2 random variables - kahrbjy.files.wordpress.com · ee360 random signal analysis chapter 2: random...

EE360RandomSignalanalysis

Chapter 2: Random Variables 1'

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EE360-RandomSignalA

nalys

is-E

lect

rical

Engi

neering Department - JUST.

EE360-RandomSignalAnalysis-

Ele

ctr ical Engineering Department - JUST.

EE360-RandomSignal

Anal

ysis

-Ele

ctric

alEn

gine

ering Department - JUST.

EE360-Electrica

lEng

inee

ring

Depa

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2 Random Variables

2.1 The Random Variable Concept

A real r.v. is a real function of the elements of a sample space S.

(Random variable=chance variable=stochastic variable)

(Chung page 75) A r.v. is numerically valued function X of ω with domain Ω.

ω ∈ Ω : ω → X(ω) is called a r.v. on Ω.

Jordan University of Science and Technology - Electrical Engineering Abdel-Rahman Jaradat



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Ex. 2.1-1

Flip a coin, roll a die. H → 1, T → (-2). Outcome of the experiment number from die

× value of H/T.

Example 2.1.1: A random variable mapping of a sample space




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Ex. 2.1-2

Spin a wheel having marks 1 to 12. Sample space S = 0 < s ≤ 12.

Define a r.v. X = X(S) = s2. This maps elements in S onto the real line as the

set 0 < x ≤ 144

Fig. 2.1-2 Mapping applicable to Example 2.1.2




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Conditions for a function to be a r.v.:

1. every point in S must correspond to only one value of a r.v.

2. set X ≤ x shall be an event for any real number x.

PX ≤ x=sum of the probs. of all elementary events corresponding to

X ≤ x.

3. PX = −∞ = PX = ∞ = 0




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y = f(v)

• v: domain of a function = set of values the argument v may take.

• y: range of a function = set of corresponding values of the function.

X=r.v.

• domain of X is S.

• range of X is infinite (−∞ < x < ∞)

P = probability measure

• domain of P is all subsets defined on S.

• range of P is 0 ≤ P ≤ 1




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Discrete and Continuous Random Variables

• Discrete r.v. can take discrete values.

Sample space for a discrete r.v. can be discrete, continuous, or mixed.

• Cont. r.v. is one having a cont. range of values.

Sample space for cont. r.v. can’t be discrete.

• Mixed r.v. is one for which some of its values are discrete and some are

continuous.




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Ex. 2.1-3

S = 1, 2, 3, 4 and a r.v. X = X(s) = s3. Elements in S map to discrete

points x in the set 1, 8, 27, 64.

If the probabilities of the elements of S are

P (1) = 424 , P (2) = 3

24 , P (3) = 724 , P (4) = 10

24 , then

the probabilities of the r.v.’s values become

P (X = 1) = 424 , P (X = 8) = 3

24 , P (X = 27) = 724 , P (X = 64) = 10

24 ,

this is so because of the one-to-one mapping.




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Ex. 2.1-4

Prob. of occurrence of any discrete value of a continuous r.v. = 0

If the continuous r.v. T can take the values from -60 to 120, then the probability that

this r.v. fall in a small region dt centered at t in the given range will have a

probability = dt120−(−60) = dt

180 .

To get the probability that the r.v. = t we take the limit when dt → 0 and hence the

prob. =0




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2.2 Distribution Function

Cumulative probability distribution function (CPDF) defined as

PX ≤ x = probability of the event X ≤ x = function of x

= FX(x) with (−∞ ≤ x ≤ ∞)

Since FX(x) is a probability measure, then the following properties hold:

1. FX(−∞) = 0

2. FX(+∞) = 1

3. 0 ≤ FX(x) ≤ 1

4. FX(x1) ≤ FX(x2) if x1 < x2

5. Px1 < X ≤ x2 = FX(x2) − FX(x1)

6. FX(x+) = FX(x) function continuous from the right




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For the discrete case, one can write CPDF as follows

FX(x) =

N∑

i=1

PX = xiu(x − xi) with u(x) =

1, x ≥ 0

0, x < 0

or

FX(x) =

N∑

i=1

P (xi)u(x − xi)

Note that P (X = xi) = P (xi) is a shorthand notation.

Properties 1,2, 4, and 6 used to verify if GX(x) is a valid distribution function.




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Example 2.2-1

Let X have the discrete values in the set −1,−0.5, 0.7, 1.5, 3. The

corresponding probabilities are assumed to be 0.1, 0.2, 0.1, 0.4, 0.2.

Now PX < −1 = 0 because there are no sample space points in the set

X < −1. Only when X = −1 do we obtain one outcome.




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%Example 2.2.1: Discrete distribution and density functions




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Example 2.2-2

Let X have the continuous values in the set 0 < X ≤ 12 with FX(X) = 112X .

Example 2.2.2: Continuous (Uniform) distribution and density functions




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Density Function

Probability Density Function (pdf) is defined as

fX(x) =dFX(x)

dx

and for the discrete case

fX(x) =

N∑

i=1

P (X = xi)δ(x − xi)

• Existence If the derivative of FX(x) exists, then fX(x) exists. Note the

following:

φ(x0) =

∫ x0

−∞

φ(x)δ(x − x0)dx

δ(x) =du(x)

dx




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u(x) =

∫ x

−∞

δ(ξ)dξ

From the above, for the discrete case

fX(x) =

N∑

i=1

P (xi)δ(x − xi)

• Properties of Probability Density Functions (pdf)

1. 0 ≤ fX(x),∀x

2.∫

∞

−∞fX(x)dx = 1

3. FX(x) =∫ x

−∞fX(ξ)dξ

4. Px1 < X ≤ x2 =∫ x2

x1fX(x)dx = FX(x2) − FX(x1)

Properties (1) and (2) must be satisfied for a function to be pdf.




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Example 2.3-1

Let us test the function gX(s) shown in Figure 2.3-la to see if it can be a valid

density function. It obviously satisfies property 1 since it is nonnegative. Its area is

aα = 1, (must equal unity to satisfy property 2). Therefore, a = 1/α is necessary

if gX(x) is to be a density. Suppose a = 1/α. To find the applicable distribution

function we first write

gX(x) =

0 x0 − α > x ≥ x0 + α

1α2 (x − x0 + α) x0 − α ≤ x < x0

1α − 1

α2 (x − x0) x0 ≤ x < x0 + α




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GX(x) =

0 x0 − α > x∫

x

x0−αgX(ξ)dξ = 1

2α2 (x − x0 + α)2 x0 − α ≤ x < x0

1

2+

∫

x

x0gX(ξ)dξ = 1

2+ 1

α(x − x0) − 1

2α2 (x − x0)2 x0 ≤ x < x0 + α

1 x0 + α ≤ x




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%Example 2.3.1: A possible pdf and Cpdf




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Example 2.3-2

Suppose a r.v. is known to have the triangular pdf of the preceding example with

x0 = 8, α = 5, a = 1α = 1

5 . From previous example formulas:

fX(x) =

0 3 > x ≥ 13

(x − 3)/25 3 ≤ x < 8

0.2 − (x − 8)/25 8 ≤ x < 13

P4.5 < X ≤ 6.7 =

∫ 6.7

4.5

[(x − 3)/25]dx =1

25[x2

2− 3x]|6.7

4.5 = 0.2288




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Example 2.3.3

Given

FX(x) = u(x)[1 − e−x2

b ], b > 0 is a constant

its density function

fX(x) =dFX(x)

dx= u(x)

d

dx[1 − e−x2/b] + [1 − e−x2/b]

du(x)

dx

= (1 − e−x2/b)δ(x) + u(x)2x

be−x2/b = u(x)

2x

be−x2/b

since at x = 0, (1 − e−x2/b) = 0.




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The Gaussian Random Variable

gaussian r.v. has the density function

fX(x) =1

√

2πσ2X

e−

(x−aX )2

2σ2X

and CPDF

FX(x) =1

√

2πσ2X

∫ x

−∞

e−

(ξ−aX )2

2σ2X dξ




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The Normalized Gaussian Random Variable

The normalized gaussian r.v. has mean=0 and unit variance, aX = 0, σX = 1 and

the density function

f(x) =1√2π

e−x2

2

and CPDF

F (x) =1√2π

∫ x

−∞

e−(ξ)2

2 dξ

F (−x) = 1 − F (x)

From general gaussian to normalized we subtract the mean and divide by standard

deviation:

FX(x) = F (x − aX

σX)




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%pdf and cpdf of a gaussian r.v.




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%plot Gaussian density function, mean=3, stdev (sigma)=1

ax=3; sigma=1; %range for x from -3 to +3

x=-1:0.1:7;

y=1/sqrt(2 * pi * sigmaˆ2) * exp(-(x-ax).ˆ2/(2 * sigmaˆ2));

plot(x,y); ylabel(’Gaussian pdf’);

xlabel(’mean=3, sigma=1’); grid;

print -deps pdfgaussian.eps

−4 −2 0 2 4 6 8 100

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

Gau

ssia

n pd

f

mean=3, sigma=1




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Example 2.4.1

Find PX ≤ 5.5 for a gaussian r.v. having aX = 3 and σX = 2.

First normalize your data using the relation

FX(x) = F (x − aX

σX) =

∫ x

−∞

fX(ξ)dξ

, here (x − aX)/σX = (5.5 − 3)/2 = 1.25 and hence

PX ≤ 5.5 = FX(5.5) = F (1.25) and using table in appendix B we get

F (1.25) = 0.8944




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Example 2.4.2

X = Height of a cloud is gaussian r.v., aX = 1830m, σX = 460m.

Find PX > 2750m?

PX > 2750m = 1 − PX ≤ 2750m = 1 − FX(2750) =

1 − F ( 2750−1830460 ) = 1 − F (2.0) = 1 − 0.9772 = 0.0228

Example 2.4.3

gaussian r.v., aX = 7, σX = 0.5, find P (X ≤ 7.3) =?

P (X ≤ 7.3) = FX(7.3) = F ( 7.3−70.5 ) = F (0.6) = 0.7257


EE360 Random Signal analysis

Chapter2:RandomVariables27

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1.00001.0000

1.0000

Gaussian

Distribution

Function:

FX

(x)

=∫

x−∞

fX

(ξ)dξ

=F

(

x−

ax

σX

)

Norm

alizedgaussian

r.v:V

aluesofF

(x)

for

0≤

x≤

3.8

9in

stepsof0.01

JordanUniversityofScienceandTechnology-ElectricalEngineeringAbdel-RahmanJaradat



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Other Distribution and Density Examples: Binomial

Let 0 < p < 1, N = 1, 2, · · · , then the function

fX(x) =N

∑

k=0

(

N

k

)

pk(1 − p)N−kδ(x − k)

is called the binomial density function. The quantity(

Nk

)

is the binomial coefficient(

Nk

)

= N !k!(N−k)! and the binomial cpdf is given by

FX(x) =N

∑

k=0

(

N

k

)

pk(1 − p)N−ku(x − k)




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%Binomial pdf and cpdf: N = 6, p = 0.25




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%binomial density function

N=6; p=0.25; q=1-p; y=[];cy=[];

for k=0:1:N

t=pˆk * qˆ(N-k) * prod(N:-1:N-k+1)/prod(1:1:k);

y=[y t]; cy=[cy sum(y)];

end

plot([0:1:N],y,’ * ’, [0:1:N],cy,’o’);

ylabel(’ * =PMF and o=CPDF for Binomial’);

xlabel(’Binomial N=6, p=0.25’); grid;

print -deps pmfbinomial.eps




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%0 1 2 3 4 5 6

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

*=P

MF

and

o=

CP

DF

for

Bin

omia

l

Binomial N=6, p=0.25




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Other Distribution and Density Examples: Poisson

the Poisson r.v. X has a pdf and cpdf given by

fX(x) = e−b∞∑

k=0

bk

k!δ(x − k)

PX = k = e−b bk

k!

FX(x) = e−b∞∑

k=0

bk

k!u(x − k)

with b > 0 is a real constant b = λT where λ = average rate and T is the time

interval of interest.




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Example 2.5.1

Cars arrive at the rate of 50 cars per hour (5060 per minute). If service time = 1

minute, what is the probability of forming a queue?

Here PX = k = e−b bk

k! . T = 1 minute, b = λT = 56 × 1 = 5

6 Probability of

a waiting line = probability of two or more cars arrive in 1 minute interval= 1-

probability of 0 or 1 cars arrive in 1 minute interval =

1 − PX = 0 − PX = 1 = 1 − e−56 (1 + 5

6 ) = 0.2032




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Other Distribution and Density Examples: Uniform

The uniform pdf and cpdf are defined by:

fX(x) =

1b−a a ≤ x ≤ b

0 elsewhere

FX(x) =

0 x < a

x−ab−a a ≤ x < b

1 b ≤ x




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Uniform pdf and cpdf




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Other Distribution and Density Examples: Exponential

The exponential pdf and cpdf are given by:

fX(x) =

1b e−(x−a)/b x > a

0 x < a

FX(x) =

1 − e−(x−a)/b x > a

0 x < a

for real numbers −∞ < a < ∞ and b > 0.




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%Exponential pdf and cpdf




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Example 2.5.2

Given fP (p) = 1P0

e−p

P0 u(p).

PP > P0 = 1 − PP ≤ P0 = 1 − FP (P0) = 1 − (1 − e−P0/P0) =

e−1 = 0.368




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Other Distribution and Density Examples: Rayleigh

The Rayleigh pdf and cpdf are:

fX(x) =

2b (x − a)e−(x−a)2/b x ≥ a

0 x < a

FX(x) =

1 − e−(x−a)2/b x ≥ a

0 x < a

for real numbers −∞ < a < ∞ and b > 0.




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Ray

leig

hpd

fand

cpdf




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Example 2.5.3

Find the value x = x0 of a Rayleigh r.v. for which

PX ≤ x0 = PX > x0 = FX(x0) = 0.5. This value is called the median

of the r.v.

FX(x0) = 1 − e−(x0−a)2/b = 0.5, hence x0 = a +√

b ln(2).




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Conditional Distribution and Density Functions

P (A|B) =P (A ∩ B)

P (B)

Conditional Distribution: Let A be an event X ≤ x for a r.v. X . The conditional

distribution function of X which is denoted by

FX(x|B) = PX ≤ x|B =PX ≤ x ∩ B

P (B)

the joint event consists of all outcomes s such that

X(s) ≤ x, and s ∈ B




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Example 2.6.1

Two boxes have red, green, and blue balls in them.

xi Ball Color Box 1 Box 2 Total

1 Red 5 80 85

2 Green 35 60 95

3 Blue 60 10 70

Totals 100 150 250

Table 1: Numbers of colored balls in two boxes




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Let the event B1 select box 1, P (B1) = 0.2, while the event B2 select box 2,

P (B2) = 0.8. B1, B2 are mutually exclusive events with

P (B1 ∪ B2) = P (B1) + P (B2) = 1.0

Define X a r.v. to have the values 1, 2, 3 if a red, green, blue ball is selected.

Let B be the event of selecting a box B1 or B2.

P (X = 1|B = B1) = 5100 , P (X = 1|B = B2) = 80

150

P (X = 2|B = B1) = 35100 , P (X = 2|B = B2) = 60

150

P (X = 3|B = B1) = 60100 , P (X = 3|B = B2) = 10

150

Table 2: Conditional probabilities: example 2.6.1




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the conditional pdf fX(x|B1) = 5100δ(x − 1) + 35

100δ(x − 2) + 60100δ(x − 3)

and by integration we get the conditional cpdf:

FX(x|B1) = 5100u(x − 1) + 35

100u(x − 2) + 60100u(x − 3)

Compute

P (X = 1) = P (X = 1|B1)P (B1) + P (X = 1|B2)P (B2)

=5

100× 2

10+

80

150× 8

10= 0.437

P (X = 2) =35

100× 2

10+

60

150× 8

10= 0.390

P (X = 3) =60

100× 2

10+

10

150× 8

10= 0.173

from which fX(x) = 0.437δ(x − 1) + 0.390δ(x − 2) + 0.173δ(x − 3)

and FX(x) = 0.437u(x − 1) + 0.390u(x − 2) + 0.173u(x − 3)




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Dis

trib

utio

nsan

dde

nsiti

esfo

rE

xam

ple

2.6.

1




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Methods of Defining Conditioning Event

1. Event B may depend on some r.v. other than X (Chapter 4).

2. Event B is defined in terms of the r.v. X : let B = X ≤ b where b real

number −∞ < b < ∞.

Substituting we get

FX(x|X ≤ b) = PX ≤ x|X ≤ b =PX ≤ x ∩ X ≤ b

PX ≤ bfor all events X ≤ b for which PX ≤ b 6= 0

Two cases

(a) b ≤ x: here the event X ≤ b is a subset of the event X ≤ x, so

X ≤ x ∩ X ≤ b = X ≤ b and

FX(x|X ≤ b) = PX ≤ x|X ≤ b =PX ≤ bPX ≤ b = 1, b ≤ x




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%

(b) x < b: the event X ≤ x is a subset of the event X ≤ b, so

X ≤ x ∩ X ≤ b = X ≤ x and

FX(x|X ≤ b) = PX ≤ x|X ≤ b =PX ≤ x ∩ X ≤ b

PX ≤ b

=PX ≤ xPX ≤ b =

FX(x)

FX(b), x < b

combining the last two expressions we get

FX(x|X ≤ b) =

FX(x)FX(b) x < b

1 b ≤ x

fX(x|X ≤ b) =

fX(x)FX(b) = fX(x)

∫

b

−∞

fX(x)dxx < b

0 b ≤ x




&

$

%

Note that

FX(x|X ≤ b) ≥ FX(x) and fX(x|X ≤ b) ≥ fX(x), x < b

the main result can be extended to the case B = a < X ≤ b see problem

2.6.2




&

$

%Dis

trib

utio

nsan

dde

nsiti

esap

plic

able

toa

cond

ition

ing

even

tB=

X≤

b



Chapter 2: Random Variables 52

FX(x|X ≤ b) =

FX(x)FX(b)

x < b

1 b ≤ x

1.0

b x

FX(x|X ≤ b)

FX(x)




Example 2.6.2

Target is a circle of 50 m radius. Landing point measured from target center is

Rayleigh distributed r.v. with b = 800 and a = 0. Find the probability that landing

occurs at 10m from the center of target?

FX(x) = [1 − e−x2/800]u(x)

P (within 10m|on target) =FX(10)

FX(50)=

1 − e−100/800

1 − e−2500/800= 0.1229




SUMMARY

This chapter is concerned with two things:

(1) the definition of a random variable; it is always a numerical quantity, regardless

of the random experiment from which it derives; and

(2) the various functions that describe the probabilistic behavior of a random

variable.

The various types of random variables (discrete, continuous, mixed) were defined.

The concepts of probability density and cumulative probability distribution functions

were introduced to define the probabilistic behavior of a random variable.

The important gaussian random variable was discussed in detail, while others

(binomial, Poisson, etc.) were defined.

Finally, conditional distribution and density functions were discussed to demonstrate

how probabilities that are associated with a random variable can depend on some

random event.


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