Basic notions of probability theory: continuous probability distributions. Piero Baraldi

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1 Basic notions of probability theory: continuous probability distributions Piero Baraldi

2 Probability distributions for reliability, safety and risk analysis: discrete probability distributions continuous probability distributions

3 Probability functions (continuous random variables) Let be a random variable which takes continuous values in R Its cumulative distribution is F x = P( x) F (x) x

4 Probability functions (continuous random variables) Let be a random variable which takes continuous values in R Its cumulative distribution is F x = P( x) F (x) Let us consider a small interval dx: P x < x + dx = F x + dx F (x) The probability density function f (x) is defined by: f ( x) lim dx0 F ( x dx) F dx ( x) df dx f (x) Notice that: f (x) is not a probability but a probability per unit of x (probability density) f (x) 0 + f x dx = 1 x x

5 Summary measures:percentiles, median, mean, variance Distribution Percentiles (x α ): F ( x ) 100 F (x) x 10 x 90 x

6 Summary measures:percentiles, median, mean, variance Distribution Percentiles (x α ): F ( x ) 100 Median of the distribution (x 50 ): The probability to be below or above is equal F ( x 50 ) 0.5 Mean Value (Expected Value): Where the probability mass is concentrated on average? E[ ] x p ( discrete random variables) i i i1 n xf ( x) dx ( continuous random variables) Variance (var[]): It is a measure of the dispersion of the values around the mean ( x ) p ( discrete random variables) i i i ( x ) f ( x) dx ( continuous random variables)

7 Exercise 6 Suppose that a random variable is described by a PDF of the form f x x x elsewhere 1. Find the value of α for which f (x) is a PDF?. What is P( > 5)? 3. Compute the following: Mean of Variance of Standard Deviation of Median of

8 Exercise 6: Solution In order to satisfy all the properties of a PDF, we must have: 10 0 x dx 1.0 (10) f x 3/10 α = (3/1000) 0 10

9 Exercise 6: Solution 5 3 3x 5 P( 5) 1 P 5 1 dx Mean of x E( ) x( ) dx x 7.50 Variance of x Var( ) ( x 7.5) ( ) dx x 15 x (7.5) x dx Standard Deviation of Standard Deviation= Var( ) 3.75 = 1.94

10 Exercise 6: Solution Median of : The modal value is x m x dx 3 xm x m = 7.94

11 Reliability T = Time to failure of a component (random variable) f T Probability density function (pdf) at time t: f T (t) Cumulative distribution function (cdf) at time t = probability of having a failure before t : F T (t) = P(T<t) Reliability at time t = Probability that the component does not fail up to t: t R(t)=1-F T (t) P t T 1 t F T t 1 R(t) t1 t t t

12 Probability density function: interpretation We start out a new item at time t = 0 and at time t=0, we ask: «What is the probability that the item will fail in the interval [t, t+δt]?» t T t t f ( t) t P T

13 Hazard Function We start out a new item at time t = 0 and at time t=0, we ask: «What is the probability that the item will fail in the interval [t, t+δt]?» t T t t f ( t) t P T We started out a new item at time t = 0; the item has survived until time t, we ask: «What is the probability that the item will fail in the next interval [t, t+δt]?» Hazard function t T t t T t h ( t) t P T

14 Hazard Function and Reliability h T ( t) dt P( t T t dt T P( t t) T t dt) P( T t) ft ( t) dt R( t) f t = df(t) dt = d(1 R(t)) dt = dr(t) dt h T t dt = dr(t) R(t) t න h T t dt = ln R(t) 0 R t = e t 0 ht t dt f t = h t R t = h(t)e 0 t ht t dt

15 Hazard Function: the Bath-Tub Curve Usually, the hazard function shows three distinct phases: i. Decreasing - infant mortality or burn in period: Failures due to defective pieces of equipment not manufactured or constructed properly (missing parts, substandard material batches, damage in shipping,...) The items are tested at the factory before they are distributed to the users much of the infant mortality is removed before the items are delivered for use. (i) (ii) (iii) HUMAN ANALOGY Congenital defects Car accidents, Loss of bone mass Arterial hardening,

16 Hazard Function: the Bath-Tub Curve Usually, the hazard function shows three distinct phases: i. Decreasing - infant mortality or burn in period: ii. iii. Failures due to defective pieces of equipment not manufactured or constructed properly (missing parts, substandard material batches, damage in shipping,...) Constant - useful life Random failures due to unavoidable loads coming from without (earthquakes, power surges, vibration, temperature fluctuations,...) Increasing ageing Aging failures due to cumulative effects such as corrosion, embrittlement, fatigue, cracking, (i) (ii) (iii) HUMAN ANALOGY Congenital defects Human accidents Loss of bone mass femur fracture Arterial hardening,

17 Univariate continuous probability distributions: 1) exponential distribution ) Weibull distribution 3) Normal distribution

18 Continuous Distributions: Exponential Distribution T=failure time h T (t)=λ constant 0 t P{T>t}=P{no failure in (0,t)}= =Poisson(k=0;(0,t),λ) = λt 0 0! e λt =e λt F T t = 1 P T > t = 1 e λt R t = 1 F T t = e λt f T (t) = λe λt It is the only distribution characterized by a constant failure rate Piero Baraldi

19 Exponential Distribution and bath tub curve (i) (ii) (iii) Piero Baraldi

20 Exponential Distribution moments Integration by parts E T = 0 + tf t dt = 0 + tλe λt dt = 1 λ =MTTF Var T = 1 λ Piero Baraldi

21 Exercise 7 A rotary pump has a constant failure rate λ = hours -1 (data from OREDA 00). You are required to find: the probability that the pump survives 1 month (730 hours) the pump mean time to failure suppose that the pump has been working without failures for two months (1460 hours), which is the probability that the pump will survive another month? Piero Baraldi

22 Exercise 7: Solution A rotary pump has a constant failure rate λ = hours -1 (data from OREDA 00). You are required to find: the probability that the pump survives 1 month (730 hours) the pump mean time to failure (MTTF) suppose that the pump has been working without failures for two months (1460 hours), which is the probability that the pump will survive another month? P T > 730 = R 730 = e λ 730 = 0.73 MTTF=E T = 1 = 336 hours 3, month λ P(T > 190) P T > 190 T > 1460 = P(T > 1460) = e λ 160 e λ 1460 = e λ( ) = == e λ(730) = 0.73 Piero Baraldi

23 Exercise 7: Solution A rotary pump has a constant failure rate λ = hours -1 (data from OREDA 00). You are required to find: the probability that the pump survives 1 month (730 hours) the pump mean time to failure (MTTF) suppose that the pump has been working without failures for two months (1460 hours), which is the probability that the pump will survive another month? P T > 70 = R 730 = e λ 730 = 0.73 MTTF=E T = 1 = 336 hours 3, month λ P(T > 190) P T > 190 T > 1460 = P(T > 1460) = e λ 160 e λ 1460 = e λ( ) = == e λ(730) = 0.73 memorylessness Piero Baraldi

24 Exponential distribution: memorylessness A component with constant failure rate, λ, is found still operational at a given time t 1 (age of the component). What is the probability that it will fail in the next period of time of length τ? P T t 1 + τ T > t 1 = F t 1+τ F t 1 R(t 1 ) = e λt 1 e λ(t 1+τ) e λt 1 = P(t 1<T t 1 +τ) P(T>t 1 ) = 1 e λ(t 1+τ) 1 e λt 1 e λt 1 = 1 e λτ =F(τ) Still exponential with failure rate λ! The probability that it will fail in some period of time of lengths τ does not depend from the component age t 1 (the component is always as good as new) = Piero Baraldi

25 Statistical distribution of the failure times of N components with constant failure rate N 0 = 1000; λ = 0.01, N t = Number of Component Working at t E[F] Expected Number of failures 00 between (t 5 T < t) MTTF= t Piero Baraldi

26 Statistical distribution of the failure times of N components with constant failure rate N 0 = 1000; λ = 0.01, E[F] Expected Number of failures 00 between (t 5 T < t) 150 N t = number of component working at t MTTF= F=number of failure in t 5 T < t = Random variable t Piero Baraldi

27 Statistical distribution of the failure times of N components with constant failure rate N 0 = 1000; λ = 0.01, E[F] Expected Number of failures 00 between (t 5 T < t) 150 N t = number of component working at t MTTF= F=number of failure in t 5 T < t = Random variable Binomial Distribution t E F = N t P{failure in t, t + Δt = N t λδt Piero Baraldi

28 Statistical distribution of the failure times of N components with constant failure rate N = 1000 λ = 0.01 Number of failures between (t T < t + 5) MTTF= E[Number of failures between 0 and 5 ] λ 5 N = [Number of failures between 5 and 50] λ 5 N surv = E[Number of failures between 50 and 75] λ 5 N surv = t Piero Baraldi

29 Univariate continuous probability distributions: 1) exponential distribution ) Weibull distribution 3) Normal distribution

30 Continuous Distributions : the Weibull Distribution The age of a component influences its failure process so that the hazard rate does not remain constant throughout the lifetime: h t = λαt α 1, t > 0 F t = P T < t = 1 e t 0 ht t dt = 1 e t 0 λαt α 1 dt = 1 e λtα f T t = df dt = λαtλtα E[ T ] 1 ; Var[ T ] 1 1 k1 x ( k) x e dx k 0 0 Piero Baraldi 30

31 Weibull Distribution and bath tub curve α<1 α=1 α>1 (i) (ii) (iii) Piero Baraldi

32 Univariate continuous probability distributions: 1) exponential distribution ) Weibull distribution 3) Normal distribution

33 Continuous Distributions: Normal (or Gaussian) Distribution Probability density function: Expected value and variance: ] [ ] [ Var E 0 ;, 1 ), ; ( 1 x x e x f ), ( ~ N It is the only distribution with a symmetric bell shape! f (x)

34 Transformations of random variables Random variables: ~f x Y, y = g(x) Monotonically increasing How to find the pdf of Y: f Y y? y g(x) x Piero Baraldi

35 Transformations of random variables Random variables: ~f x Y, y = g(x) Monotonically increasing How to find the pdf of Y: f Y y? y y + dy y g(x) x x + dx x P x < x + dx = P y Y y + dy f x dx = f Y y dy Piero Baraldi f Y y = f x dx dy = f x 1 dg(x) dx

36 Standard Normal Variable ~N μ, σ = 1 πσ e S 1 x μ σ Standard Normal Variable What is the pdf of S?

37 Standard Normal Variable ~N μ, σ = 1 πσ e S 1 x μ σ Standard Normal Variable What is the pdf of S, f S (s)? f S s = f x dx ds = f x 1 d x μ σ dx = σf x = σ 1 πσ e 1 x μ σ = 1 π e 1 s = N(0,1) f S (s) F S s = s 1 π න e 1 ξ dξ from tables s

38 Table of Standard Normal Probability 1 1 x x e d x (x) x (x) x (x)

39 Table of Standard Normal Probability x (x) x (x) x (x)

40 Table of Standard Normal Probability x (x) x (x) x 1-(x) E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E-14

41 Exercise 8 Suppose, from historical data, that the total annual rainfall in a catch basin is estimated to be normal (gaussian) N(60cm, 15 cm ), What is the probability that in the next year the annual rainfall will be between 40 and 70 cm?

42 Exercise 8 (Solution) Suppose, from historical data, that the total annual rainfall in a catch basin is estimated to be normal (gaussian) N(60cm,15cm), What is the probability that in the next year the annual rainfall will be between 40 and 70 cm? Solution: Let be the normally distributed random variable denoting the total annual rainfall. The probability that will be between two values a and b is: By doing the substitution, we get: 1 x - 1 P( a b) e dx S s P(a b) e ds b a b a

43 Exercise 8: Solution b s P(a b) e b a ds a The probability that the variable will be between 40 and 70 is P(40 70) From Tables and and we therefore obtain for the probability: P(40 70)

44 Standard Normal Variable 1 x - 1 P( a b) e dx b a S S N(0,1) b s P(a b) e ds a b s P(a b) e b a ds a

45 Central limit theorem For any sequence of n independent random variable i, their sum = σ n i=1 i is a random variable which, for large n, tends to be distributed as a normal distribution

46 Other Properties of Normal Variables If i are independent, identically distributed random variables with mean μ and finite variance given by σ S n = σ n i=0 n i N(μ, σ n ) If i are independent normal random variables with mean μ i and finite variance given by σ i, and b i εr are constants n Q n = i=1 n b i i N b i μ i, i=1 n b i σ i=1 Piero Baraldi

47 Exercise 9 The daily concentration of a certain pollutant in a stream has the exponential distribution 1. If the mean daily concentration of the pollutant is mg/10 3 liter, determine the constant c in the exponential distribution.. Suppose that the problem of pollution will occur if the concentration of the pollutant exceeds 6mg/10 3 liter. What is the probability of a pollution problem resulting from this pollutant in a single day? 3. What is the return period (in days) associated with this concentration level of 6 mg/10 3 liter? Assume that the concentration of the pollutant is statistically independent between days. 4. What is the probability that this pollutant will cause a pollution problem at most once in the next 3 days? 5. If instead of the exponential distribution, the daily pollutant concentration is Gaussian with the same mean and variance, what would be the probability of pollution in a day in this case?

48 Exercise 9: Solution Verify the normalization of the probability density function 0 ce cx dx cx 1 e 0 1 Then, from the expected value of the exponential distribution: E[] = 1/c = c = x [ ] ( )0.5 8 E x e dx 0 E[ ] E [ ] 8 4 4

49 Exercise 9: Solution P(pollution) P( 6) p 1 P( 6) x 0.5x e dx 1 e E[ T 6] 0days p P pollution at most once in 3 days ( ) p (1 p ) 1 k 0 3 k 3k k P( > 6) = 1 P( 6) = 1 P(υ ) = 1 Φ() = = 0.03

50 Objectives of These Lectures What is a random variable? What is a probability density function (pdf)? What is a cumulative distribution function (CDF)? What is the hazard function and its relationship with the pdf and CDF? The bath-tub curve Binomial, Geometric and Poisson Distribution Exponential Distribution and its memoryless property Weibull Distribution Gaussian Distribution and the central limit theorem Piero Baraldi

51 Lecture 1,,3,4: Where to study? Slides Red book ( An introduction to the basics or reliability and risk analysis, E. io): 4.1, 4.,4.3 (no 4.3.4),4.4,4.5,4.6, Exercises on Green Book ( basics of reliability and risk analysis Workout Problems and Solutions, E. io, P. Baraldi, F. Cadini) All problems in Chapter 4 If you are interested in probabilistic approaches for treating uncertainty, you can refer to: Uncertainty in Risk Assessment The Representation and Treatment of Uncertainties by Probabilistic and Non-Probabilistic Methods Chapter Piero Baraldi

52 THE END (probably) Thank you for the attention!!!!

53 Univariate continuous probability distributions: 1) exponential distribution ) Weibull distribution 3) Normal distribution 4) Lognormal distribution

54 Univariate Continuous Distributions Log-normal Distribution Probability density function: 0, 1 1 ), ; ( ln 1 x x e x x g Notice that σ and μ are not the expected value and standard deviation of

55 Univariate Continuous Distributions Log-normal Distribution Probability density function: Expected value and variance: 1) ( ] [ ] [ e e Var e E 0, 1 1 ), ; ( ln 1 x x e x x g N normal Log, ~ ln ), ( ~ Notice that σ and μ are not the expected value and standard deviation of

56 Univariate Continuous Distributions Log-normal Distribution Probability density function: Expected value and variance: 1) ( ] [ ] [ e e Var e E 0, 1 1 ), ; ( ln 1 x x e x x g N normal Log, ~ ln ), ( ~ Note: if σ = ln 1 + σ μ μ = ln μ σ + μ

57 Example: With reference to previous Example, assume that the total annual rainfall is lognormally distributed (instead of normally) with the same mean and standard deviation of 60 cm and 15 cm, respectively. What is the probability that in future years the annual rainfall will be between 40 and 70 cm, under this assumption? Solution: Recall that if the distribution of a random variable is log-normal, then the distribution of the variable ln is normal N( z, z). The probability density function of the log-normal random variable is: 1 1 f ( x) e x z 1z ln x z

58 Example: Solution σ = ln 1 + σ μ = ln μ μ σ + μ First we compute the values of the two parameters z, z of the distribution of the normal variable. With the data of the previous Example for the values of and and equations, we have: x z z Now, the probability that the annual rainfall will be between 40 cm and 70 cm, is ln(70) 4.06 ln(40) 4.06 P(40 70) P(40 70) x