2.1 random variable concept

2.1 Random Variable Concept

• Given an experiment defined by a sample space S with elements s, we assign a real number to every s according to some rule as follows;

– X : a function that maps all elements of the sample space into points on the real line

)(sX

Example 2.1-1

• Rolling a die and flipping a coin– Sample space {H,T} x {1,2, ..., 6}– Let the r.v. be a function X as follows

1. A coin head (H) outcome -> positive value shown up on the die

2. A coin tail (T) outcome -> negative and twice value shown up on the die

Example 2.1-2

• The pointer on a wheel of chance is spun– The possible outcomes

are the numbers from 0 to 12 marked on the wheel

– Sample space {0 < s 12}

– Define a r.v. by the function

– Points in S map onto the real line as the set {0 < x 144}

2)( ssXX


• Conditions for a Function to be a Random Variable– Every point in S must correspond to only one value of the

r.v.– The set { X x } shall be an event for any real number x

• The set {X(s) x} corresponds to those points s in the sample space for which the X(s) does not exceed the number x

This means that the set is not a set of numbers but a set of experimental outcomes

– The probabilities of the events {X=} and {X=- } shall be 0:• This condition does not prevent X from being either - or for

some values of s; it only requires that the prob. of the set of those s be zero

- -2 -1 0 1 2

... ...The function that maps outcomes in the following manner can not be a r.v.


• Discrete and Continuous Random Variables– Discrete r.v.:Having only discrete values

• Example 2.1-1 : discrete r.v. defined on a discrete sample space

– Continuous r.v. : Having a continuous range of values• Example 2.1-2 : continuous r.v. defined on a continuous sample

space– Mixed r.v. : Some of its values are discrete and some are

continuous

2.2 Distribution Function

• Cumulative Probability Distribution Function – CDF– The probability of the event {X x} :– Depend on x– A function of x– Distribution function of X– Properties

)()()6(

)()(}{)5()()()4(

1)(0)3(1)()2(0)()1(

1221

2121

xFxF

xFxFxXxPxxifxFxF

xFFF

XX

XX

XX

X

X

X

}{)( xXPxFX

; consider a discrete distribution


• Discrete Random Variable– If X is a discrete r.v., : stair-step form– Amplitude of a step : probability of occurrence of the value of X whe

re the step occurs– If the values of X are denoted xi, we may write

where u(·) is the unit-step function

– Using

)(xFX

N

iiiX xxuxXPxF

1)(}{)(

0001

)(xx

xu

}{)( ii xXPxP

N

iii

N

iiiX xxuxPxxuxXPxF

11

)()()(}{)(


Fx(3)=Fx(3+)=1 by the property (6)

Example 2.2-1

• Let X have the discrete values in the set {-1, -0.5, 0.7, 1.5, 3}

• The corresponding probabilities {0.1, 0.2, 0.1, 0.4, 0.2}

)()(

ii xx

dxxxdu

Example 2.2-2

• Wheel-of-chance experiment

2.3 Density Function• Probability Density Function (PDF) of r.v. X

• Existence– If the derivative of FX(x) exists, then fX(x) exists– If there may be places where dFX(x)/dx is not defined (abrupt

change in slope), FX(x) is a function with step-type discontinuities such as in ex. 2.2-2.

dxxdF

xf XX

)()(

2.3 Density Function– Discrete r.v.

• Stair-step form distribution function• Description of the derivative of FX(x) at stairstep points

– Unit-impulse function, (t), used

N

iiiX

N

iiiX

xxxPxf

xxuxPxF

1

1

)()()(

)()()(

2.3 Density Function• Unit-Impulse Function (t)

– Definition by its integral property

(x) : any continuous function at the point x = x0

(t) – Can be integrated as a “function” with infinite amplitude,

area of unity, and zero duration• The relationship of unit-impulse and unit-step functions

• The general impulse function– Shown symbolically as a vertical arrow occurring at the

point x=x0 and having an amplitude equal to the amplitude of the step function for which it is the derivative

dxxxxx )()()( 00

)()()()( xudordx

xduxx

2.3 Density Function• A discrete r.v.

– The density function for a discrete r.v. exists

– Using impulse functions to describe the derivative of FX

(x) at its stair step points

N

iiiX

N

iiiX

xxxPxf

xxuxPxF

1

1

)()()(

)()()(

2.3 Density Function• Properties of Density Functions

2

1)(}{)4(

)()()3(

1)()2(

)(0)1(

21x

x X

xXX

X

X

dxxfxXxP

dfxF

dxxf

xallxf

Example 2.3-1• Test the function gX(x) if it can be a valid density function

– Property 1 : nonnegative– Property 2: Its area a=1 a=1/

Example 2.3-3

• Find its density function

0for 21)()(22

xe

bxe

dxd

dxxdFxf b

xbx

XX

01)()(2

bwhereexuxF b

x

X

)(2)(2

xuebx

xf bx

X

2.4 Gaussian r.v.• A r.v. X is called Gaussian if its density function has the form

XX

ax

X

X aandwhereexf X

x

02

1)(2

2

2)(

2

x

a

X

X dexF X

x

2

2

2)(

22

1)(

2.4 Gaussian r.v.• Numerical or approximation methods for Gaussian r.v.

– Tables many tables according to various– Only one table according to normalized – Consider

– For a negative value of x ; – From FX(x), consider

XX a,

XX a,0,1 XX a

xx

a

X

X dexFdexF X

x

22)(

2

2

2

2

21)(

2

1)(

)(1)( xFxF

XXXX

X auddua

u

,

X

XX

axFxF

)(

Example 2.4-1• Find the probability of the event {X5.5} for Gaussian r.v. havin

g aX = 3 and X = 2

8944.0)25.1()5.5(}5.5{

)(

)5.5(}5.5{

FFXP

axFxF

FXP

X

X

XX

X

Example 2.4-2• Assume that the height of clouds at some location is Gaussian r.

v. X with aX = 1830m and X = 460m• Find the probability that clouds will be higher than 2750m

)(1)( xFxF

0228.09772.00.1

)0.2(1460

183027501

)2750(1}2750{1}2750{

F

F

FXPXP

X

2.4 Gaussian r.v.• Evaluations of F(X) by approximation

• Ex) 2.4-3 Gaussian r.v. with aX = 7, X = 0.5

x

dexQwherexQxF

2

2

21)()(1)(

02)1(

121)(

2

22

22

xe

bxaxadexQ

x

x

510.5,339.0 ba

7264.0251.5)6.0(339.0)6.0(661.0

11)6.0(1)6.0(

7257.0)6.0(5.0

73.7)3.7(}3.7{

2/)6.0(

2

2

eQF

FFFXP X

Table B-1 : F(0.6)=0.7257

2.5 Other Distribution and Density• Binomial distribution

– Density function & distribution function

– Bernoulli trial– For N=6 and p=0.25

N

k

kNkX kxpp

kN

xf0

)()1()(

N

k

kNkX kxupp

kN

xF0

)()1()(

,2,110 Nandp

2.5 Other Distribution and Density• Poisson distribution

– Density function & distribution function

– Quite similar to those for the binomial r.v.– If N and p0 for the binomial case in such a way that Np=b, the

Poisson case results– Applications

1. The number of defective units in sample taken from a production line2. The number of telephone calls made during a period of time3. The number of electrons emitted from a small section of cathode in a gi

ven time interval– If the time interval of interest has duration T, and the events being c

ounted are known to occur at an average rate and have a Poisson distribution, then b = T.

0b

00

)(!

)( ,)(!

)(k

kb

Xk

kb

X kxukbexFkx

kbexf

2.5 Other Distribution and Density

• Uniform distribution

– The quantization of signal samples prior to encoding in digital communication systems

The error introduced in the round-off process uniform distributed– Figure 2.5-2

elsewhere0

)(1

)( bxaabxf X

xb

bxabxax

axxFX

1

0)(

2.5 Other Distribution and Density

• Exponential

ax

axebxf

bax

X

0

1)(

)(

ax

axebxF

bax

X

0

11)()(

2.5 Other Distribution and Density• Rayleigh

– The envelope of one type of noise when passed through a bandpass filter

ax

axeaxbxf

bax

X

0

)(2)(

2)(

axaxexF b

ax

X

01)(

2)(

2.6 Conditional Distribution and Density Functions

• Conditional Probability– For two events A and B where P(B)0, the conditional

probability of A given B

• Conditional Distribution– A : identified as the event {X x} for the r.v. x– Conditional distribution function of X

– = the joint event– This joint event consists of all outcomes s such that

– The conditional distribution : discrete, continuous, or mixed random variables

)()()|(

BPBAPBAP

)|()(

}{}|{ BxFBP

BxXPBxXP X

}{ BxX BxX }{

BsandxsX )(


• Properties of Conditional Distribution

)|()|()6(

)|()|(}|{)5()|()|()4(

1)|(0)3(1)|()2(0)|()1(

1221

2121

BxFBxF

BxFBxFBxXxPxxifBxFBxF

BxFBFBF

XX

XX

XX

X

X

X


• Conditional Density– Conditional density function of the r.v. x

• The derivative of the conditional distribution function

• If FX(x|B) contains discontinuities, impulse response are present in fX(x|B) to account for the derivatives at the discontinuities

– Properties

dxBxdFBxf X

X)|()|(

2

1

)|(}|{)4(

)|()|()3(

1)|()2(

)|(0)1(

21

x

x X

x

XX

X

X

dxBxfBxXxP

dBfBxF

dxBxf

xallBxf


• Example 2.6-1– Red, green, and blue balls in two boxes as shown in the following

table– Select a box first, and then take a ball from the selected box– The 2nd box is larger than the 1st box causing the 2nd box to be

selected more frequently – Event selecting the 2nd box : B2, P(B2)=8/10

– Event selecting the 1st box : B1, P(B1)=2/10

– Event selecting a red ball : r.v. x1

– Event selecting a green ball : r.v. x2

– Event selecting a blue ball : r.v. x3 Box

Xi Ball color 1 2 Totals

1 Red 5 80 85

2 Green 35 60 95

3 Blue 60 10 70

Totals 100 150 250


Box

Xi Ball color 1 2 Totals

1 Red 5 80 85

2 Green 35 60 95

3 Blue 60 10 70

Totals 100 150 250

15010

210060

1

15060

210035

1

15080

21005

1

)|3()|3()|2()|2(

)|1()|1(

BBXPBBXPBBXPBBXP

BBXPBBXP

)3()2()1()|()3()2()1()|(

10060

10035

1005

1

10060

10035

1005

1

xuxuxuBxFxxxBxf

X

X


– total probability

173.0)()|3()()|3()3(

390.0)()|2()()|2()2(

437.0)()|1()()|1()1(

108

15010

102

10060

2211

108

15060

102

10035

2211

108

15080

102

1005

2211

BPBXPBPBXPXP

BPBXPBPBXPXP

BPBXPBPBXPXP

)3(173.0)2(390.0)1(437.0)()3(173.0)2(390.0)1(437.0)(

xuxuxuxFxxxxf

X

X


• Event B : }{ bXB

)(}{}|{)|(

)|()(

}{}|{

bXPbXxXPbXxXPbXxF

BxFBP

BxXPBxXP

X

X

)()()( bXbXxX

1)()(

)(}{}|{)|(

bXPbXP

bXPbXxXPbXxXPbXxFX

)()()( xXbXxX

)()(

)(}{}|{)|(

bXPxXP

bXPbXxXPbXxXPbXxFX

Case 1. bx :

Case 2. b>x :


– Conditional distribution

– From the assumption that the conditioning event has nonzero prob.,

– Similarly,

bx

bxdxxf

xfbFxf

bXxf

bx

bxbFxF

bXxF

b

X

X

X

X

X

X

X

X

0)(

)()()(

)|(

1)()(

)|(

1)(0 bFX )()|( xFbXxF XX

bxxfbXxf XX )()|(


• Example 2.6-2– Sky divers try to land within a target circle– Miss-distance from the point has the Rayleigh distribution wit

hb=800m2 and a=0

– Target : a circle of 50m radius with a bull’s eye of 10m radius– The prob. of sky diver hitting the bull’s eye, given that the la

nding is on the target ?Sol) x = 10, b = 50

)(1)( 800

2

xuexFx

X

1229.0)1()1(

)50()10(

)50|10(8002500

800100

ee

FF

XFX

XX

Random Poisson Points• Papoulis pp. 117

– Points in nonoverlapping intervals• P{ka in ta, kb in tb}

)()|()( BPBAPBAP

where A = {ka in ta}, B = {kb in tb}

ta tb

0 T

bb knb

kb

b Tt

Tt

kn

BP

1)(

Random Poisson Points

baa kkn

b

a

k

b

a

a

b

tTt

tTt

kkn

BAP

1)|(

bababkkn

b

akn

b

k

b

ak

b

a

b

b tTt

Tt

tTt

Tt

kkn

kn

BAP

11)(

baba kknba

kb

ka

baba Tt

Tt

Tt

Tt

kknkkn

1

)!(!!!

bb knb

kb

b Tt

Tt

kn

BP

1)(

Homework• Prob. 2.3-1, 2.3-5, 2.3-9, 2.3-12, 2.3-13, 2.4-

3, 2.4-4, 2.5-7• Due : next Tuesday

2.1 random variable concept

Documents

function x

value of x

values of x

point x

continuous function

x existenceif

values of s

sample space s