Mohammed Jahanzeb Choudary ipment for finding the wave height and wind speed. The City College of
Figure Anemometer Figure WaveRider Bouy
City University of New York lculation to find the future estimation of the location's behavior and characteristics,
9.15
Assume, First measurement L=N(2.15, )
Second measurement L=N(2.20, 2)
and Third measurement L=N(2.18, 3)
Applying Eqs. 9.14 and 9.15 and considering First and Second measurements,
L' ' =
2.20 2 + 2.15 (2 ) 2
= 2.16km
2 + (2 ) 2
and,
'
L' =
2 (2 )
9.17
is the parameter to be updated with the prior distribution parameter
'
= 3
'
= 0.2 3 = 0.6
(a) From the observation we get
t = N(
2 + 3 0 .5
,
) = N ( 2.5, 0.354)
2
2
From the relationship given in Table 9.1,
0.354 2 3 + 0.62 2.5
=
= 2.629
0.62 +
9.18
(a) The parameter needs to be updated is p. Its suitable to set a conjugate prior for the
problem. Thus, suppose p is Beta distributed with parameter p and q. With the information
presented in the problem, we can get
E '( p ) =
q'
q'r'
= 0.5 Var '( p
9.16
Assume the prior distribution of p to be a Beta distribution, then,
E ' ( p) =
q'
= 0.1 - (1)
q'+ r '
and
Var ' ( p) =
a' r '
= 0.06 2 -(2)
2
(q'+ r ' ) (q '+ r '+1)
From Equations 1 and 2, we get,
q = 2.4 and r = 9 2.4 = 21.6
From Table 9.1, for a b
9.19
Using conjugate distributions, we assume Gamma distribution as prior distribution of .
From Table 9.1, for an exponential basic random variable,
E ' ( ) =
k'
= 0.5;
v'
Var ' ( ) =
k ' 0 .5
=
= (0.5 0.2) 2
2
v'
v'
So, v = 50
and, k = 25
Now, v = v +
x
9.20
is the parameter to be updated. Let X denote the crack length.
The probability of crack length larger than 4 is:
P( X > 4) = 1 P ( X < 4) = e 4
The probability of crack length smaller than 6 is:
P ( X < 6 ) = 1 e 6
Likelihood function is
L( ) = e
9.21
(a) The occurrence rate of the tornado is estimated from the historical record as 0.1/year.
The probability of x occurrences during time t (in years) is
P( X = x ) =
(0.1t ) x 0.1t
e
x!
The probability that the tornado hits the town during the next 5
9.22
h is the parameter to be updated.
(a)
10
10
0
0
E ( x ) = E ( x | h ) f ( h )dh = 0.5hi0.003h 2 = 3.75
where
h
h
0
0
E ( x | h ) = xf ( x | h )dx =
x
dh = 0.5h
h
(b)
(i) the range of h is 4<h<10 upon the observation
(ii) likelihood:
1
h
L( h | x = 4
9.23
(a) Let x denote the fraction of grouted length. Assumptions: the grouted length is normally
distributed; the mean grouted length is normally distributed; the variance of fraction of length
grouted to the constant and can be approximated by the obser
9.24
(1) Prior
Non-informative prior is used as (P.M. Lee. Bayesian Statistics: An Introduction. Edward
Arnold, 1989)
2
f '( T , T )
1
2
T
(2) Likelihood
, tn | T , T )
p(t1 , t2 ,
(t1 T ) 2
(t2 T ) 2
(tn T ) 2
1
1
=
exp
i
exp
ii
exp
2
2
2
2 T 2
9.25
Let x denotes the rated value, and y denotes the actual value.
2
Based on 9.24 to 9.27, = 0.512 , = 0.751 , 2 = 1.732 , s x = 34.167
From equation 9.26, E (Y | x ) = 0.512 + 0.751x
The variance can be calculated by equation 9.34 as
Var (Y | x ) =
6 1
9.13
(a) T is N( , 10)
Sample mean = t = 65 min.
n=5
So the posterior distribution of ,
f ' ' ( ) = N (t ,
) = N (65,
10
n
) = N (65,4.47) min.
5
(b) We know,
f ( ) = N(65,4.47) min.
L( ) = N (60,
10
) = N(60, 3.16) min.
10
From Equation 8.14 and 8.15,
'
9.14
(a) The sample mean =
s =
2
1
(32 o 04'+31o 59'+32 o 01'+32 o 05'+31o 57'+32 o 00' ) = 32 o 01'
6
1
cfw_(3' ) 2 + (2' ) 2 + 0 + (4' ) 2 + (4' ) 2 + (1' ) 2 = 9.2
6 1
s = 3.03' = 0.05 o
The actual value of the angle is N (32 o 01' ,
0.05 o
)
6
or,
N(
9.10
is the parameter to be updated with the prior distribution of:
P '( = 1 5) = 1 / 3
P '( = 1 10) = 2 / 3
(a) Let =accidents were reported on days 2 and 5. The probability to observe is
P( ) = P( | = 1/ 5) P '( = 1/ 5) + P( | = 1/10) P '( = 1 /10)
1
2
9.3
P(HAHF) = P(HALF) = 0.3, P(LAHF) = P(LALF) = 0.2
(a) P( H A H F ) = P( H A H F HAHF) P(HAHF) + P( H A H F HALF) P(HALF) + P( H A H F LAHF)
P(LAHF) + P( H A H F LALF) P(LALF)
= 0.30.3 + 0.40.3 + 0.20.2 + 0.250.2
= 0.3
(b) P(HAHF H A H F ) = P( H A H F
9.7
Since,
where
f ( ) = kL( ) f ( )
L( ) =
e
12
(
12
1!
)1
, f ' ( ) =
0.271
; 0.5 20
= 0, elsewhere
So,
f ( ) = k e
12
12
0.271
;
0.5 20
= 0, elsewhere
To determine the constant k,
20
0.271 12
k
12 e d = 1.0 k = 4.79
0.5
So the posterior distribution o
9.8
(a) P(p > 0.9) = 0.7 and P(p > 0.9) = 0.3
(b) f (p) = k L(p) f (p)
3
3
L(p) = ( ) p3 (1-p)0 = p3
0 p 0.9
0.9 p 1.0
1
3
8
6
4
To find k,
0.9
10
f"(p)
So, f (p) = k p3 1/3;
= k p3 7;
=0, elsewhere
12
1.0
0
2
0.9
k p dp +
3
0
3
k 7 p d p = 1 .0
0
or,
p
9.12
Let X = compression index
X is N( , 0.16)
Sample mean x =
1
(0.75 + 0.89 + 0.91 + 0.81) = 0.84
4
(a) From Equation 8.13,
f ( ) = N ( x,
)
n
= N(0.84, 0.16/2) = N(0.84, 0.08)
'
'
(b) f ( ) = N( , ) = N(0.8, 0.2)
L( ) = N(0.84, 0.08) = N ( x,
)
n
From
9.11
M is the parameter to be updated. Its convenient to prescribe a conjugate prior to the Poisson
process. From the information given in the problem, the mean and variance of the gamma
distribution of M is:
E '( ) =
k'
k '/ v '2
= 10 '( ) =
= 0 .4
v'
k
Set-1
CE26400
Quiz 2
Name: _
1. Suppose an offshore platform is designed against the 100-year wave (i.e. a
wave height corresponding to a return period of 100 years) but is intended to
operate for 25 years only. What is the probability that it will be sub