Estadística, Profesora: María Durbán 1 Bernouilli distribution Binomial distribution Geometric distribution 1 Poisson Process Poisson distribution Exponential distribution 3 Normal distribution 4 Normal distribution as an approximation of other distributions 1 Bernouilli Process Probability models Estadística, Profesora: María Durbán 2 1 Bernouilli Process 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 population and remains constant regardless of the sample observed quantity The observations are independent Pr( ) Pr( ) 1 D p A q p = = = − Estadística, Profesora: María Durbán 3 1 Bernouilli Process When an experiment has the following characteristics: There are only toew possible results: Aceptable (A) Defective (D) population The observations are independent Estadística, Profesora: María Durbán 4 1 Bernouilli Process 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
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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
Estadística, Profesora: María Durbán2
1 Bernouilli Process
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
== = −
Estadística, Profesora: María Durbán3
1 Bernouilli Process
When an experiment has the following characteristics:
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
The observations are independent
Pr( )Pr( ) 1
D pA q p
== = −
Estadística, Profesora: María Durbán4
1 Bernouilli Process
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
Estadística, Profesora: María Durbán5
1 Bernouilli Process
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
= = × − + × =
= = − − + − = −
Estadística, Profesora: María Durbán6
1 Bernouilli Process
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
Estadística, Profesora: María Durbán7
The probability function is:
1 Bernouilli Process
( ) (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
1 Bernouilli Process
Estadística, Profesora: María Durbán9
1 Bernouilli Process
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 =
Estadística, Profesora: María Durbán10
1 Bernouilli Process
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?
X = Number of defective circuits among the 40
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
Estadística, Profesora: María Durbán11
1 Bernouilli Process
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?
X = Number of defective circuits among the 40
~ (40,0.01)X B
0 4040Pr( 0) 0.01 (1 0.01) 0.669
0X ⎛ ⎞
= = − =⎜ ⎟⎝ ⎠
Estadística, Profesora: María Durbán12
1 Bernouilli Process
Geometric distribution
The experiment has the following characteristics:
There are only two possible results
The probability of success is constant
The observations are independent
The experiment is repeated until the first success
X = Number of times that an experiment is repeated until a successoccurs.
The arrivals of clients at a service point are stable and independent. on average, one client arrives per minute.
Estadística, Profesora: María Durbán25
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
Estadística, Profesora: María Durbán26
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
Estadística, Profesora: María Durbán27
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
Estadística, Profesora: María Durbán28
2 Poisson Process
0.1( ) 0.1 xf x e−=
0.5( ) 0.5 xf x e−=
2( ) 2 xf x e−=
Estadística, Profesora: María Durbán29
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)=
The arrivals of clients at a service point are stable and independent. on average, one client arrives per minute.
What is the probability that no clients arrive in 3 minutes?
X = Number of clients per minute
Y = Time between clients
Estadística, Profesora: María Durbán30
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−> > = > = − =
Estadística, Profesora: María Durbán31
1 Bernouilli ProcessBernouilli distributionBinomial distributionGeometric distribution
2 Poisson ProcessPoisson distributionExponential distribution
3 Normal distribution
4 Normal distribution as an approximation of other distributions
Normal distribution
Probability models
Estadística, Profesora: María Durbán32
3 Normal distribution
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
Estadística, Profesora: María Durbán33
3 Normal distribution
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
Estadística, Profesora: María Durbán34
It is bell shaped and symmetric with respect to the mean
μ
( )f x
3 Normal distribution
0.5 0.5
Mean, median and mode are the same.
Estadística, Profesora: María Durbán35
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)?
3 Normal distribution
It is a scalefactor
It is a translationfactor
Estadística, Profesora: María Durbán36
The probability is the areaunder the curve
c dX
f(X)
3 Normal distribution
Pr(c d)X≤ ≤It is not possible to compute the probability of an interval, simply by integrating the density function
Estadística, Profesora: María Durbán37
μ
σ
Density of X
Density of de X-μ
0
Density of (X-μ)/σ
1
3 Normal distribution
All Normal distributions can be transformed into N(0,1)
XX Z μσ−
→ =
Estadística, Profesora: María Durbán38
3 Normal distribution
~ (3, 2)X N
Pr( 6)X ≤3 6
0 1.56 3Pr Pr( 1.5)
2Z Z−⎛ ⎞≤ = ≤⎜ ⎟
⎝ ⎠
Samearea
~ (0,1)Z N
Estadística, Profesora: María Durbán39
3 Normal distribution
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
−
−
≤ ≥
< ≥
Estadística, Profesora: María Durbán40
3 Normal distribution
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
−
−
≤ ≥
< ≥
Estadística, Profesora: María Durbán41
3 Normal distribution
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> =
Estadística, Profesora: María Durbán42
3 Normal distribution
Example
6000 7000Pr( 6000) Pr Pr( 1.66)600
X Z Z−⎛ ⎞< = < = < −⎜ ⎟⎝ ⎠
1.66
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?
Estadística, Profesora: María Durbán43
3 Normal distribution
Example
6000 7000Pr( 6000) Pr Pr( 1.66)600
X Z Z−⎛ ⎞< = < = < −⎜ ⎟⎝ ⎠
1.66
1 Pr( 1.66)Z= − ≤
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?
Estadística, Profesora: María Durbán44
1 Pr( 1.66)Z= − <
3 Normal distribution
Example
1 0.95150.0485
= −=
What is the probability that a semiconductor fails before6000 hours?
Estadística, Profesora: María Durbán45
3 Normal distribution
Example
7000Pr( ) 0.9505 Pr 0.9505600
aX a Z −⎛ ⎞> = → > =⎜ ⎟⎝ ⎠
a
0.95
The lifetime of a semiconductor follows a Normal distribution with mean equal to7000 hours and standard deviation 600 hours
What is the lifetime exceeded by 95.05% of semiconductors?
Estadística, Profesora: María Durbán46
3 Normal distribution
Example
7000Pr( ) 0.9505 Pr 0.9505600
aX a Z −⎛ ⎞> = → > =⎜ ⎟⎝ ⎠
-b
0.95
-b Negative value
The lifetime of a semiconductor follows a Normal distribution with mean equal to7000 hours and standard deviation 600 hours
What is the lifetime exceeded by 95.05% of semiconductors?
Estadística, Profesora: María Durbán47
3 Normal distribution
Example
( 7000)Pr( ) 0.9505 Pr 0.9505600
aX a Z − −⎛ ⎞> = → < =⎜ ⎟⎝ ⎠
b
0.95
b
The lifetime of a semiconductor follows a Normal distribution with mean equal to7000 hours and standard deviation 600 hours
What is the lifetime exceeded by 95.05% of semiconductors?
Estadística, Profesora: María Durbán48
( 7000)Pr( ) Pr 0.9505600
aX a Z − −⎛ ⎞> = < =⎜ ⎟⎝ ⎠
What is the lifetime exceeded by 95% of semiconductors?
Example
3 Normal distribution
( 7000) 1.65600
6010
a
a
− −=
⇓=
95.05% of semiconductorslast more than 6010 hours
Estadística, Profesora: María Durbán49
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
3 Normal distribution
More Examples of how to compute probabilities
≤
Estadística, Profesora: María Durbán50
Normal distribution is important since, although some r.v. do not follow aNormal distribution, some statistics/estimators follow a Normal distribution.
3 Normal distribution
Estadística, Profesora: María Durbán51
3 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
Estadística, Profesora: María Durbán52
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
3 Normal distribution
Estadística, Profesora: María Durbán53
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
3 Normal distribution
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
Estadística, Profesora: María Durbán54
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
3 Normal distribution
2(0,1)i
i
YN
μ
σ
−≈∑
∑
1 2 nY X X X= + + +K
( )2~ ,i iY N μ σ∑ ∑the distribution of
Estadística, Profesora: María Durbán55
Central Limit Theorem
3 Normal distribution
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.
Estadística, Profesora: María Durbán56
1 Bernouilli ProcessBernouilli distributionBinomial distributionGeometric distribution
2 Poisson ProcessPoisson distributionExponential distribution
3 Normal distribution
4 Normal distribution as an approximation of other distributions
Probability models
Estadística, Profesora: María Durbán57
4 Normal as an approximation of other distributions
The Binomial is teh sum of Bernouilli variable tha take values equal to 0 o 1.