reviewch3 discrete random variables

1

241-460 Introduction to QueueingNetworks : Engineering Approach

Chapter 3 Discrete RandomChapter 3 Discrete Random Variables

Assoc. Prof. Thossaporn Kamolphiwong

Centre for Network Research (CNR)

Department of Computer Engineering, Faculty of Engineering

Prince of Songkla University ThailandPrince of Songkla University, ThailandEmail : [email protected]

Outline

• Random variables• Types of Random variables• Types of Random variablesDiscrete Random variablesContinuous Random variables

• Probability Mass function (PMF)

Chapter 3 : Discrete Random Variables

2

Random Variables

• Outcome of random experiment need not be a numbernumber

• However, interested in measurement numerical attribute of outcome.Total number of headsExecution time of job

• Outcome are random measurement will be• Outcome are random measurement will be random (random variable)


Definitions

Random variable X is a function that assigns a real

number X() to each outcome in the samplenumber ,X(), to each outcome in the sample

space of a random experiment.

Outcomes

Random VariableX()= x = 2

S


0 1 2-1-2

SX

Sx = {-2, -1,0, 1,2}

3

Name of random variable

Name of random variable :uses Capital letter,X

X can be a set of possible valuesSX : range of random variable XSY : range of random variable Y


Example

Coin is tossed three times and the sequence of heads and tails is noted.

Let X : the number of heads in three coin tosses.

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT , , , , , , ,X() : 3 2 2 1 2 1 1 0

Sx = {0, 1, 2, 3}


4

Types of Random Variables

Two types of Random Variables• Discrete random variableX is a discrete random variable if the range of X is

countableSX = {x0, x1, x2,…}

X is a finite random variable if all values with nonzero probability are in the finite setp y

SX = {x0, x1, x2,…,xn}

• Continuous random variable


Probability Mass Function

• Probability mass function (PMF) is a function that gives the probability that a discretefunction that gives the probability that a discreterandom variable is exactly equal to some value.


5


• DefinitionThe probability mass function (PMF) of theThe probability mass function (PMF) of the

discrete random variable X is

Random variable X

fX (x) = P[ X() = x ]


Event of all outcome in S

PMF of random variable X

PMF Example

Observe three calls at a telephone switch where voice call (v) and data calls (d) are equally likely.voice call (v) and data calls (d) are equally likely. Let X denote the number of voice calls, Y the number of data calls and the corresponding values of the random variables X, Y

ddd ddv dvd dvv vdd vdv vvd vvv

1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8

Outcomes()

P[]


1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8

0 1 1 2 1 2 2 3

P[]

X

3 2 2 1 2 1 1 0Y

6

(Continue)

ddd ddv dvd dvv vdd vdv vvd vvvOutcomes()

1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8P[]

• Find PMF of random variable X

1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8P[]

0 1 1 2 1 2 2 3X

SX = { 0, 1, 2, 3 }

fX(x) = P[X() = x ] = ??


(Continue)


1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8P[]

f (1) = P[X() = 1]

1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8P[]

0 1 1 2 1 2 2 3X

X denote the number of voice calls,


fX(1) P[X() 1]

7

(Continue)


P[] 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8

fX(0) = P[X(ddd) = 0] = P[X = 0] = 1/8

fX(1) = P[X(ddv) = 1] + P[X(dvd) = 1] + P[X(vdd) = 1]

= P[X = 1] = 1/8 + 1/8 + 1/8 =3/8

P[]

X

1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8

0 1 1 2 1 2 2 3

fX(2) = P[X(dvv) = 2] + P[X(vdv) = 2] + P[X(vvd) = 2]

= P[X = 2] = 1/8 + 1/8 + 1/8 =3/8

fX(3) = P[X(vvv) = 3] = P[X = 3] = 1/8


(Continue)

• PMF of random variable X

x

x

x

x

xfX

3

2

1

0

8/1

8/3

8/3

8/1


otherwise0

8

(Continue)

• Find PMF of Y

P[Y=0] = P[{vvv}] = 1/8[ ] [{ }]

P[Y=1] = P[{dvv}]+P[{vdv}]+P[vvd}] = 1/8 + 1/8 + 1/8 = 3/8

P[Y=2] = P[{ddv}]+P[{dvd}]+P[{vdd}]= 1/8 + 1/8 + 1/8 = 3/8

P[Y=3] = P[{ddd}] = 1/8


0.4

Y PY(y)

0 1/8


0

0.1

0.2

0.3

0 1 2 3

f Y(y

)

y

1 3/8

2 3/8

3 1/8

Properties of PMF

• For any x, fX(x) > 0

• xSx fX(x) = 1

• For any event B SX, the probability that X is

in the set B is Probability Mass Function

0 4

P[B] = fX(x)


00.10.20.30.4

0 1 2 3

f X(x

)

XxB

9

Cumulative distribution function

• Definition

Cumulative distribution function (CDF) of a randomCumulative distribution function (CDF) of a random variable X is defined as the probability of the event {X < x}:

FX(x) = P[X < x]


0.10.20.30.4

f X(x

)


X( ) [ ]0

0 1 2 3X

FX(2) = P[X < 2]

Example

Find CDF of Y


P[]

Y

1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8

3 2 2 1 2 1 1 0

Y PY(y)

0 1/8


0.4


1 3/8

2 3/8

3 1/80

0.1

0.2

0.3

0 1 2 3

f Y(y

)

y

10

(Continue)

F[Y=0] = fY(0) = 1/8

F[Y=1] = f (0) + f (1) = 1/8 + 3/8 =4/8 = 1/2

Y PY(y)

0 1/8

1 3/8

2 3/8

3 1/8F[Y=1] = fY(0) + fY(1) = 1/8 + 3/8 =4/8 = 1/2

F[Y=2] = fY(0) + fY(1) + fY(2) = 1/8 + 3/8 + 3/8 = 7/8

F[Y=3] = fY(0)+fY(1)+fY(2)+fY(3)=1/8+3/8+3/8 +1/8 = 1

Cumulative Distribution Function

0 75

1Y FY(y)

0 1/8


0

0.25

0.5

0.75

0 1 2 3No. of data calls

FY(y

)

y

0 1/8

1 1/2

2 7/8

3 1

Compare : PMF & CDF

• Probability Mass Function• Cumulative Distribution FunctionDistribution FunctionCumulative Distribution Function

0

0.25

0.5

0.75

1


FY(y

)

y


0

0.1

0.2

0.3

0.4


f Y(y

)

y


y 0 1 2 3

fY(y) 1/8 3/8 3/8 1/8

y 0 1 2 3

FY(y) 1/8 1/2 7/8 1

11

Properties of CDF

• 0 < FX (x) < 1

•

•

0)()(lim XX

xFxF

1)(lim XX

xFxF


Properties of CDF (2)

• FX(x) is a nondecreasing function of x, that is, if a < b, then FX (a) < FX (b)b, then FX (a) FX (b)

Let

a = 1,

b = 3


0

0.25

0.5

0.75

1

0 1 2 3

FX(x

)

x

FX (1) < FX (3)


No. of data calls

12


• For all a < b, then FX (b) - FX (a) = P[a < X < b]


0

0.25

0.5

0.75

1

0 1 2 3No. of voice calls

FX(x

)

x


0

0.1

0.2

0.3

0.4


f X(x

)

x

Let a = 1, b = 3

FX (3) - FX (1) = P[1 < X < 3]



• P[X > x] = 1 - FX(x)


0

0.25

0.5

0.75

1

0 1 2 3

FX(x

)

x


0

0.1

0.2

0.3

0.4


f X(x

)

x


No. of voice calls

FX(2) = 7/8

13

Example

Consider choosing a child family at random and looking at the sexes of the children from firstlooking at the sexes of the children from first born to last born. They will stop when they have a child of each sex, or stop when they have 3 children. Let X denote the number of girls.

• Find the PMF of XFind the PMF of X

• Find the CDF of X


Solution

S = {GGG, GGB, GB, BG, BBG, BBB}

S = {3 2 1 0}SX = {3, 2, 1, 0}

• Probabilities of OutcomesA three child family

S = {GGG,GGB,GBG,BGG,GBB,BGB,BBG,BBB}S {GGG,GGB,GBG,BGG,GBB,BGB,BBG,BBB}

P{} =1/8A two child family

S = {GG, GB, BG, BB} P{} =1/4


14

(Continue)

P[{GGG}] = P[{BBB}] = 1/8

P[{BBG}] = P[{GGB}] = 1/8P[{BBG}] = P[{GGB}] = 1/8

P[{GB}] = P[{BG}] = ¼

Find PMF of X

P[X = 0] = P[{BBB}] = 1/8

P[X = 1] = P[{BBG}] +P[{GB}] + P[{BG}]

= 1/8 + ¼ + ¼ = 5/8

P[X = 2] = P[{GGB}] = 1/8

P[X = 3] = P[{GGG}] = 1/8


(Continue)

• Find CDF of X

P[X < 0] = P[X = 0] = 1/8

P[X < 1] = P[X = 0] + P[X = 1]

= 1/8 + 5/8 = 6/8 = 3/4

P[X < 2] = P[X = 0] + P[X = 1] + P[X = 2]

= 1/8 + 5/8 + 1/8 = 7/8

P[X < 3] = P[X = 0] + P[X = 1] + P[X = 2] + P[X = 3]

= 1/8 + 5/8 + 1/8 + 1/8 = 1


15

(Continue)

PMF f X CDF f X

2

1

0

8/1

8/5

8/1

x

x

x

xf X

2

1

0

8/7

4/3

8/1

x

x

x

xFX

PMF of X CDF of X


38/1 x 31 x

(Continue)

Cumulative DistributionFunction

Probability Mass FunctionFunction

0

0.25

0.5

0.75

1


FX(x

)

x00 1 2 3

No. of data calls

f X(x

)

x

0.25

0.5

0.75

1


x 0 1 2 3

fX(x) 1/8 5/8 1/8 1/8

x 0 1 2 3

FX(x) 1/8 3/4 7/8 1

16

Some important Discrete Random Variable

Discrete Random Variables• Bernoulli Random Variable• Bernoulli Random Variable• Binomial Random Variable• Pascal Random Variable• Geometric Random Variable• Poisson Random Variable


Bernoulli Random Variable

• Definition

A : an event relate to outcomes of random experimentA : an event relate to outcomes of random experiment

XA : Random variable,

XA is called Bernoulli random variable if

success in if

fail in not if

1

0

A

AX A

SX = {0, 1}


otherwise0

1

01

p

p

PX

where 0 < p <1P[A] = p : probability of

“success”

17

Bernoulli RV Example

Suppose you test one circuit. With probability p, the circuit is rejected. Let X be the number ofthe circuit is rejected. Let X be the number of rejected circuits in one test. What is PX(x)?

Sol There are two outcomes,X = 1 with probability p andX = 0 with probability 1-p


otherwise0

1

01

xp

xp

xf X

Bernoulli Random Variable

• If there is a 0.2 probability of a reject, then 080 x

otherwise0

12.0

08.0

x

x

xXPxf X

PMF of Bernoulli RV

1

x)


f X(x

)

0

0.5

0 1x

f X(x

18

Binomial Random variable

Binomial Experiment

Th i t i t f f• The experiment consists of a sequence of ntrials, where n is fixed in advance of the experiment.

• The trials are identical, and each trial can result in one of the same two possible outcomes,

hi h d t d b (S) f il (F)which are denoted by success (S) or failure (F).• The trials are independent.• The probability of success is constant from trial

to trial: denoted by p.Chapter 3 : Discrete Random Variables

Binomial Random variable

Binomial random variable

Gi bi i l i t i ti f• Given a binomial experiment consisting of ntrials

• Let X = the number of S’s among n trials

SX = {0,1,2,…, n}X is binomial rv if the PMF of X has the formX is binomial rv if the PMF of X has the form


nkppk

nkXP knk ..., 0,for 1

19

Binomial RV Example

Suppose you test 24 circuits and you find k rejects. With probability p = 0.2, the circuit is rejected.With probability p 0.2, the circuit is rejected.

Let X be the number of rejected circuits in one test. What is PMF of X?

Sx = {0, 1, 2, …, 24}


24 ..., 0,for 8.02.024 24

k

kkXP kk

PMF of Binomial RV

0.15

0.2

n = 24 p = 0 2Probability mass

0.15

0.2

n = 24, p = 0.5

0

0.05

0.1

1 3 5 7 9 11 13 15 17 19 21 23 25

n 24, p 0.2

function of binomial random variable

P[X k] is maximum at


0

0.05

0.1

1 3 5 7 9 11 13 15 17 19 21 23 25

P[X=k] is maximum at kmax = (n+1)p

20

Example

Transmitting binary digits through a communication channel, the number of digitscommunication channel, the number of digits received correctly, Cn, out of n transmitted digits. Find

The probability of exactly i errors Pe[i]

The probability of an error-free transmission


Solution

p : probability of success to transmit binary digitE : # of digits received erroneouslyE : # of digits received erroneously

The probability of exactly i errors Pe[i]

Th b bilit f f t i i

nippi

niEP iin

e ..., 0,for 1

The probability of an error-free transmission


nne pp

nEP

00

21

Example

• Suppose that student guesses on each question of a multiple-choice quiz. If each question hasof a multiple choice quiz. If each question has four different choices, find the probability that student gets one or more correct answers on a 10-item quiz.


Solution

• Let X = the correct number of answers

• each question has four different choices then p• each question has four different choices, then p= 0.25

• PMF of X is

10

where x = 0, 1, 2, …, 10


xx

xxXP

1025.0125.0

10

22

(Continue)

• The probability of at least one correct answer is

P(X ≥ 1) = 1 - P(X=0)

056.075.025.00

100 100

XP

P(X ≥ 1) = 1 - 0.056 = .944


Example

• Suppose that student guesses on each question of a multiple-choice quiz. If the quiz consists ofof a multiple choice quiz. If the quiz consists of three questions, question 1 has 3 possible answers, question two has 4 possible answers, and question 3 has 5 possible answers, find the probability that student gets one or more correct answers.


23

Solution

• Let X = the correct number of answersX is not binomial !X is not binomial !Xi is the number of correct answers on question i

X = X1 + X2 +X3

(note that the only possible values for Xi are 0 and 1, with 0 representing an incorrect answer and 1representing a correct answer)


(Continue)

• Since each Xi are independent, the probability of at least one correct answer isat least one correct answer is

P(X ≥ 1) = 1 - P(X=0)

= 1 – [P(X1 = 0)P(X2 = 0)P(X3 = 0)]

= 1 – (⅔)(3/4)(4/5)


24

Pascal Random Variable

• The experiment consists of a sequence of independent trials.independent trials.

• Each trial can result in a success (S) or a failure (F).

• The experiment continues until a total of ksuccesses have been observed, where k is a specified positive integerspecified positive integer.

• The probability of success is constant from trial to trial: denoted by p.


Pascal Random Variable

The PMF of the Pascal r.v. X = number of trials until we observe the kth success, withuntil we observe the k success, withparameters k = number of successes, is

kxk ppk

xxXP

11

1

where x = k, k + 1, k + 2,…


25

Binomial & Pascal

Pascal : test circuit until find k successes.

Binomial : test n circuit, find k successes.


Pascal RV Example

• If there is a 0.2 probability of a reject and we seek four detective circuits, the random variableseek four detective circuits, the random variable L is the number of tests necessary to find the four circuits.

44 8.02.03

1][

ll

lLP

PMF of Pascal RV

0 04

0.06


3

][

0

0.02

0.04

0 10 20 30 40l

f L(l

)

l = 4, 5, …

26

Geometric Random Variable

• The experiment consists of a sequence of independent trials.independent trials.

• Each trial can result in a success (S) or a failure (F).

• The experiment continues until first success.• Probability of success, p, is the same for each

bobservation



X : the number of independent Bernoulli trials

until the first occurrence of a successuntil the first occurrence of a successX is called geometric random variable if the PMF of

X has form

xpp x 211 1

fail


otherwise

xppxXP

,...2,1

0

1][

27

PMF of Geometric RV

0.3

0.4

0.5

0

0.1

0.2

0.3

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

p = 0.5

0.4

Probability mass function of thegeometric random variable


0

0.1

0.2

0.3

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

p = 0.3

CDF of Geometric RV

• The CDF of X evaluated at integer values can be shown i j 1shown

k

j

jpqkXP1

1

1

0

k

i

iqp

i = j -1

k k

i qq

1

1

1

,Where q = 1 - p


kk

qq

qpkXP

11

1

i q

q0 1

28

Modified Geometric RV

• Interested X' = X -1, the number of failure before a success occurssuccess occurs

• Modified geometric rv : Defined as total no. of failed trials before first success., X= X' + 1. Then X' is a modified geometric random variable with PMF:

P[X' = k] = P[X - 1 = k] = P[X = k + 1]

= (1-p)kp ,k = 0,1,2,…



• Geometric distribution is the only discrete

distribution that satisfies MEMORYLESS

property.

• Future outcomes are independent of the past

eventsevents.


29

Memoryless property

• LetX : # of trials up to the first successp j : # of all failures k : additional trials up to the first success,

X = j + k

F il i lX


First successkj

Fail trial

Memoryless property

“Memoryless property:”

• If a success has not occurred in the first j trials, then the probability of having to perform at least k more trials is the same as the probability of initially having

1, allfor | kjkXPjXjkXP

to perform at least k trials

• Thus, each time a failure, the system “forgets” and begin a new as if it were performing the first trial


30

Example

In a test of integrated circuits there is a probability p = 0.2 that each circuit is rejected. Let Y equalp 0.2 that each circuit is rejected. Let Y equal the number of tests up to and including the first test that discovers a reject.

What is the PMF of Y?


Solution

Let a : accepted circuit, r : reject circuitY = 1r r rY = 2 Y = 3

P[Y=1] = p,

p

a1- p

p

a1- p a1- p

p

P[Y=2] = p(1-p),

P[Y=3] = p(1-p)2


31

(Continue)

• p = 0.2

yy 218020 1

otherwise

yppyYP

y ,...2,1

0

1][

1


otherwise

yyYP

y ,...2,1

0

8.02.0][

PMF of Geometric RV

otherwise0

,...2,18.02.0 1 yyf

y

Y

PMF of Geometric RV

0 1

0.2

Y(y

)

otherwise0


0

0.1

0 10 20Y

f Y

32

Geometric and Pascal Distributions

• Let X1, X2, … Xn, … (independent and identically distributed)distributed)

1 2 3 4 5 6 7 8 9 10 11 12

X1 X2 X3

X = X1 + X2 + X3

Trials

• Pascal random variable represented as a sum of geometric random variables.


: Indicates a trial that results in a “success”

Poisson Random Variable

• Interested in counting the number of occurrences (N) of an event in a certain timeoccurrences (N) of an event in a certain time period or in a certain region in space

• The PMF for Poisson random variable is given by

...,2,1,0 !

kek

kNPk

where is the average number of event occurrences in a specified time interval


33

PMF of Poisson RV

Probability mass function of Poisson0.3

0.4

0.5

= 0.75 function of Poisson random variable

0

0.1

0.2

1 3 5 7 9 11 13 15 17 19 21 23

0 15

0.2

0.25

3

0.1

0.15 = 9


0

0.05

0.1

0.15

1 2 3 4 5 6 7 8 9 1011 12 131415 1617 181920 212223 24

= 3

0

0.05

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Poisson RV in physical situations

• Sequence of Bernoulli trials take place in time• Condition hold• Condition holdOne event can occur in subintervalOutcome in different subintervals are

independentProbability of an event occurrence in subinterval

is p = /n is average number of eventsis p /n, is average number of events observed in T –second interval


34

(Continue)

t

Event occurrences in n subintervals of [0,T]

M : # of trials until the occurrences of an event is a geometric random variable with p = /n

0 Tt


0 Tt

MT/n

Poisson RV Example

Requests for telephone connections arrive at a telephone switching office at a rate of calls per second. It is known that the number of requests in a time period is a Poisson random variable.

Find the probability that there are no call requests in t second.

Fi d th b bilit th t thFind the probability that there are n or more requests.


35

Solution

• The average number of call requests in a t-second period is = t.second period is t.

• N(t) : the number of requests in t second a Poisson random variable with = t.

• PMF of Poisson random variable is given by

k


ek

ktNP!

Solution

• Probability that there are no call requests in tsecondsseconds

• Probability that there are n or more requests

tt eet

tNP !0

00

P[N(t) > n]


1

0 !1

n

k

tk

ek

t

= 1 - P[N(t) < n]

36

Poisson RV Example

The number of hits at a Web site in any time interval is a Poisson random variable. Ainterval is a Poisson random variable. A particular site has on average = 2 hits per second.

What is the probability that there are no hits in an interval of 0.25 seconds?

What is the probability that there are no moreWhat is the probability that there are no more than two hits in an interval of one second?


(Continue)

• H : number of hits at a Web site in any time intervalinterval

• average = 2 hits per second.

• = T = (2 hits/s) x (t) = 0.5t hits

210!50 5.0 hhet th


otherwise0

,...2,1,0!5.0 hhethtHP

37

Solution

• What is the probability that there are no hits in an interval of 0.25 seconds?an interval of 0.25 seconds?

= T = (2 hits/s) x (0.25s) = 0.5 hits

The PMF of no hits is

P[H=0]=fH(0) = 0.50e-0.5/0! = 0.607


Solution

• What is the probability that there are no more than two hits in an interval of one second?than two hits in an interval of one second?

In an interval of 1 second,= T = (2 hits/s) x (1s) = 2 hits.The PMF of H is


otherwise0

,...2,1,0!12 2 hhehf

h

H

38

Solution

{H < 2} = {H = 0} {H = 1} {H = 2}

P[H < 2] = P[H = 0] + P[H = 1] + P[H = 2]

= PH(0) + PH(1) + PH(2)

= e-2 + 21e-2 /1! + 22e-2/2! = 0.677


Poisson RV Example

The number of database queries processed by a computer in any 10-second interval is a Poissoncomputer in any 10 second interval is a Poisson random variable, K, with = 5 queries.

(a) What is the probability that there will be no queries processed in a 10-second interval?

(b) What is the probability that at least two queries will be processed in a 2-second interval?


39

Solution

The PMF is5k

(a) The probability that there will be no queries processed in a 10-second interval is

,...2,1,0 !

5 5

kk

ekf

k

K


50 ef K

Solution

(b) = T, = 5, T = 10 = 0.5 queries/second

N : number of queries processed in a 2-second interval, = 2 = 1

Therefore

,...2,1,0 !1 nnenf N

Therefore,

P[N ≥ 2] = 1 – PN(0) – PN(1) = 1 – e-1 – e-1 = 0.264


40

Example

Calls arrive at random times at a telephone switching office with an average of = 0.25switching office with an average of 0.25 calls/second. What is the PMF of the number of calls that arrive in any T = 20-second interval?


(Continue)

• Let j = number of calls that arrive

t 0 25 20 5 ll• = t = 0.25x20 = 5 calls

otherwise

jjejf

j

J

,...2,1,0

0

!/5 5

0.1

0.2

f J(j

)


00 5 10 15j

41

Poisson → Binomial

• Poisson probabilities can be approximate th bi i l b biliti if i l dthe binomial probabilities if n is large and pis small, then = np.

...,2,1,0 !

1

ke

kpp

k

nkf

kknk

N


Poisson RV Example

The probability of a bit error in a communication line is 10-3.line is 10 .

Find the probability that a block of 1000 bits has five or more errors.


42

Solution

• Each bit transmission corresponds to a Bernoulli trial with a “success” corresponding to a bit errortrial with a success corresponding to a bit error in transmission

• The probability of k errors in 1000 transmission is given by the binomial probability with n = 1000and p = 10-3


Solution

• The Poisson approximation to binomial probabilities uses = np = 1000(10-3) = 1probabilities uses np 1000(10 ) 1

515 KPKP

4

0 !1

k

k

ek

111111 1


!4

1

!3

1

!2

1

!1

111 1e

00366.05 KP

43

Summary of Discrete RV

RV PMF

Bernoulli BernoulliSuccess/Fail

BinomialTest n time until

k success

1

01

x

x

p

pxXP

knk ppk

nkXP

1


PascalTest until k

success kxk pp

k

xkXP

11

1

Summary of Discrete RV

RV PMF

Geometric RV 1 1xGeometric RVTest until success

Poisson RVOccurrence in a period

of time

0

1 1x

X

ppxP

ek

kNPk

!


44

Average

• The average is a statistic of the collection of numerical observationsnumerical observations

• Use to describe the entire collection• Averages:MeanMedianModeMode


Average

• adding up all numbers in the collection andMean :

adding up all numbers in the collection anddividing by the number of terms in the sum

• number in the middle of the set of numbersn is odd (n+1)/2 th

i th f /2 d /2 +1 th

Median :


n is even the average of n/2 and n/2 +1 th

• most common number in the collection ofobservations

Mode :

45

Example

For one quiz, 10 students have the following grades (on a scale of 0 to 10)

Sum of the ten grades is 68.4, 5, 5, 5, 7, 7, 8, 8, 9, 10 mean = 68/10 = 6.8

)9, 5, 10, 8, 4, 7, 5, 5, 8, 7

Find the mean, the median, the mode------------------------------------------------------------


4, 5, 5, 5, 7, 7, 8, 8, 9, 10 median = (7+7)/2 = 7(four scores below 7 and four score above 7)

4, 5, 5, 5, 7, 7, 8, 8, 9, 10 mode = 5 (score occurs more often than any other)

Expected Value of Random Variable

• A measure of the center of the distribution• The sum of all possible values for a random• The sum of all possible values for a random

variable, each value multiplied by its probability of occurrence

XSx

XX xxPXE Probability Distribution

0

0.1

0.2

0.3

0.4

f X(x

)


00 1 2 3 4 5x

Expected value

46

Expected Value of Random Variable

x(1) = -2, x(2) = 0x(3) = 0, x(4) = 1Outcomes

RVX()=x

• Experiment RV X

• Experiment Trial n independent

x(5) = 2, x(6) = -1 ect.SX = {-2, -1, 0, 1, 2}

0 1 2-1-2SX

• x(i) ith trial• x(1), x(2), …, x(n) set of n sample values of X


Expected Value of RV

n

in ix

nm

1

1• Average

xSx

xn xNn

m1

• n trials, assume each x SX occur Nx times

• Relative frequency of probability


n

NxP x

nX

lim

47

Expected Value

• Expected value

N1

lili

xSxx

nn

nxN

nm limlim

xSx

x

n n

Nx lim

X xxP

1

2

3


XSx

X

X

XExxP

xSx

x P(x) xP(x)

0 0 3 Probability Distribution0 00

Example

0 0.31 0.252 0.23 0.154 0.05

0

0.1

0.2

0.3

0.4

0 1 2 3 4 5x

f X(x

)

0.000.250.400.450.20

5 0.05

)(xP )]([ xxP

Total 1.00Center of distribution

Expected Value

0.251.55


48

Bernoulli Expected value

1

01 xpxPXPMF, SX = {0, 1}

1xpX

Expected Value

pppxxPXEXSx

X

110

, X { , }


The Bernoulli random variable X has expected value E[X] = p

PMF,

Geometric Expected Value

,...3,2,11 1 xppxP xX

SX = {1, 2,…}

Expected Value

ppxpXEx

x 111

1

x 1


The geometric random variable X has expected value E[X] = 1/p

49

Infinite Geometric Series

x

xpxpXE1

11 qqdq

dpXE 1 ,

1

1

x

x

x

x

x

qd

p

qdq

dp

xqp

0

0

1

p

qpq

p

qq

1

1 1

12

n

x

x

n

x

qdq

dp

qdq

p

0

0

limq

qn

x

x

n

1

1lim

0

Infinite Geometric Series


Poisson Expected Value

0 !

x

ex

xXE 0 !x x

!11

!

1

xxx

xx

1

1

!1x

x

xeXE

1 !x

x

ex

x


0 !m

m

meXE

Let m = x - 1 e

ee

50

Expected Value of other RV

RV E[X]RV E[X]Binomial RV

Test n time until ksuccess

Pascal RV

np


Test until k success k/p

Derived Random Variables

• Use sample values of a random variable to compute other quantitiescompute other quantities

• ExampleMeasure the power level of received signal in

millwatts(x) convert to decibels by calculating y = 10log10x


51

Derived Random Variables

OutcomesS

RVX()=X

0 1 2-1-2

SX

RV Y

0 1 2 3 4


0SY

3

Derived random variable

Example

• Random Variable X = # of pages in colourprinting machinep g

• Probability model PX(x) for # of pages in each colour printing machine

• Charging plan for printing:1st page = 10 Baht2nd page = 9 Baht 3rd 8 B ht

Find a function Y = g(X) f th h f l3rd page = 8 Baht

4th page = 7 Baht5th page = 6 Baht6 – 10 pages = 50 Baht


for the charge for colourprinting

52

Solution

RV XOutcomesS Outcomes

3 4 1021

SX

OutcomesS Outcomes


: Outcomes of colour printing machineX : # of pages in colour printing machine

Solution

RV X

OutcomesS Outcomes

3 4 1021

SX

RV Y


10 19 27 50

SY

Derived RV

106

51

50

5.05.10 2

X

XXXY

53

Solution

X

1

Y

10123456

1019273440


78910

50

PMF of Derived RV

For discrete random variable X,

th PMF f Y (X) ithe PMF of Y = g(X) is

yxgx

XY xPyP:


54

Example

• Suppose the probability model for # of pages Xof a printing machine is

otherwise

x

x

xPX 8,7,6,5

4,3,2,1

0

1.0

15.0

of a printing machine is

• Find the PMF of Y : the charge for printing


106

51

50

5.05.10 2

X

XXXXgY

PY(y)

Solution

X PX(x) Y

PY[10] = PX[1] = 0.15

PY[19] = PX[2] = 0.15

PY[27] = PX[3] = 0.15

PY[34] = PX[4] = 0.15

PY[40] = PX[5] = 0.1

1 0.152 0.153 0.154 0.155 0.10

1019 273440

0.150.150.150.150.1

PY[50] = PX[6]+PX[7]+PX[8]

= 0.1+0.1+0.1 = 0.3


6 0.107 0.108 0.10

50 0.3

55

Example

The amplitude V (volts) of a sinusoidal signal is a random variable with PMF

otherwise0

3,...,2,37/1 vvPV

0

0.1

0.2

-5 0 5v

P[V

=v]


Let Y = V2/2 watts denote the average power of the transmitted signal. Find PY(y)

v

Solution

Y = V2/2

S = {-3 -2 -1 0 1 2 3} vPYP SV = {-3, -2, -1, 0, 1, 2, 3}

SY = {0, 0.5, 2, 4.5}

Y = 0, PY(0) = PV(0) = 1/7

Y = 0.5, PY(0.5) = PV(1) + PV(-1)

yvgvVY vPYP

:

= 1/7 + 1/7 =2/7

Y = 0.5, 2, 4.5, PY(y) = 2/7


56

Solution

3237/1 v

542507/2

07/1

y

y

yPY

otherwise0

3,...,2,37/1 vvPV


otherwise0

5.4,2,5.07/2 yyPY

Expected Value of a Derived RV

Given a random variable X with PMF PX(x) and the derived random variable Y = g(X) the

XSx

XY xPxgYE

the derived random variable Y g(X), the expected value of Y is


ySy

YY yyPYE

57

Example

• What is E[Y]?

otherwise

x

x

xPX 8,7,6,5

4,3,2,1

0

1.0

15.0


106

51

50

5.05.10 2

X

XXXYXg

Solution

8

1xX xPxgYE

E[Y] = 0.15[(10.5-0.5) + (10.5(2)-0.5(2)2 )+ (10.5(3) - 0.5(3)2) + (10.5(4)-0.5(4)2)]

4 (0 1)(50)(3)

1x

8

6

24

1

2 1.0501.055.055.1015.05.05.10xx

XXYE

+ 4 + (0.1)(50)(3)

= 0.15[10 +19 + 27 + 34] + 4 + 15

= 32.5


58

Property of Expected value

For any random variable X,

E[Y] = E[X - µX] = 0

E[Y] = E[aX + b] = aE[X] + b


A linear transformation

Example

• Let V = g(R) = 4R + 7 and

,

2

0

0

4/3

4/1

otherwise

r

r

rPR


• Find E[R], E[V]

59

Solution

• E[R] = 0(1/4) + 2(3/4) =3/2

• V = 4R + 7 S = {7 15}• V = 4R + 7 SV = {7, 15}

• PV(7) = ¼ ,PV(15) = ¾

E[4R+7] = 4E[R] + 7 = 4(3/2) + 7

E[V] = (VPV)= 7(1/4)+15(3/4)


4(3/2) + 7= 13

7(1/4)+15(3/4)= 13

Variance and Standard Deviation

• VarianceDescribes the difference between RV X and its

expected value Y = X – E[X]

How far from average?Because E[Y] = E[X-µX] = 0, it is better to know E[|Y|]

E[Y2] = E[(X - µX)2] Variance


Var[X] = E[(X - µX)2]

60

Variance

• Theorem

2222][Var XEXEXEX X

Var[aX + b] = a2Var[X]


[ ] [ ]

Standard Deviation

• Standard DeviationA measure of dispersion expressed in terms

XX Var

A measure of dispersion expressed in terms of the original units


61

x P(x) xP(x)

0 0 05 0x2 x2P(x)

0 0

Example

Probability Distribution

0.3

0.4

0 0.05 01 0.2 0.22 0.2 0.43 0.1 0.34 0.3 1.2

0 01 0.24 0.49 0.316 1.2

-

x

PX(x

)

0

0.1

0.2

0.3

0 1 2 3 4 5

+

E[X] = 2.855 0.15 0.75 25 0.75

10.45Total 1.00 2.85Var[X] = 10.45 – 2.852

= 2.33X = 1.5


)]([ 2 xPx )(xP )]([ xxPE[X2]

E[X] 2.85

Theorem

(a) If X is Bernoulli (p), then Var[X] = p(1-p)

(b) If X is Geometric (p), then Var[X] = (1-p)/p2

(c) If X is Binomial (p), then Var[X] = np(1-p)

(d) If X is Pascal (p), then Var[X] = k(1-p)/p2

(e) If X is Poisson (p), then Var[X] =


62

References

1. Alberto Leon-Garcia, Probability and Random Processes for Electrical Engineering, 3rd Ed., Addision-Wesley Publishing, 1994

2. Roy D. Yates, David J. Goodman, Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineering, 2nd, John Wiley & Sons, Inc, 2005

3. Jay L. Devore, Probability and Statistics for Engineering and the Sciences, 3rd edition, Brooks/Cole Publishing Company, USA, 1991.


(Continue)

4. Robert B. Cooper, Introduction to QueueingTheory, 2nd edition,Theory, 2nd edition,

5. Donald Gross, Carl M. Harris, Fundamentals of Queueing Theory, 3rd edition, Wiley-Interscience Publication, USA, 1998.

6. Leonard Kleinrock, Queueing Systems Vol. I: Theory, A Wiley-Interscience Publication,Theory, A Wiley Interscience Publication, Canada, 1975


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