Common Distributions in Stochastic Analysis and Inverse Modeling Presented by Gordon A. Fenton
63 Common Discrete Distributions
Bernoulli Trials: - form the basis for 6 very common distributions (binomial, geometric, negative binomial, Poisson, exponential, and gamma) Fundamental Assumptions; 1. Each trial has two possible outcomes (0/1, true/false, success/failure, on/off, yes/no, etc) 2. Trials are independent (allows us to easily compute probabilities) 3. Probability of “success” remains constant from trial to trial
Note: if we relax assumptions 1 and 2 we get Markov Chains, another suite of powerful stochastic models 64 Bernoulli Trials
Let Xii =1 if the 'th trial is a "success" = 0 if not then the Bernoulli probability distribution is
P1[X i = ] = p
P[X i = 0] =1 − p = q
65 Bernoulli Trials
1
E[X i ] =∑ jXP[ i = j] =+= 0( q ) 1( p ) p j=0 1 2= 2= =+=22 EXii∑ jXP[ j] 0( q ) 1() pp j=0 =22 − 2 =−= Var[X i ] EEXi [ Xi ] pp pq
66 Common Discrete Distributions
Bernoulli Family of Distributions
Discrete Trials Every “Instant” Becomes a Trial 1) Binomial: 4) Poisson:
N n = number of “successes” N t = number of “successes” in n trials in time t 2) Geometric: 5) Exponential: (continuous)
T 1 = number of trials until T 1 = time to the first the first “success” “success” 3) Negative Binomial: 6) Gamma: (continuous)
T k = number of trials until T k = time to the k’th the k’th “success” “success”
If N n = number of “successes” in n trials, then n Nn = X12+XX ++ n = ∑ X i i=1 since X i = 1 if the trial results in a “success” and 0 if not.
nn n = = = = Mean: E[Nn ] EE∑∑Xii[ Xp] ∑np ii=11= i= 1 Variance: nn = = Var[Nn ] Var ∑∑XXi Var[ i ] (since independent) ii=11= n =∑ pq = npq i=1 68 Binomial Distribution
Consider P2 [N5 = ] =P {SSFFF} ∪{ SFSFF} ∪∪ { FFFSS} =PP[SSFFF] +[ SFSFF] ++ P[ FFFSS] = PPP[SSFFF] [ ] [ ] P[ ] P[ ] ++ PP[ SFSF] [ ] PP[ ] [ ] P[F] =pp23qq +23 ++ p 23 q
5 23 = pq 2
In general the binomial distribution is
n k nk− P[Nn = k] = pq k
69 Binomial Distribution
n k nk− P[Npn = kq] = k n = 10 p = 0.4
70 Binomial Distribution: Example
Suppose that the probability of failure, p f , of a slope is investigated using Monte Carlo simulation. That is, a series of n = 5000 independent realizations of the slope strength are simulated from a given probability distribution and the number of failures counted to be
n f = 120 . What is the estimated failure probability and what is the ‘standard error’ (standard deviation) of this estimate?
Solution: each realization is an independent trial with two possible outcomes, success or failure, having constant probability of failure ( p f ). Thus, each realization is a Bernoulli trial, and the number of failures is a realization of N n , which follows a binomial distribution.
71 Binomial Distribution: Example
We have nn= 5000 and f = 120
The failure probability is estimated by number of failures 120 pˆ = = = 0.02 4 f number of trials 5000
Nn In general pˆ f = n N 1 so that σ 2 =Varpnˆ = Var n = Var[ ] pˆ f f 2 nn npq pq = = nn2 which, for pp ˆ f = 0.024 gives us 0.024( 1− 0.024) σ = 0.0022 pˆ f 5000 72 Geometric Distribution
Let T 1 = number of trials until the first “success”
3 Consider PP[T1 = 4] =[ FFFS] = pq
k−1 In general P[T1 = k] = pq
∞ k−1 1 E[T1 ] =∑kpq = k=1 p
∞ − 1 q =2 −2 =21k −= Var[T1 ] EET1 [T1 ] ∑ k pq 2 2 k=1 p p
73 Geometric Distribution
Because all trials are independent, the geometric distribution is “memoryless”. That is, it doesn’t matter when we start looking at a sequence of trials – the distribution of the number of trials to the next “success” remains the same. That is, ∞ m−1 P[T >+jk] ∑ = ++ pq P[Tq>+j kT| > j] = 1 = mjk1 = k 1 1 ∞ n−1 P[T1 > j] ∑ nj= +1 pq and ∞ ∞∞ mm−1 km k1 k P[T1 > k] =∑ pq = p ∑∑ q = pq q = pq = q mk=+==1 mk m 0 1− q
74 Geometric Distribution
k−1 P[T1 = k] = pq
p = 0.4
75 Geometric Distribution: Example
A series of piles have been designed to be able to withstand a certain test load with probability p = 0.8 . If the resulting piles are sequentially tested at that design load, what is the probability that the first pile to fail the test is the 7’th pile tested?
Solution: If piles can be assumed to fail independently with constant probability, then this is again a sequence of Bernoulli trials. We thus want to compute
76−1 P[T1 = 7] =pq = (0.8)(0.2) = 0.00005
76 Negative Binomial Distribution
Let T k = the number of trials until the k’th “success”
Consider
PP[T3 = 5] =[{ SSFFS} ∪{ SFSFS} ∪∪ { FFSSS}] =PP[SSFFS] +[ SFSFS] ++ P[ FFSSS]
4 32 = p q 2
m −1 k mk− In general P[Tk = m] = pq k −1
77 Negative Binomial Distribution
The negative binomial distribution is a generalization of the geometric distribution (the k’th “success”). Its mean and variance can be obtained by realizing that T k is the sum of k independent T 1 's . That is
TTT= + ++ T k 1112 1k so that kk k E[T ] =EETT =[ ] = k ∑∑1j 1 jj=1 =1 p and kk Vra [T ] = Var TT= Var[ ] (since independent) k ∑∑1j 1 jj=1 =1 kq =kVar[T1 ] = 2 p 78 Negative Binomial Distribution
m −1 k mk− P[Tk = mp] = q k −1
p = 0.4 m = 3
79 Negative Binomial Distribution: Example
A series of piles have been designed to be able to withstand a certain test load with probability p = 0.8 . If the resulting piles are sequentially tested at that design load, what is the probability that the third pile to fail the test is the 7’th pile tested?
Solution: If piles can be assumed to fail independently with constant probability, then this is again a sequence of Bernoulli trials. We thus want to compute
k −−1 m km− 71 37−3 P7[T3 = ] = pq = (0.8) (0.2) m −−1 31
6 34 = (0.8) ( 0.2) = 0. 00082 2 80 Poisson Distribution
We now let every instant in time (or space) become a Bernoulli trial (i.e., a sequence of independent trials)
“Successes” can now occur at any instant in time.
PROBLEM: in any time interval, there are now an infinite number of Bernoulli trials. If the probability of “success” of a trial is, say, 0.2, then in any time interval we get an infinite number of “successes”. This is not a very useful model! SOLUTION: we can no longer talk about probability of “success” of individual trials, p (which becomes zero). We must now talk about the mean rate of “success”, λ .
81 Poisson Distribution
Governs many rate dependent processes (arrivals of vehicles at an intersection, ‘packets’ through an internet gateway, earthquakes exceeding magnitude M, floods, load extremes, etc.)
Let N t = number of “successes” (arrivals) in time interval t
The distribution of N t is arrived at in the limit as the number of trials goes to infinity assuming the probability of “success” is proportional to the number of trials (see the course notes for details). We get (λt)k P[Nk=] = e−λt t k!
82 Poisson Distribution
t = 4.5 λ = 0.9
(λt)k P[Ne= k] = −λt t k!
83 Poisson Distribution: Example Many research papers suggest that the arrivals of earthquakes follow a Poisson process over time. Suppose that the mean time between earthquakes is 50 years at a particular location. 1. How long must the time period be so that the probability that no earthquakes occur during that period is at most 0.1?
Solution: λ =1/ E[T1 ] = 1/ 50 = 0.02 1. Let N t be the number of earthquakes over the time interval t. We want to find t such that 0 (λt) −−λλ P[Ne= 0] =tt = e ≤ 0.1 t 0!
this gives t ≥−ln( 0.1) / λ = −ln( 0.1) / 0.02= 115 years
84 Poisson Distribution: Example
2. Suppose that 50 years pass without any earthquakes occurring. What is the probability that another 50 years will pass without any earthquakes occurring?
Solution:
PP[NN100 =∩=00N50 ] [ 100 = 0] P0[N100= |0N 50 = ] = = PP[NN50 = 00] [ 50 = ] −100λ e −−λ = =ee50 = 1 = 0.368 e−50λ
Note that due to memorylessness,
−−510λ P[N50 = 0] =ee = = 0.368
85 Simulating a Poisson Distribution Since the Poisson process derives from an infinite number of independent and equilikely Bernoulli trials over time (or space), its simulation is simple: uniformly distributed (equilikely) on any interval.
86 Common Continuous Distributions
The Bernoulli Family extends to two continuous distributions when every instant in time (space) becomes a Bernoulli trial;
1. Exponential: T 1 = time until the next “success” (arrival). This is the continuous time analog to the geometric distribution.
2. Gamma: T k = time until the k’th success (or arrival). This is the continuous time analog to the negative binomial distribution.
Consider the Poisson Distribution which governs the number of “successes” in time interval t when every instant is an independent Bernoulli trial. We know that the Poisson distribution specifies that (λt)k P[Ne= k] = −λt t k! What is the probability that the time to the first “success” is greater than t? If the first “success” occurs later than t, then the number of successes in time interval t, must be zero, i.e. (λt)0 P[T > t] =P[Ne= 0] =−λt = e−λt 1 t 0!
88 Exponential Distribution 0 (λt) −λt −λt If P[T1 > t] =P[Net = 0] = = e 0! then Ft( ) =P T ≤ 1 = 1− e−λt T1 [ 1 ] dF( t) so that f (t) =T1 = λe−λt , t ≥ 0 T1 dt
1 E[T ] = 1 λ 1 Var[T ] = 1 λ 2
89 Exponential Distribution
Again, by virtue of the fact that every instant in time (space) is an independent Bernoulli trial, the exponential distribution is also memoryless. That is,
PP[TT111>+∩tsT>tt] [ >+s] P|[T1 >+ tsTt1 >] = = PP[TT>>tt] [ ] 11 −+λ (ts) e −λs = = e e−λt
−λs but since P [ Ts 1 > ] = e it is apparent that it doesn’t matter when you start looking. I.e., the probability that the next “success” is time s in the future is independent of t (when we start looking).
90 Exponential Distribution
The “memoryless” property of the exponential distribution means that it is a common “lifetime” model (where “success” is defined as “failure”!) in systems that neither improve nor degrade with time – failure probability remains constant with time.
However, some systems improve with time (e.g. early strength of concrete) and some systems degrade with time (e.g. fatigue), which leads to the Weibull distribution to be discussed shortly.
91 Exponential Distribution: Example Let us assume that earthquakes in a certain region occur on average once every 50 years and that the number of earthquakes in any time interval follows a Poisson distribution. Under these conditions, what is the probability that less than 30 years will pass before the next earthquake occurs?
Solution: Let T 1 be the time to the next earthquake. Then, since the number of earthquakes follow a Poisson distribution, the time between earthquakes follows an exponential distribution.
Thus, T 1 follows an exponential distribution with λ = 1/ 50 = 0.02 earthquakes per year (on average), and
−30(0.02 ) P[T1 < 30 years] =−=1 e 0.549
92 Gamma Distribution
The last of the Bernoulli Family of distributions is a generalization of the exponential distribution.
Let T k = time until the k’th “success” which is evidently the sum of the times between the first k “successes”, i.e.
TT= +TT ++ k 1112 1k which leads to the following moments k E[T ] =kTE[ ] = k 1 λ k Var[T ] =kTVar[ ] = k 1 λ 2 93 Gamma Distribution
To determine the Gamma distribution, lets consider an rd example: P [ Tt 3 > ] means the probability that the 3 “success” occurs after time t. For this to be true, we must have had no more than 2
“successes” within time interval t. That is, the event Tt 3 > is equivalent to the event Nt ≤ 2 In other words:
P[Tt3 > ] =P[NNt≤===2] = P[ t 012] ++ PP[NN tt] [ ] (λλtt)1 ( )2 = e−−λλtt+ ee+ − λt 1! 2! 2 (λt) j = e−λt ∑ j=0 j!
94 Gamma Distribution
2 j −λt (λt) If P[Tte3 >=] ∑ j=0 j! j 2 (λt) then Ft()=P[T ≤ t] = 1− e−λt T3 3 ∑ j=0 j!
In general k−1 (λt) j Ft()= P[T ≤ t] = 1− e−λt Tk k ∑ j! j=0 is the Gamma distribution
95 Gamma Distribution
λ = 1 k = 3
96 Gamma Distribution: Example As in the previous example, let us assume that earthquakes in a certain region occur on average once every 50 years and that the number of earthquakes in any time interval follows a Poisson distribution. Under these conditions, what is the probability that less than 150 years will pass before two or more earthquakes occur?
Solution: Let T 2 be the time to the occurrence of the second earthquake. Then, since earthquakes occur according to a
Poisson process, T 2 must follow a Gamma distribution with k =2 and λ = 1/ 50, which gives us 1 (150 / 50) P[T < 150] ==−+=F (150) 1e−150/50 1 0.801 2 T2 1!
97 Gamma Distribution: Example
Note that since this is a Poisson process, the same result is obtained by noticing that
PP[TN2 <≥150] = [ 150 2] so that
P2[N105 ≥ ] =−=−−1P[N150 <==2] 1PP[ NN150 01] [ 150 ] 1 (150 / 50) =−−1 ee−−150/50 150/50 1! = 0.081 which is one way of determining the CDF for the Gamma distribution (as we saw earlier). 98 Weibull Distribution • generalization of the exponential distribution • now has two parameters for more flexibility in distribution shape • common lifetime model • improving system if β < 1 • degrading system if β > 1 • constant failure probability (exponential) if β = 1
β β β f ()x= (λ xe) −()λx X x −()λx β FxX ()= 1− e
99 Weibull Distribution
β β β fx( ) = (λ xe) −()λx X x −()λx β FX (1x) = − e
100 Weibull Distribution: Example The time to 90% consolidation of a sample of a certain clay has a Weibull distribution with β = 0.5 . A significant number of tests have shown that 81% of clay samples reach 90% consolidation in under 5516 hours. What is the median time to attain 90% consolidation?
Solution: Let X be the time until a clay sample reaches 90% consolidation. Then we are told that X follows a Weibull distribution with β = 0.5 . We first need to compute the other Weibull parameter, λ . To do this we make use of the fact that we know P [ X ≤ 5516 ] = 0.81 , and since P [ XF≤ 5516] = X (5516) we have 0.5 Fe(5516) =−= 1 −(5516λ ) 0.81
0.5 e−(5516λ ) = 0.19 λ = 1/2000 101 Weibull Distribution: Example
We are now looking for the median, x , which is the point which divides the distribution in half. That is, we want to find x such that Fx ( ) = 0.5 . In other words,
0.5 x 1− exp − =0.5 ⇒= x 960.9 hours 2000
102 Uniform Distribution 1 fx()= αβ ≤≤x UNIFORM X βα−
The uniform distribution is useful in representing random variables which have known upper and lower bounds and which have equal likelihood of occurring anywhere between these bounds.
The distribution makes no assumptions regarding preferential likelihood of the random variable since all possible values are equally likely.
fxX () Let αβ= 3 and = 7 αβ+ E[X ] =µ = X 2 0.25 2 alternative notation 2 (βα− ) Var[X ] =σ = X 12 x 0 1 2 3 4 5 6 7 8 9 103 Normal Distribution
The best known Probability Density Function is the Normal or Gaussian distribution.
Let X be a normally distributed random variable with mean and standard deviation given by µσXX and . In this case the PDF is given by:
2 11x − µX E[X ] = µX fxX ( )= exp − σπ 2 σ Var[X ] = σ 2 X 2 X X
104 NORMAL DISTRIBUTION
µσXX=100 = 50
Mean,median and mode fxX ()
The area under the distribution is unity
inflection points are at µσ±
x 105 NORMAL DISTRIBUTION
To compute probabilities: 2 a a 1x−µX − 1 2 σ ≤== X P[ Xxa] ∫ f X ( ) dx ∫ e dx −∞ σ X 2π −∞ which has no closed form solution. It must be evaluated numerically provide tables for probabilities.
Problem: we’d need one table for each possible value of µ X
and σ X . Solution:
X − µµXXaa−− µX PP[ XZ≤≤a] = = P≤ σσXX σX a − µ = Φ X σ X
106 STANDARD NORMAL DISTRIBUTION
X − µ Where Z = X is the standard normal. σ X
X − µX 1 E[Z ] =E =E[ X −=µX ] 0 σσXX X − µ 1 Var[ X ] =X = −=µ = Var[Z ] Var 22Var[ X X ] 1 σσXX σX
Now we need one table for the standard normal, µσZZ=0 and = 1 and we standardize:
X − µµXXaa−− µX PP[ X ≤≤≤a] = = PZ σσXX σX a − µ =Φ=X Standard Normal CDF σ X 107 STANDARD NORMAL DISTRIBUTION
The Standard Normal distribution is so important that it is commonly given its own symbols:
X − µ Standardized variable: Z = X σ X
112 Density function: φZ (z) =exp − z 2π 2
a − µ Cumulative Distribution: P[ X ≤ a] = Φ=ΦX (z) σ X
108 STANDARD NORMAL DISTRIBUTION
µσZZ=0 = 1
φZ ()z
The area under the distribution is unity
inflection points are now at ± 1
z
109 Standard Normal Cumulative Distribution Function (CDF)
Φ()z
Standard tables cover z>0
z
110 Standard Normal Function CDF
Table gives Φ(z) for z≥ 0
111 Standard Normal Function CDF For z < 0 Φ(z) = 1- Φ(-z)
112 Example calculations using the Standard Normal Function CDF
Example 1:
Permeability measurements have indicated that k is normally distributed −−88 with the properties: µσkk=4.1×= 10 m/s and 1× 10 m/s − What is the probability that k >×4.5 108 m/s ?
fkK ( )
We want this area
k
…but tables only give area to the left of a given point… 113 fkK ( ) …so estimate this area and subtract from 1
k
Pk[>× 4.5 10−−88 ] =−≤× 1Pk [ 4.5 10 ] 4.5× 10−−88 −× 4.1 10 = −Φ 1 −8 1× 10 =1 −Φ( 0.4) =1 − 0.65542
= 0.3446 (34%) 114 The Reliability Index The Reliability Index is a measure of the margin of safety in “standard deviation units”.
For example, if dealing with a normally distributed Factor of Safety (where FS=1 implies failure), the reliability Index is given by: µ −1 β = FS σ FS If the Factor of Safety is lognornal, the reliability Index is given by: µ β = ln(FS ) σ ln(FS )
For normally distributed random variables, the “reliability index” (β) is uniquely related to the “probability of failure” (pf) through the expression:
p f =1 − Φ(β ) 115 0.5
f p Probability ofFailure:
Reliability Index: β
Probability of Failure vs. Reliability Index for a Normal Distribution 116 Consider a normal distribution of the Factor of Safety (FS)
fFS
µFS =1.5
σ FS = 0.21
pf is this area
FS
β is this distance ÷ the standard deviation
(1.5 −1) β = = 2.38 p f =1 − Φ( 2.38) = 1 − 0.991343 = 0.0087 0.21 117 Lognormal Distribution
Let X be a lognormally distributed random variable with a mean and standard deviation given by µσXX and .
σ X Let the Coefficient of Variation vX = µX
If XX is lognormally distributed, this means that ln is normally distributed. Let the mean and standard deviation of this underlying normal distribution of lnX be given by µσln XX and ln respectively.
118 The PDF for a lognormal distribution is given by:
2 11ln x − µln X f X ()x = exp − σπ 2 σ x ln X 2 ln X If the mean and standard deviation of the lognormal random variable x are µσXX and , then the mean and standard deviation of the underlying normal distribution of lnx are given by:
1 2 µµln XX=−+ln ln{ 1 v X} 2 2 σ ln XX=ln{ 1 + v }
119 Lognormal Distribution
To compute probabilities: 1. convert to normal by taking logarithms on both sides of the inequality, 2. standardize by subtracting mean and dividing by standard deviation on both sides of inequality, 3. look up probability in standard normal tables
lnXa−−µµln XX ln ln P[ Xa≤≤] = P[ ln Xln a] = P ≤ σσln XXln
ln aa−−µµln XX ln ln = P Z ≤ = Φ σσln XX ln
120 Going in the other direction....
1 2 µX=exp µσln XX + ln 2
2 σµXX= exp( σln X) − 1
Further relationships involving the lognormal distribution:
MedianXX= exp(µln )
2 ModeX= exp(µσln XX − ln )
121 LOGNORMAL DISTRIBUTION
µσXX=100 = 50 0.012 Mode= 71.6 fxX () = 0.01 Median 89.4 Mean= 100 0.008
0.006 The area under the distribution is unity
0.004
0.002
0 x
0 50 100 150 200 250 300 122350 SIGNIFICANCE OF THE MEDIAN IN A LOGNORMAL DISTRIBUTION
µσXX=100 = 50 fx() X Median= 89.4
The area under the distribution is unity
Area = 50% Area = 50%
x123 Example 2:
Permeability measurements have indicated that k is lognormally distributed −−88 with the properties: µσkk=4.1×= 10 m/s and 1× 10 m/s − What is the probability that k >×4.5 108 m/s ? First find the properties of the underlying normal distribution of ln k
2 2 σ k 1 σ ln k =+=+=ln 1 ln 1 0.2404 µ 4.1 k 11 µ=−ln µσ2 =ln(4.1 ×− 10− 82 ) (0.2404) =− 17.0386 ln kk22ln k
124 ln(4.5×=− 10−8 ) 16.92 fkln K (ln )
We want this area
ln k
…but tables only give area to the left of a given point…
125 ln(4.5×=− 10−8 ) 16.92 fkln K (ln )
…so estimate this area and subtract from 1
ln k
Pk[>× 4.5 10−−88 ] =−≤× 1Pk [ 4.5 10 ] −16.92 −− ( 17.0386) =1 −Φ 0.2404 =1 −Φ( 0.5075) ≈1 −Φ( 0.51) =1 − 0.69497 = 0.3050 (31%) (was 0.3446 using the normal distribution) 126 COMPARISON OF NORMAL AND LOGNORMAL DISTRIBUTIONS
σ X µXX=100 (Mean) v = (Coefficient of Variation) µX
fxX () vX = 0.1
vX = 0.3 vX = 0.5
x
127 ....not much difference for vX < 0.3 Calculations involving a single random variable (SRV)
Example of Bearing Capacity
q
2 φuu= 0,c kN/m
What is the relationship between the Factor of Safety (FS) of a conventional bearing capacity calculation (based on the mean strength) and the probability of failure (pf)?
128 First perform a deterministic calculation
Conventional bearing capacity calculations typically involve high factors of safety of at least 3.
The bearing capacity of an undrained clay is given by the Prandtl equation:
qcuu=(2 +π )
= 5.14cu where cu is a design “mean” value of the undrained shear strength.
2 If cu = 100kN/m , and FS = 3 this implies an allowable bearing pressure of:
5.14× 100 q = =171kN/m2 all 3
129 Now perform a probabilistic calculation
If additional data comes in to indicate that the undrained strength is a lognormal random variable with µσ= 100kN/m22 and = 50kN/m what is the probability ccuu of the actual bearing capacity being less than the factored deterministic value Pq [u < 171] ?
qu is a linear function of the random variable cquu from = 5.14 c u hence E[qc ]= 5.14E[ ] thus µµ =5.14 =514 u u qcuu and Var[qc ]= 5.142 Var[ ] thus σσ =5.14 =257 u u qcuu
1 (Note that vv= = ) quuc 2 130 First find the properties of the underlying normal distribution of ln qu
2 2 1 σ ln qq=ln{ 1 +=v } ln 1 + = 0.47 uu 2 11 µ=−=ln µσ22ln(514) − (0.47) = 6.13 ln qquu22ln q u ln171− 6.13 Pq[u <=Φ 171] 0.47 =Φ−( 2.10) =1 −Φ( 2.10) =1 − 0.982
= 0.018 (1.8%) 131 Example of Slope Stability
Dimensionless strength 1 parameter H µ 2 C = cu DH γ sat H φ= 0, σγ , o u cu sat β = 26.57
assume D = 2
If cu is treated as a random variable with mean µσ and standard deviation ccuu what is the relationship between
FS (based on the mean) and the probability of failure p f ? 132 Deterministic calculation
For a slope with β = 26.5 7o and a depth ratio D = 2, FS= 5.88 C µ where C = cu γ sat H
133 Probabilistic calculation 1) Note that FS∝= C or FS KC ,
hence µµFS= K C and vv FS= C 2) If C is lognormal, then so is FS .
Compute the underlying normal properties σµln FS and ln FS 3) Find the probability P[ FS < 1] from standard tables.
If FS is lognormal, P[ FS<= 1] P[ ln FS < ln1]
ln1− µµln FS ln FS =Φ =Φ− σσln FS ln FS
1− µFS If FS is normal, P[ FS <=Φ 1] σ FS
134 σ C vC = µC
High!
The MEAN strength may be too optiminstic….
135 Extreme Value Distributions
Most systems fail due to a combination of extremes in load and/or resistance, so extreme value distributions are very useful in assessing system reliability.
Let Ynn= max( XX12 , ,… , X)
and Y1= min( XX 12 , ,… , Xn )
We are interested in determining the distributions of YYn and 1
136 Exact Extreme Value Distributions: Maximum
Since Y n is the maximum of the X’s, then the event Y n ≤ y implies that all of the X’s are also less than y. In other words,
PP[YXnn≤y] = [ 12 ≤∩ yX≤∩ y ∩ X ≤ y]
If it is assumed that the X’s are independent and identically distributed, then Fy( ) =P[ Y≤ y] = PP[ X≤≤ yy] [ X ] P[X ≤ y] Ynn 1 2 n n = [FyX ()]
dF( y) n−1 f( y) =Yn = nfy( yF)[ ()] Yn dy XX
137 Exact Extreme Value Distributions: Example
Suppose that fissure lengths, X, in a rock mass have an − x exponential distribution with fx X ( ) = e . What, then, does the distribution of the maximum fissure length, Y n , look like for n = 1, 5, and 50 fissures?
Solution:
If n = 1, then Y n is the maximum of one observed fissure, which of course is just the distribution of the single fissure length.
Thus, when n = 1, the distribution of Y n is just the exponential distribution;
f y =fy = e− y Y1 ( ) X ( )
138 Exact Extreme Value Distributions: Example When n = 5, we have 5 Fy=PPY≤y = X ≤≤yyP X P X ≤ y= Fy Y5 ( ) [ 51] [ ] [ 25] [ ] X ( )
− y 5 = 1− e
dyF ( ) 4 fy( ) =Y5 = 51ee−−yy− Y5 dy
Similarly, when n = 50, we have
50 50 F ( y) =P Y≤ y = Fy( ) = 1− e− y Y50 [ 50 ] X
dyF ( ) 49 f ( y) = Y50 = 50e−−yy 1− e Y50 dy
139 Exact Extreme Value Distributions: Example
140 Exact Extreme Value Distributions: Minimum
Since Y 1 is the minimum of the X’s, then the event Y 1 > y implies that all of the X’s are also greater than y. In other words,
PP[YX1 > y] =[ 12>> yX∩ > y∩∩ Xn y] If it is assumed that the X’s are independent and identically distributed, then n P[Y1 > y] =>>PP[ X1 yy] [ XX2 ] P[ n >=y] [1− FX ()y ]
n so that F y = 11−−F y Y1 ( ) X ( )
dF ( y) n−1 f ( y) = Y1 = nf( y)[1− F ()y ] Y1 dy X X
141 Exact Extreme Value Distributions: Example
A series of 5 soil samples are taken at a site and their shear strengths determined. Suppose that a subsequent design is going to be based on the minimum shear strength observed out of the 5 samples. If the shear strengths of the individual samples are exponentially distributed with parameter λ = 0.025 m 2 /N k then what is the distribution of the design shear strength?
Solution: 5 5 5 Fy( ) =−−11 Fy( ) =−−− 111e−−λλyy =− 1 e YX1 ( )
−5λ y =1 − e which is also exponential with ‘rate’ n λ
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