Module 9 Modeling Uncertainty: THEORETICAL PROBABILITY MODELS

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Module 9 Modeling Uncertainty:

THEORETICAL PROBABILITY MODELS

Topics• Binomial Distribution• Poisson Distribution• Exponential Distribution• Normal Distribution• Beta Distribution

Introduction

• Module 7:– basic probability– use in decision problems

• Module 8:– subjective probability– modeling for decision analysis

• Module 9:– theoretical distributions– application to decision analysis

• Module 9 software tutorial

Theoretical Probability ModelsLearning Objectives

• Refresh knowledge:– Binomial distribution– Poisson distribution– Exponential distribution– Normal distribution

• Gain knowledge:– Beta distribution

Binomial Distribution(a discrete distribution)

Model characteristics:

• Dichotomous outcomes– Two possible outcomes– One outcome can occur

• Constant probability of “success”

• Independence

Binomial Distribution

Mathematical model:

• PB(R= r | n, p) = [n! / r! (n - r)!] pr(1-p)n-r

• B subscript = binomial probability

• R = binomial random variable

• r = number of successful outcomes

• n = number of events or trials

• p = probability of successful outcome

• E (R) = = np

Var (R) = = np(1- p)

Probability mass function notation:• f (x) = ( ) px qn-x

• X = binomial random variable• x = number of successful outcomes• n = number of events or trials• p = probability of a successful outcome• q = 1 – p = probability of an unsuccessful

outcome • E (X) = µ = np

V (X) = σ2 = µq

Cumulative distribution function:

• Probability of k or fewer “successful” outcomes

• F (x ≤ k) = ∑ ( ) px qn-xk

• Frequent application situations:– Quality control– Reliability– Survey sampling

• Approximation by other models:– Poisson: p → 0 and n → ∞– Normal: p ≤ 0.5 and np > 5, OR

p > 0.5 and nq > 5

Example application:Items are manufactured in large lots, from each of which

twenty units are selected at random. The lot is accepted if the

sample contains three or fewer defectives. If the production

process yields, on the average, ten percent defectives, what is

the probability of lot acceptance?

Formulation:

Determine the probability of three or fewer

“successes” in 20 independent trials, each

having 0.1 probability of success.

F (x ≤ 3) = ∑ ( ) (0.1)x (0.9)20-x = 0.8673

Poisson Distribution(a discrete distribution)

• Time interval or spatial region

• Probability of an event is small

• Events are independent

• Events take place at a constant rate

Poisson Distribution

Mathematical model:

• PP (X = k | m) = (e –m m k) / k !• P subscript = Poisson probability

• X = Poisson random variable

• k = number of events that occur

• e = natural logs base, 2.718…

• m = mean number of event occurrences

• E (X) = µ = m

Var (X) = σ2 = m

Probability mass function notation:• f (x) = (e – λ λ x ) / x !

• x = number of event occurrences• λ = mean number of event occurrences• e = natural logs base, 2.718…

• E (X) = µ = λV (X) = σ2 = λ

• Probability of k or fewer occurrences within a temporal or spatial interval

• F (x ≤ k) = ( e – λ λ x ) / x !∑X = 0

• Frequent application situations:– Quality control– Reliability– Queuing– Physical properties

• Approximation by another model:– Normal: λ > 5

Example application:The probability that a person will have a negative

reaction to the injection of a certain serum is 0.001.

Determine the probability that two or more of 1,000

people will have a negative reaction to the injection.

Formulation:Determine the probability that zero or one

people will have a negative reaction

when λ = np = 1, and take the complement.

1 – F (x ≤ 1) = 1 - e-11x / x! = 0.264∑x= 0

Exponential Distribution(a continuous distribution)

• Time or distance between two outcome occurrences

• Poisson model characteristics required

• Uses Poisson mean ( m or λ )

Exponential Distribution

Mathematical model:

• f E ( T = t | m ) = me-mt

• E subscript = exponential probability

• T = exponential random variable

• t = interarrival period

• m = mean of Poisson distribution

• e = natural logs base, 2.718..

• E (T) = μ = 1 / m

Var (T) = σ2 = 1 / m2

Probability density function notation:

• f (x) = λ e -λx

• x = length of interval between

occurrences

• λ = mean of Poisson distribution

• e = natural logs base, 2.718…

• E (X) = μ = 1 / λ

V (X) = σ2 = 1 / λ2

• Probability that the interval between two occurrences is of length k or less

• F ( x ≤ k ) = ∫ λ e – λ x d x = 1 – e – λ k0

• Frequent application situations:– Quality control– Reliability– Queuing– Physical phenomena

• Applicable when underlying process is Poisson

Example Application:The life of a certain brand of light bulb can be approximated

by the exponential probability density function. If the mean

life of the light bulb is 1,000 hours, determine the probability

that the bulb will last more than 1,000 hours

Formulation:

• Determine the probability that a light bulb will have a life of 1000 or fewer hours, when µ = 1000, and take the complement

• µ = 1/λ = 1000 so λ = 1/1000 = 0.001

• 1 – F(x ≤ 1000) = 1 - ∫ 0.001 e- 0.001x dx

= 1 - {1- e-0.001(1000) } = 1 - {1 – e-1 } = 0.3680

Normal (Gaussian) Distribution(a continuous distribution)

• “Bell–shaped” curve

• Effective for measured phenomena

• Effective for multiple sources of uncertainty

• Two parameters, µ and σ

Normal Distribution

Mathematical model:• fn ( y | µ,σ2 ) = (2πσ2)-1/2 exp [ -(y-µ)2 / 2σ2]

• N subscript = normal probability• y = value(s) of random variable Y• π = 3.14159…• e = natural logs base 2.718…• µ = the distribution mean• σ = the distribution standard deviation

• E (Y) = µ Var (Y) = σ2

Normal Distribution

Probability density function:• f (x) = (2πσ2)-1/2 exp [ -(x-µ)2 / 2σ2]

- ∞ < x < ∞

- ∞ < µ < ∞

- σ > 0

- π, e are constants

- µ, σ are parameters

• E (X) = µ

Var (X) = σ2

Normal Distribution

Cumulative distribution function:• Probability that a value of x or less will occur• F (x) = (2πσ2)-1/2 ∫ exp [ -(v-µ)2 / 2σ2] dv• P (a < X ≤ b) = F (b) – F (a)

= (2πσ2)-1/2∫ exp [ -(v-µ)2 / 2σ2] dv

Normal Distribution

• Frequent application situations:– measured phenomena

– measured results of multiple additive phenomena

– approximation of other distributions

– numerous inferential statistical methods

Normal Distribution

Example application:The ages of employees at a company are

normally distributed with a mean of 50 years

and a standard deviation of 5 years. Determine

the percentage of employees whose ages are

likely to be between 50 and 52.5 years.

Normal Distribution

Formulation:• Determine the probability of age ≤ 50 years, and determine the

probability of age ≤ 52.5 years. Take the difference, and multiply the

result by 100 to obtain the percentage.

• Using the standard normal variate, Z = (x –μ) / σ, and cumulative distribution function probabilities from either tabled values or appropriate software, we obtainF(x ≤ 50) = F(Z ≤ 0) = 0.5000 and

F(x ≤ 52.5) = F(Z ≤ 0.5) = 0.6915

• ThenP(50 < x ≤ 52.5) = 0.6915 – 0.5000 = 0.1915 or 19.15 %

Beta Distribution

• Variable bounded at both ends

• Numerous distribution shapes

• Events are independent

• Models proportions

Beta Distribution

Generalized beta probability density function:

• f (x | λ1, λ2, ω1, ω2)

where - ω1 ≤ x ≤ ω2

- λ1, λ2, ≥ 0 - Γ(λ) = (λ – 1)! when λ Є I+, OR

____ 1

ω2-ω1

_________(λ1+λ2)Γ

Γ (λ1) Γ (λ2) (______x – ω1ω2 – ω1)

λ1-1(1 - _____x – ω1ω2 – ω1)

= ∫ xλ-1 e-x dx otherwise∞

Beta Distribution

More frequently,

set ω1 = 0 and ω2 =1 • So f (x | λ1, λ2) = xλ

1-1 (1-x)λ

Where 0 ≤ x ≤ 1 λ1, λ2 ≥ 0 Γ (λ) = (λ-1)! Where λ Є I+, OR

__________Γ (λ1+ λ2)

Γ (λ1) Γ (λ2)

∫ xλ-1 e-x dx otherwise∞

Beta Distribution

• When λ1 and λ2 are positive integers:

• Probability density function

f (x | λ1, λ2) = x λ1-1 (1 –

x)λ2-1

where the random variable X = x

and 0 ≤ x ≤ 1

_____________(λ1+ λ2 – 1)!

(λ1 – 1)! (λ2 – 1)!

Beta Distribution

• Let n = λ1 + λ2 and r = λ1

• f β ( q | r,n ) = q r-1 ( 1 – q )n-r-1

• β subscript = beta probability

• Q = Beta random variable

• q = proportion value between 0 and 1

• r = number of “successes”

• n = number of trials

• E (Q) = µ = r / n

Var (Q) = σ2 = r ( n – r) / [ n2 ( n + 1) ]

( r – 1) ! ( n – r – 1) !________________(n – 1)!

Beta Distribution

Cumulative distribution function:• F (Q ≤ q | r,n) = ∫ υr-1 (1 – υ) dυ

• Called Incomplete Beta Function– Equivalence to binomial cumulative distribution function

• Values from tables or software

______________(n – 1)!(r – 1)! (n – r – 1) 0

Beta Distribution

• Frequent application situations:– Proportions or percentages

– Physical variables in restricted intervals

– Tolerances in quality control and reliability

– PERT networks (generalized form)

– Bayesian analysis (informative a priori)

• Approximated by difference between two normal distributions when λ1+ λ2 ≥ ~ 30

Beta Distribution

Example Application:• Suppose that the percentage of employees who submit

medical benefits claims each year is a beta random variable. As the benefits coordinator for a small firm with 40 employees, you currently expect that about 30 percent will most likely submit claims this year. Determine the probability that more than 16 employees will submit claims.

Beta Distribution

Formulation:• Determine the probability that 16 or fewer of 40

employees will submit claims, and take the complement.

• μ = r/n = 0.30 = r = (0.30)(40) = 12

q = 16/40 = 0.40

• From table values or software,

Fβ (Q ≤ 0.40 | 12, 40) = 0.91

• Thus Pβ (Q > 0.40) = 1 – 0.91 = 0.09

Summary

• Theoretical distributions to model uncertainty• Five distributions:

– Binomial– Poisson– Exponential– Normal– Beta

• Each distribution:– Mathematical formulation– Mean and variance– Graphical representations– Application situations– Example problem

Module 9 Modeling Uncertainty: THEORETICAL PROBABILITY MODELS

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