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INSE 6320 --Week 3

Risk Analysis for Information and Systems Engineering

Descri tive Statistics Discrete Probability Distributions Continuous Probability Distributions Stochastic Processes

Dr. A. Ben Hamza Concordia University

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Random Variables and Probability Density Functions

Arandom variable is a quantity whose value is not known exactly but its probability distribution is known. The

value of the random variable will vary from trial to trial as the experiment is repeated. The variables

probability density function(PDF) describes how these values are distributed (i.e. it gives the probability that

the variable value falls within a particular interval).

Smallest values

are most likely

f(x)All values between 0

and 1 are equally likely

f(x)

Continuous PDFs

xExponential distribution

(e.g. event rainfall)

0

0 31 2 4 x

f(x)

Discrete distribution

(e.g. number of severe storms)

Only discrete

values (integers)

are possible

Probability thatx = 20.2

0.3

0.250.15

0.1

0 1 xUniform distribution

(e.g. soil texture)

A Discrete PDF

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Mean of a Random Variable

The relationship for determining the Mean or Expected Value is the same as therelationship for finding centroidof a geometric shape. According to this similarity,

graphical methods used to determine centroid could be used to find the MeanValue for some simple density functions.

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Variance of a Random Variable

The (population) variance of random variable (RV) gives an idea of how

widely spread the values of the RV are likely to be. It is the second moment of

the distribution, indicating how closely concentrated around the expected

value of the distribution is. The variance is defined by

The variance is a measure of risk. The variance examines the differences

2 2 2( ) ( ) ( ( ))Var X E X E X

( ) is referred to as the standard deviationVar X

.

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Poisson Probability Distribution

The Poisson distribution is

Where the parameter >0 is the mean number of successes in the interval.

The mean and variance of the Poisson distribution are

( ) 0,1, 2,...!

xe

f x xx

2

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The number of typographical errors in new editions of textbooks varies

considerably from book to book. After some analysis an instructor concludes

that the number of errors is Poisson distributed with amean of 1.5 per 100

pages. The instructor randomly selects 100 pages of a new book. What is the

probability that there are no typos?

Poisson Distribution: Example

a s, w a s = w en = .

T here is about a 22% chance of f ind ing zero error s

1.5 01.5(0) ( 0) 0.2231

! 0!

xe e

f P Xx

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Normal Probability Distribution

The normal probability distributionis themost important distribution for describing

a continuous random variable.

It has been used in a wide variety ofapplications:

Heights and weights of people

Test scores

Scientific measurements

The normal distribution is

with mean and variance

The normal distribution is:

The visual appearance of the normal

distribution is a symmetric, unimodal or-

2

2

( )

21

( )2

x

f x e x

2

2( , )X N

Amounts of rainfall

It is widely used in statistical inference

.

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Calculating Normal Probabilities

We can use the following function to convert any normal random variable to a

standard normal random variable

Some advice: alwaysdraw a picture!

0

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Calculating Normal Probabilities

P(45 < X < 60) ?mean of 50 minutes and a

standard deviation of 10 minutes

0

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Examples:

(0.76) 0.776373

(1.3) ?( 3) 1 (3) ?

(3.86) ?

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Lognormal Distribution Probability Density Function

A random variable X is said to have the Lognormal Distribution with

parameters and, where > 0 and > 0, if the probability densityfunction of X is:

for x >0

2

2

1ln

21x

,

, for x 02

0

x

f(x)

0

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Lognormal Distribution - Probability Distribution Function

If X ~ LN(,),

then Y= ln (X) ~ N(,)

where F(z) is the cumulative probability distribution function of N(0,1)

xFxXPxF

n)()(

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Lognormal Distribution

Mean or Expected Value of X

2

2

1

)(

eXEX

e an o

Standard Deviation of X

median e

2

1

12

e2

2eX

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Lognormal Distribution - Example

A theoretical justification based on a certain material failure mechanism

underlies the assumption that ductile strength X of a material has a

lognormal distribution.

If the parameters are =5 and =0.1 ,

Find:(a) x and x

(c) P(110 X 130)

(d) The median ductile strength(e) The expected number having strength at least 120, if ten different

samples of an alloy steel of this type were subjected to a strength test.

(f) The minimum acceptable strength, If the smallest 5% of strength

values were unacceptable.

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Lognormal Distribution Example Solution

(a)2

2

)(

eXEX005.5e

16.149

)1(22

2 eeX

223

933.14

16


(b))120(1)120( XPXP

)1.0

0.5120ln(1 ZP

9834.0

0166.01

.

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(C)

)1.0

0.5130ln

1.0

0.5110ln()130110(

ZPXP

)99.2()32.1(

)32.199.2(

FF

ZP

(d)

092.0

..

41.1485

5.0 eemedianX

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(e) Let Y=number of items tested that have strength of at

least 120 y=0,1,2,,10

)120( XPp)120(1 XP

983.0

0170.01

)12.2(1

)1.0

0.5120ln(1

F

ZP

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f) The value of x, say xms, for which is

83.9

983.010)(

)983.0,10(~

npYE

BY

05.0)( msxXPe ermne as o ows:

and ,

,

so that

,

therefore 964.125

64.11.0

0.5ln

05.0)64.1(

05.0)1.0

0.5ln(

ms

ms

ms

x

x

ZP

xZP

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Exponential Distribution

A random variableXis defined to be exponential random variable (or

sayXis exponentially distributed) with positive parameter if its

probability density function is given by:

if 0, 0( )

0 if 0

xe x

f xx

00Note: ( ) 1x xf x dx e dx e

Thus, f(x) is a probability density function.

The cumulative distribution function:

0

00

( ) ( ) ( ) 0

For 0, ( ) 0 0

For 0, ( ) 1

x

x xt t x

F x P X x f t dt

x F x dt

x F x e dt e e

1 if 0

( )0 if 0

xe x

F xx

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Exponential Distribution

Expectation:

0 0

0 0 0

0

[ ] ( )

Integration by part:

1 1[ ] ( )

x x

x x x x

E X xf x dx x e dx x de

E X xe e dx e dx e

Variance:

2 2 2 2

0 0

2 2

20 0 0 0

2

2 2

2 2

[ ] ( )

Integration by part:

2 2 1 2[ ] ( )2 2

2 1 1Var [ ] [ ] ( [ ])

x x

x x

E X x f x dx x e dx x de

E X x e e x dx xe dx x e dx

X E X E X

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Exponential Distribution: Example

The lifetime of an alkaline battery (measured in hours) is exponentiallydistributed with = 0.05. Find the prob ability a battery will last b etween 10 & 15hours

(10 15)

15 10

P X

F F

T here is about a 13%

chance a batter y w i l l only

last 10 to 15 hours

(10 15)P X (0.05)(10) ( 0.05)(15)

0.5 0.75

0.1341

e e

e e

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Gamma Distribution

A continuous random variableXis said to have a Gamma Distribution, if

the probability density function ofXis

,0)(

1 1

xforex

x

),;( xf0

where >0 and >0

The Standard Gamma Distribution has = 1

The parameter is called the scale parameter because values other than

1 either stretch or compress the probability density function.

Important applications in waiting time and reliability analysis. Special cases

include exponential and chi-square distributions

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Gamma FunctionDefinition

For , the Gamma Function is defined by

Properties of the gamma function:

0

0

1)( dxex x

)(

(1) For any

(2) For any positive integer,

(3)

)1()1()(,1

)!1()(, nnn

2

1

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Gamma Density Functions

0.4

0.6

0.8

1),;( xf 5.0,2

1,1

0 2 4 6 80

0.2

x

,

1,2

If X~G(, ), then

Mean or Expected Value:

Standard Deviation:

)(XE

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Stochastic Process - Introduction

Stochastic processes are processes that proceed randomly in time.

Rather than consider fixed random variablesX, Y, etc. or even

sequences of i.i.d. random variables, we consider sequencesX0,X1,X2, ., whereXtrepresent some random quantity at time t.

In general, the valueX might depend on the quantityX- at time t-1,or even the valueXs for other times s < t.

Example: simple random walk .

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Stochastic Process - Definition

A stochastic process is a family of time indexed random variables Xtwhere tbelongs to an index set. Formal notation, whereIis

an index set that is a subset ofR.

Examples of index sets:1) I = (-, ) or I = [0, ]. In this case X

tis a continuous time

ItXt :

.

2) I = {0, 1, 2, .} or I = {0, 1, 2, }. In this case Xtis a discrete

time stochastic process.

We use uppercase letter {Xt} to describe the process. A time series,{xt} is a realization or sample function from a certain process.

We use information from a time series to estimate parameters andproperties of process {Xt}.

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Probability Distribution of a Process

For any stochastic process with index setI, its probability

distribution function is uniquely determined by its finite dimensional

distributions.

The kdimensional distribution function of a process is defined by

for any and any real numbersx1, ,xk .

The distribution function tells us everything we need to know about

the process {Xt}.

kttkXX kktt,...,,..., 11,..., 11

Ittk,...,

1

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Moments of Stochastic Process

We can describe a stochastic process via its moments, i.e.,

We often use the first two moments.

The mean function of the process is

The variance function of the process is

etc.,, 2sttt XXEXEXE

.ttXE

.2ttXVar

The covariance function betweenXt ,Xs is

The correlation function betweenXt ,Xs is

These moments are often function of time.

ssttst XXEXX ,Cov

22

,Cov,

st

st

st

XXXX

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Counting process

A stochastic process {N(t) : t 0} is a counting process ifN(t) representsthe total number of events that occur by timet.

eg, # of persons entering a store before time t, # of people who were

born by time t, # of goals a soccer player scores by timet.

N t should satisf :

N(t)>0

N(t) is integer valued

Ifs

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The Poisson Process: Example

For some reason, you decide everyday at 3:00

PM to go to the bus stop and count the number

of buses that arrive. You record the number of

buses that have passed after 10 minutes

time

X1=1 min X2=2 min

1st Bus

Arr iv al

2nd Bus

Arr iv al

X3=4 min

4th Bus

Arr iv al

5th Bus

Arr iv al

S1 = 1 mint=0 S2 = 3 min S3 = 7 min S5 = 15 min

X4=2 min

3rd Bus

Arr iv al

S4 = 9 min

X5=6 min

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For some reason, you decide everyday at 3:00

PM to go to the bus stop and count the number

of buses that arrive. You record the number of

buses that have passed after 10 minutes

time

X1=10 min

2nd Bus

Arr iv al

t=0 S1 = 10 min S2 = 16 min

1st Bus

Arr iv al

X2=6 min

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Given thatXi follow an exponential distribution thenN(t=10) follows

a Poisson Distribution

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