (X in this example). It
may increase continuously as in the figure below, or by discrete jumps, or both, but it cannot
decrease (monotonically nondecreasing). We will generally refer to this function as the
“distribution function.” It is bounded between 0 and 1, i.e., it cannot be less than 0 or more
than 1.
Probability
1
0
X (random variable value)
1.15 Discrete Random Variables
A discrete random variable is one whose outcome values are from a finite set. All discrete
random variables have a Probability Mass Function, pX(t), (pmf) where:
p X (t )  FX (t )  FX (t  )  P[ X  t ]
and
FX (t ) 
p
ti t
X
(t i )
p X ( ti )  0
1.16 Continuous Random Variables
A continuous random variable is one whose outcome values are from an infinite set. All
continuous random variables have a Probability Density Function, fX(t) (pdf), where
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f X (t ) 
d
FX (t )
dt
and
t
FX (t ) 
f
X
( x)dx

Note that all random variables have a cumulative probability distribution function (PDF).
Not all have a pmf or pdf.
1.17 Expectation
The expected value of a random variable is defined as
E[X]   tf x (t ) dt
and
E[X] 

Px ( ti ) 0
if X is a continuous random variable
t i Px (t i )
if X is a discrete random variable
Now the Reimann integral you learned about in calculus is

b
a
g ( x)dx  lim
n 
n
 g ( x )[ x
'
i
i
 xi 1 ]
i=1
where xi is some point between xi 1 and xi . A more general form of integration uses the
Stieltjes integral which is

b
a
g(x)dH(x) 
lim
n
n
 (g(x
'
i
)[H(x i )  H(x i 1 )]
i=1
Now, if H(x) is continuous and differentiable then this can be simplified to

b
a
b
g ( x)dH ( x)   g ( x)(
a
dH ( x)
)dx
dx
which is a Reimann iintegral.
Consider the following function, H(x)
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H(x)
x
X2
X1
Assume we want: E[g(x)]
1) Decompose H(x) into two functions L(x) and K(x) where L(x) represents the discrete
portion of H(x) and K(x) the continuous portion.
L(x)
x
X1
X2
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  
 
L( x) =  H x+  H x 
which is a discrete staircase function.
K(x)
X1
x
X2
K(x) is some piece-wise continuous function
If I represents the integral of interest then
b
b
a
a
I   g ( x)dH ( x)   g ( x)[

dK ( x)
]dx   g ( xi' )[ L( xi )] where L X i   H ( xi )  H ( xi' )
dx
L ( xi )  0

Assuming g(x) is continuous and differentiable, then

I   g (u)dFX (u)
-
Consider the case where g(x) = x. In this case I is the E[X]. Formally


-
udFX (u) =

0
-

udFX (u) -  ud[1  FX (u)]
0
Call the first integral on the right hand side A and the second B. In B we have used the fact
that dFX (u)=-d[1-FX (u)] . Now integrating each of these by parts yields
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0
0
A :  uFX (u ) |0   FX (u )du  0 (if E[ X ] exists)   FX (u )du


B := (u1  FX (u ) |0   [1  FX (u )]du )  0 (if E[ X ] exists)   [1  FX (u )]du


0
0
Thus




0
0

udFX (u)  [1  FX (u)]du  FX (u)du
Finally, if X is a positive random variable


0
0
E[ X ]   1  FX (u ) du   R X (u )du
where RX (u)  1  P[ X  u]  PX  u is known as the Reliability Function of X.
1.18 Jointly Distributed Random Variables
We may be interested in systems involving more than one random variable. In the case of
two random variables, X and Y, we talk about the joint probability distribution function, i.e.,
FXY (s, t )  P[ X  s, Y  t ]
Now, if X and Y are continuous, then the joint density function exists and is
f XY ( s, t ) 
2
FXY ( s, t )
st
However, if X and Y are discrete, then the joint mass function exists and is
p XY (s, t )  FXY (s, t )  FXY (s  , t  )
Consider continuous random variables X and Y where we know that Y=y.
P[ X  s, Y  y ]
FX ( s Y  y ) 
P[Y  y]
Now if we let y  Y  y+y and then and take the limit as y0 we get
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FX ( s Y  y ) 

lim P[ X  s, y  Y  y  y ]
y  0
P[ y  Y  y  y ]
lim FXY ( s, y  y )  FXY ( s, y ) y
y  0 FY ( y  y )  FY ( y )
y
 FXY ( s, y  y )  FXY ( s, y ) 

y



y  0 
y
 FY ( y  y )  FY ( y ) 

FXY ( s, y )
y

f Y ( y)

f X (s Y  y) 
lim
f ( s, y )

FX ( s Y  y )  XY
2
f Y ( y)
y
If X, Y are independent then FXY (s, t )  FX (s) FY (t ) .
1.19 Some Important Discrete Distributions
1.19.1 Bernoulli Distribution
A Bernoulli distribution is associated with an experiment consisting of a single trial where
there can be one of two possible outcomes: success with probability p or failure with
probability q = (1-p). Let X be a Bernoulli random variable then
 0(failure )
1  p if k  0
X 
and p X (k )  
if k  1
1(success )
p
Examples:
- single toss of a coin
- operating state of a component (operating or failed) at some time
1.19.2 Binomial Distribution
A Binomial distribution is associated with an experiment of n independent Bernoulli trials.
Let X represent the number of successes, then
 n
pX (i)    pi q ni
i  0,1,2,..., n
 i
1.19.3 Geometric Distribution
A Geometric distribution is associated with an experiment of independent Bernoulli trials
that are performed until a success occurs. In this case
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p X (i )  pq i 1
i  1,2,3,..., 
1.19.4 Poisson Distribution
A Poisson distribution is associated with an experiment in which the number of occurrences
of some outcome are counted over some time period T such that  represents the rate of
occurrences over the time period. In this case
( T ) i  T
p X (i ) 
e
i!
The Poisson distribution can be used to approximate the binomial distribution when n is large
and p is small. Let
  np  p 

n
Using the binomial mass function, we know
n!
n!      
PX (i ) 
p i (1  p ) n i 
  1  
(n  i )!i!
(n  i )!i!  n   n 
i
n i
 
1 
i 
(n)( n  1)( n  2).....( n  i  1)   n 

i!
ni   i
1  
 n
n


i
n(n  1).....( n  i  1)    
  

ni
 i!  


n
   
1



 n 
i 
  
1   
 n 
Now if n is large and p is small then
 

1    e
n


n(n  1).....( n  i  1) n i
 i 1
ni
n
i

 e
i
 e 
Therefore, p x (i)  (1)
which is the Poisson mass function.
i! 1
i!
n
1.20 Some Important Continuous Distributions
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1.20.1 Uniform Distribution
A Uniform distibution is associated with an experiment in which the outcome ranges
continuously over the values from a to b.
axb
1
f X  x    ba
otherwise
 0
xa
FX x  
ba
which has the symbol U (a, b)
1.20.2 Exponential Distribution
An Exponential distribution has the following probability density function and probability
distribution function.
f X ( x)  e x
FX ( x)  1  e
0x
x
Suppose we are measuring the length of time until a certain event occurs, e.g., failure of a
light bulb. Let the random variable X be the time (measured from time 0) until the event
occurs. We say X has an exponential distribution (or less formally, "the event is
exponential") if
FX (t )  Pr[ X  t ]  1  e  t , t  0
where  > 0 is the parameter, or mean rate of occurrence of the event. Recall that  = 1/E(X)
= (mean time until event occurs)-1, which motivates the use of the term "mean rate."
There are five very important properties of the exponential distribution that we will exploit in
helping us model various situations.
1.20.2.1 Property 1: Lack of Memory
Given that the light bulb has not yet failed by time t, what is the probability that it fails in the
next s time units?
Answer: The same as the probability that a new light bulb fails in the first s time units.
Interpretation: the propensity to fail of a component with an exponentially distributed
lifetime is completely random and not a function of the age of the component.
To verify this property mathematically, note that:
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P[failure in next s time units/no failure by time t] is
 P[ X  t  s| X  t ]

 
 

 P[t  X  t  s] / P[ X  t ]  1  e   ( t  s )  1  e  t / 1  (1  e  t )


 e   t  e   (t s) / e   t


 e  t 1  e  s / e  t
 1  e  s
 P[ X  s]
 P[A new light bulb fails in first s time units]
Note that there are (at least) two equivalent ways of verifying the key step above:
(i) P[t  X  t  s]  P[( X  t ) and not ( X  t  s)]
 P[ X  t ] - P[ X  t  s]
 e  t  e  ( t + s )
A second way is based upon the fact that the event {X > t + s} is contained in (implies) the
event {X > t}. Considering the figure below we are interested in the probability of the
shaded event, but we see that
P[ X  t ]  P[t  X  t  s]  P[ X  t  s]
Therefore,
P[t  X  t  s]  P[ X  t ]  P[ X  t  s]
(ii)
t s
  e  x dx
t
 x
= -e
t s
t
 e  t  e  ( t  s )
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{X>t+s}
{t<X<t+s}
{X>t}
1.20.2.2 Property 2: Constant Failure Rate:
Let t be small then P[ X  t ]  1  e t . Then using the Taylor Series expansion of ex

 t 2  ....
1  e  t  1  1  t 


2


 t
(The last step involves neglecting higher order terms of t.)
This impiles that, no matter how long the component has been in operation, the probability of
failure in next small time interval of length t is approximately t.
1.20.2.3 Property 3: Combined failure rate from more than one component
Suppose two components are put into service at the same time:
X = life time of first component,
Y = life time of second component,
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where X and Y are independent and exponentially distributed with mean rates  and ,
respectively. Thus,
P[Y  t] = 1 - e-t, t  0
P[Y  t] = 1 - e-t, t  0
P[X > t] = e-t
P[Y > t] = e-t
What is the probability distribution of the time until first failure (of either one component or
the other)?
Let U = the time until either fails, i.e., U = min(X,Y). Then
P[U > t] = P[min ( X , Y ) > t]
= P[ X > t] P[Y > t], since X and Y independen t
= e  t e   t
= e (    ) t
 P[U  t] = 1 - e (    )t , t  0
i.e., U has an exponential distribution with mean rate =  +  = sum of rates of components.
 P[U  t  t | U  t ]  (   )t
(from Property 2)
What is the probability that the first failure is component 1?
Assume that at time t, component 1 fails but component 2 has not yet failed, i.e., the event is
that X=t and Y>t. Now, since X and Y are independent, then
P[ X  t, Y  t ]  P[ X  t ]P[Y  t ]  f X (t )dt (1  FX (t )). Integrating over all possible values
of t leads to

P[ X  Y ] 

 f X (t )(1  FY (t ))dt   e
0
Component 2?


 t  t
e
0

dt    e  (  )t dt 
0


1.20.2.4 Property 4: Relation to Poisson Distribution
Suppose that when the light bulb fails, we replace it with an identical bulb. When that bulb
fails, we replace it with an identical bulb, and so forth.
Define X1 = life time of first bulb
X2 = life time of second bulb


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
Xk = life time of kth bulb
The assumption that the light bulbs are identical (i.e., all come from the same manufacturing
process with the same specifications) leads to the natural assumption that the random
variables X1, X2, ...., Xk, have a common distribution function. It is also reasonable to
assume that the failure of one bulb does not influence the failure another since thay are put
into operation sequentially. Therefore,
P[ X k  t | X k 1  x k 1 ,..., X 1  x1 ]  P[ X k  t ]
 1  e  t
and
FX k (t )  1  e  t
The process of successive failures of the light bulbs can be represented graphically as
follows:
N(t)
5
4
3
2
1
0
x1
x1
first
failure
x1
second
failure
t
t
x1
third
failure
fourth
failure
Note that if we wish to describe completely the process of successive failures, there are two
completely equivalent ways to do so:
i) Specify the times at which the first, second, etc., failures occurred, or equivalently
specify the successive times between failures,
X1, X2, X3, ...
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ii) Specify for each time point t  0, the number of failures (denoted N(t)) that occurred
between time 0 and time t.
The collection of random variables {N(t), t  0} is called the counting process for the
successive events (failures) where N(t) = number of events (failures) between 0 and t. The
relation between {N(t), t  0} and {X1, X2, X3, ...} is the following:
N (t )  max{ n | X 1  X 2  ...  X n  t}
OR
{N (t )  k}  { X 1  ...  X k  t  X 1  ...  X k 1 }
From this it follows that
P[ N (t )  k ]  P[ X 1  ...  X k  t  X 1  ...  X k 1 ]
(If the two events are the same, then they must, of course, have the same probability.) The
probability on the right can be calculated using the known facts that each Xi, i=1, 2,...k+1,
has the same exponential distribution, with parameter , and that they are independent. The
calculation is a bit involved (convolution integral) but the result is
P[ N (t )  k ]  e
 t
( t ) k
, k  0,1,...
k!
This should be recognized as the Poisson probability mass function. Thus, for each fixed
time t, N(t), the number of events that have occurred up to t has a Poisson distribution with
parameter t. Note that:
E[ N (t )]   t , or
E[ N (t )]
 ,
t
so that the mean number of events per unit time =  , another reason for calling  the mean
rate at which events are occurring. The values of P[N(t) = k] for any particular t and , can
be found from a Poisson table with parametert.
1.20.2.5 Property 5: Detected and undetected failures.
Suppose that when a failure occurs, it has a probability p of being detected and 1-p of not
being detected. Suppose the number of failures is governed by a Poisson process, {N(t), t >
0} (i.e., the times between failures are identically and exponentially distributed) with mean
failure rate . Then what can we say about the number of detected failures?
Define Nd(t) = number of detected failures in [0, t].
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( pt ) k
, k  0,1,...
k!
That is, the number of detected failures is also governed by a Poisson process, {Nd(t), t > 0},
Fact: P[ N d (t )  k ]  e  p t
with mean rate t. In particular,
E[Nd(t)] = pt = pE[N(t)],
i.e., the mean number of detected failures = p times the mean number of actual failures which
is plausible. The implication of this is that a Poisson stream can be split apart into Poisson
streams where the sum of the rates of the spilt off streams equals the rate of the original
stream.
This property has many applications to systems where Poisson events are split into two or
more different classes of events. For example, assume jobs enter a job shop according to a
Poisson process with mean rate  = 12 per hour. A fraction p = 1/3 of the entering jobs go to
machine center #1, and 1 - p = 2/3 to machine center #2. Then the job-arrival processes at
machine centers #1 and #2 are both Poisson, with mean rates p = 1/3 x 12 = 4 per hour, and
(1 - p) = 2/3 x 12 = 8 per hour, respectively.
1.20.3 Erlang Distribution
Let X1 .... Xn be mutually independent exponentially distributed random variables (stages)
with rates 1... n then
n
Y   Xi
i 1
has an n-stage general Erlang distribution. If 1 = 2... n =,, i.e., all the stage rates are the
same, then Y is an n-stage special Erlang. This was shown earlier under Property 3 of the
Exponential Distribution. The density function in this case is f Y (t ) 
(t ) n 1  t
e
.
(n  1)!
1.20.4 Normal Distribution
A Normally distributed random variable has the following distribution function:
FX ( x) 
1
 1  t    2 
 
x  
 2    
e
dt . There is no closed form expression for this function.

 2  
1.21 Convolution Integral
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Often we are interested in a random variable which a function of two or more other random
variables. We know something about the original random variables (distribution or density
functions) and we want to determine the same for the new random variable. For example, let
X1 and X2 be independent random variables and let Y = X1 + X2. Due to independence we
know that
F X1 X 2 ( x1, x2 )  FX ( x1) FX ( x2 )
1
2
Therefore, using the Law of Total Probability we get
P[Y  z ]  P[ X 1  X 2  z ]   P[X 1  X 2  z | X 1  x1 ]dFX 1 ( x1 )
x1
This combined with knowing that
P[ X 1  X 2  z | X 1  x1 ]  P[ X 2  z  x1 ]  FX 2 ( z  x1 )
leads to
FY ( z )   FX 2 ( z  x1 )dFX 1 ( x1 )
x1
Then differentiating with respect to z yields
dFY ( z )   dFX 2 ( z  x1 )dFX 1 ( x1 )
x1
For the case where both X1 and X2 are discrete we can show that this yields
z 1
pY ( z )   p X 2 ( z  i ) p X1 (i )
i 1
Proof:
P[Y  z ]  P[Y  z ]  P[Y  z  1]
or equivalently
pY ( z)  FY ( z)  FY (Z  1)
Now, from earlier we know that
z 1
FY ( z )   FX 2 ( z  i ) p X1 (i )
i 1
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Then,
z 1
z 11
i 1
i 1
pY (z)   FX 2 ( z  i ) p X1 (i ) 
F
X2
( z  1  i ) p X1 (i )
z 2
z 2
i 1
i 1
 FX 2 ( z  ( z  1)) p X1 ( z  1)   FX 2 ( z  i ) p X1 (i )   FX 2 ( z  1  i ) p X1 (i )
z 2
 FX 2 ( z  ( z  1)) p X1 ( z  1)   [( FX 2 ( z  i )  FX 2 ( z  1  i )) p X1 (i )]
i 1
z 2
 FX 2 (1) p X1 ( z  1)   [ p X 2 ( z  i ) p X1 (i )]
i 1
Now if X2 = 1, 2, … then
1
FX 2 (1)   p X 2 ( j )  p X 2 (1)
j 1
Therefore,
z 1
pY ( z )   p X 2 ( z  i ) p X1 (i )
i 1
Intuitively, this should make sense. For example, let X1 and X2 be independent and
geometrically distributed with identical trial success probability p. Let Xi be the number of
flips of a coin required to get a heads and let Y = X1 + X2 be the number of flips of the first
coin to get a heads added to the number of flips required to get a heads on the second
(identical) coin. If Y = z and X1 = x1 then X2 = z-x1 and since both X1 and X2 must be
greater than or equal to 1 (you must flip the coin at least once to get a heads) there are z-1
different ways (outcomes of X1 and X2) for Y = z. Therefore, in order to determine P[Y = z]
we simply sum up P[X2=z-x1] .P[X1=x1] for all possible values of X1, i.e.,
z 1
P[Y  z ]   P[ X 2  z  x1 ] P[ X 1  x1 ]
x1 1
and then
z 1
z 1
x1 1
x1 1
P[Y  z ]   P[ X 2  z  x1 ] P[ X 1  x1 ]   FX 2 ( z  x1 ) p X1 ( x1 )
Plugging in the distribution and mass functions for the geometrically distributed random
variables we get
z 1
P[Y  z ]   pq z  x1 1 pq x1 1  p 2 q z  2 ( z  1)
x1 1
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P[Y  z ]  p
z
2
q
i 2
(i  1)
i 2
Now if X1 is a discrete random variable then
FY ( z )   FX 2 ( z  x1 )dFX 1 ( x1 )   FX 2 ( z  x1 )P[ X 1  x1 ]   FX 2 ( z  x1 ) p X 1 ( x1 )
x1
x1
x1
and if X1 is continuous then
FY ( z )   FX 2 ( z  x1 )dFX 1 ( x1 )   FX 2 ( z  x1 ) f X 1 ( x1 )dx1
x1
x1
Example 1:
Let Z = X/Y where X and Y are geometrically distributed with P[success]=p.
Since X and Y are discrete and positive then the possible values of Z can be written as a ratio
of integers a and b where a represents the value of X and b represents the value of Y. For a
given value of Z, there are an infinite number of combinations of a and b, thus

P[ Z  a / b]   pq ia1 pq ib1
i 1



2
p
q2
2
p
q2
2
p
q2

q
( a b ) i
i 1

 (q
( a b ) i
)  q ( a b ) 0
i 0


1

 1
( a b )
1 q

Example 2:
Let M = N1 + N2 where N1, N2 are Poisson counting processes with rates and ,
respectively. We know
p N1 (k ) 
p N 2 (k ) 
k
k!
k
k!
e  k  0,1,2,..., 
e   k  0,1,2,..., 
Now
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m
p M (m)   p N 2 (m  k ) p N1 ( k )
k 0
m

 m k
k  0 ( m  k )!
e
(    )
m
e 
k
k!
 m k
e 
k
 (m  k )! k !
k 0

e
(    )
m !  m k k

m! k 0 (m  k )! k !
m
e (    )

(   ) m
m!
(    ) m (    )

e
m!
Notice that M also has a Poisson mass function with rate equal to the sum of the individual
rates, i.e., the merging of two Poisson streams yields a Poisson stream.
Example 3:
Consider Y = X1 + X2 where Xi has an exponential distribution with rate .
z
FY ( y )   FX 2 ( z  x)dFX1 ( x)
0
y
  (1  e  ( y  x ) )e x dx
0
 e x | 0y xez | 0y
 1  e y  ye y
then
f Y ( y)  2 ye y
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04/29/17