chapter 03 random variables part a
TRANSCRIPT
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Definition of a Random Variable
Random Variable [m-w.org]
: a variable that is itself a function of the result of a statistical
experiment in which each outcome has a definite
probability of occurrence
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Definition of a Random Variable
A random variable is a mapping from an outcome s of a
random experiment to a real number
: X X S S → ⊂
domainrange
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Definition of a Random Variable
:X S S →
head
an om
ExperimentSample Space Random
VariableX(s)
tail0 1 R
S x
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Definition of a Random Variable
:X S S → ⊂
X (s)
1 2 R3 4 5 6
Random
Sample Space, S
Random Variable
S x
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Definition of a Random Variable
More than one outcomes can be mapped to the same real
number:X S S →
X(s)
Random
0 1
S x
xper menSample Space
Random Variable
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Types of Random Variables
Discrete random variables: have a countable (finite or infinite)image S x = {0, 1}
S x = {…, -3, -2, -1, 0, 1, 2, 3, …}
Continuous random variables: have an uncountable image S x = (0, 1]
S x = R
Mixed random variables: have an image which contains
continuous and discrete parts S x = {0} U (0, 1]
We will mostly focus on discrete and continuous random
var a es
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screte an om ar a es
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Probability Mass Function
The Probability Mass Function (pmf ) or the discrete probability
density function provides the probability of a particular point in
For a countable S X ={a0, a2, …, an}, the pmf is the set of
probabilities
( ) { }Pr , 1,2, ,X k k p a X a k n = = = …
p X (ak )
S X ={a0=0, a1=1, …, a5 =5},
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X
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Properties of a PMF
P1: ( )0 1X k p a ≤ ≤
P2: ( ) 1k X
X k a S p a ∈ =∑
X k
S X ={a0=0, a1=1, …, a5 =5},
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Cumulative Distribution Function (CDF) of Discrete
Random Variable
The Cumulative Distribution Function (CDF) for a discrete rv is
defined as:
( ) { } ( )PrX X
x t F t X t p x
≤= ≤ = ∑
p X ( x )
mf
F X ( x )
CDF
X 1 2 3 4 X 1 2 3 4
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Cumulative Distribution Function (CDF) of Discrete
Random Variable
CDF can be used to find the probability of a range of values in a
rv’s image:
{ } { } { }( ( )
Pr Pr Pr
X X
a X b X b X a F b F a
< ≤ = ≤ − ≤= −
p X ( x ) F X ( x )
pmf CDF
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X 1 2 3 4 X 1 2 3 4
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Properties of a CDF
F1: ( )0 1X F x x ≤ ≤ ∀ − ∞ < < ∞
F2: ( )X X a a ⇒
p X ( x )
pmf
F X ( x )
X 1 2 3 4 X 1 2 3 4
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Properties of a CDF
F3:( )lim 0x X F x →−∞ =
F4:
mx X x →∞ =
( ) ( ) ( )1 1X i X i X i F x F x p x + += +
p X ( x ) F X ( x )
pmf CDF
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X 1 2 3 4 X 1 2 3 4
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Expected Value of a Discrete Random
Variable The expected value, expectation or mean of a discrete rv is the
“average” value of the random variable
What is the average value of a random variable whose image is
S = 1 6 7 9 13 ?
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Expected Value of a Discrete Random
Variable What is the average value of the following random variable
whose image is S X ={1, 6, 7, 9, 13}?
If your answer is 7.2 then you assumed that all of the values in the
rv’s ima e have e ual wei hts
( )
1 1 1 1 1 1
1 6 7 9 13 1 6 7 9 13 7.25 5 5 5 5 5× + × + × + × + × = + + + + =
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Expected Value of a Discrete Random
Variable Mathematically, the expected value of a discrete random
variable is:
{ } { }Prk X
X k k a S
X a X a µ∈Ε = = =∑
,
In such cases, we say that the expected value does not exist
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Variance of a Random Variable
Variance of a rv is a measure of “the amount of variation of a rv
around its mean”
Intuitively, which of the following discrete rvs has a higher
variance?
X
0.5E { X }=3.87
X
0.4E { X }=5.2
0.033
.
0.2
0.1
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1 6 7 9 13 X 1 6 7 9 13 X
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Variance of a Random Variable
Intuitively, which of the following discrete rvs has a higher
variance?
q X ( x ) has a higher variance because it varies more around its mean
than p X ( x )
p X ( x )
E X =3.87
q X ( x )
E { X }=5.20.5
0.4
0.2
0.4
1 6 7 9 13
0.033
X 1 6 7 9 13
0.1
X
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Variance of a Random Variable
Mathematically, the variance of a discrete rv is defined as:
2var r
k X
X k k
a S
a a σ
∈
= = − =
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Variance of a Random Variable
x x
0.5
0.4
E { X }=3.87 0.4 E { X }=5.2
var { X }=15.36
0.033
0.2
0.1
var { X }=9.91
1 6 7 9 13 X 1 6 7 9 13 X
{ } ( )2var 1 3.87 0.5X = − × + { } ( )2
var 1 5.2 0.4X = − × +
( ) ( )
( ) ( )
2 2
2 2
6 3.87 0.4 7 3.87 0.033
9 3.87 0.033 13 3.87 0.033
− × + − × +
− × + − ×
( ) ( )
( ) ( )
2 2
2 2
6 5.2 0.2 7 5.2 0.2
9 5.2 0.1 13 5.2 0.1
− × + − × +
− × + − ×
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.= .=
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Standard Deviation of a Random Variable
In many scenarios, we use the square root of the variance
called its standard deviation
{ }varX X σ =
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Standard Deviation of a Random Variable
p X ( x )
0.5E { X }=3.87
q X ( x )
0.4E { X }=5.2
0.4
0.2σ
X =3.14
σ X =3.92
1 6 7 9 13
0.033
X 1 6 7 9 13
.
X
{ }var 15.36 3.92X X σ = = ={ }var 9.91 3.14X X σ = = =
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Discrete Uniform Random Variable
A discrete uniform rv, D, has a finite image and all the elements
of the image have equal probabilities
Pr{D=k }
1/n
What do you think are the expected value and standard
k x1 x2 x3 xn
deviation of this random variable?
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Bernoulli Random Variable
A Bernoulli Random Variable is defined on a single event A
This rv is based on an experiment called a Bernoulli trial
The experiment is performed and the event A either happens or does not
happen
Thus the sam le s ace of a Bernoulli rv is binar
Sample Space
Bernoulli Random
VariableX sB
A=head
er n o ul l i T c
o
Ac
= Not head= tail
0 1 Ri al : T o s s
i n
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S x
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Bernoulli Random Variable
A Bernoulli Random Variable is defined on a single event A
This rv is based on a the experiment called a Bernoulli trial
The experiment is performed and the event A either happens or does not
happen
Thus the sample space of a Bernoulli rv is binary
Bernoulli Random
VariableX sP
A=Pak wins
R
k i s t an c r i
pl a y s A u
Ac =Pak losses 0 1
Sample Space
k e t t e am
s t r al i a
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Bernoulli Random Variable
A Bernoulli Random Variable is defined on a single event A
This rv is based on a the experiment called a Bernoulli trial
The experiment is performed and the event A either happens or does not
happen
Thus the sample space of a Bernoulli rv is binary
Bernoulli Random
Of course the Pr{A} = 0 for this
ex eriment
A=Pak wins
VariableX(s)
P ak i s t a
pl a y
Ac =Pak losses0 1
R c r i c k e t t
A u s t r al i
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Sample Space
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images.google.com
am
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Bernoulli Random Variable
Typical examples of Bernoulli rvs in communication:
Transmit a bit over a wireless channel
0−
> bit is received error-free 1 −> bit received is not received error-free => bit is received with errors
Outcomes:
0 −> packet is received
1 −> acket is not received => acket is lost en-route due to con estion
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Bernoulli Random Variable
Sample space of a Bernoulli rv, I, is binary
Both outcomes are mapped to real numbers,
Traditionally: I( A) = 1 and I( Ac) = 0 are used to represent a Bernoulli rv’s
outcomes
The pmf of I is:
Pr{I = 1} = p
Pr{I = 0} = 1 − p
Pr{I =k }
p
I
1- p
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0 1
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Bernoulli Random Variable
Example:
Consider the experiment of a fair coin toss. What is the expected
value and the variance of this Bernoulli rv?
Pr{I =k }
0.5
I 0 1
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Discrete Random Variables
Example:
Consider the experiment of a fair coin toss. What is the expected
va ue an e var ance o s ernou rv
Since the coin toss is fair: Pr{I = 1} = 0.5 and Pr{I = 0} = 0.5
. . .
var{I} = (1−0.5)2x(0.5) + (0−0.5)2x(0.5) = 0.25
Pr{I =k }
=
σ I =0.5
0.5
.
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Bernoulli Random Variable
Example:
Consider a binary symmetric channel with probability of bit-error
0.1. What is the expected value and the variance of this Bernoulli
rv?
Pr{I =k }
0.9
I 0 1
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Bernoulli Random Variable
Example:
Consider a binary symmetric channel with probability of bit-error 0.1. What is the
ex ected value and the variance of this Bernoulli rv?
The event of interest here is a bit-error:
=> Pr{I = 1} = 0.9 and Pr{I = 0} = 0.1
= . . = .
var{I} = (1−0.9)2x(0.9) + (0−0.9)2x(0.1) = 0.09
Note that the variance of this pmf is smaller than the variance of the coin tosspm
Pr{I =k }
0.9σ I =0.3
0.1
µ I =0.9
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I 0 1
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Binomial Random Variable
Consider a collection of n independent Bernoulli trials
A Binomial Random Variable is the total number of occurrences
of an event A in this independent Bernoulli collection
Send n bits count the number of bits that are received with errors
Send n packets, count the number of packets that are not lost
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Binomial Random Variable
If I j ( A)=1 and I j ( Ac)=0, j =1,2,…,n, are used to represent the
outcomes of the Bernoulli trials then the Binomial Random
ar a e, , s
n
X I = ∑
So what is the ima e of ?
1 j =
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Binomial Random Variable
If I j ( A)=1 and I j ( Ac)=0, j =1,2,…,n, are used to represent the
outcomes of the Bernoulli trials then the Binomial Random
, ,
n
j X I = ∑
So what is the image of X ?
1 j =
S X = {0, 1, 2, 3, …, n}
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Binomial Random Variable
Pr{I j (A)=1} = p and Pr{I j (A)=0} = 1- p
Then a Binomial rv X is defined as:
n
X I =1 j =
An t e pm o a nom a rv s
n k k
n −
r p pk = = −
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Binomial Random Variable
The pmf of a Binomial rv X is:
n −
; , rn p p p
k
= = = −
s pm g ves e pro a y a exac y ou o a o a o n
Bernoulli trials were successes
Be very careful about the definition of a success
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Binomial Random Variable
Pr{ X =k }
k
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Binomial Random Variable
Example:
Consider a binary symmetric channel with probability of bit-error p.
n e pro a y a a pac e o n s s rece ve w one
or more errors?
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Binomial Random Variable
Example:
Consider a binary symmetric channel with probability of bit-error p.
n e pro a y a a pac e o n s s rece ve w one
or more errors?
Pr{I j = 1} = 0.1 and Pr{I j = 0} = 0.9, j =1,2,…,n
Then the probability that a packet is received with errors is
( ) ( ) ( ) ( ){ }Pr 1 2 3X X X X n = ∪ = ∪ = ∪ ∪ =…
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Binomial Random Variable
Example:
Consider a binary symmetric channel with probability of bit-error p.
n e pro a y a a pac e o n s s rece ve w one
or more errors?
{ } ( ) ( ) ( ) ( ){ }
{ } { } { } { }
Pr pkt with errs Pr 1 2 3
Pr 1 Pr 2 Pr 3 Pr
X X X X n
X X X X n
= = ∪ = ∪ = ∪ ∪ =
= = + = + = + + =
…
…
( ) ( ) ( ) ( )1 2 31 2 3
1 1 1 11 2 3
n n n n nn n n n n
p p p p p p p pn
− − − − = − + − + − + + −
…
( )1
1
n
n i i
i
p pi −
= = − ∑
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Binomial Random Variable
Example:
Consider a binary symmetric channel with probability of bit-error p.
n e pro a y a a pac e o n s s rece ve w one
or more errors?n
n i n
−
There is an easier way to compute the same probability by noting
1
r p w errsi
p pi
=
= −
that:
{ } { } { } { }Pr pkt with errs Pr 0 1 Pr 0 1 Pr 0X X X = > = − < = − =
( )00
1 10
n n p p− = − −
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p= − −
45
Connection Between Bernoulli and Binomial
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Connection Between Bernoulli and Binomial
RVs
I ( A)=1, I ( A)=0n Bernoulli trials
A
Pr{ A}
I
Pr{I =1}= p
X
{ } ( )Pr 1
n k k
n
X k p pk −
= = −
Pr{I =0}=1- p
Bernoulli Trial Bernoulli RV Binomial RV
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d bl
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Geometric Random Variable
A Geometric Random Variable, M, is the number of Bernoulli
trials until the first occurrence of an event A
The experiment is stopped as soon as event A is observed
The image of a Geometric random variable is infinite but
countable
M = , , , …
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G i R d V i bl
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Geometric Random Variable
The pmf of a Geometric Random Variable, Z , is
1k −= = = −
Al s o c al
g e o
OR
e d
t h em
m e
t r i c p
Depending on whether the success trial is included in the total
rZ p p p= = = − o d i f i e d
f
count or not
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G t i R d V i bl
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Geometric Random Variable
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Image courtesy of Wikipedia article on Geometric Distribution
Connection between Geometric and Binomial
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Connection between Geometric and Binomial
RVs Can we find the probability of a Geometric random variable
using the Binomial random variable?
{ } ( )Pr 1k
Z k p p= = − Geometric
{ } ( )Pr 1n k k
n Z k p p
k
− = = −
Binomial
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Connection between Geometric and Binomial
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RVsCan we find the probability of a Geometric random variable using
the Binomial random variable?
11 k k k
− −
{ } ( ) { }1
1 1
Pr 1 Prk
X k p p Z k −
= = − = =
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Connection between Geometric and Binomial
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RVs In k Bernoulli trials, there are k ways in which you can have 1
success and k -1 failures
Since the Binomial random variable counts successes andfailures it sums and considers all the k outcomes to ether
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Connection between Geometric and Binomial
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RVsFor k =4, a Binomial random variable X jointly considers the
outcomes 0001, 0010, 0100, 1000
Pr{ X = 1} = Pr{0001} + Pr{0010} + Pr{0100} + Pr{1000}Since the underlying Bernoulli trials are independent:
Pr{ X = 1} = (k )Pr{one success in k trials}
,one of these outcomes, 0001
=> Pr{ Z = 1} = Pr{ X = 1} / k
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Memoryless Property
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Memoryless Property
It can be shown that the Geometric rv satisfies the memoryless
property
The memoryless property is satisfied when:
{ } { }Pr PrZ j k Z k Z j = + ≥ = =
This property is also called the Markov Property
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Memoryless Property
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Memoryless Property
The memoryless property is satisfied when:
{ } { }Pr PrZ j k Z k Z j = + ≥ = =
For the Geometric rv, RHS of the above equation is:
r p p= = −
property, we need to show that
Pr 1j
Z k Z k = = −
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Memoryless Property
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Memoryless Property
To prove that the Geometric rv satisfies the memoryless
property, we need to show that
Let’s expand the LHS using the definition of conditional
{ } ( )Pr 1j
Z j k Z k p p= + ≥ = −
probability
{ }Pr Z j k Z k
= + ≥∩{ }
{ }
Pr
Pr
Z k
Z j k
≥
= +=
( )1 2
1
1 1 1
j k
k k k
p p+
+ +
−=
− − − …
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Memoryless Property
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Memoryless Property
Continued from last page
j k +−
( ) ( ) ( )( )
1 2Pr
1 1 1
1
k k k
j k
Z j k Z j p p p p p p
p p
+ +
+
= + ≥ =
− + − + − +−
…
( ) ( ) ( ) ( )( )
( )
0 1 21 1 1 1
1
k
j k
p p p p p p p
p p
+
=− − + − + − +
−
…
( )
( )
1
1
k
j
p
p p
−
= −
Summation over all possible values
of the Geometric pmf
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Memoryless Property
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Memoryless Property
It can also be shown that the Geometric rv is the only discrete
random variable that satisfies the memoryless property
Because of the memoryless property, the Geometric rv can bethought of as the number of failures between two successes
OR the inter-arrival time between successes
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