Presentation is loading. Please wait.

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

## Presentation on theme: "Chapter 4 Functions of Random Variables Instructor: Prof. Wilson Tang CIVL 181 Modelling Systems with Uncertainties."— Presentation transcript:

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

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

Consider M = 4X + 10 Observation: the distribution of y depends on (1) Distribution of X (2) g(X) X 654 0.6 0.2 M 343026 0.6 0.2 X654 M343026 wider distribution

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)

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

if X 1 and X 2 are Poisson with mean rates 1 and 2 respectively  Z is Poisson with z = 1 + 2 (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

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

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 =  0.3 2 + 0.8 2 + 0.7 2 =1.1 a) P(S > 18) = 1 –  = 1 –  (3.55) = 0.000193 Assume D, L, W are s.i.

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 = 21.15  R = 0.15  R = N(21.15, 3.17)  X = -14.1 + 21.15 = 7.05  X =  1.1 2 + 3.17 2 = 3.36 P(F) =  =  (-2.1) = 0.018 X = R – S  R – design capacity 1.5 – design safety factor, SF

If the target is P(F) = 0.001   R = ? and assume  R = 0.15 Recall:  X =  R - 14.1  X =  (0.15  R ) 2 + 1.1 2  =  -1 (0.001) = – 3.09   0.0225  R 2 +1.21 = 20.8 – 2.95  R + 0.105  R 2   0.0825  R 2 – 2.95  R +19.59 = 0  R = 8.812 or 26.9 Set: P(F) =  = 0.001 Since  R should be larger than 21.5  R = 26.9 Check  R = 8.812 P(F) =  =  (3.09)  1.0

E4.9 Start Framing X1X2X3 X4 X5 23 4 51 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

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 = 20 + 60 = 80  M =  5 2  X 2 + 10 2  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) 2 + 10 2 (1.2) 2 + 96 = 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

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 = 20 2. ? T

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

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

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

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.

E4.10 (Cont’d)

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.

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) = ?

Mean rates for each summer month, S = Similarly, W = Z = X + Y Poisson with parameters m Z = m X + m Y = 2.7  2 + 0.58  1 = 5.98 Recall Poisson Distribution X

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

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

(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 = 41.33  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) 

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

Case 3: If R is discrete, S is continuous Example on Case 3 S = N(5, 1) r P(R = r) 567 0.1 0.3 0.6

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 

Example: S = N(5, 2) R = N(  R, 1)  R = ? Require P f = 0.001 or P S = 0.999  =  -1 (0.999) = 3.1  Design  R = 5 + 3.1  2 2 + 1 2 = 11.93

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. ?

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

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

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

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

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

Example If  K = 400 ;  K = 200 If M = 100 

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

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 )

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

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

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 )

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

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

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

Similar presentations

Ads by Google