reviewch4-continuous random variables

1

241241--460 460 Introduction to Introduction to QueueingQueueingNetworks : Engineering ApproachNetworks : Engineering Approach

Chapter 4 Continuous RandomChapter 4 Continuous Random

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

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

Chapter 4 Continuous Random Chapter 4 Continuous Random VariablesVariables

Email : [email protected]

Outline

Continuous Random Variables• Cumulative Distribution Function (CDF)• Cumulative Distribution Function (CDF)• Probability Density Function (PDF)• Some Useful Continuous RV• Expected Value• Derived random variable• Conditioning a Continuous RV

Chapter 4 : Continuous Random Variables

2

Continuous RV

Continuous Random variable Y is a random variable where the data can take infinitely many values.where the data can take infinitely many values.

Outcomes

Random VariableY() = y

S


0 1 2-1-2

SY

SY = {y|-2 y 2}

Random Variables

Discrete & Continuous RV

Discrete

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

ContinuousS

OutcomesS

0 1 2-1-2SX

Continuous

SY = {y| -2 < y < 2}


OutcomesS

0 1 2-1-2SY

3

Continuous Sample Space

A continuous set of numbers, sometimes referred to as an interval, contains all ofreferred to as an interval, contains all of the real numbers between two limits.

(x1,x2)= {x | x1 x x2}

[x1,x2]= {x | x1 x x2}

[x x )= {x | x x x }[x1,x2) {x | x1 x x2}

(x1,x2]= {x | x1 x x2}


Continuous Random Variables

• A continuous random variable is one for which

the outcome can be any value in an interval of

the real number line.

• Don’t calculate P[X = x] but calculate

b h d b l bP[a < X < b], where a and b are real numbers

• For a continuous random variable P[X = x] = 0


4

Example

a wheel of circumference 1m mark a point on the perimeter

X

mark a point on the perimeter at the top of the wheel.

spinning the pointer in center of wheel

X : distance, 0 X 1

Fo a gi en hat is the For a given x, what is the probability P[X = x]?


Y=nY=2

Y=1

Solution

• Find PMF of Y• Y = # of arc in which pointer

Y

Y=3

Y 2

,...,2,1/1 nyn

f pstops

• SY = {1, 2, …, n} PY(y) = ??

otherwise0

,...,2,1/1 nynyPY


5

(Continue)

• X = distance from marked point to pointer stopsY=1

Y=2Y=n

X

to pointer stops

• P[X = x] < P[Y = nx] = 1/n

• If n ,

Y 2

Y=3

a : smallest integer > a 0

1limlim

nnxYPxXP

nn

• P[X=x] < 0,


g

Y = nx,

Y = 12x0.2 = 2.4 = 3P[X=x] = 0

Cumulative Distribution Function

The cumulative distribution function (CDF)if d i bl X iif random variable X is

FX(x) = P[X < x]


6

CDF Theorem

For any random variable X,

FX(-) = 0

FX() = 1

P[x1 X x2] = FX(x2) – FX(x1)[ 1 2] X( 2) X( 1)


Example

Find CDF of X (P[X < x]), 0 < X <1

Y=1Y=2

Y=3

Y=n

X


7

Solution

Let n = 10 and x = 0.17

{Y = 1}< {X = 0 17} < {Y =2}Y=1

Y=2Y=10

X

{Y = 1}< {X = 0.17} < {Y =2}

{Y < nx - 1}{X < x}{Y < nx}{Y < 10x0.17 - 1}={X < 0.17}={Y < 10x0.17}{Y < 1.7 - 1}{X < 0.17}{Y < 1.7}

Y=3

{Y < 1} {X < x} {Y < 2}

FY(1) < FX(0.17) < FY(2)


(Continue)

• FY(1) < FX(0.17) < FY(2)

nk

y

nkynk

y

nkyFY ,...,2,1 ,

1

1

0

1

0


8

(Continue)

• FY(nx-1) < FX(x) < FY(nx)

n

nxxF

n

nxX

1

xn

nxn

lim n

nxxF

n

nxn

Xn

lim1

lim


nn

n

nxxF

nn

nxn

Xnn

lim1

limlim FX(x) = x

Solution

• The CDF of X is

10

00 x

F

1

10

1 x

xxxFX

0.8

1

1.2

X(x

)


0

0.4

0 0.2 0.4 0.6 0.8 1 1.2

x

FX

9

Probability Density Function

p1 = P[x1 < X < x1+]

F ( +) F ( )2

FX(x)

= FX(x1+) - FX(x1)

p2 = P[x2 < X < x2+]

= FX(x2+) - FX(x2)x1

x2

p 1

p 2 x

Ffd i til

,0Average slope


)()(

)()(][

11

1111

xFxF

xFxFxXxP

XX

XX

dx

xdFxf

xFofderivativeslope

XX

X

)(


The probability density function (PDF) of a continuous random variable X iscontinuous random variable X is

d

xdFxf X

X Slop at point x


dx

10

Theorem

For a continuous random variable X with PDF f (x)

fX(x) 0 for all x,

,)( duufxFx

XX

PDF fX(x),


1)( dxxf X

PDF and CDF

FX(x2) – FX(x1)

fX(x)

The probability of observing X in an interval is the area under the PDF graph between the

x1 x2x

dxxfxXxPx

x X )(][2

121


two end points of the interval

11

Some thoughts

• The probability of any individual outcome is zero.zero.

• The probability mass function (PMF) does not apply for a continuous random variable.

• For a continuous random variable, the probabilities apply to intervals.


PMF versus PDF

Discrete Continuous

A finite set of number All numbers in an interval

x0, x1, x2, …, xn

Probability Mass Function, fx(x)

P[X=x] = fx(x)

Probability Density Function, fx(x)

P[a<x<b] = [area under the graph of fx over [a,b] ]

fX(x)fX(x) Probability given by

Probability given by height


a b

fX

xx

given by area.

12

Discrete Continuous

A finite set of number All numbers in an interval

Comparison of CDF

x0, x1, x2, …, xn

Cumulative Distribution Function, Fx(x)

P[X<x] = Fx(x)

Cumulative Distribution Function, Fx(x)

P[X < x] = Fx(x)

FX(x)

x

FX(x)

x


Summary PDF and PMF

• Discrete random variable, the probabilitydistribution is referred as a probability massfunction (PMF)function (PMF).

• Continuous random variable, the probabilitydistribution is referred as a probability densityfunction (PDF).

• The defining property of a PDF is that the totalarea under the curve is equal to one.


13

Example

The CDF of X is 00 x

Find• The PDF of X

1x1

10

00

xx

x

xFX

• Probability of the event {1/4 X 3/4}.


Solution

10

00

xx

x

xF

• fX(x) = ?

• fX(x) = dFX(x)/dx

1x1

10 xxxFX

101

0

x

otherwisexf X

1

x)


101 x 0.5

x0 0.5 1 1.5

f X(x

14

P[1/4 X 3/4]

= F[3/4] – F[1/4]

Solution

101

0 otherwisexf X [ ] [ ]

= ¾ - ¼ = ½

Or

P[1/4 X 3/4]

101 x

f X

14/34/3

dxdxxf X

1

X(x

)


21

41

43

4/14/1

0.5

x0 0.5 1 1.5

f X(

Note : Interval

Note:When we work with continuous random variable itWhen we work with continuous random variable, it

is usually not necessary to be precise about specifying whether or not range of numbers includes the endpoints. This is because individual numbers have probability zero. The following intervals have the same probability.g p y

(¼, ¾) (¼, ¾]

[¼, ¾) [¼, ¾]


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Expected Values

The expected value of a continuous random variable X isvariable X is

For Discrete Random Variable :

dxxxfXE XX


xSx

X xxPXE

Expected Values of Derived RV

• Expected value of continuous random variable

• Expected value of continuous function

( ) ( )XE X xf x dx


( ) ( ) ( )XE g X g X f x dx

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Theorem

[ ] 0For any random variable X, E[X - X] = 0

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

Var[X] = E[X2] - 2X

Var[aX + b] = a2Var[X]


Example

The probability density function of the random variable Y isvariable Y is

Sketch the PDF and find the following

otherwise

yyyfY

11

0

2/3 2

(1) The expected value E[Y](2) The variance Var[Y](3) The standard deviation Y


17

(Continue)

• Sketch the PDF

h

yyyfY

112/3 2

otherwiseyfY

0

1.5

3

PDF


1 1.50.5-0.5-1-1.5y

Solution

• The expected value3

PDF

dyyyfYE Y

1

1

3

23 dy

y

1

1 1.50.5-0.5-1-1.5

1.5

y


08

3

8

3

8

31

1

4

y

18

(Continue)

• The variance 22Var YYEY

dyyyYE 222

2

3

1

1

5

10

3 y The Variance

0


10

5

3

Var[X] = 3/5

(Continue)

• The standard deviation Y

5/3Var YY


19

Some Useful Continuous RV

Some Random Variables• Uniform Random Variable• Uniform Random Variable• Exponential Random Variable


Uniform Random Variable

• Uniform Random VariableThe uniform random variable arises in situationsThe uniform random variable arises in situations

where all values in an interval of the real line are equally likely to occur


20

Uniform Random Variable

Definition Uniform Random Variable

X is a uniform (a,b) random variable on the interval

otherwise

bxaabxf X 0

/1

[a,b] if the PDF of X is

where the two parameters and b > a


where the two parameters and b > a

Uniform RV Theory

The CDF of X is

If X is a uniform (a, b) random variable

The CDF of X is

bx

bxaab

axax

xFX

1

0

The expected value of X is E[X] = (b+a)/2

The variance of X is Var[X] = (b-a)2/12


21

CDF of Uniform RV

( ) duufxF

x

XX

u

a b duab

xFx

X

1

x

a

uab

1

x

duab

x

a

1

aab

xab

11

0


bx

bxaab

axax

xFX

1

0

ab

ax

Exponential Random Variables

• Exponential Random Variablearises in the modeling of the time between occurrence

0

00

x

x

exf

xX

The probability density function is

The cumulative distribution function is

of events


is the rate at which events occur

0

0

1

0

x

x

exF

xX

22

CDF of Exponential RV

x

uX duexF

x

XX duufxF

x

uedu

xxu eee 0

0


xX exF 1

PDF and CDF

• = 2

0 1

0.20.3

0.40.50.6

fX(x)


0 20.40.6

0.81.0

1.2

FX(x

)

Cumulative Distribution Function

xe 1xe


0.1

-3 0 3 6 9 12 15x

0.2

-3 0 3 6 9 12 15x

23

Memoryless Property of an Exponential RV

• Memoryless property:

P[X > t + h | X > t] P[X > h]

Waiting t second

Waiting more h second

Waiting h second

P[X > t + h | X > t] = P[X > h]

Waiting more h second

The probability of waiting at least an additional h seconds is the same regardless of how long one has already been waiting


Prove : Memoryless Property

tXhtXP

• Prove : P[X > t + h | X > t] = P[X > h]

tXP

tXhtXPtXhtXP

|

X > t

X > t + h

{X > t + h } { X > t} = {X > t + h }


t t + htime

24

Prove : Memoryless Property(2)

tXP

tXhtXPtXhtXP

| tXP tXP

htXP

ht

udue


t

udue

ht

th

ee

e

Prove : Memoryless Property(3)

hXPhXP 1

h

udue h

ue

P[X > h ] = P[X > t + h | X > t]


he

25

Example

The transmission time X of messages in a communication system obeys the exponentialcommunication system obeys the exponential probability law with parameter , that is,

P[X > x] = e-x , x > 0.

Findh f• the CDF of X.

• P[T < X < 2T], where T = 1/.


Solution

• CDF of X

F ( ) P[X < ] 1 P[X > ]

P[X > x] = e-x , x > 0.

FX (x)

0

0

1

0

x

x

exF

xX

= P[X < x] = 1 – P[X > x]

• P[T < X < 2T]


= P[T < X < 2T] = FX(2T) – FX (T)

= 1 – e-2 – (1-e-1) 0.233

26

Solution(2)

fX (x)FX (x)

1

x

e-x

x

1 – e-x

00 x

F 00 x

1


01 xe

xFxX

0

0

x

x

exf

xX

Example

The waiting time X of a customer in a queueingsystem is zero if he finds the system idle, and ansystem is zero if he finds the system idle, and an exponentially distributed random length of time if he finds the system busy. The probabilities that he finds the system idle or busy are p and 1 – p, respectively.

Find the CDF of X


28

(Continue)

FX(x) = P[X < x|idle]p + P[X < x|busy](1-p)

= p + (1 p)(1 e-x)= p + (1-p)(1 – e x)

1

FX(x)

00 x


0 xp

0

0

11

0

x

x

eppxF

xX

Example

The probability that a telephone call lasts no more than t minutes is often modeled as an

otherwise

tetF

t

T

0

0

1 3/

exponential CDF.

What is the PDF of the duration in minutes of a


What is the PDF of the duration in minutes of a telephone conversation?

What is the probability that a conversation will last between 2 and 4 minutes?

29

Solution

1

FT(t

)

te

tFt 01 3/

• The PDF of T 0.4

t)

0.5

t4 9

F

otherwisetFT

0

tdFtf T

T


0.2

t3 8

f T(t

0

dt

tfT

otherwise

te t 0

0

3/1 3/

Solution

• P[2 T 4] = ?

• Using CDF

P[2 T 4] = F4(4) – F2(2) = e-2/3 – e-4/3 = 0.25

• Using PDF

4

3/3/1 tT etf

3/1 tT etF


dteTP t 3/4

2 3

142

34324

2

3/ eee t

30

Probability Models of Derived RV

• Discrete : Determine fY(y) from g(X) and fX(x))If Y = g(X)

yxgx

XY xPyP:

Discrete : Determine fY(y) from g(X) and fX(x))

• Continuous Derived RV


Find CDF FY(y) = P[Y < y] Compute PDF fY(y) = dFY(y)/dy

Example

Find PDF of Y, Y = 100X

• Y = location of the pointer Y=1Y=2

Y=n

X

• Y = location of the pointer on 1-meter circumference of circle

• X = location of the pointer after spinning

Y=2

Y=3

00 x


1

10

0

1

0

x

x

x

xxFX

31

Solution

Y 100X

• FY(y) = P[Y < y]

• Y = 100X

• FY(y) = P[100X < y]

= P[X < y/100]


Solution(2)

• FY(y) = P[X < y/100]P[X < x]

00 y

1

10

0

1

0

x

x

x

xxFX

P[X < x]


100

1000

1

100/

y

y

y

yyFY

32

Solution(3)

• Find PDF of Y, Y = 100X

1000100/1

0

y

otherwiseyfyF

dy

dYY


Derived Random Variable

• If Y = aX, a > 0

• FY(y) = P[Y < y] = P[aX < y]

= P[X < y/a] = FX(y/a)


ayfady

aydFyf X

XY /

1/

33

Derived Random Variable (2)

• If Y = X + b, FY(y) = P[Y < y]

FY(y) = P[X + b < y]

= P[X < y - b] = FX(y-b)

bdF


byfdy

bydFyf X

XY

Example

Let X have the triangular PDF

Find

otherwise

xxxf X

10

0

2

• PDF of Y = aX

• Sketch the PDF of Y for a = ½, 1, 2


34

Solution

Y = aX, FY(y) = FX(y/a)

• 0 < x < 1

• fY(y) = (1/a)fX(y/a)

fY(y) = (1/a)fX(y/a), a > 0

2

3

4

f Y(y

)

a=1/2

a=12

0 < y < a

= (1/a)(2y/a)

= 2y/a2


00 1/2 1 2

1

yf

a=2

Conditioning a Continuous RV

Conditional PDF given an Event

F d i bl X ith PDF f ( ) d• For a random variable X with PDF fX(x) and an event B SX with P[B] > 0, the conditional PDF of X given B is

Bxxf X ,


otherwise

BPxf BX

0|

35

Conditioning a Continuous RV

Theorem :Gi t {B } d th diti l• Given an event space {Bi} and the conditional

PDFs fX|Bi(x),

• If {xB}, the conditional expected value of X

ii

BXX BPxfxfi |

f { } p f

is


dxxxfBXE BX ||

Example

Suppose the duration T (in minutes) of a telephone call is a exponential (1/3) random variable:

otherwise

tetf

t

T

,0

0

31 3

0

0.2

0.4

0 5 10t

f T(t

)

• For calls that last at least 2 minutes, what is the conditional PDF of the call duration?

• What is the condition Expected Value


36

Solution

• The probability of the condition event T > 2 is

• The conditional PDF of T given T > 2 is

ttfT

2

2TP

2

dttfT 32

2

331 edte t


otherwise

TP

tf

tfT

TT

0

22|

Solution (2)

te t 21 32

/T>

2(t)

0 2

0.4

otherwise

tetf TT

2

032|


t

f T/

0

0.2

0 2 4 6 8 10

37

Solution (3)

• The conditional expected value is

2| TTE

32

2

32 dtete tt

2

3231 dtet t


minutes

2

2

532

References

1. Alberto Leon-Garcia, Probability and Random Processes for Electrical Engineering, 3rd Ed.,Processes for Electrical Engineering, 3 Ed., Addision-Wesley Publishing, 2008

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

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


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