Probability models 1 Bernouilli Process - UC3MGeometric distribution 1 Poisson Process Poisson distribution Exponential distribution 3 Normal distribution 4 Normal distribution as

Estadística, Profesora: María Durbán1

1 Bernouilli ProcessBernouilli distributionBinomial distributionGeometric distribution

1 Poisson ProcessPoisson distributionExponential distribution

3 Normal distribution

4 Normal distribution as an approximation of other distributions

1 Bernouilli Process

Probability models



When an experiment has the following characteristics:

There are only two possible results: Acceptable (A)Defective (D)

The proportion of A and D is constant in the populationand remains constant regardless of the sample observed quantity

The observations are independent

Pr( )Pr( ) 1

D pA q p

== = −




There are only toew possible results: Aceptable (A)Defective (D)

The proportion of A and D is constant in the populationand remains constant regardless of the sampleobserved quantity


Pr( )Pr( ) 1

D pA q p

== = −



Examples

Toss a coin

Observe if a manufactures piece is defective or not

Sex on a newborn baby

Transmission (right or wrong) of a bit in a digital channel



Bernouilli distribution

0 if A occurs 1 Pr( 0) 1 if A does not ocurr Pr( 1)

q p XX

p X→ = − = =

=→ = =

The probability function is:

1( ) (1 ) 0,1x xp x p p x−= − =

[ ]

[ ] 2 2

0 (1 ) 1

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

E X p p p

Var X p p p p p p

μ

σ

= = × − + × =

= = − − + − = −



Binomial distribution

X = Number of successes in n trials

X takes values 0,1,2,…,n

When a Bernoulli experiment with parameter p is repeated a fixed number of times, n, the number of successes follows a el Binomial distribution with parameters (n,p).

~ ( , )X B n p




( ) (1 ) , 0,1, ,r n rnP X r p p r n

r−⎛ ⎞

= = − =⎜ ⎟⎝ ⎠

K

[ ][ ] (1 )

E X np

Var X np p

=

= −Estadística, Profesora: María Durbán

8

n=5

n=25 p=0.75 p=0.5 p=0.2




Example

An electronic device has 40 integrated circuit. The probability that a circuit is defective is, and the circuits are independent. The device will work if there areNo defective circuits.

What is the probability that the device works?

X = Number of defective circuits among the 40

Pr( 0)X =



Example

An electronic device has 40 integrated circuit. The probability that a circuit isdefective is, and the circuits are independent. The device will work if there areNo defective circuits.

What is the probability that that the device works?


Pr( 0)X =Experiment: Observe if a circuit is defective or not. It is repeated 40 times

The circuits are independent

The probability of being defective is constant, 0.01



Example

An electronic device has 40 integrated circuit. The probability that a circuit isdefective is, and the circuits are independent. The device will work if there areNo defective circuits.

What is the probability that that the device works?


~ (40,0.01)X B

0 4040Pr( 0) 0.01 (1 0.01) 0.669

0X ⎛ ⎞

= = − =⎜ ⎟⎝ ⎠



Geometric distribution

The experiment has the following characteristics:

There are only two possible results

The probability of success is constant


The experiment is repeated until the first success

X = Number of times that an experiment is repeated until a successoccurs.

~ ( )X Ge p




1,..... are Bernouilli iX i n=

1 2 3 4

1 0 0 0 1 Pr( 1)0 1 0 0 2 Pr( 2)0 0 1 0 3 Pr( 3)0 0 0 1 4 Pr( 4)

X X X X X

X X pX X qpX X qqpX X qqqp

↓ ↓ ↓ ↓ ↓⇒ = = =⇒ = = =⇒ = = =⇒ = = =

L

L

L

L

L


1( ) (1 ) , 1, 2,rP X r p p r−= = − = KEstadística, Profesora: María Durbán

14



[ ][ ] 2

1/

(1 ) /

E X p

Var X p p

=

= −

1 2 3 4

1 0 0 0 1 Pr( 1)0 1 0 0 2 Pr( 2)0 0 1 0 3 Pr( 3)0 0 0 1 4 Pr( 4)

X X X X X

X X pX X qpX X qqpX X qqqp

↓ ↓ ↓ ↓ ↓⇒ = = =⇒ = = =⇒ = = =⇒ = = =

L

L

L

L

L

1,..... are Bernouilli iX i n=






Example

The probability that a bit transmitted through a digital channel is received As an error is 0.1. If the transmissions are independent,

What is the average number of transmissions that have to be observed until the first error occurs?

X = Number of transmissions until the first error occurs

[ ] 1/ 1/ 0.1 10E X p= = =






2 Poisson Process

Probability models


2 Poisson Process


We observe the occurrence of events in an interval

The probability that this happens in an interval

is the same for intervals of the same lengthis proportional to the length of the interval

The events occur independently. The number of events in an interval is independent of the number of events in another interval


2 Poisson Process

X = Number of events in an interval of fixed length

Poisson distribution can be obtained as a limit of the Binomial distribution when

Poisson distribution

y 0n p→ ∞ →

npλ = → Average number of events in that interval

lim 1n

en

λλ −⎛ ⎞− =⎜ ⎟⎝ ⎠n → ∞

n → ∞



( ) , 0,1,!

reP X r rr

λλ−

= = = K

[ ] [ ]

[ ]

1

0 1

1 2 1 2

! ( 1)!

~ ( ) ~ ( ) independient ~ ( )

r reE X E X r er r

Var XX P Y P X Y P

λλλ λλ λ λ

λλ λ λ λ

− −∞ ∞−= → = = =

−

=

+ +

∑ ∑


2 Poisson Process


2 Poisson Process



2 Poisson Process

Examples

Number of faults in a millimeter of cable.

Number of phone calls in a switchboard per hour.

Number of mistakes in a page of a document


2 Poisson Process

Example

What is the probability that no clients arrive in 3 minutes?

X = Number of clients per minute

Y = Number of clients in 3 minutes

~ ( 1)X P λ→ =

~ ( 3)Y P λ→ =

3 033Pr( 0)

0!eY e

−−= = =

The arrivals of clients at a service point are stable and independent. on average, one client arrives per minute.


2 Poisson Process

Example

Opening the service point 8 hours, cost 6000 euros per day. What should be the minimum charge per client so that the service point is profitable?

Y = Number of clients in 8 hours ~ ( 60 8 480)Y P λ→ = × =

Profit = Charge x Y -6000

[ ]Expected Profit = Charge 6000 0 = Charge 480 6000 0

E Y× − >

× − >Charge > 12.5



2 Poisson Process

The Exponential distribution is used to model

Time between phone callsTime between arivals at a service point Lifetime of a lamp

When the number of events follows a Poisson distribution, the time between events follows an Exponential distribution.

M

Distribución de exponencial


2 Poisson Process

Exponential distribution

X = Number of events per time unit

T = Time until the first even occurs

We can calculate the distribution function:

0 0( ) (zero events in (0,t ))P T t P> =

~ ( )X P λ

X= Number of events per time unitY = Number of events in (0,t0) 0~ ( )Y P tλ

00( ) Pr( 0) tP T t Y e λ−> = = =

00 0( ) ( ) 1 tF t P T t e λ−= ≤ = −

~ ( )X P λ


2 Poisson Process

~ ( )X P λ

( )( ) , 0tdF tf t e tdt

λλ −= = ≥

[ ][ ] 2

1/

1/

E X

Var X

λ

λ

=

=

If there are λ events, on average, per time unitThe average time between events is 1/λ

X = Number of events per time unit

T = Time until the first even occurs

Exponential distribution


2 Poisson Process

0.1( ) 0.1 xf x e−=

0.5( ) 0.5 xf x e−=

2( ) 2 xf x e−=


2 Poisson Process

Example

~ ( 1)X P λ→ =

~ ( 1)T Exp λ→ =

( )1 3 3Pr( 3) 1 Pr( 3) 1 (3) 1 1T T F e e− × −> = − ≤ = − = − − =

Pr(there are no clients in 3 minutes)=


What is the probability that no clients arrive in 3 minutes?

X = Number of clients per minute

Y = Time between clients


2 Poisson Process

Property

1 2 1 2Pr(T > t +t / T > t ) = Pr( T > t )

1 22

1

( t +t )1 2 1 1 2

t1 1

Pr(T > t +t T > t ) Pr( T > t +t ) = Pr( T > t ) Pr( T > t )

te ee

λλ

λ

−−

−= =I

If there hasn’t been any clients in 4 minutes, what is the probability that there are no clients in the next 3 minutes?

3Pr( 7 | 4) Pr( 3) 1 (3)Y Y Y F e−> > = > = − =






Normal distribution

Probability models



Normal or Gaussian distribution describe a large amount of random processes

Measurement errorsNoise in a digital signalElectric current …

In many occasions others distributions can be approximated to a Normal distribution

It is the base of statistical inference



It is characterized by two parameters: the mean, μ, and the standard deviation, σ.

It can take any real value.

Its density function is:

2

2( )2

2

1( ) e2

[ ] [ ]

x

f x x

E X Var X

μσ

πσμ σ

− −

= − ∞ < < ∞

= =

( , )N μ σ


It is bell shaped and symmetric with respect to the mean

μ

( )f x


0.5 0.5

Mean, median and mode are the same.


The effect ofThe effect of μμ andand σσ

How does the standard deviation affect the shape of f(x)?σ= 2

σ =3σ =4

μ = 10 μ = 11 μ = 12How does the mean affect the position of f(x)?


It is a scalefactor

It is a translationfactor


The probability is the areaunder the curve

c dX

f(X)


Pr(c d)X≤ ≤It is not possible to compute the probability of an interval, simply by integrating the density function


μ

σ

Density of X

Density of de X-μ

0

Density of (X-μ)/σ

1


All Normal distributions can be transformed into N(0,1)

XX Z μσ−

→ =



~ (3, 2)X N

Pr( 6)X ≤3 6

0 1.56 3Pr Pr( 1.5)

2Z Z−⎛ ⎞≤ = ≤⎜ ⎟

⎝ ⎠

Samearea

~ (0,1)Z N



The distribution function of the standardized Normal has its ownnotation:

There exists certain boundaries to the function Q, that are used to calculate error boundaries of probabilities in communication systems

( ) Pr( ) ( )( ) ( ) 1 ( )

F x X x xQ x Pt X x x

φφ

= ≤ == > = −

( ) 1 ( )Q x Q x− = −

2

2

2

2

1( ) 02

1( ) 02

x

x

Q x e x

Q x e xxπ

−

−

≤ ≥

< ≥



The distribution function of the standardized Normal has its ownnotation:

There exists certain boundaries to the function Q, that are used to calculate error boundaries of probabilities in communication systems

( ) Pr( ) ( )( ) ( ) 1 ( )

F x X x xQ x Pt X x x

φφ

= ≤ == > = −

( ) 1 ( )Q x Q x− = −

2

2

2

2

1( ) 02

1( ) 02

x

x

Q x e x

Q x e xxπ

−

−

≤ ≥

< ≥



Example

The lifetime of a semiconductor follows a Normal distribution with mean equal to7000 hours and standard deviation 600 hours

What is the probability that a semiconductor fails before 6000 hours?

What is the lifetime exceeded by 95.05% of semiconductors?

Pr( 6000)X <

Pr( ) 0.9505X a> =



Example

6000 7000Pr( 6000) Pr Pr( 1.66)600

X Z Z−⎛ ⎞< = < = < −⎜ ⎟⎝ ⎠

1.66





Example

6000 7000Pr( 6000) Pr Pr( 1.66)600

X Z Z−⎛ ⎞< = < = < −⎜ ⎟⎝ ⎠

1.66

1 Pr( 1.66)Z= − ≤




1 Pr( 1.66)Z= − <


Example

1 0.95150.0485

= −=

What is the probability that a semiconductor fails before6000 hours?



Example

7000Pr( ) 0.9505 Pr 0.9505600

aX a Z −⎛ ⎞> = → > =⎜ ⎟⎝ ⎠

a

0.95





Example

7000Pr( ) 0.9505 Pr 0.9505600

aX a Z −⎛ ⎞> = → > =⎜ ⎟⎝ ⎠

-b

0.95

-b Negative value





Example

( 7000)Pr( ) 0.9505 Pr 0.9505600

aX a Z − −⎛ ⎞> = → < =⎜ ⎟⎝ ⎠

b

0.95

b




( 7000)Pr( ) Pr 0.9505600

aX a Z − −⎛ ⎞> = < =⎜ ⎟⎝ ⎠

What is the lifetime exceeded by 95% of semiconductors?

Example


( 7000) 1.65600

6010

a

a

− −=

⇓=

95.05% of semiconductorslast more than 6010 hours


Pr( -0.6 < Z < 1.83 )=

Pr( Z < 1.83 ) - Pr( Z -0.6 )

Pr ( Z <-0.6) =Pr ( Z >0.6 ) =1 - Pr (Z < 0.6 ) =1 – 0.7257 =

0.2743

Pr( Z < 1.83 ) =0.9664

= 0.7257 - 0.0336= 0.6921

1.83-0.6


More Examples of how to compute probabilities

≤


Normal distribution is important since, although some r.v. do not follow aNormal distribution, some statistics/estimators follow a Normal distribution.




50 55 60 65 70

010

2030

4050

60

x

Example

Let X be an Uniform r.v. in [50,70].We have a sample of size 2000.

The sample has mean 59.9 andstandard deviation 17

The histogram does not resemble a Normal distribution with the same mean and standard deviation


We randomly choose groups of 10 observations.

Compute the sample mean for each group

Means of each group are, more or less, similar to the mean of the original variable.

556970

565766

656263

595351

545954

695151

655354

605866

696060

596359

3ª2ª1ªGroups

59.4 58.5 61.1



55.22075556.160009

57.09926458.038518

58.97777359.917028

60.85628261.795537

62.73479263.674046

64.613301

aa$x

0

10

20

30

40

a


The distribution of the sample means follow,approximately a Normal distribution.

The mean of this new variable is very similar to the mean of the original variable.

The observations of the new variable are closer to each other. The standard deviation is smaller, in this case, it is 1.92

xxx


Let suppose that we have n in dependent variables Xi with means (μι) and standard deviations (σi) and they follow any distribution.

When n increases,

Central Limit Theorem


2(0,1)i

i

YN

μ

σ

−≈∑

∑

1 2 nY X X X= + + +K

( )2~ ,i iY N μ σ∑ ∑the distribution of


Central Limit Theorem


It doesn’t matter what we measure, when we average over a large sample, we will have Normal distribution

Let suppose tah we have n in dependent variables Xi withmeans (μι) and standard deviations (σi) and they follow anyDistribution.






Probability models


4 Normal as an approximation of other distributions

The Binomial is teh sum of Bernouilli variable tha take values equal to 0 o 1.

Binomial-Normal

1 2 nY X X X= + +K [ ][ ] (1 )

i

i

E X pVar X p p

=

= −C.L.T.

( ), (1 )Y N np np p≈ − 305

nnpq

>>


5.000 7.625 10.250 12.875 15.500 18.125 20.750 23.375 26.000x

0.00

0.04

0.08

0.12


Binomial-Normal

( )15, 10.5N

50 0.310.5

n pnpq

= ==



Example

A semiconductors manufacturer admits that 2% of the chips produced are defective. The chips are packed in lots of 2000.

A buyer will reject the lot if there 25 o more defective chips

What is the probability of rejecting the lot? Pr( 25)

25 40Pr6.26

Pr( 2.4)Pr( 2.4) 0.9918

X

Z

ZZ

≥

↓

−⎛ ⎞≥ =⎜ ⎟⎝ ⎠

≥ − =≤ =

~ (2000,0.02)30

40(1 ) 39.2

X Bnnpnp p

>=

− =

(40,6.26)X N≈



Poisson-Normal

Poisson distribution appears as the limit of a Binomial distribution when the number of experiments tends to infinity.

We will approximate to a Normal distribution when λ is large (λ > 5)

( )~ ( )

,

X P

X N

λ

λ λ≈



Poisson-Normal



Example

The number of defects on a metal surface per squared meter follows a Poisson distribution with mean 100.

If we analyze a squared meter of surface. What is the probability of finding 95 or more defects?

Pr( 95)

95 100Pr Pr( 0.5)10

Pr( 0.5) 0.6915

X

Z Z

Z

≥

↓

−⎛ ⎞≥ = ≥ − =⎜ ⎟⎝ ⎠

≤ =

Documents

Probability models 1 Bernouilli Process - UC3MGeometric distribution 1 Poisson Process Poisson distribution Exponential distribution 3 Normal distribution 4 Normal distribution as