Full CHapter 4 Solutions

8/8/2019 Full CHapter 4 Solutions

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4 Functions of Random Variables

1. Derive the probability density function for Y, given that Y = X4; for thecases

(i) fX (x) =12 ; 0 x 2

(ii) fX (x) = c exp(x); x 0:Plot all density functions.

Solution:

g(x) = X4

g0(x) = 4X3:

Mathematically there are four roots given by (obtain these easily by den-

ing Y = Z2

and Z = X2

and performing successive square roots):

x1 = 4p

y; x2 = 4py;x3 = i 4

py; x4 = i 4py:

(i) For

fX (x) =1

2; 0 x 2;

the general transformation considering only the positive root is

fY (y) =fX (x1)

jg0 (x1)j

=

1

4y3=4

1

2

=1

8y3=4; 0 y 16:

Note that Z160

1

8y3=4dy =

18 41 y1=416

0

= 1:

The density plots are shown in Figures 27 and 28.

(ii) GivenfX (x) = c exp(x); x 0;

rst nd the value of c :

Z10

c exp(x)dx = 1:0=) c = 1:0:

As x has a positive range, the negative roots should be neglected forcalculations.

42


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0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4

0.0

0.2

0.4

0.6

x

f(x)

Figure 27: Density Function fX (x) = 1=2; 0 x 2

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 160

1

2

3

4

5

y

f(y)

Figure 28: fY(y) =1

8y3=4; 0 y 16

43


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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 160

1

2

3

4

5

y

f(y)

Figure 30: fY(y) =1

4y3=4exp( 4py); y 0

density transformation:

g(x) = aX2

g0(x) = 2aX

fY (y) =

fX (x1)

jg0 (x1)j :

(i) Given that fX (x) = c exp(x); x 0; rst nd the value of c :Z10

c exp(x)dx = 1:0=) c = 1:0:

Then,

fY (y) =fX (x1)

jg0 (x1)j

= 1

2a

ry

a

exp(rya

)

=1

2p

ayexp(

ry

a); y 0:

45


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0.040.060.080.100.120.140.160.180.200.220.240.260.280.300.320.340.360.380.400.420

1

2

3

4

5

x

y

12p

yexp(py)

For the case a = 1; the plot of density function fX (x) is shown in Figure31.

0 1 2 3 4 50.0

0.2

0.4

0.6

0.8

1.0

x

f(x)

Figure 31: Density Function fX (x) = exp(x); x > 0

(ii) If fX (x) is a Rayleigh density, it is given by

f(x) =x

2exp

x

2

22

; x 0:

46


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As there is only one root, we have

fY (y) = fX (x1)jg0 (x1)j

=1

2p

ay

ry

a

1

2exp

y

2a2

=1

2a2exp

y

2a2

; y > 0:

For the case = 1 and a = 1; the plots of these density functions areshown in Figures 32 and 33.

0 1 2 3 4 50.0

0.1

0.2

0.3

0.4

0.5

0.6

x

f(x)

Figure 32: Density Function fX(x) = x expnx22

o; x > 0

(iii) The lognormal density is given by

fX (x) =1

xp

2exp

(1

2

ln x

2); 0 < x < 1:

So fY (y) is found as

fY (y) =fX (x1)

jg0 (x1)j

= 12ap

ya

1pya

p2

exp8 0For mean value = 1, standard deviation = 1; and a = 1; the densityfunctions are plotted in Figures 34 and 35.

3. Derive the probability density function for Y; given Y = a + X3: Considertwo cases:

(i) fX (x) = c exp(x); x 0;(ii) fX (x) is a lognormal density.

Plot all density functions.Solution:

(i) Given the density fX (x) = c exp(x); x 0; rst nd the value of c :Z10

c exp(x)dx = 1 =) c = 1:

ForX3 = Y a;

X has three equal roots,3p

Y a:

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0 2 4 6 8 10 12 14 16 18 20 22 24

0.00

0.05

0.10

0.15

0.20

x

y

Figure 34: Density Function fX(x) =1

xp

2exp

(1

2

ln x 1

1

2); x > 0

0 1 2 3 4 50.00

0.05

0.10

0.15

0.20

y

f(y)

Figure 35: Density Function fY(y) =1

2yp

2exp

8

0

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Performing the density transformation:

x1 = 3py a; x2 = 3py a; x3 = 3py ag(x) = a + X3

g0(x) = 2X2

fY (y) =fX (x1)

jg0 (x1)j +fX (x2)

jg0 (x2)j +fX (x3)

jg0 (x3)j=

3

2(y a) 23 exp(3p

y a); y 0:

Density functions for a = 1 are shown in Figures 36 and 37.

0 1 2 3 4 50.0

0.2

0.4

0.6

0.8

1.0

x

f(x)

Figure 36: Density Function fX (x) = exp(x); x > 0

(ii) The lognormal density is given by

fX (x) =1

xp

2exp

(1

2

ln x

2); 0 < x < 1:

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0 1 2 3 4 50

5

10

15

20

25

x

y

Figure 37: Density function fY(y) =3

2(y a) 23 exp(3p

y a); y > 0

Performing the density transformation:

x1 =3p

y a; x2 = 3p

y a; x3 = 3p

y ag(x) = a + X3

g0(x) = 2X2

fY (y) =fX (x1)

jg0 (x1)j +fX (x2)

jg0 (x2)j +fX (x3)

jg0 (x3)j

= 12(y a) 23

33p

y ap2 exp(

12

ln

3py a

2)

=3

2p

2(y a) exp(

12

ln 3

py a

2):

Density functions for = 1; = 1 and a = 1 are shown in Figures 38 and39.

4. Derive the probability density function for Y; given Y = X3 for

fX (x) = 1=c2; 2 < x < 4:

Sketch all density functions.

Solution: Integrating the density function over the domain yields:Z42

1

c2dx = 1

c2 = 2:

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0 1 2 3 4 5 6 7 8 9 10

0.00

0.05

0.10

0.15

0.20

0.25

x

f(x)

Figure 38: Density Function fX (x) =1

xp2

expn12

ln x1

1

2o; x > 0

1 2 3 4 50.0

0.5

1.0

1.5

2.0

2.5

y

f(y)

Figure 39: Density Function fY(y) =3

2p

2(y 1) exp(

12

ln 3p

y 1 11

2)

52


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Then, x1 = y1=3 and

g(X) = X3g0(X) = 3X2 = 3Y2=3

fY(y) =fX (x1)

jg0(x1)j =1=23y2=3 = 16y2=3 ; 8 < y < 64:

5. The random variables X and Y are related by the equation

Y = eX :

(i) Suppose that X is uniformly distributed as 0 X 5: Find fY (y) :Sketch this density function.

(ii) Suppose now that X is governed by the density function fX = cex;

0 X 5; where c is a constant. Find fY (y) : Sketch this densityfunction.Solution:

(i) From the given information fX (x) = 1=5: Performing the density trans-formation:

x1 = ln y

g(x) = exp(x)

g0(x) = exp(x)

fY (y) =fX (x1)

jg0 (x1)j =1=5

exp(ln y)

=

1

5y ; 1 y 148:4fY (1) = 1=5

fY (148:4) = 1=(5 148:4) = 1:3477 103:The density functions are shown in Figures 40 and 41.

(ii) Suppose now that X is governed by the density function

fX = cex; 0 X 5;

where c is a constant. Find fY (y) : Sketch this density function.

First nd the value of c :R50

c exp(x)dx = 1 =) c = 1:0068:

fY (y) = fX (x1)jg0 (x1)j = 1:0068 exp( ln y)exp(ln y) = 1:0068y2 ; 1 y 148:4fY (1) = 1:0068

fY (148:4) = 1:0068= (148:4)2 = 4: 5717 105:

The density functions are shown in Figures 42 and 43.

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0 1 2 3 4 5 6

0.05

0.10

0.15

0.20

x

f(x)

Figure 40: Density function fX (x) = 1=5; 0 x 5

0 20 40 60 80 100 120 1400.00

0.05

0.10

0.15

0.20

y

f(y)

Figure 41: Density function fY(y) =1

5y; 1 y 148:4

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0 1 2 3 4 50.0

0.2

0.4

0.6

0.8

1.0

x

f(x)

Figure 42: Density Function, fX (x) = 1:0068ex; 0 x 5

0 20 40 60 80 100 120 1400.0

0.2

0.4

0.6

0.8

1.0

y

f(y)

Figure 43: Density Function, fY(y) =1:0068

y2; 1 y 148:5

55


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6. Derive the probability density functions for Y; given Y = jXj ; and

fX (x) =

8>:

14

if 2 x 0;1

2exp(x) if x 0:

Solution: Performing the density transformation:

x1 = y; x2 = yg(x) = X

g0(x) = 1

fY (y) =fX (x1)

jg0 (x1)j +fX (x2)

jg0 (x2)j

= 14 + 12 exp(y) + 14 + 12 exp(y)

=1

2+

1

2(exp(y) + exp (y)) ; y > 0:

The plots for the density function are shown in Figures 44 and 45.

-3 -2 -1 0 1 2 3 4 5

0.2

0.4

0.6

x

f(x)

Figure 44: fX (x)

7. Given the following: Y = 3X

4 and fX (x) = 0:5; where

1

x

1;nd fY(y) and plot both density functions.

Solution: For any density function fX (x) ; the general transformationfor one root is:

fY (y) =fX (x1)

jg0 (x1)j ;

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0 1 2 3 4 5

10

20

30

40

50

60

70

y

f(y)

Figure 45: Density Function fY =12 +

12 (exp(x) + exp (x)) ; y > 0

If g(x) = 3X 4; then g0(x) = 3 and

fY (y) =0:5

3= 0:16667; 7 y 1:

The plots for the density functions are shown in Figures 46 and 47.

8. Given the following: Y = 3X 4 and fX(x) = N(0; 0:33); nd fY(y) andplot both density functions.

Solution: As X is governed by the Gaussian density, the probabilitydensity function is given by

fX (x) =1

p

2exp

(1

2

x

2); 1 < x < 1:

In this case

x1 = (y + 4)=3;

g(x) = 3x 4g0(x) = 3

fY (y) = 13

1p

2exp

0BBB@(Y + 4)

3

2

221CCCA ; over all x;

fY (y) =1

3

1

(0:33)p

2exp

(Y + 4)

2

18

!for = 0; = 1:

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-1.4 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

0.1

0.2

0.3

0.4

0.5

0.6

x

f(x)

Figure 46: Density Function fX =12 ; 1 x 1

-8 -7 -6 -5 -4 -3 -2 -1 0 1

0.05

0.10

0.15

0.20

x

y

Figure 47: Density Function fY =16 ; 7 x 1

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The density function plots are given in Figures 48 and 49.

-1.0 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1.0

-1.0

-0.5

0.5

1.0

x

y

Figure 48: Density Function fX (x) =1

(1)p

2exp

1

2

x1

2; 1 0 1:

9. Given the uid drag equation: FD = CDV2; where CD is a constant and

fV(v) = 0:1; for the range 10 v 20; nd fFD and plot both densityfunctions.

Solution: For this case there are two roots

v = pFD=CD; and dvdFD = 1

2pFDCD :So the general transformation is given by

fFD (FD) =1

2p

CDFD

"fV

rFDCD

!+ fV

r

FDCD

!#u (FD) ;

where u() is the unit step function. However, since v has a positive rangewe must drop the negative root. Therefore,

fFD (FD) =1

2p

CDFD

1

10

u (FD)

=

1

20pCDFD u (FD) :Suppose, CD = 2:0: Then,

fFD (FD) =1

20p

2FD; 200 FD 800:

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-14 -12 -10 -8 -6 -4 -2 0 2 4 6

0.02

0.04

0.06

0.08

0.10

0.12

x

y

Figure 49: Density Function fY (y) =1

3

1

(1)p

2exp

(y + 4)

2

18(1)2

!; 1 0

1

0 2 4 6 8 10 12 14 16 18 20 220.00

0.02

0.04

0.06

0.08

0.10

0.12

v

f(v)

Figure 50: Density Function fV(v) =110 ; 10 v 20

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0 2 4 6 8 10 12 14 16 18 20 220.000

0.005

0.010

0.015

Fd

f(Fd)

Figure 51: Density Function fFD =1

20p

2FD; 200 FD 800

The density plots are shown in Figures 50 and 51.

If 10 v 20 and 10 v 20; then fV(v) = 0:05 and we wouldretain both positive and negative roots.

10. For the same drag equation, FD = CDV2; V is standard normal N(0; 1):

Note that FD 0: Derive fFD and sketch both density functions.Solution:

fV (v) =1

p2exp

v2

2 :

The roots are v = p

FD=CD; and here, since V is governed by a Normaldensity, it can be positive or negative and we retain both roots. Thus,

dv

dFD= 1

2p

FDCD

fFD (FD) =1

2p

CDFD

"fV

rFDCD

!+ fV

r

FDCD

!#

=1p

2CDFDexp

FD

2CD

; FD > 0:

The density function fV(v) is shown in Figure 52.

For CD = 2:0; the density function fFD (FD) is shown in Figure 53.

11. For the function Y = a tan X; a > 0; derive the general relation for fY(y):Then, assume X is uniformly distributed over [; ] and nd fY(y): Thenplot density functions for X and Y:

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-1.0 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1.0

-1.0

-0.5

0.5

1.0

v

f(v)

Figure 52: Density Function fV (v) =1p2

exp

v

2

2

; 1 0 1

Solution: The general transformation is found as follows:

xn = arctan (y=a) ; n = :::; 1; 0; 1;:::

g0(x) =a

cos2 x=

a2 + y2

a

fY (y) =a

a2 + y2

1Xn=1

fX (xn) :

We have used the geometrical relation implied by the function y = a tan x:That is, tan x = y=a and therefore,

cos x = a=p

a2 + y2:

For the range [; ] ; x = arctan(y=a); there is one root. Then,

fY (y) =a

a2 + y2

1

2

=

a

2(a2 + y2):

See Figures 54 and 55.

12. For the function Z = XY ; nd fZ (z) for the cases:

(i) fXY (x; y) = [(b a)(d c)]1; a x b; c y d: Plot.(ii) fXY (x; y) = Cexp[(x + y)]; a x b; c y d: Plot.Solution: Using the general method, we dene a variable W;

W = X:

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0 1 2 3 4 5

0.05

0.10

0.15

0.20

0.25

0.30

Fd

f(Fd)

Figure 53: Density Function fFD (FD) =1p

2CDFDexp

FD

2CD

; FD > 0

Solving for X and Y,

X = W = g1

Y =Z

W= g2:

The Jacobian is given by

J = @g1=@z @g1=@w@g2=@z @g2=@w

=

0 11=W Z=W2 = 1W:

fXY (x; y) is dened on the rectangle dened by a < X < b and c < Y < d;Since X = W; we know a < W < b: The range for Z can be derived fromthat ofY: It is given that c < Y < d: Writing Y in terms of Z and W;

c < Y < d ! c < ZW

< d:

Solving for Z;cW < Z < dW:

Z ranges between two lines: Z = aW and Z = dW: Therefore, fW Z (w; z)is dened in a domain that resembles a quadrilateral, as shown in Figure56 and 57.

In order to obtain the marginal density, the joint distribution fZ (z) needsto be integrated over W: The range ofW depends on Z: Assuming ad < cb

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-1.4 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1.0 1.2 1.4

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

Figure 54: a tan X for a = 1

0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16

-5

-4

-3

-2

-1

0

1

2

3

4

5

x

y

Figure 55: fY (y) =a

2(a2 + y2)for a = 1

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a b

c

d

X

Y

Figure 56: Original Domain

Z

W

bd

da

ba

Z = cW

Z = dW

ca

cb

Figure 57: Transformed Domain

65


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as shown in Figure 57, W ranges on

a < W < Zc , ca < Z < ad

Z

d< W 0 and a1 > a22. a1a2 > 0 and a1 < a2

3. a1a2 < 0 and a1 > a2

4. a1a2 < 0 and a1 < a2:

Here, let us consider the case where a1 and a2 are of the same sign. Inaddition, let us assume that a1 > a2: Then,

FY (y) =

Z Za1~x1+a~x2


32/44


33/44

Solution:

18. Derive Equation 4.26.Solution:

19. Find the general expression for the rst order approximation for and for the following:

(i)

=P L

AE;

with all variables uncorrelated and

P = 1000 P = 10 L = 35 L = 15A = 0:1 A = 0:01 E = 10 106 E = 0:01 106:

(ii) solve (i) except that LA = 0:5 and all other correlations zero.(iii)

V = L3 + ar3;

where and a are constants and

L = 100 L = 5 r = 40 r = 2:

(iv) solve (iii) for the cases: Lr = 1; 1; 0: Discuss.Solution :

(i)

= g(P;L;A;E) =P L

AE:

The approximate mean is given by

' g(P; L; A; E )

+1

2

@2g

@P22P +

@2g

@L22L +

@2g

@A22A +

@2g

@E22E

+ 2@2g

@P@LP LPL + 2

@2g

@P@AP APA + 2

@2g

@P@EP E PE

+2@2g

@L@ALALA + 2

@2g

@L@ELELE + 2

@2g

@A@EAEAE

;

'

PL

AE+

1

20 + 0 + 2 PL

3AE2

A+ 2

PL

A3E2

E

+ 21

AEP LPL 2

L2AE

P APA 2L

A2E

P E PE

2 P2AE

LALA 2P

A2E

LELE + 2PL2A

2E

AEAE

;

74


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and variance is

V arfYg '

@g@P

22P +

@g@L

22L +

@g@E

22E +

@g@A

22A

+ 2@g

@P

@g

@LP LPL + 2

@g

@P

@g

@EP E PE

+ 2@g

@P

@g

@AP APA + 2

@g

@L

@g

@ELELE

+ 2@g

@L

@g

@ALALA + 2

@g

@E

@g

@AEAEA:

V arfg ' L

AE 2

2P + P

AE 2

2L + LP2AE

2

2A + LPA

2E

2

2E

+ 2LP2A

2E

P LPL 22LP3A

2E

P APA 22LP2A

3E

P E PE

2 L2P

3A2E

LALA 2L

2P

2A3E

LELE + 22L

2P

3A3E

AEAE:

The rst term approximation of the mean is

' g(P; L; A; E)=

PLAE

=1000(35)

0:1(10

106)

= 0:035:

As all the variables are uncorrelated, the variance is approximately givenby

V arfg '

LAE

22P +

P

AE

22L +

LP

2AE

22A +

LP

A2E

22E

=

35

0:1 (10 106)2

(10)2 +

1000

0:1 (10 106)2

(15)2

+

1000(35)

(0:1)2 (10 106)2

(0:01)2 +

1000(35)

0:1 (10 106)22

(0:01 106)2

= 2:3737 104

= 1:5407 102

:

(ii) When LA = 0:5 and all other correlations zero, the rst approxima-

75


35/44

tion of the mean remains the same whereas the variance is given by

V arfg =

LAE

2

2P +

PAE

2

2L +

LP2AE

2

2A +

LPA

2E

2

2E

2 L2P

3A2E

LALA

= 2:3737 104 2 35 10002

(0:1)3 (10 106)2 (0:5) (15) (0:01)

= 1:8487 104 = 1:3597 102:

(iii) Given:

V = L3 + ar3 = g(L; r):

The approximate mean is given by

v = g(L; r) +1

2

@2g

@L22L +

@2g

@r22r + 2

@2g

@L@rLrLr

= 3L + a3r +

1

2

6L

2L + 6ar

2r

:

The rst order approximation is

v = 3L + a

3r;

and the variance is given by

V arfYg '

@g

@L

22L +

@g

@r

22r + 2

@g

@L

@g

@rLrLr

=

32L2

2L +

3a2r2

2r + 2

32L

3a2r

LrLr:

(iv) The rst order approximation of the mean is

v = 3L + a

3r

= (100)3 + a(40)3

= 3:2056 106 for a = 1;

and the variance for a = 1 and Lr = 1; 1 and 0 are

V arfYgLr

=1 =

32L2 2L +

3a2r

2 2r + 2

32L

3a2r

LrLr

=

3 (100)22 (5)2 + 3 (40)22 (2)2+2 3 (100)2 3 (40)2 (1) (5) (2)

= 2:3121 1011;

76


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37/44

fore,

E(R) = 2

gsin2 + 12

2g

sin2 2 + 4g cos2

+1

2

4

2

gsin2

2

=3002

9:8sin(2

6) +

1

2

2

9:8sin

2

6

(35)2

+4 300

9:8cos

2

6

0:5

6

(0:10)

35

+1

2

4 300

2

9:8sin(2

6)

6

(0:10)2

= 7953:3 + 108:25 + 56:100 43:609= 8074:0:

In this case, the percent dierence is

% =8074:0 7953:3

8074:0 100 = 1:4949%:

The variance is given by

V arR =

2

gsin2

22 + 2

2

gsin2

2

2g

cos2

+

2

2g

cos2

22

=2 3009:8 sin(2 6 )

2

352

+2

2 300

9:8sin(2

6)

2 3002

9:8cos(2

6)

0:5

6

(0:10)

35

+

2 3002

9:8cos(2

6)

2 6

(0:10)2

= 3:4439 106 + 8:9236 105 + 2:3122 105= 4:5675 106

R =p

4:5675 106 = 2137:2

R1 =

p4:5675 106

8074:0= 0:26470

R2 = p4:5675 106

7953:3= 0:26872:

21. Obtain the mean and variance ofY as expanded in the Taylor series Equa-tion 4.27.

Solution:

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Table 2: Sample Realizations of V and M

pi Wi Vi = 30 25Wi Mi = 750 + (312:5)Wi0:42742 97:53 2408:31 29728:840:25981 91:83 2265:84 27947:980:95105 115:3 2853:40 35292:410:28134 92:57 2284:14 28176:740:13554 87:61 2160:21 26627:610:32111 93:92 2317:94 28599:290:31008 93:54 2308:57 28482:10:14649 87:98 2169:52 26743:960:79250 109:95 2718:63 33607:810:99164 116:72 2887:89 35723:68

Table 3: Sample Statistics for V and M

V (lb) M(lb-ft)

2437:44 30093:04 276:55 3456:86 0:1134 0:1148

26. For the cantilever beam of Figure 4.18, both parameters R and L arerandom variables. L is Gaussian with L = 25 ft and L = 0:1 ft; R is auniform random variable between the values 9; 800 10; 200 lb. Estimatethe mean and standard deviation of the shear V and bending moment M:

Solution: From Example 4.16, W is 2500 lb. The shear and moment atthe xed end are, respectively,

V = R W L= R 2500L lb;

M = RL + W L

L

2

= RL + 1250L2 lb-ft:

The shear and moment are functions of R and L both of which are randomvariables. R is a uniform random variable with lower limit a = 9; 800 lband the upper limit b = 10; 200 lb. L is a Gaussian with mean L = 25 ftand standard deviation L = 0:1 ft.

First, we can use Table of random uniform numbers to generate realiza-tions of R by using the relation Ri = a + (b a)pi: Ten values of Ri areshown in Table 4.

Next the realizations of L can be obtained in by one of the two followingmethods:

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Table 4: Sample Realizations of R

pi Ri0:42742 9470:80:25981 8803:10:95105 115570:28134 8888:90:13554 8308:00:32111 9047:30:31008 9003:40:14649 8351:60:79250 109250:99164 11719

Table 5: Sample Realizations of Lpi si Li = Li + Lisi

0:32111 0:67 24:9330:31008 0:65 24:9350:14649 1:11 24:8890:7925 0:73 25:073

0:99164 2:11 25:2110:62948 3:01 25:3010:55292 0:11 25:0110:8825 1:11 25:111

0:70974 0:44 25:0440:12102 1:21 24:879

Method 1: Using random numbers from the Standard Uniform Table4.2.

Each value of pi is used as the entry point into the standard normal tableto nd the corresponding approximate value of si. Once si is found, therealizations of L can be obtained by using the relation Li = Li + Lisi.The values are tabulated in Table 5.

Once both Ri and Li are found, the realizations of mean and varianceare found using

Vi = Ri 2500Li lb;Mi = RiLi + 1250L2i lb-ft:

These are given in the Table 6.

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Ri Li Vi = Ri 2500Li Mi = RiLi + 1250L2i

9470:8 24:933 52861:7 540932:658803:1 24:935 53534:4 557687:4811557 24:889 50665:5 486685:738888:9 25:073 53793:6 562947:778308:0 25:211 54719:5 585040:169047:3 25:301 54205:2 571270:019003:4 25:011 53524:1 556753:618351:6 25:111 54425:9 578485:8710925 25:044 51685 510396:7211719 24:879 50478:5 482148:80

Table 7: Sample Statistics for V and MV (lb) M(lb-ft)

52989:34 543234:88 1534:32 37371:22 0:029 0:0688

These mean and the variance values can be averaged using

x

=1

n

nXi=1

xi

x =

vuut

n

Xi=1

(xi x)2

n1

;

to estimate their respective means and standard deviations, shown in Table7.

Method 2: Use the standard normal random numbers (Table 4.6).

Pick 10 random numbers from Table 4.6. These are the standard normalvariables si. Once si is found, the realizations of L can be obtained byusing the relation

Li = Li + Lisi:

The values are tabulated in Table 8.

Then the standard procedure is followed. The realizations of mean and

variance are found using

Vi = Ri 2500Li lb;Mi = RiLi + 1250L2i lb-ft:

These are given in the Table 9

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Table 8: Sample Realizations of Lsi Li = Li + Lisi

0:58885 25:061:3286 24:87

4:1952 102 24:10:46021 24:950:48187 24:95

1:5813 25:160:77828 25:080:52954 25:05

0:50652 24:95

0:23290 24:98


Ri Li Vi = Ri 2500Li Mi = RiLi + 1250L2i9470:8 25:06 53176:41 547606:968803:1 24:87 53364:75 554060:3911557 24:1 50932:51 492111:318888:9 24:95

53496:05 556562:91

8308:0 24:95 54071:53 570941:559047:3 25:16 53848:03 563551:239003:4 25:08 53691:17 560336:108351:6 25:05 54280:79 575330:8810925 24:95 51448:37 505515:8311719 24:98 50722:78 487092:99:

Table 10: Sample Statistics for V and MV (lb) M(lb-ft)

52903:24 541311:02 1340:39 33280:30 0:0253 0:0615

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The respective means and standard deviations are listed in Table 10.

27. Draw the density function Equation 4.38. Generate an additional threevalues of xi:

Solution:

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