reviewch5-multiple of random variable

8/3/2019 ReviewCh5-Multiple of Random Variable

http://slidepdf.com/reader/full/reviewch5-multiple-of-random-variable 1/51

1

241-460 Introduction to Queueing

Netw orks : Engineering Approach

Assoc. Prof. Thossaporn KamolphiwongCentre for Network Research (CNR)

Department of Computer Engineering, Faculty of EngineeringPrince of Songkla University, Thailand

apter u t p e o an om Variables

Email : [email protected]

Outline

Multiple of Random Variable

Events and Probabilities

Joint Probability Mass Function

Joint Cumulative Distribution Function

• Part II

Joint Pro a i ity Density Function

Marginal PMF & PDF

Conditional joint PMF & PDF

Chapter 5 : Multiple of Random Variables



2

Part I



Random Variable

Experiment

Outcomes

Observe

Math. model Assign number


Probability Random Variable

Math. model



3

Multiple Random Variable

Experiment

Outcomes

ObserveInterest >one thing in

Experiment

Assign vector numberto each outcomes


Random Variable Multiple RVPair of RV

Vector Random Variables

Vector random variables is a function that

ass gns a vec or o rea num ers o eac

outcome ( ) in S and several quantities

are of interest




4

Example

A random experiment consist of selecting a

’ .

Let = outcome of this experiment,

H ( ) = height of student in inches,

W ( ) = weight of student in ponds, and

A = a e of student in ears.

The vector ( H ( ), W ( ), A( )) is a vector random

variable


Events and Probabil ities

Consider the two-dimensional variable Z = ( X,Y ).

Find the re ion of the lane corres ondin to the events

A = { X + Y < 10}

= + <




5

Event and P robability(2)

y

(10,0)

(0,10)

x A

(10,0)

(0,10)

xC

X+Y =10

X 2+Y 2 = 100


A = { X + Y < 10} C = { X 2 + Y 2 < 100}

Probability

A2={ x1< X < x2} y

y2

A1={Y < y2}

x1 x2

A = { x1 < X < x2} {Y < y2}

A2 A1

x

A

• =

• P[ A] = P[{ X in A2} {Y in A1}]




6

Probability(2)

B = { x1 < X < x2} { y1 < Y < y2}

y

B

y2

y1


x

P[ B] = P[{ x1 < X < x2} { y1 < Y < y2}]

x1 x2

Probability(3)

For n-dimensional random variable X = ( X 1, X 2, …, X n)

Event

A = { X 1 in A1} { X 2 in A2} … { X n in An}

The probability of events:

P[ A] = P[{ X 1 in A1}{ X 2 in A2}…{ X n in An}]


P[ A] = P[ X 1 in A1,…, X n in An]



7


• Probability Mass functionRandom variable X

• Joint probability mass function

Event of all outcome X () in S

PMF of random variable X

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


S X,Y = {( x,y)|P X,Y ( x,y) > 0}

P X,Y ( x,y) = P[ X = x,Y = y]

Example

• Test two integrated circuits one after the other.On each test the ossible outcomes are a(accept) and r (reject). Assume that all circuits

are acceptable with probability 0.9 and that the

outcomes of successive tests are independent.Count the number of acceptable circuits X andcount the number of successful tests Y before

you observe the first reject . (If both tests aresuccessful, let Y = 2.) Draw a tree diagram for

the experiment and find the joint PMF of

X and Y




8

Solution

• aa 0.81

a

0.9 a X = 2, Y = 2

0.09

0.09

0.01

• ar

• ra

• rr

.

0.1r

0.1 r

0.9

0.1 r

a

X = 1, Y = 1

X = 1, Y = 0

X = 0, Y = 0


P[aa] = 0.81, P[ar ] = P[ra] = 0.09, P[rr ] = 0.01

S = {aa, ar, ra, rr }

Solution(2)

Let

into the pair of RV ( X,Y )

g(aa) = (2,2), g(ar ) = (1,1)g(ra) = (1,0), g(rr ) = (0,0)

• P X,Y ( x,y) = P[ X= x, Y= y]

P X,Y (2 ,2) = P[ X= 2 , Y= 2] = P[aa] = 0.81

P X,Y (1 ,1) = 0.09, P X,Y (1 ,0) = 0.09

P X,Y (0 ,0) = 0.01




9

Solution(3)

x

y x

11

2,2

09.0

81.0

otherwise

y x

y x y xP Y X

0,0

0,1

0

01.0

09.0,,

P X,Y ( x,y) y = 0 y = 1 y =2


x = 0 0.01 0 0

x = 1 0.09 0.09 0

x = 2 0 0 0.81

Solution(4)

y

1

2

x

.81

.09

.09.01 0

1

2

0

0.3

0.6

0.9

01

2

0.010.09

0.81

y


P X,Y ( x,y)



10

Joint PMF Properties

,,

X Y S x S y

Y X

0,, X Y S x S y

Y X y xP


Joint PMF Theorem

For discrete random variables X and Y and any set B

in the X, Y plane, the probability of the event

B y x

Y X y xP BP

,, ,

{( X,Y ) B} is

X

P X ,Y


B={ X 2 + Y 2 < 4}



11

Example

• Find the probability of the event B that X , thenumber of acce table circuits e uals Y the

number of successful tests before observing thefirst failure

P X,Y ( x,y) y = 0 y = 1 y =2

x = 0 0.01 0 0


x = 1 0.09 0.09 0

x = 2 0 0 0.81

Solution

X : # of acceptable circuits

observing the 1st failure

P X,Y

( x,y) y = 0 y = 1 y = 2

x = 0 0.01 0 0

x = 1 0.09 0.09 0 X

Y P X ,Y

S X,Y = {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0),

(2,1), (2,2)}


.



12

(Continue)

B = {( x,y)| X = Y }Y P X ,Y

B S X,Y = {(0,0), (1,1), (2,2)} X

B={( x,y) | X = Y }


B = {(0,0), (1,1), (2,2)}

Solution(2)

B = {(0,0), (1,1), (2,2)} P X,Y ( x,y) y = 0 y = 1 y =2

P[ B] = ??

P B = P 0 0 + P 1 1 + P 2 2

x = 0 0.01 0 0

x = 1 0.09 0.09 0

x = 2 0 0 0.81

, , ,

= 0.01 + 0.09 + 0.81 = 0.91




13

Marginal PMF

• Experiment with 2 RVs, X, Y

,other ( X )

P X ( x) or PY ( y)

For discrete random variables X and Y with

joint PMF P X,Y ( x,y) ,


P X ( x) = P X,Y ( x,y)

PY ( y) = P X,Y ( x,y)

ySY

xS X

Example

P X,Y ( x,y) y = 0 y = 1 y =2

x = 0 0.01 0 0

Find Marginal PMF of X and Y

x = 1 0.09 0.09 0

x = 2 0 0 0.81

Solution

Marginal PMF for X

•


, ,

2

0

, , y

Y X X y xP xP

P X (0) = 0.01

P X (1) = 0.09 + 0.09 = 0.18

P X (2) = 0.81



14

Solution

P X,Y ( x,y) y = 0 y = 1 y =2

x = 0 0.01 0 0

• Marginal PMF for Y P X ( x)

0.01

x = 1 0.09 0.09 0

x = 2 0 0 0.81

• PY (0) = 0.01 + 0.09

= 0.1

0

, , x

Y X Y y xP yP

PY ( y) 0.1 0.09 0.81

0.18

0.81

1


• PY (1) = 0.09

• PY (2) = 0.81

Solution (2)

x

x

1

0

18.0

01.0

otherwise

x xP X

2

0

81.0

y 01.0


otherwise

y

y yPY

2

0

81.0

.



15

Example

The number of bytes N in a message has aeometric distribution with arameter 1- and

range S N = {0,1,2 ,…}.

Suppose that messages are broken into packets of maximum length M bytes.

Let Q be then number of full packets in a messageand let R be the number of b tes left over.

Find the joint PMF PQ,R(q,r ) and the marginal PMF’s of Q and R


Solution

• Joint PMF

N : messa e b tes

M : packets of maximum length

Q : number of full packets { Q = 0,1,…}

R : number of bytes left over { R = 0,1,…, M -1}

N = qM + r N bytes


M bytes R bytes



16

Solution(2)

PMF of Geometric RV with parameter p :

• first success on the kth attempt

P X [ X = k ] = p(1 - p)k -1 , k = 1,2,3,…

• k failures before the first success

PY [Y = k ] = p(1 - p)k , k = 0,1,2,…

(modified Geometric RV)


Solution(3)

PMF of Geometric RV with parameter 1- p :

(modified Geometric RV)

P[ N = n] = pn(1- p), n = 0,1,2,…




17

Solution(4)

P[ N = n] = pn(1- p) N = qM+r

= P[ N = qM+r ]

= P[Q = q, R = r ] = pn(1- p)


- p

P[Q = q, R = r ] = (1-p) pqM+r

Solution(5)

• Marginal PMF of Q

Q : number of full packets

R = 0 N = qM

R = 1 N = qM + 1

…

R = M – 1 N = qM + M -1




18

Solution(6)

P[ N = n] = P[Q = q, R = r ] = (1-p) pqM+r

Marginal PMF of Q

P[Q = q] = P[ N in {qM, qM +1 ,…, qM+( M-1)}]

1

1 M

r qM p p


0r

1

0

1 M

k

k qM p p p

Solution(7)

P[Q = q]

1

1 M

r qM p p p

,...2,1,0 1

11 q p

p p p

M

qM

0r


P[Q = q] = (1 - p M ) pqM



19

Solution(8)

• Marginal PMF of R

R : number of bytes left over { R = 0,1,…, M -1}

Q = 1 N = r

Q = 2 N = M + r

Q = 3 N = 2 M + r

…Q = q N = qM + r

…


Solution(9)

• Marginal PMF of R

P R = r

0 1q

r qM

p p

, , ,…,

1,...,2,1,0 1

M r p r


1 p



20

Joint CDF

• Cumulative Distribution function

F X ( x) = P[ X < x]

• Joint Cumulative Distribution

function


F X,Y ( x,y) = P[ X < x, Y < y]

(Contiue)

=

( x,y)

Y

y

, , ,


F X,Y ( x,y) X x



21

Theorem : Joint CDF

Theorem : For any pair of random variables X, Y,

0 < F X,Y ( x,y) < 1

F X ( x) = F X,Y ( x,) = P[ X < x, Y < ]

F Y ( y) = F X,Y (, y) = P[ X < , Y < y]

F (-, y) = F ( x, -) = 0 , ,

If x < x1 and y < y1 then F X,Y ( x,y) < F X,Y ( x1 ,y1)

F X,Y (, ) = 1


Summary


Events and Probabilities


Joint Cumulative Distribution Function




22

Joint Probability Density Function

• Definition: The Joint PDF of the continuousrandom variable X and Y is a function f X,Y ( x,y)

x y

Y X Y X dvduvu f y xF ,, ,,

with the property


Single RV X, f X ( x) measure of probability/unit length

Two RV X and Y f X,Y ( x,y) measure probability/unit area

Joint PDF

Theorem :

y x

y xF y x f

Y X

Y X

,,

,

2

,

P[ x < X < x+dx , y < Y < y + dy] = f X,Y ( x,y)dxdy




23

Joint PDF : Theorem

P[ x1< X < x2, y1<Y < y2] = F X,Y ( x2 ,y2) – F X,Y ( x2 ,y1)

– X,Y 1 , 2 X,Y 1 , 1

( x2 ,y2)

Y


( x1 ,y1) X

Joint PDF : Theorem

• f X,Y ( x,y) > 0 for all ( x, y)

1,, dxdy y x f Y X •

• Y X dxdy y x f AP ,,


A



25

Solution

Case 1 completely insidethe region

Y

1

y ( x, y)

• x > 1 and y > 1

F X,Y ( x,y) = 1

Case 2 outside the region

• x < 0 or y < 0

Y

1

X 1 x

F X,Y ( x,y) = 0


X 1

y

x

,

49

( x, y)

x < 0

y < 0

Solution(2)

Case 3 ( x,y) is inside the area of nonzero probability

1

Y

X ( x,y) y

f X,Y ( x, y) = 2

y x

y x

Y X Y X

dudv

dudvvu f y xF

2

,,,,


X 1

y x

v

Y X dudv y xF 0

, 2, xv



26

Solution(3)

y x

dudv xF 2,

Y

f X,Y ( x, y) = 2

y

dvv x0

2

22 y xy

v0

,1

X 1

X ( x,y) y

x


Solution(4)

Case 4 ( x,y) above the triangle

,

1

Y

X ( x,y) y

f X,Y ( x, y) = 2

x x

vY X dudv y xF

0, 2,

x

dvv x2


X 1 xv

0

2 x



27

Solution(5)

Case 5 ( x,y) is the right of

triangleY

y

v

Y X dudv y xF 0

1

, 2,

• 0 < y < 1, x > 1 1

X

X ( x,y) y

x

X,Y ,

v y

dvv12


0

22 y y

Solution

The resulting CDF is

1,1

1,10

10,0

10

0or0

1

2

2

0

,2

2

2

,

y x

x y

x y x

x y

y x

y y

x

y xy

y xF Y X




28

Marginal PDF

• Interested only in one random variable

Ignore Y and observe only X

Ignore X and observer only Y

Theorem : If X and Y are random variables with joint PDF f X,Y ( x,y),

dx y x f y f

dy y x f x f

Y X Y

Y X X

,

,

,

,


Example

2

• The PDF of X and Y is

otherwise

y x x y y x f Y X

,

0,,

• Find The marginal PDFs f X ( x) and f Y ( y)




29

(Continue)

y x x y 1 ,114 / 5 2

-1 < x < 1

otherwiseY X

0,,

Y

1 y > x2


y <

X

-1 1

0.5

0

Solution

• Find f X ( x)

1

Y X=x

0.6 )

dy y

dy y x f x f Y X X

4 / 5

,, 0.5

-1 0 1

X x2

1

4 / 52

dy y x

0.2

.

0-1.5 -1 -0.5 0 0.5 1

x

f X ( x


58

44

18

5

22

1

4

5 x

x



30

Solution

• Find f Y ( y)1

Y

2

3

)

dx y

Y X Y

4 / 5

,,

-1 0 1 X

y1/2

Y=y

-y1/2

y

ydx y 4 / 5

51

-1 -0.5 0 0.5 1 x

f Y (


2

5

423

y

y y

Functions of Two Random Variable

• Use two random variable to computer a newrandom variable

• Example

X : Amplitude of signal transmitted by radio station

Y : attenuation of the signal as it travels to theantenna of a moving car

W : amplitude of the signal at the radio receiver in

W = X/Y




31

Theorem : For discrete random variables X

PMF of Function

and Y, the derived random variable W =

g( X,Y ) has PMF

w y xg y x

Y X W y xPwP,:,

, ,


(Continue)

xPwP , P P X,Y X,Y (( x,y x,y)) w y xg y x ,:,

,


w = g( x,y)



32

A firm sends out two kinds of promotional facsimiles. Onekind contains only text and requires 40 seconds to

Example

.pictures that take 60 seconds per page. Faxes can be1, 2, or 3 pages long. Let random variable L : length of afax in pages. S L = {1, 2, 3}. Let the random variable T :time to send each page. ST = {40,60}. After observingmany fax transmissions, the firm derives the following


P L,T (l,t ) t = 40 sec t = 60 secl = 1 page 0.15 0.1

l = 2 pages 0.3 0.2

l = 3 pages 0.15 0.1

(Continue)

• Let D = g( L,T ) = LT be the total duration in

second of a fax transmission. Find the ran e S D, the PMF P D(d ), and the expected value E [ D]


64



33

Solution

PMF P D(d ) P L,T (l,t ) t = 40 sec t = 60 sec

l = 1 a e 0.15 0.1

• S D = {40, 60, 80, 120, 180}

• P D(40) = P L,T (1,40) = 0.15

• P D(60) = P L,T (1,60) = 0.1

• P D(80) = P L,T (2,40) = 0.3

. .

l = 2 pages 0.3 0.2

l = 3 pages 0.15 0.1

• P D(120) = P L,T (2,60)+P L,T (3,40) = 0.35

• P D(180) = 0.1

• P D(d ) = 0;d 40, 60, 80, 120, 180


(Continue)

P l t t = 40 sec t = 60 sec D (second) P D( d ) ,

l = 1 page 0.15 0.1

l = 2 pages 0.3 0.2

l = 3 pages 0.15 0.1

40 0.15

60 0.1

80 0.3

120 0.35


180 0.1



34

Solution (cont.)

Expected value E [ D]

DSd

D d dP D E

= 40(0.15)+60(0.1)+80(0.3)+120(0.35)+180(.1)

= 96 secP L,T (l,t ) t = 40 sec t = 60 sec


l = 1 page 0.15 0.1

l = 2 pages 0.3 0.2

l = 3 pages 0.15 0.1

Theorem : For continuous random

PDF of Function

,

W = g( X,Y ) is

w y xg

Y X W dxdy y x f wW PwF ,

, ,




35

(Continue)

Y X W dxdy y x f wW PwF , ,

P P X,Y X,Y (( x,y x,y))

w y xg ,


w = g( x,y)

Example

Y x

• X and Y have the joint PDF

otherwise y x f Y X

,

0,,

Y

Y=wX

• Find the PDF of W = Y / X


X



36

Solution

• F W (w) = P[W w]

= w = w

Y

Y=wX Y< wX


X

= w

Solution

For w < 0, F W (w) = 0,

w > 0, CDF is

0 0

dxdyewX Y P y x Y

Y=wX Y< wX

0 0

dxdye

wx

y x


0 0

dxdyee

wx

y x X

= w



37

(Continue)

dxdyeewX Y P

wx

y x 0 0

w

dxee wx x

1

10


(Continue)

0

01 w

w

wX Y PwF W

00

wwdF w w


02 ww

dw



38

Theorem : For variable X and Y , the expectedvalue of W = g( X,Y ) is

Expected Value of Function

Discrete:

X Y S x S y

Y X y xP y xgW E ,, ,

Continuous :


dxdy y xP y xgW E Y X ,,

,

Example

P L,T (l,t ) t = 40

sec

t = 60

sec

• Compute E [ D] directlyfrom P L,T (l,t )

l = 1 page 0.15 0.1

l = 2 pages 0.3 0.2

l = 3 pages 0.15 0.1

• g(l,t ) = lt

3

L T Sl St

T L t lPt lg D E ,, ,


T L

l t

,,

1 60,40

1.060315.04032.06023.04021.060115.0401

sec 96 D E



39

Theorem:

Expected Value of Function

E [g1( X,Y ) +…+ gn( X,Y )] = E [g1( X,Y )] +…+ E [gn( X,Y )]

E [ X+Y ] = E [ X ] + E [Y ]


Conditioning by an Event

Definition : For discrete random variable X and , ,

conditional joint PMF of X and Y given B is

P X,Y | B( x,y) = P[ X = x, Y=y| B]


EventPair of RV



40


• Know joint of PMF orPDF

• Given Event B

• Want to know condition joint Probability when

P P X,Y X,Y (( x,y x,y))

EventEventBB

probability that ( x,y) B]



• Theorem : Condition Joint PMF of

the X,Y plane with P[ B] > 0 ,

otherwise

B y x BP

y xP

y xP

Y X

BY X

,

0

,

,

,

|,

• Theorem: Conditional Joint PDF of

P( x, y)

P[ B]


,

otherwise

B y x BP

y x f

y x f

Y X

BY X

,

0

,

,

,

|,



41

1/16

y

P X,Y ( x,y)

Example

• Random variables X any Y

have the oint PMF P x

1/8

1/8

1/12

1/12

1/12

1/16

1/16

1/16

1

2

3

1 2 3 4

1/4

x

,

as shown. Let B denote theevent X + Y < 4. Find theconditional PMF of X and Y

given B

Chapter 5 : Multiple of Random VariablesChapter 5 : Multiple of Random Variables

(Continue)

y P X,Y ( x,y) B = {( x,y)| X + Y < 4}

1/8

1/8

1/12

1/12

1/12

1/16

1/16

1/16

1

2

3

41/16

1/4

x

= , , , , , , ,

P[ B] = P X,Y (1,1) + P X,Y (2,1) +

P X,Y (2,2) + P X,Y (3,1)= ¼ + 1/8 + 1/8 + 1/12

= 7/12


X+Y = 4

X+Y < 4



42

Solution

y xP

y xPY X

BY X

,,

,

|,

• P X,Y|B(1,1) = (¼)/(7/12) = 3/7

• P X,Y|B(2,1) = (1/8)/(7/12) = 3/14

• P X,Y|B(3,1) = (1/12)/(7/12) = 1/7 1

2

3

4

3/14

y

P X,Y|B( x,y)

3/7 3/14 1/7

• P X,Y|B(2,2) = (1/8)/(7/12) = 3/14


1 2 3 4 x

• Theorem : For random variable X and Y and an

event B o nonzero robabilit the conditional

Conditional Expected Value

expected value of W = g( X,Y ) given B is

Discrete :

X Y S x S y

BY X y xP y xg BW E ,,| |,

Continuous:


dxdy y x f y xg BY X ,, |,



43

Example

Find the conditional expected value of W = X + Y

given the event B = { X + Y < 4}

B = {(1,1), (2,1), (3,1), (2,2)}

P X,Y|B( x,y) = {3/7, 3/14, 1/7, 3/14}

1/4

1/8

1/8

1/12

1/12

1/12

1/16

1/16

1/16

1

2

3

41/16

y P X,Y ( x,y)

BY X y xP y x BW E |, ,|

14

41

14

34

7

14

14

33

7

32


1 2 3 4 x

,

Conditioning by Random Variable

• Outcome of experiment ( x,y) B

Derive new robabilit model for ex eriment conditioning by event

• Change B = { X = x} or B = {Y = y} RVKnowledge of Y derive new probability model

of X conditioning by RV

Con itioning y RV

Conditional PMF of X given Y

Conditional PDF of X given Y




44

Condition PMF

• Definition : For any event Y = y such that PY ( y) >

0 the condition PMF o X iven Y = is

P X|Y ( x|y) = P[ X=x|Y=y]

• Theorem : For random variables X and Y with joint

PMF P X,Y ( x,y) and x and y such that P X ( x) > 0 and

Y , X,Y , X|Y Y Y|X X


• Definition : For y such that f Y ( y) > 0 , the conditional

PDF of X given Y {Y = y} is

Condition PDF

• Theorem : f X,Y ( x,y) = f X|Y ( x|y) f Y ( y) = f Y|X ( y|x) f X ( x)

y f

y x f y x f

Y

Y X

Y X

,|

,

|

x f

y x f x y f

X

Y X

X Y

,|

,

|




45

Example

Random variable X and Y

have the oint PMF1/16

y

P X,Y ( x,y), as shown. Findthe condition PMF of Y

given X = x for each x S x

1/8

1/8

1/12

1/12

1/12

1/16

1/16

1/16

1

2

3

1 2 3 4

1/4

x y xP xP

Y X ,|

,


xP X

(Continue)

X = 1 P( X = 1) = ¼

1/16

y

x

x

2,

1,

8 / 18 / 1

4 / 1

1/8

1/8

1/12

1/12

1/12

1/16

1/16

1/16

12

3

1 2 3 4

1/4

x


otherwise

x

X

,

4,

,

0

16 / 116 / 116 / 116 / 1



46

(Continue)

1/16

y

x

x

2,

1,

4 / 1

4 / 1

1/8

1/8

1/12

1/12

1/12

1/16

1/16

1/16

1

2

3

1 2 3 4

1/4

x

otherwise

x

x xP X

,

4,

3,

0

4 / 1

4 / 1

X = 1, Y = 1

1

1,11|1

,

|

X

Y X

X Y P

PP


141

41

1

1,11|1

,

| X

Y X

X Y P

PP

(Continue)

1/16

y

x

x

2,

1,

4 / 1

4 / 1

2

1

41

81

2

1,22|1

,

| X

Y X

X Y P

PP

1/8

1/8

1/12

1/12

1/12

1/16

1/16

1/16

12

3

1 2 3 4

1/4

x

otherwise

x

x xP X

,

4,

3,

0

4 / 1

4 / 1


2

1

41

81

2

2,22|2

,

| X

Y X

X Y P

PP

otherwise

y yP X Y

2,1

0

2 / 12||



47

(Continue)

From Theorem

y xP Y X ,,

otherwise

y yP X Y

2,1

0

2 / 12||

Then

otherwise

y yP X Y

1

0

11|

|

xP X

X Y |


otherwise

y yP X Y

4,3,2,1

0

4 / 14||

otherwise

y yP

X Y

3,2,1

0

3 / 13||

Example

A server cluster has two servers labeled A and B.Incoming jobs are independently routed by thefront end equipment (called server switch) toserver A with probability p and to server B with

probability (1- p). The number of jobs, X , arrivingper unit time is Poisson distributed withparameter .

of jobs, Y , received by server A, per unit time.




48

Solution

A PY [k ] = ?? p

erverSwitch

B

Poisson()

n o s

1- pServer cluster Server cluster

Let X = # of jobs arrive per unit time at servercluster

Y = # of jobs arrive per unit time at server A


(Continue)

Y X Y k nPk P ,,4

X

X

X is Poisson Distribution

X

Sn

X X Y Y nPnk Pk P ||

1

2

3

1 2 3 4Y

!n

enP

n

X




49

(Continue)

• Jobs occur as a sequence of n independent

Bernoulli trials

k nk

X Y p pk

nnk P

1||

• Condition probability that [Y=k ] given [ X=n] is


Binomial PMF

(Continue)

X X Y Y nPnk Pk P ||

k n

k nk

k n

nk nk

Y

k n

p

k

e p

n

e p p

k

nk P

!

1

!

!1

X Sn


p

k

Y ek

e pk P

1

!

e (1- p)



50

• Theorem : Discrete

Conditional Expected Value of a

Function

• Theorem : Continuous

X S x

Y X ,, |

X S x

Y X y x xP yY X E || |


dx y xP y xg yY Y X g E Y X |,|, |

dx y x xP yY X E Y X

|| |

Independent Random Variables

Definition : Random variables X and Y are independent

i and onl i

Discrete : P X,Y ( x,y) = P X ( x)PY ( y)

Continuous : f X,Y ( x,y) = f X (x)f Y (y)




Example

• Are Q and R independent?

r

M

q M M p p

p p pr RPqQP

1

11

r RP

p pr Mq

1

r P ,

Therefore Q and R are independent


,

References

1. Alberto Leon-Garcia, Probability and RandomProcesses for Electrical En ineerin 3rd Ed. Addision-Wesley Publishing, 2008

2. Roy D. Yates, David J. Goodman, Probabilityand Stochastic Processes: A FriendlyIntroduction for Electrical and ComputerEngineering, 2nd, John Wiley & Sons, Inc, 2005

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

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