Basic notions of probability theory • Discrete Random Variables · 2019. 3. 4. · Probability...

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Piero Baraldi Piero Baraldi Basic notions of probability theory Discrete Random Variables
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Transcript of Basic notions of probability theory • Discrete Random Variables · 2019. 3. 4. · Probability...

  • Piero BaraldiPiero Baraldi

    Basic notions of probability theory

    • Discrete Random Variables

  • Piero Baraldi

    Contents

    o Basic Definitionso Boolean Logico Definitions of probabilityo Probability lawso Random variableso Probability Distributions

  • Piero Baraldi

    Random variables

  • Piero Baraldi

    Random variablesExperiment: 𝜀𝜀Sample space: ΩGeneric outcome: 𝜔𝜔

    Ω𝜔𝜔

  • Piero Baraldi

    Random variablesExperiment: 𝜀𝜀Sample space: ΩGeneric outcome: 𝜔𝜔

    Ω𝜔𝜔

    X(ω) random variable in ℛ

    ℛ𝑋𝑋(𝜔𝜔)

    Univocal mapping

  • Piero Baraldi

    Random variable - Example

    Ω1 2

    X(ω) in ℜ

    Univocal mapping

    34 5 6

    1 2 3 4 5 6

    Experiment: 𝜀𝜀 = {throwing a die}Sample space: Ω={1,2,3,4,5,6}Generic outcome: 𝜔𝜔

  • Piero Baraldi

    Random variable - EventExperiment: 𝜀𝜀 = {throwing a die}Sample space: Ω={1,2,3,4,5,6}Event: 𝐸𝐸1= {1,2,3,4}

    𝐸𝐸2= ∅

    Ω1 2

    X(ω)

    𝐸𝐸1={X

  • Piero Baraldi

    Random variables

    X(ω) random variable in ℛ

    General mathematical models of random behaviours (It is not necessary to speak of the physical process)

    They apply to different physical phenomena which behave similarly

    Experiment: 𝜀𝜀Sample space: ΩGeneric outcome: 𝜔𝜔

  • Piero Baraldi

    Probability distributions for reliability, safety and risk

    analysis

  • Piero Baraldi

    Probability functions (I)

    o 𝐹𝐹𝑋𝑋 𝑥𝑥 = 𝑃𝑃 𝑋𝑋 ≤ 𝑥𝑥o Properties:

    0)(lim =−∞→

    xFXx

    1)(lim =+∞→

    xFXx

    • Cumulative Distribution Function (cdf)

  • Piero Baraldi

    Probability functions (I)

    o 𝐹𝐹𝑋𝑋 𝑥𝑥 = 𝑃𝑃 𝑋𝑋 ≤ 𝑥𝑥o Properties:

    • FX(x) is a non-decreasing function of x• The probability that X takes on a value in the interval [a,b] is:

    FX(x)

    x

    0)(lim =−∞→

    xFXx

    1)(lim =+∞→

    xFXx

    { } )()( aFbFbXaP XX −=≤<

    • Cumulative Distribution Function (cdf)

  • Piero Baraldi

    • discrete probability distributions

    Probability distributions for reliability, safety and risk analysis:

    • continuous probability distributions

  • Piero Baraldi

    • 𝑋𝑋 – random variable takes discretevalues 𝑥𝑥𝑖𝑖 , 𝑖𝑖 = 1, … ,𝑛𝑛 Probability mass function:

    𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)=P{X= xi}=𝑝𝑝𝑖𝑖

    x

    0.096

    0.384

    0 1 2 3

    0.008

    pX(x) 0.512

    • Probability Mass Function (pmf)

    Probability functions (II, discrete random variables)

    𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)

    �𝑖𝑖=1

    𝑛𝑛

    𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖) = 1

  • Piero Baraldi

    • 𝑋𝑋 – random variable takes discretevalues 𝑥𝑥𝑖𝑖 , 𝑖𝑖 = 1, … ,𝑛𝑛 Probability mass function:

    𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)=P{X= xi}=𝑝𝑝𝑖𝑖

    Cumulative distribution function:x

    0.096

    0.384

    0 1 2 3

    0.008

    pX(x)

    0.104

    0.616

    1.00

    0 1 2 3

    0.008

    FX(x)

    0.512

    • Probability Mass Function (pmf)

    Probability functions (II, discrete random variables)

    𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)

    𝐹𝐹𝑋𝑋 𝑥𝑥 = 𝑃𝑃 𝑋𝑋 ≤ 𝑥𝑥 =

    = �𝑖𝑖: 𝑥𝑥𝑖𝑖≤𝑥𝑥

    𝑓𝑓(𝑥𝑥𝑖𝑖)

    �𝑖𝑖=1

    𝑛𝑛

    𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖) = 1

  • Piero Baraldi

    Summary measures: median, variance, …

    • Mean Value (Expected Value):

    𝜇𝜇𝑋𝑋 = 𝐸𝐸 𝑥𝑥 = �𝑖𝑖=1

    𝑛𝑛

    𝑥𝑥𝑖𝑖𝑝𝑝𝑖𝑖

    • Variance:

    It is a measure of the dispersion of the values around the mean

    Where the probability mass is concentrated on average?

    𝑉𝑉𝑉𝑉𝑉𝑉 𝑋𝑋 = 𝜎𝜎𝑋𝑋2 = �𝑖𝑖=1

    𝑛𝑛

    𝑥𝑥𝑖𝑖 − 𝜇𝜇𝑋𝑋 2𝑝𝑝𝑖𝑖

  • Piero Baraldi

    Exercise. 1 (Probabilty Mass Function)

    Ω‘head’ ‘tail’

    𝑥𝑥0 1𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)

    Experiment: 𝜀𝜀 = {tossing a fair coin}

    Questions:1. 1. Draw the probability mass function2. 2. Draw the cumulative distribution

  • Piero Baraldi

    Ω‘head’ ‘tail’

    𝑥𝑥0 1

    𝑥𝑥0 1

    Probability mass function:𝑓𝑓𝑋𝑋(0) = 𝑃𝑃 𝑋𝑋 = 0 = 0.5𝑓𝑓𝑋𝑋(1) = 𝑃𝑃 𝑋𝑋 = 1 = 0.5

    𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)

    0.5

    1

    Experiment: 𝜀𝜀 = {tossing a coin}Exercise. 1 (Probabilty Mass Function)

  • Piero Baraldi

    Ω‘head’ ‘tail’

    𝑥𝑥0 1

    𝑥𝑥0 1

    𝑥𝑥0 1

    Probability mass function:𝑓𝑓𝑋𝑋(0) = 𝑃𝑃 𝑋𝑋 = 0 = 0.5𝑓𝑓𝑋𝑋(1) = 𝑃𝑃 𝑋𝑋 = 1 = 0.5

    Cumulative distribution

    𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)

    𝐹𝐹𝑋𝑋(𝑥𝑥)

    0.5

    0.5

    1

    1

    Experiment: 𝜀𝜀 = {tossing a coin}Exercise. 1 (Probabilty Mass Function)

  • Piero Baraldi

    Exercise 2: Probability Mass Distribution

    In the district of the ‘big red lakes’, there are 5 oil wells. Let 𝑋𝑋 indicates the number of operating oil wells. According to some data, the government assumes that the following equation properly describes the number of operating wells.

    𝑓𝑓𝑋𝑋(𝑖𝑖) = 𝑃𝑃 𝑋𝑋 = 𝑖𝑖 = 𝐴𝐴 � 𝑖𝑖 with 𝑖𝑖 = 0,1, … , 5

    You are required to:a) Compute Ab) What is the probability that only three or less oil wells are operative?c) What is the mean and standard deviation of X?

  • Piero Baraldi

    Exercise 2: Probability Mass Distribution (Solution)

    In the district of the ‘big red lakes’, there are 5 oil wells. Let X indicates the number of operating oil wells. According to some data, the government assumes that the following equation properly describes the number of operating wells.

    𝑓𝑓𝑋𝑋(𝑖𝑖) = 𝑃𝑃 𝑋𝑋 = 𝑖𝑖 = 𝐴𝐴 � 𝑖𝑖 with 𝑖𝑖 = 0,1, … , 5

    You are required to:a) Compute A

    From ∑𝑖𝑖=05 𝑓𝑓𝑋𝑋(𝑖𝑖) = 1 𝐴𝐴 0 + 1 + 2 + 3 + 4 + 5 = 1 𝐴𝐴 =115

  • Piero Baraldi

    In the district of the ‘big red lakes’, there are 5 oil wells. Let X indicates the number of operating oil wells. According to some data, the government assumes that the following equation properly describes the number of operating wells:

    𝑓𝑓𝑋𝑋(𝑖𝑖) = 𝑃𝑃 𝑋𝑋 = 𝑖𝑖 = 𝐴𝐴 � 𝑖𝑖 with 𝑖𝑖 = 0,1, … , 5

    You are required to:a) Compute Ab) What is the probability that only three or less oil wells are operative?

    𝑃𝑃 𝑋𝑋 ≤ 3 = 𝑓𝑓𝑋𝑋 0 + 𝑓𝑓𝑋𝑋 1 +𝑓𝑓𝑋𝑋 2 + 𝑓𝑓𝑋𝑋 3 = 𝐴𝐴 0 + 1 + 2 + 3 = 6𝐴𝐴 =6

    15

    Exercise 2: Probability Mass Distribution (Solution)

  • Piero Baraldi

    In the district of the ‘big red lakes’, there are 5 oil wells. Let X indicates the number of operating oil wells. According to some data, the government assumes that the following equation properly describes the number of operating wells:

    𝑓𝑓𝑋𝑋(𝑖𝑖) = 𝑃𝑃 𝑋𝑋 = 𝑖𝑖 = 𝐴𝐴 � 𝑖𝑖 with 𝑖𝑖 = 0,1, … , 5

    You are required to:a) Compute Ab) What is the probability than only three or less oil wells are operative?c) What is the mean and standard deviation of X?

    𝜇𝜇𝑋𝑋 = 𝐸𝐸 𝑥𝑥 = �𝑖𝑖=0

    𝑛𝑛

    𝑖𝑖 � 𝑝𝑝𝑖𝑖 = �𝑖𝑖=0

    𝑛𝑛

    𝑖𝑖2 � 𝐴𝐴 = 𝐴𝐴 1 + 4 + 9 + 16 + 25 = 𝐴𝐴 � 55 =5515

    = 3,66

    𝜎𝜎𝑋𝑋2 = �𝑖𝑖=0

    𝑛𝑛

    𝑖𝑖 − 𝜇𝜇𝑋𝑋 2𝑝𝑝𝑖𝑖 = �𝑖𝑖=0

    𝑛𝑛

    𝑖𝑖 − 𝜇𝜇𝑋𝑋 2 � 𝐴𝐴 � 𝑖𝑖 = ⋯ = 1,55 → 𝜎𝜎𝑋𝑋 = 1.25

    Exercise 2: Probability Mass Distribution (Solution)

  • Piero Baraldi

    Univariate discrete probability distributions

  • Piero Baraldi

    Exercise 3A contractor is planning the purchase of equipment, including bulldozers,

    needed for a new project in a remote area. Suppose that from hisprevious experience, he figures there is a 80% chance that eachbulldozer can last at least 6 months without any breakdown.

    • If he has purchased 3 bulldozers, what is the probability that there will be only 1 bulldozer left operative in 6 months?

  • Piero Baraldi

    Exercise 3: Solutiono G = event where a Bulldozer is in good condition after

    6 months.o B = event where a Bulldozer is in bad condition after

    6 month.

    o The possible statuses of the three bulldozers would be: {GGG, GGB, GBG, GBB, BGG, BGB, BBG, BBB}

    o The probability of having only 1 bulldozer left operative is:

    𝑃𝑃 𝑜𝑜𝑛𝑛𝑜𝑜𝑜𝑜 1 𝑏𝑏𝑏𝑏𝑜𝑜𝑜𝑜𝑏𝑏𝑜𝑜𝑏𝑏𝑏𝑏𝑉𝑉 𝑖𝑖𝑖𝑖 𝑜𝑜𝑝𝑝𝑏𝑏𝑉𝑉𝑉𝑉𝑜𝑜𝑖𝑖𝑜𝑜𝑏𝑏 = 𝑃𝑃 𝐺𝐺𝐺𝐺𝐺𝐺 + 𝑃𝑃 𝐺𝐺𝐺𝐺𝐺𝐺 + 𝑃𝑃 𝐺𝐺𝐺𝐺𝐺𝐺= 3 ∗ 0.8 � 0.2 � 0.2 = 0.096

  • Piero Baraldi

    Exercise 3 (continue)• Let X be the random variable whose values represent the number of

    good bulldozers after 6 months. The probability that a bulldozer will remain operational after 6 months is p = 0.8. Using the above information, you are required to:

    1. plot the probability mass function (PMF) as well as the cumulative distribution function (CDF) of X.

    2. Compute the following quantities:• Mean of X• Variance of X• Standard deviation of X

  • Piero Baraldi

    x

    0.096

    0.384

    0 1 2 3

    0.008

    pX(x)

    0.104

    0.488

    1.00

    0 1 2 3

    0.008

    FX(x)

    Exercise 3: Solution

    • X = {0, 1, 2, 3}• p=P{no failure in (0,6 months)}=0.8

    • PX (0) = (1-p)3=0.008• PX (1) = 3p(1–p)2=0.096• PX (2) = 3p2(1–p)=0.384• PX (3) = p3 = 0.512

    0.512

  • Piero Baraldi

    Exercise 3: Solution

    o Mean of X

    E(X) = 0(0.008) + 1(0.096) + 2(0.384) + 3(0.512) = 2.40

    o Var(X)

    Var(X) = 0.008(0 – 2.4)2 + 0.096(1 – 2.4)2 + 0.384(2 – 2.4)2 + 0.512(3 –2.4)2 = 0.48

    o Standard deviation

    Standard deviation = Var(X)1/2=0.481/2= 0.69

    …= 3 � 𝑝𝑝

    … = 3 � 𝑝𝑝 �(1- 𝑝𝑝)

  • Piero Baraldi

    Univariate discrete probability distributions:

    1) binomial distribution2) geometric distribution3) Poisson distribution

  • Piero Baraldi

    Univariate Discrete Distributions: Binomial Distribution (I)

    Y = discrete random variable with only two possible outcomes:

    • Y=1 (success) with P{Y=1}=p

    • Y=0 (failure) with P{Y=0}=1-p

    We perform n different trials of the experiment, 𝑌𝑌1, … ,𝑌𝑌𝑛𝑛

    X= discrete random variable counting the number of success out of the n trial (independentlyfrom the sequence with which successes appear):

    𝑋𝑋 = ∑𝑖𝑖=1𝑛𝑛 𝑌𝑌𝑖𝑖 Ω = 0,1,2, … ,𝑛𝑛

    The probability mass function:

    𝑛𝑛𝑥𝑥 =binomial coefficient=

    𝑛𝑛!𝑛𝑛−𝑥𝑥 !𝑥𝑥!

    x=0,1, 2,…,n

    Bernoulli process

    𝑏𝑏 𝑥𝑥;𝑛𝑛,𝑝𝑝 = 𝑛𝑛𝑥𝑥 𝑝𝑝𝑥𝑥 1 − 𝑝𝑝 𝑛𝑛−𝑥𝑥 with

  • Piero Baraldi

    Univariate Discrete Distributions: Binomial Distribution (II)

    )1(][][

    pnpXVarnpXE

    −==

    x=0,1, 2,…n𝑏𝑏 𝑥𝑥;𝑛𝑛,𝑝𝑝 = 𝑛𝑛𝑥𝑥 𝑝𝑝𝑥𝑥 1 − 𝑝𝑝 𝑛𝑛−𝑥𝑥 with

  • Piero Baraldi

    Exercise 320% of the items produced by a machine are defective. You are required to find:• the probability distribution of the number of healthy items before a

    defective item is found. • What is the probability of producing more than 2 consecutive healthy

    items?

  • Piero Baraldi

    Exercise 3 (Solution)20% of items produced by a machine are defective. You are required to find:• the probability distribution of the number of healthy items produced

    before a defective item is found.

    Probability that an item is defective: 𝑝𝑝 = 0.2 Let T be the number of healthy items produced before a defective item

    g 𝑜𝑜 = 0; 𝑝𝑝 = 𝑃𝑃 first item produced is defective = 𝑝𝑝

    T1

    Defective item

  • Piero Baraldi

    Exercise 3 (Solution)20% of items produced by a machine are defective. You are required to find:• the probability distribution of the number of healthy items produced

    before a defective item is found.

    Probability that an item is defective: 𝑝𝑝 = 0.2 Let T be the number of healthy items produced before a defective item

    g 𝑜𝑜 = 0; 𝑝𝑝 = 𝑃𝑃 first item produced is defective = 𝑝𝑝

    g 𝑜𝑜 = 1; 𝑝𝑝 = 𝑃𝑃 first item produced is healthyand second item produced is defective = (1 − 𝑝𝑝)𝑝𝑝

    T1

    T1 2

  • Piero Baraldi

    Exercise 3 (Solution)20% of items produced by a machine are defective. You are required to find:• the probability distribution of the number of healthy items produced

    before a defective item is found.

    Probability that an item is defective: 𝑝𝑝 = 0.2 Let T be the number of healthy items produced before a defective item

    g 𝑜𝑜 = 0; 𝑝𝑝 = 𝑃𝑃 first item produced is 𝑏𝑏𝑏𝑏𝑓𝑓𝑏𝑏𝑑𝑑𝑜𝑜𝑖𝑖𝑜𝑜𝑏𝑏 = 𝑝𝑝

    g 𝑜𝑜 = 1; 𝑝𝑝 = 𝑃𝑃 first item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜and second item produced is 𝑏𝑏𝑏𝑏𝑓𝑓𝑏𝑏𝑑𝑑𝑜𝑜𝑖𝑖𝑜𝑜𝑏𝑏 = (1 − 𝑝𝑝)𝑝𝑝

    T1

    T1 2

    T

    g 𝑜𝑜;𝑝𝑝 = 𝑃𝑃

    first item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜second item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜

    …𝑜𝑜 − 1 − th item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜𝑜𝑜 − th item produced is healthy𝑜𝑜 + 1 itemproduced is 𝑏𝑏𝑏𝑏𝑓𝑓𝑏𝑏𝑑𝑑𝑜𝑜𝑖𝑖𝑜𝑜𝑏𝑏

    = 1 − 𝑝𝑝 𝑡𝑡 𝑝𝑝

    1 2 t t+1

  • Piero Baraldi

    Exercise 320% of item produced by a machine are defective. You are required to find:• the probability distribution of the number of healthy items before a

    defective item is found. • What is the probability of producing more than 2 consecutive healthy items?

    g 𝑜𝑜; 𝑝𝑝 = 𝑃𝑃

    first item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜second item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜

    …𝑜𝑜 − 1 − th item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜𝑜𝑜 − th item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜

    𝑜𝑜 + 1 − th item produced is 𝑏𝑏𝑏𝑏𝑓𝑓𝑏𝑏𝑑𝑑𝑜𝑜𝑖𝑖𝑜𝑜𝑏𝑏

    = 1 − 𝑝𝑝 𝑡𝑡 𝑝𝑝

    𝑃𝑃 more than 2 consecutive healthy items are produced =

    = �𝑡𝑡=3

    +∞

    𝑔𝑔 𝑜𝑜; 𝑝𝑝 = 1 −�𝑡𝑡=0

    2

    𝑔𝑔 𝑜𝑜; 𝑝𝑝 = 1 − 𝑝𝑝 + 1 − 𝑝𝑝 𝑝𝑝 + 1 − 𝑝𝑝 2𝑝𝑝 = 0.51

  • Piero Baraldi

    Univariate discrete probability distributions:1) binomial distribution2) geometric distribution3) Poisson distribution

  • Piero Baraldi

    Univariate Discrete Distributions, Geometric Distribution

    𝑝𝑝 = P{failure}

    T= trail of the first experiment whose outcome is “failure”

    The probability mass function:

    ppptg t 1)1();( −−= t=1, 2,…

    Expected value:

    pppppppptTE t

    t

    1)]1(1[

    ...])1(3)1(21[)1(][ 221

    1=

    −−=+−+−+=−= −

    =∑

  • Piero Baraldi

    Univariate Discrete Distributions, Geometric Distribution

    T= trail of the first failure (or number of trials between two successive occurrences of failure);𝑝𝑝 = P{success}

    The probability mass function:

    ppptg t 1)1();( −−= t=1, 2,…

    Expected value (return period):

    pppppppptTE t

    t

    1)]1(1[

    ...])1(3)1(21[)1(][ 221

    1=

    −−=+−+−+=−= −

    =∑

  • Piero Baraldi

    Exercise 4The Christmas tree of New York has 25000 light bulbs. The probability that a single light bulb fails during the first month of service is p=0.004. You are required to find:• the probability that less than 100 light bulbs fail during the first month of

    service

  • Piero Baraldi

    Exercise 4The Christmas tree of the city has 25000 light bulbs. The probability that a single light bulb fails during the first month of service is p=0.004. You are required to find:• the probability that less than 100 light bulbs fail during the first month of

    service

    x=0,1, 2,…n𝑏𝑏 𝑥𝑥;𝑛𝑛,𝑝𝑝 = 𝑛𝑛𝑥𝑥 𝑝𝑝𝑥𝑥 1 − 𝑝𝑝 𝑛𝑛−𝑥𝑥 with

    Binomial distribution:

    𝑃𝑃 X

  • Piero Baraldi

    Exercise 4The Christmas tree of the city has 25000 light bulbs. The probability that a single light bulb fails during the first month of service is p=0.004. You are required to find:• the probability that less than 100 light bulbs fail during the first month of

    service

    x=0,1, 2,…n𝑏𝑏 𝑥𝑥;𝑛𝑛,𝑝𝑝 = 𝑛𝑛𝑥𝑥 𝑝𝑝𝑥𝑥 1 − 𝑝𝑝 𝑛𝑛−𝑥𝑥 with

    Binomial distribution:

    𝑃𝑃 X

  • Piero Baraldi

    Approximation of the binomial distribution in the case of: 𝑝𝑝 → 0 𝑛𝑛 → ∞

    It depends from only one parameter: 𝜇𝜇 = 𝑛𝑛𝑝𝑝=100 = E[X]

    which can be interpreted as the average number of faulty light bulbs in 1 month.

    𝑏𝑏 𝑥𝑥;𝑛𝑛, 𝑝𝑝 =𝑛𝑛𝑝𝑝𝑥𝑥!

    𝑥𝑥

    𝑏𝑏−𝑛𝑛𝑛𝑛

    𝑏𝑏 𝑥𝑥;𝑛𝑛, 𝑝𝑝 =𝑛𝑛𝑝𝑝𝑥𝑥!

    𝑥𝑥

    𝑏𝑏−𝑛𝑛𝑛𝑛 → 𝜋𝜋 𝑥𝑥; 𝜇𝜇 =𝜇𝜇𝑥𝑥!

    𝑥𝑥

    𝑏𝑏−𝜇𝜇 𝑝𝑝 → 0 𝑛𝑛 → ∞

    From the binomial to the poisson distribution

  • Piero Baraldi

    Univariate discrete probability distributions:1) binomial distribution2) geometric distribution3) Poisson distribution

  • Piero Baraldi

    k=0,1, 2,…

    Univariate Discrete Distributions, Poisson Distribution

    Stochastic events that occur in a (continuous) period of time (e.g. failures, earthquakes,…):

    • Rate of occurrence, λ, is constant

    • Discrete Random Variable:

    𝐾𝐾 = number of events in the period of observation (0, 𝑜𝑜)

    • Probability mass function: tk

    ekttkp λλλ −=!)()),,0(;(

    tKVartKEλ

    λ=

    =][

    ][

  • Piero Baraldi

    Exercise 5

    The occurrences of flood may be modelled by a Poisson process with constant rate of occurrence υ. Let p(k; t, υ) denote the probability of kflood occurrences in t years.

    If the mean occurrence rate of floods for a certain region A is once every 8 years, you are required to find:• the probability of no floods in a 10-year period• of 1 flood in a 10-year period• of more than 3 floods in a 10-year period.

  • Piero Baraldi

    Exercise 5: Solution

    P{k = 0 in 10 years}= e-1.25 = 0.286P{k = 1 in 10 years}= 1.25(e-1.25) = 0.3525P{k > 3 in 10 years}= 1–P{k ≤ 3 in 10 years}=

    = 0.0394

    2 331.25 1.25

    0

    1.25 1.251 ( ; 10, 0.125) 1 0.286 0.35752! 3!k

    p k t e eυ − −=

    = − = = = − − − −∑

    Mean occurrence rate of floods: once every 8 yearsBeing 𝐸𝐸 𝑘𝑘 = 𝜈𝜈 � 𝑜𝑜 → 1 = 𝜈𝜈 � 8 𝑜𝑜 → 𝜈𝜈 = 1

    8𝑜𝑜−1

    0.3525

  • Piero Baraldi

    Exercise 5 (continue)

    The occurrences of flood may be modelled by a Poisson process with rate υ. Let p(k; t, υ) denote the probability of k flood occurrences in t years.

    If the mean occurrence rate of floods for a certain region A is once every 8 years, you are required to find• the probability of no floods in a 10-year period• of 1 flood in a 10-year period• of more than 3 floods in a 10-year period.

    Consider a condenser of an electricity production plant in region A. If a flood occurs, the probability that it fails is 0.05. Compute:• the probability that the condenser will not fail over the 10-year

    period.

  • Piero Baraldi

    Exercise 5 (continue): SolutionP{condenser fails | flood} = 0.05 P{condenser survives | flood} = 0.95

    condenser survives over 10 𝑜𝑜 =

    = �𝑘𝑘=0

    +∞

    (condenser survives over 10 y) AND (𝑘𝑘 floods over 10 years)

    P{condenser survives AND 0 flood}= P{0 flood} P{condenser survives | 0 flood}= 0.286 (1) = 0.286

    P{condenser survives AND 1 flood}= P{1 flood} P{condenser survives | 1 flood}= (0.3525)(0.95)= 0.3396

    P{condenser survives AND k independent floods}=P{k floods}P{condenser survives|k independent floods}=

    P{condenser survives over 10 years} =

    kk

    ek

    )95,0(!

    25.1 25.1−

    9394.0!

    1875.1!

    1875.1)95,0(!

    25.1

    1875.1

    0

    25.125.1

    0

    25.1

    0

    25.1

    ===

    ==

    ∑∑∞+

    =

    −−

    +∞

    =

    −+∞

    =

    eek

    e

    ek

    ek

    k

    kk

    k

    k

    kk

    幻灯片编号 1幻灯片编号 2幻灯片编号 3幻灯片编号 4幻灯片编号 5幻灯片编号 6幻灯片编号 7幻灯片编号 8幻灯片编号 9幻灯片编号 10幻灯片编号 11discrete probability distributions�幻灯片编号 13幻灯片编号 14幻灯片编号 15Exercise. 1 (Probabilty Mass Function)Exercise. 1 (Probabilty Mass Function)Exercise. 1 (Probabilty Mass Function)Exercise 2: Probability Mass DistributionExercise 2: Probability Mass Distribution (Solution)Exercise 2: Probability Mass Distribution (Solution)Exercise 2: Probability Mass Distribution (Solution)Univariate discrete probability distributions� 幻灯片编号 24幻灯片编号 25幻灯片编号 26幻灯片编号 27幻灯片编号 28Univariate discrete probability distributions:��1) binomial distribution�2) geometric distribution�3) Poisson distribution� 幻灯片编号 30幻灯片编号 31幻灯片编号 32幻灯片编号 33幻灯片编号 34幻灯片编号 35幻灯片编号 36Univariate discrete probability distributions:�1) binomial distribution�2) geometric distribution�3) Poisson distribution� 幻灯片编号 38幻灯片编号 39幻灯片编号 40幻灯片编号 41幻灯片编号 42幻灯片编号 43Univariate discrete probability distributions:�1) binomial distribution�2) geometric distribution�3) Poisson distribution� 幻灯片编号 45幻灯片编号 46幻灯片编号 47幻灯片编号 48幻灯片编号 49