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Chapter 4 Functions of Random Variables Instructor: Prof. Wilson Tang CIVL 181 Modelling Systems with Uncertainties

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Functions of Random Variable (R.V.) In general, Y = g(X) Y = g(X 1, X 2,…, X n ) If we know distribution of X distribution of Y? M = 4X + 10 X 5 22 M 1) X b b = Xtan 2) cost of delay = aX 2 3) where X – length of delay

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Consider M = 4X + 10 Observation: the distribution of y depends on (1) Distribution of X (2) g(X) X M X654 M wider distribution

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YXX X Y X Eq. 4.6 E4.1 See E4.1 in text for details X X where Y X For monotonic function g(X)

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E4.2 When g(X) is not linear shape may change Consider Y = ln(X) y 0 f(y) x 1 0 e f(x) LN(, ) x = 0 y = – x = 1 y = 0 x = e y = x = y = y = N(, ) See E4.2 in text for details X

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if X 1 and X 2 are Poisson with mean rates 1 and 2 respectively Z is Poisson with z = (see E4.5 on p. 175) if X is N( , ) Y is N(a + b, a ) if X is N( , ) Y is N(a , a ) if X is LN(, ) Y is LN(ln a +, ) Summary of Common Results

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Summary of Common Results (Cont’d) if X 1 and X 2 are N( , ) and N( , ) respectively Z is N( z, z ) where if X 1 and X 2 are s.i. 12 = 0 if X i = N( i, i ) ; i = 1 to n Z is N( z, z ) where

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E4.8 S = D + L + W Column with capacity, R D = 4.2, D = 0.3, D = 7% L = 6.5, L = 0.8, L = 12% W = 3.4, W = 0.7, W = 21% S = total load = D + L + W S = D + L + W = 14.1 S = =1.1 a) P(S > 18) = 1 – = 1 – (3.55) = Assume D, L, W are s.i.

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E4.8 (Cont’d) P(failure) = P(R < S) = P(R – S < 0) = P(X < 0) R = N( R, R R ) where R = 1.5 S = 1.5 x 14.1 = R = 0.15 R = N(21.15, 3.17) X = = 7.05 X = = 3.36 P(F) = = (-2.1) = X = R – S R – design capacity 1.5 – design safety factor, SF

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If the target is P(F) = R = ? and assume R = 0.15 Recall: X = R X = (0.15 R ) = -1 (0.001) = – 3.09 R = 20.8 – 2.95 R R 2 R 2 – 2.95 R = 0 R = or 26.9 Set: P(F) = = Since R should be larger than 21.5 R = 26.9 Check R = P(F) = = (3.09) 1.0

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E4.9 Start Framing X1X2X3 X4 X X 1 = Excavation X 2 = Ftg. Const. X 3 = Wall Const. X 4 = Sub-assembly X 5 = Delivery to site P(Framing at least 8 days from start) = P(T > 8) = P(T 1 > 8 T 2 > 8) = P(T 1 > 8) + P(T 2 > 8) – P(T 1 > 8) P(T 2 > 8) where T 1 = X 1 + X 2 +X 3 T 2 = X 4 + X 5

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YX M W = N(10, 2) 523 M X N(4, 0.8) Y N(6, 1.2) M = 5X + 10Y M is Normal with M, M M = 5 X + 10 Y = = 80 M = 5 2 X y 2 + corr. term where corr. term = 2a 1 a 2 1 2 = 2 5 10 1 0.8 1.2 = 96 M = 5 2 (0.8) (1.2) = 16 Alternative way: M = 5X + 10Y = 5(2W/5) + 10(3W/5) = 2W + 6W = 8W M = 8 W = 80 M = 8 W = 16 Assume = 1

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W = weight of a truck = N(100, 20) We are interested in the total weight of 2 trucks or 1. Total weight = T = 2W Normal with T = 2 W = 200 T = 2 W = 40 Total weight = T = W 1 + W 2 Normal with T = 200 T = = if s.i. and 1 = 2 = ? T

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E4.10 Sand Footing P S Sand Property – MFooting Property – B and I Assume P, B, I, and M are s.i. and log-normal with parameters P, B, I, M and P, B, I, M, respectively

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E4.10 (Cont’d) Observe: Y X1X1X1X1 X2X2X2X2 X3X3X3X3 X4X4X4X4 Normal Y or lnS is normal S is lognormal LN( S, S )

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E4.10 (Cont’d) (a) S = ? and c.o.v. of S, S = ?

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E4.10 (Cont’d) (b) Maximum allowable settlement = 2.5 in., reliability against excessive settlement = ? (c) If an expense of $100 is spent so that the c.o.v. of M can be reduced to 5% for getting better information. Would you spend this money? Assume the damage cost is $50,000 for the exceedance of the maximum allowable settlement.

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E4.10 (Cont’d)

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Assume C 0 is the initial cost of construction. Then, the expected cost of the first design (Not spending $100) is The expected cost of the second design is E(C 1 ) < E(C 2 ) It is better for not spending the money to gather more information on M.

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The occurrence of thunderstorms in a town in Texas follow Poisson process during each season: Winter: Oct – March Summer: April – Sept 21 – year record 73 storms in winters 340 storms in summers 340 storms in summers Determine: P(4 storms during three month of March, April & May) = ?

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Mean rates for each summer month, S = Similarly, W = Z = X + Y Poisson with parameters m Z = m X + m Y = 2.7 1 = 5.98 Recall Poisson Distribution X

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Central Limit Theorem S will approach a normal distribution regardless the individual probability distribution of X i if N is large enough

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P4.27 f E (e) 0+1 1/2 e (a) E(E) = 0

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(c) T = total mean = E 1 + E 2 +…+ E 124 E(T) = 124 0 = 0 Var(T) = Var(E 1 ) + Var(E 2 ) +…+ Var(E 124 ) = 124Var(E) = 124 1/3 = T = 6.43 Distribution of T Normal (C.L.T) P( T < 0.01% of 2 miles) (b) P( E < 0.01% of length of segment) (area under rectangular curve) Assume s.i. By making more measurements, the % error is decreasing P(within given tolerance limit)

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Reliability Computation Suppose R denotes resistance or capacity S denotes load or demand Satisfactory Performance = {S < R} P S = P(S < R) and P f = 1 - P S Case 1: If R, S are normal where Z = S – R and Z = S 2 + R 2 Case 2: If R, S are lognormal where Z = S – R and Z = S 2 + R 2

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Case 3: If R is discrete, S is continuous Example on Case 3 S = N(5, 1) r P(R = r)

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Reliability – Based Design Observe for Case 1 in which R and S are both Normal If P S Reliability = Reliability index = -1 (P S ) Design R = S + S 2 + R 2

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Example: S = N(5, 2) R = N( R, 1) R = ? Require P f = or P S = = -1 (0.999) = 3.1 Design R = = 11.93

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Moments of function of R.V. In most cases, we can’t derive f y (y) easily. Let us try to determine mean and Var of Y. Let us try to determine mean and Var of Y. ?

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Y = aX + b E(Y) = aE(X) + b Why? E(X) 1 For linear case

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Y = a 1 X 1 + a 2 X 2 E(Y) = a 1 E(X 1 ) + a 2 E(X 2 ) = a 1 2 a 2 2 a 1 a 2 1 2 if X 1 and X 2 are s.i. = 0 Var(Y) = a 1 2 a 2 2 2 2 Extension ( not only true for Normal)

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Mean and Variance of general functions Recall we have formula for E(aX + b) and E(a 1 X 1 + a 2 X 2 +…+ a n X n ) but how about a general non-linear function: Y = g(X) E[g(X)] = ? Var[g(X)] = ? Rigorous method However computations may be complex, also we may not have f X ( x ). X X X

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First order approximation: g( X ) g’( X ) (X – X ) X g(X) XX X Taylor Series Approximation x - x Var(X) …

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Observe validity of linear approx depends on: 1) Function g is almost linear, i.e. small curvature 2) x is small, i.e. distribution of X is narrow

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Example If K = 400 ; K = 200 If M = 100

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K M M What if M is also random? M = 100; M = 20 then

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Observe contribution of uncertainties from 2 components: Sensitivity factor + Var(X i ) How to invest money for research? a) Identify R.V. X i with major contributions to uncertainties b) Reduce sensitivity factor c) Reduce Var(X i )

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In general Y = g(X 1, X 2, …, X n ) E(Y) = g( 1, 2, …, n ) Propagation of uncertainties

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Uses 1. Easy calculations 2. Compare relative contributions of uncertainties – allocation of resource 3. Combine individual contributions of uncertainties

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Comparison of 1 st and 2 nd Order Approximation Observe 2 nd order approximation take the 2 nd order term Requires: 1. Small Var of X i ’s, e.g. in surveying measurements 2. Function approx. linear 2 nd order term Compare this to 1 st order E(Y) = g( X )

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Check 2nd order term for e.g. Sufficiently close to 2.0 Despite of k 1/2 & k is fairly large

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Review on Chapter 4 1. Consider Y = g(X) or Y = g(X 1, X 2, …, X n ) 2. Given distribution of X or X i distribution of Y can be obtained by a) Theoretical derivation e.g. f Y (y) = f X (g -1 ) (dg -1 /dy) b) Formula for typical cases 3. Model of sum – Normal Model of product – Lognormal 4. C.L.T. – Generalize (3.) regardless of distribution of X or X i 5. Reliability evaluation – P(X < Y) for various cases 6. Y = g(X) or Y = g(X 1, X 2, …, X n ) Given mean & Var of X determine mean & Var of Y a) exact for linear cases b) approx. for general cases

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