Random VariablesDr. Ali Hussein Muqaibel

Ver. 2.3

1

Outline• Definition of Random Variable (RV)

• Conditions on Random Variables

• Types of RV

• Cumulative Probability Distribution Function (CDF)

• Probability Density Function (PDF)

• Gaussian Random Variable

• Other random variables:

– Discrete: Binomial, Poisson.

– Continuous: Uniform, Exponential, Rayleigh …. (more example

in Appendix F)

• Conditional density & distribution

2

Chapter 2: The Random Variable

The outcome of a random experiment may not be a number,

for example tossing a coin or selecting a color ball from a box.

However we are usually interested not in the outcome itself, but rather in some measurement or numerical attribute of the outcome.

Examples

Number of heads and not in the specific order in which

heads and tails occur.

Weight of the student.

In each of these examples, a numerical value is assigned to the outcome.

3

Definition of RV

4

Example: Consider the experiment of tossing a coin 3 times and

observing the number of “Heads” (which is a numeric value).

Example: 𝐻 ← 0, 𝑇 ← 1

A real random variable Xis a function that assigns a real number X(s) to each outcome s in the sample space.

Additional Random Variable Example

Let X be a random variable that maps the “head” outcome to the positive number which correspond to the dots on the die, and “tail” outcome map to the negative number that are equal in magnitude to twice the number which appear on the die.

Example : The experiment of rolling a die and tossing a coin

5

Example: 𝑠 is the outcome of the wheel of chance.

𝑋 𝑠 = 𝑠2

• Maps to the numbers from 0 to 144

• Not necessary that the sample space maps

uniquely

𝑋 𝑠 = 𝑠 or |𝑠2|

Conditions for a Function to be a Random Variable

1. RV can not be multivalued. The random variable function

can map more than one point in

S into same point on the real

line.

2. 𝑷 𝑿 ≤ 𝒙 = sum of the

probability of all elementary

event s that satisfies the

condition.

3. 𝑷 𝑿 = −∞ = 𝟎, 𝑷 𝑿

RVs are functions that satisfies the followings:

6

Example: Tossing a coin 3 times and observing the number of heads

{ X 1 } { HTT, THT, TTH, T TT } Equivalent

P{ X 1 } P(HTT) P(THT) P(T+ TH) P(T+ + TT)Equality

7

Types of Random Variables

Discrete

Flipping a coin 3 times and counting the number of heads.

Selecting a number from the positive integers.

Number of cars arriving at gas station A.

Continuous

Selecting a number between { 0 and 6 }

{ 0 ≤ X ≤ 6 }

8

Mixed RV

Wheel of chance 0 < 𝑠 ≤ 6 → −1 , 6 < 𝑠 ≤ 12 → 𝑠

The type is based on the event not the Sample Space. Example:

Wheel of chance 0 < 𝑠 ≤ 6 → −1 , 6 < 𝑠 ≤ 12 → +1In this example S is continuous and the event is discrete

Distribution Function

If you have a random variable X which is numeric by mapping a

random experiment outcomes to the real line

Example: Flipping a coin 3 times and counting the

number of heads

We can describe the distribution of the R.V X using the probability

9

We will define two more distributions of the random variable

which will help us finally to calculate probability.

Distribution Function

We define the cumulative probability distribution function (CDF)

( ) F x P X xX

where ,

In our flipping the coin 3 times and counting the number of heads

(2) 2F P XX

(2) 2 { 0} { 1} { 2}F P X P X P X P XX

1 3 3 7

8 8 8 8

10

1: Let {0,1,2,3} with ( 0) ( 3)

8

3 ( 1) ( 2)

Exa

8

mple X P X P X

P X P X

1(0) ( 0)

8XF P X

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

8 8 2XF P X P X P X

1 3 3 7(2) ( 2) ( 0) ( 1) ( 2)

8 8 8 8XF P X P X P X P X

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

8 8 8 8XF P X P X P X P X P X

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Distribution Function Properties

1. ( ) 0XF

2. ( ) 1XF

3. 0 ( ) 1XF x

1 2 1 24. ( ) ( ) X XF x F x if x x

1 2 2 15. ( ) ( ) X XP x X x F x F x

6. ( ) ( )X XF x F x

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• From property 4: 𝐹𝑋(𝑥) is a non decreasing function

• Properties 1,2,4, & 6 can be used as a test for the

validity of distribution functions.

A right-continuous function

1: Let {0,1,2,3} with ( 0) ( 3)

8

3 ( 1) ( 2)

Exa

8

mple X P X P X

P X P X

1(0) ( 0)

8XF P X

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

8 8 2XF P X P X P X

1 3 3 7(2) ( 2) ( 0) ( 1) ( 2)

8 8 8 8XF P X P X P X P X

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

8 8 8 8XF P X P X P X P X P X

13

Consider the experiment of tossing the coin 3 times and

observing the number of heads

The probabilities and the distribution function are shown below

The stair type distribution function

can be written as

( ) ( 0) ( 1) ( 2)

( ) ( 1) ( 2)

( ) 3) ( 3

step function step function step function

step fun on

X

cti

u x u x uF x P X P X P X x

u xP X

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𝑋 discrete => 𝐹𝑋(𝑥) stair steps,

in general 𝑭𝑿 𝒙 = 𝒊=𝟏𝑵 𝑷 𝑿 = 𝒙𝒊 𝒖(𝒙 − 𝒙𝒊)

were 𝑁 can be infinite. We may simply 𝑃 𝑋 = 𝑥𝑖 = 𝑃(𝑥𝑖)

Density Function

Probability density function (pdf) 𝑓𝑋(𝑥):the derivative of the

distribution function 𝐹𝑋(𝑥) as the

• In our discrete R.V since

1

( ) ( ) ( )N

X i i

i

F x P X x u x x

( )( ) X

X

dF xf x

dx

1

( ) ( ) ( )N

X i i

i

f x P x x x

1

( ) ( ) ( )N

X i i

i

df x P X x u x x

dx

1

( ) ( )N

i i

i

dP X x u x x

dx

1

( ) ( )N

i i

i

P X x x x

15

Additional Examples: CDF, PDF

16

Properties of Density Functions

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1. 𝑓𝑋 𝑥 ≥ 0, for all 𝑥 (Density is non-negative derivative of non-decreasing function)

2. −∞+∞

𝑓𝑋 𝑥 𝑑𝑥 = 1

3. 𝐹𝑋 𝑥 = −∞𝑥𝑓𝑋 𝑧 𝑑𝑧

From (3), 𝐹𝑋 ∞ = 1

4. 𝑃 𝑥1 < 𝑋 ≤ 𝑥2 = 𝑥1𝑥2 𝑓𝑋 𝑥 𝑑𝑥

𝑃 𝑥1 < 𝑋 ≤ 𝑥2 = 𝐹 𝑥2 − 𝐹 𝑥1 = −∞𝑥2 𝑓𝑋 𝑥 𝑑𝑥 − −∞

𝑥1 𝑓𝑋 𝑥 𝑑𝑥

• Properties (1) and (2) are used to prove if a certain function can

be a valid density function.

Additional Examples: Example 1 • A random current is described by the sample space 𝑆 = −4 ≤ 𝑖 ≤ 12}. A

random variable is defined to by

𝑋 𝑖 =

−2 𝑖 ≤ −2𝑖 −2 < 𝑖 ≤ 11 1 < 𝑖 ≤ 46 4 < 𝑖

a. Show the mapping of the random variable.

b. What type of random variable is this?

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Example 2: PDF & CDF

The sample space of an experiment is 𝑆 = 0,1,2.5,6}. The random

variable is defined to be 𝑋 = 2𝑠. Sketch the density and distribution

functions of the random variable 𝑋.

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Example 3: Properties of pdf

Determine the real constant 𝑎, for arbitrary real constants 𝑚 and 𝑏 > 0, such

that

𝑓𝑋 𝑥 = 𝑎𝑒−𝑥−𝑚𝑏

is a valid density function (called Laplace density)

• The required tests are :

1. 𝑓𝑋 𝑥 ≥ 0, for all 𝑥

(Density is non-negative derivative of non-decreasing function)

2. −∞+∞

𝑓𝑋 𝑥 𝑑𝑥 = 1

• From a sketch, 𝑓𝑋 𝑥 ≥ 0 𝑖𝑓 𝑎 > 0, 𝑎𝑛𝑑 𝑏 > 0 is necessary for non-infinite

area under 𝑓𝑋 𝑥 .

• For this assumed true

−∞+∞

𝑓𝑋 𝑥 𝑑𝑥 = 2 𝑚∞𝑎𝑒−

𝑥−𝑚

𝑏 𝑑𝑥 = 2𝑎𝑏 0+∞

𝑒−𝑧𝑑𝑧 = 2𝑎𝑏 −𝑒−𝑧 |0+∞ = 2𝑎𝑏

• We did change of variable z =𝑥−𝑚

𝑏

• Thus, we require 𝒃 > 𝟎, 𝒂 = 𝟏/𝟐𝒃 20

The Gaussian (Normal) Random Variable

2 0 X Xwhere and a

A random variable X is called Gaussian if its density function

has the form

22 2

2

( )1( )

2

X xa

x

x

Xf ex

are real constants (we will see their

significance when we discuss the

mean and variance later).

The “spread” about the point

x = ax is related to σx

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What is the value of the pdf at

𝑥 = 𝑎𝑥, 𝑥 = 𝑎𝑥 − 𝜎𝑥, 𝑥 = 𝑎𝑥 + 𝜎𝑥?

The Gaussian density is the most important of all densities.

It accurately describes many practical and significant real-world quantities

such as noise, voltage, current, electrons, and quantities which are results

from many small independent random effects

The distribution function is found from

𝐹𝑋 𝑥 = −∞𝑥𝑓𝑥 𝜁 𝑑𝜁

2 2( ) 2

2

1( )

2

X xa

x

x

XF x e d

The integral has no known closed-form solution and must be evaluated by

numerical or approximation method (tables “Appendix B”, Matlab)

What is the value of the CDF at 𝑥 = 𝑎𝑥?

22

However to evaluate numerically for a given 𝑥

2 2( ) 2

2

1( )

2

X xa

x

x

XF x e d

We need 𝜎𝑥2 and 𝑎𝑥

Example: Let 𝜎𝑥2 = 3 and 𝑎𝑥 = 5, then

𝐹𝑋 𝑥 =1

2𝜋3 −∞

𝑥

𝑒−𝜁−5 2

2 3 𝑑𝜁

We then can construct the Table for various values of 𝑥.

220( 5 2(3))1

( 20) 2 3

XF e d

26( 5 2() 3)1

(6) 2 3

XF e d

Evaluate

Numerically

Evaluate

Numerically

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Finally we will get a Table for various values of 𝑥.

However there is a problem!

The Table will only work for Gaussian distribution with

𝜎𝑥2 = 3 and 𝑎𝑥 = 5 .

We know that not all Gaussian distributions have 𝜎𝑥2 = 3 and 𝑎𝑥 = 5 .

Since the combinations of 𝑎𝑥 and 𝜎𝑥2 are infinite (uncountable infinite)

Uncountable infinite tables to be constructed

Unpractical method

24

2 2( ) 2

2

1( )

2

X xa

x

x

XF x e d

we make the variable change = ( ) in ( )X x Xu a F x

2( )21

( ) 2

X xx au

XF x e du

We will show that the general distribution function 𝐹𝑋(𝑥)

2 21( ) 0 , 1

2

x

X xF x e d a

can be found in terms of the normalize distribution 𝐹(𝑥)

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= 𝐹𝑥 − 𝑎𝑥𝜎𝑋

𝐹𝑋(𝑥) = 𝐹𝑥 − 𝑎𝑥𝜎𝑋

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Example: Evaluation of Gaussian CDF

A Gaussian r.v. 𝑋 with 𝑎𝑋 = 3 & 𝜎𝑋 = 2. Evaluate

the Prb. 𝑋 ≤ 5.5 .𝑥 − 𝑎𝑥𝜎𝑋

=5.5 − 3

2= 1.25

𝐹𝑋 𝑥 = 𝑃 𝑋 ≤ 5.5 = 𝐹𝑋 5.5 = 𝐹 1.25 = 0.8944

Using the table (appendix B)

27

Example 2:

Height of clouds above the ground is Gaussian r.v. 𝑋 with

𝑎𝑥 = 1830 𝑚, 𝜎𝑋 = 460𝑚. Find 𝑃 𝑋 > 2750}.𝑃 𝑋 > 2750 = 1 − 𝑃 𝑋 ≤ 2750 = 1 − 𝐹𝑋 2750

= 1 − 𝐹2750 − 1830

460= 1 − 𝐹 2 = 1 − 0.9772 = 0.0228

≡ 2.28%

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Q- Approximation

𝐹 𝑥 = 1 − 𝑄(𝑥),

𝑄 𝑥 =1

2𝜋 𝑥∞𝑒−

𝜁2

2 𝑑𝜁

• There is no closed form solution for 𝑄(𝑥) but it can be

approximated

𝑄(𝑥) ≈1

1 − 𝑎 𝑥 + 𝑎 𝑥2 + 𝑏

𝑒−𝑥2

2

2𝜋

• 𝑥 ≥ 0 , 𝑎 = 0.339, 𝑏 = 5.510

• 𝐸𝑟𝑟𝑜𝑟 ≤ 0.27% of 𝑄

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Example: Q Approximation

• Gaussian r. v. with 𝑎𝑋 = 7, 𝜎𝑋 = 0.5 ,

• Find the 𝑃 𝑋 ≤ 7.3

𝑃 𝑋 ≤ 7.3 = 𝐹𝑋 7.3 = 𝐹7.3 − 7

0.5= 𝐹 0.6 = 1 − 𝑄 0.6

≈ 1 −1

1 − 0.339 0.6 + 0.339 0.6 2 + 5.510

𝑒−0.6 2

2

2𝜋

= 0.7264 in the table 0.7257

• 𝐸𝑟𝑟𝑜𝑟 =0.7264−0.7257

0.7257= 0.00096 ≡ 0.096%

• MATLAB: 1 − 𝑞𝑓𝑢𝑛𝑐(0.6) = 0.7257

30

Other Distribution and Density

Examples • Discrete: Binomial, Poisson.

• Continuous: Uniform, Exponential, Rayleigh ….

(more example in Appendix F)

31

Other Distribution and Density Examples

0

( ( ) (1 ) )N

k

k

X

N kp pf kN

kxx

is called the binomial density function.

is thebinomialcoefficie!

!( t

)!n

N N

k k N k

The density can be applied to the

- Bernoulli trial experiment. (only two outcomes)

- Games of chance (win or lose).

Binomial

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

- Detection problems in radar and sonar. 32

𝑁 = 6, 𝑝 = 0.25

It applies to many experiments that have only two possible

outcomes ({H,T}, {0,1}, {Target, No Target}) on any given

trial (N).

It applies when you have N trials of the experiment of only

outcomes and you ask what is the probability of k-successes out

of these N trials.

0

( ( ) (1 ) )N

k

k

X

N kp pF kN

ku xx

Binomial distribution

N = 6 p = 0.25

33

Matlab Example: Binomial

• % Dr. Ali Muqaibel

• clear all

• close all

• % EE315 HW 3

• N=6;

• P=0.25;

• x=0:N;

• y=binopdf(x,N,P)

• y1=cdf('bino',x,N,P)

• figure (1)

• stem(x,y)

• xlabel('Power delivered by the Master Station')

• ylabel ('pdf')

• % axis ([-.1 0.9 0 0.6])

• figure (2)

• stairs(x,y1)

•

• xlabel('Power delivered by the Master Station')

• ylabel ('CDF')

• % axis ([0 0.8 0 1.2])

34

0 1 2 3 4 5 6 70

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

Power delivered by the Master Station

pdf

0 1 2 3 4 5 6 70.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Power delivered by the Master Station

CD

F

The following figure illustrates the binomial density and

distribution functions for 𝑁 = 6 and 𝑝 = 0.25.

Example: Binomial Dist.

35

(consta

0

nt)

N

N b

p

p

0

( ) ( )!

bk

X

k

f x xek

bk

0

( ) ( )!

bk

X

k

F x u xk

e kb

where b > 0 is a real constant.

Binomial Poisson

The Poisson RV X has a density and distribution

PoissonDeveloped by the French mathematician Simeon Denis Poisson in 1837

N 10

N 20

N 1000

36

The Poisson RV applies to a wide variety of counting-type

applications:

- The number of defective units in a production line.

- The number of telephone calls made during a period of time.

- The number of electrons emitted from a small section of a

cathode in a given time interval.

37

Counting type problems:

Time interval of interest 𝑇Average arrival rate 𝜆How many arrival 𝑥 ?

𝑏 = 𝜆𝑇

0

( ) ( )!

bk

X

k

f x xek

bk

0

( ) ( )!

bk

X

k

F x u xk

e kb

Example: Poisson Distribution

• At a gasoline station, cars arrive according to Poisson

distribution at a rate of 50 per hour. There is 1 gasoline

pump and it takes 1 min to obtain fuel.

• What is the probability that a waiting line will occur.

• Waiting occurs if we have more than 1 customer =1 − 𝐹𝑋(1)

• 𝜆 =50

60car per min

• 𝑇 = 1 𝑚𝑖𝑛

• 𝑏 = 𝜆𝑇 =5

6

• Probability of waiting= 1 − 𝐹𝑋 1 = 1 − 𝑒−5

6 1 +5

6= 0.2032

• Probability of waiting in a line=20.32%

38

0

( ) ( )!

bk

X

k

F x u xk

e kb

Practice: Poisson Dist.

Packets arrive at an internet router at a rate of

20 𝑝𝑎𝑐𝑘𝑒𝑡𝑠 𝑝𝑒𝑟 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 and the router needs

3 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 to serve each packet. If the router can

𝑠𝑒𝑟𝑣𝑒 4 𝑝𝑎𝑐𝑘𝑒𝑡𝑠 𝑎𝑡 𝑎 𝑡𝑖𝑚𝑒. Find the probability that

new incoming calls will not be served.

39

Ans. 3.5 × 10−3

0

( ) ( )!

bk

X

k

f x xek

bk

0

( ) ( )!

bk

X

k

F x u xk

e kb

Uniform Distribution

• 𝑓𝑋 𝑥 = 1

𝑏−𝑎𝑎 ≤ 𝑥 ≤ 𝑏

0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒

• where 𝑏 > 𝑎 ,−∞ < 𝑎 < ∞

• 𝐹𝑋 𝑥 =

0 𝑥 > 𝑎𝑥−𝑎

𝑏−𝑎𝑎 ≤ 𝑥 < 𝑏

1 𝑏 ≤ 𝑥

• Practical Importance: quantization

40

Exponential Distribution

• 𝑓𝑋 𝑥 = 1

𝑏𝑒−

𝑥−𝑎

𝑏 𝑥 > 𝑎

0 𝑥 < 𝑎

• 𝐹𝑋 𝑥 = 1 − 𝑒−𝑥−𝑎

𝑏 𝑥 > 𝑎0 𝑥 < 𝑎

• where 𝑏 > 0 ,−∞ < 𝑎 < ∞

• Used to represent the fluctuation of

signal strength in “radar”… rain drop

size

41

Example: Exponential

Distribution• 𝑃 is the received by a radar from aircraft

𝑓𝑃 𝑝 =

1

𝑝0𝑒−𝑝𝑝0 𝑝 > 0

0 𝑝 < 0

• 𝑝0 is the average amount of received power 𝑃.

• What is the probability that the received power is above the

average?

• Exponential with 𝑎 = 0, 𝑏 = 𝑝0, 𝑓𝑋 𝑥 = 1

𝑏𝑒−

𝑥−𝑎

𝑏 𝑥 > 𝑎

0 𝑥 < 𝑎

• 𝑃 𝑃 > 𝑃0 = 1 − 𝑃 𝑃 ≤ 𝑝0 = 1 − 𝐹𝑃 𝑝0 = 1 − 1 − 𝑒−𝑝0𝑝0 = 𝑒−1

≈ 0.368

42

Rayleigh

The Rayleigh density and distribution functions are

2( )2( )

( )

0

x a b

X

x a e x af x b

x a

2( )1 ( )

0

x a b

X

e x aF x

x a

for real constants and 0a b

43

Important in wireless

applications, and

measurements errors

Median & Example for Rayleigh

• Median: It is the value of 𝑋 for which the

probability is 0.5 that the values of 𝑋 does not

exceed the median.

• The value 𝑥0 such that 𝑃 𝑋 ≤ 𝑥0 = 𝑃 𝑥0 < 𝑋 ,

this 𝑥 = 𝑥0 is called the median.

• Find the median for Rayleigh distribution?

• 𝑃 𝑋 ≤ 𝑥0 = 𝐹𝑋 𝑥0 = 0.5

• 𝐹𝑋 𝑥0 = 1 − 𝑒−

𝑥0−𝑎2

𝑏 = 0.5

• 𝑥0 = 𝑎 + 𝑏𝑙𝑛 2 0.5

44

𝑚𝑒𝑑𝑖𝑎𝑛 ≠ 𝑎𝑣𝑒𝑟𝑎𝑔𝑒Find out the difference

Conditional Distribution and Density Functions

For two events 𝐴 and 𝐵 the conditional probability of event

𝐴 given event 𝐵 had occurred was defined as

𝑃 𝐴 𝐵 =𝑃 𝐴 ∩ 𝐵

𝑃 𝐵

We extend the concept of conditional probability to include

random variables

45

Conditional Distribution

Let X be a random variable and define the event 𝐴𝐴 = 𝑋 ≤ 𝑥}

We define the conditional distribution function 𝐹𝑋(𝑥|𝐵)

{X x

X

} B

AP{X x B}

F (x|B) = P{X x|B} = P(B)

46

𝑃 𝐴 𝐵 =𝑃 𝐴 ∩ 𝐵

𝑃 𝐵

Properties of Conditional Distribution

X

X

(1) F ( |B) = 0

F ( |B) = P X |B

P X B 0

= 0P(B) P(B

)

proof

X

X

(2) F ( |B) = 1

F ( |B) = P X |B

P X B P(B)

Pr

= 1P(B) P(B)

oof

X(3) 0 F (x|B) 1

1 2 X 2 X 1(5) P x < X x |B = F (x |B) F (x |B)

X 1 X 2 1 2(4) F (x |B) F (x |B) if x < x

+

X X(6) F (x |B) = F (x|B)

None Decreasing

Right continuous 47

Conditional Density Functions

We define the Conditional Density Function of the random

variable X as the derivative of the conditional distribution

function

XX

dF (x|B)f (x|B) =

dx

If contain step discontinuities as when X is discrete or

mixed (continues and discrete ) then will contain

impulse functions.

XF (x|B)

Xf (x|B)

48

Properties of Conditional Density

X(1) f (x|B) 0

X(2) f (x|B) dx = 1

x

X X(3) F (x|B) = f (ξ|B)dξ

2

1

x

1 2 Xx

(4) P x < X x |B = f (x|B)dx

49

Example: Conditional density &

distribution• Two mutually exclusive events:

𝐵1 “select the smaller box”, 𝑃 𝐵1 =2

10

𝐵2 “select the larger box”, 𝑃 𝐵2 =8

10

A random variable is defined to be 𝑋

𝑃 𝑋 = 1 𝐵 = 𝐵1 =5

100

𝑃 𝑋 = 2 𝐵 = 𝐵1 =35

100

𝑃 𝑋 = 3 𝐵 = 𝐵1 =60

100

𝑓𝑋 𝑥 𝐵1 =5

100𝛿(𝑥 − 1)+

35

100𝛿(𝑥 − 2)+

60

100𝛿(𝑥 − 3)

𝐹𝑋 𝑥 𝐵1 =5

100𝑢(𝑥 − 1)+

35

100𝑢(𝑥 − 2)+

60

100𝑢(𝑥 − 3)

50

𝒙𝟏 Ball

Color

Small

box (𝐵1)

Larger

box (𝐵𝟐)

Total

1 Red 5 80 85

2 Green 35 60 95

3 Blue 60 10 70

Total 100 150 250

Continue Example

• For comparison (non

conditional) total probability:• 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 = 0.39

• 𝑃 𝑋 = 3 = 0.173

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

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

51

Next we define the event were b is a real number B = X b

< b <

X

P X x X bF (x|X b) = P X x|X b =

P X

b

P X b 0

Cas xe 1 b

{X b} {X x }

{X x} {X b} = {X b}

X

P X x X b P X bF (x|X b) = = 1

P X b P X b

x

X

b

52

Cas xe 2 b

{X x} {X b }

{X x} {X b} = {X x}

X

X

X

P X x X b P X x F (x)F (x|X b) = =

P X b P X b F (b)

By combining the two expressions we get

X

XX

F (x)x < b

F (b)F (x|X b) =

1 b x

X

x b

53

Then the conditional distribution FX(x|X ≤ b) is never smaller

then the ordinary distribution FX(x)

X XF (x|X b) F (x)

X

XX

F (x)x < b

F (b)F (x|X b) =

1 b x

54

The conditional density function derives from the derivative

XX

dF (x|X b)f (x|X b) =

dx

X Xf (x|X b) f (x)

X X

b

XX

f (x) f (x) = x < b

F (b) f (x)dx=

0 x b

Similarly for the conditional density function

55

Example 8 Let X be a random variable with an exponential

probability density function given as

0

0 0

( )

X

xe x

x

f x

2

( )

( ) Since 2 ( | )

2

0 2

X

XX

f x

f x dxf x X

x

x

Find the probability P( X < 1 | X ≤ 2 )

( | )2X

Xf x

( )X

f x

56

= 𝑒−𝑥/(1 − 𝑒−2) 0 < 𝑥 ≤ 2

0 𝑥 < 00 𝑥 > 2

1

0

( 1| )2 2| ) (X

X XP f x dX x 1

02

1

xe dx

e

1

2 11

ee

0.7310

1

02

1

xe dxe

57

Example Conditional Rayleigh

• Given in target, find 𝑃(𝑏𝑢𝑙𝑙’𝑠 𝑒𝑦𝑒 | 𝑙𝑎𝑛𝑑𝑖𝑛𝑔 𝑜𝑛 𝑡𝑎𝑟𝑔𝑒𝑡).

𝐹𝑋 𝑥 = 1 − 𝑒−𝑥2

800 𝑢 𝑥

• P(bull’s eye | landing on target)=𝐹𝑋 10

𝐹𝑋 50

• =1−𝑒

−100800

1−𝑒−2500800

= 0.1229 ≡ 12.29%

• Check end of chapter summary 58

Target= 50𝑚Bull's-eye= 10𝑚