U5 - Problem 7
We know that ambient temperature and wind speed are important in determining the current carrying
capability of a transmission line. A transmission line owner wants to maximize its sale of capacity for a
particular line in order to maximize its profits. However, it must decide how much capacity to sell on a dayahead basis. Therefore it must predict temperature and wind. Let's denote the event "warm day" as event A
and the event "windy day" as event B. The weather forecast indicates that the probability of a windy day is
0.80, the probability of a warm day is 0.30, and the probability of a warm and windy day is 0.28
(a)
(b)
(c)
(d)
(e)
(f)
Are the events A and B independent?
What is the probability of the event "windy day will be warm"?
What is the probability of the event "warm day will be windy"?
What is the probability of the event "windy and not warm"?
What is the probability of the event "windy day will be not warm"?
What is the probability of the event "not warm day will be windy"?
Solution to problem 7
(a) No, since PA  B =0.28 but PA PB =0.8*0.3=0.24
(b) PA / B = PA  B / PB =0.28/0.80=0.35
(c) PB / A = PA  B / PA =0.28/0.30=0.93
(d) Denote the event "not warm" as a. Then Pa  B  PA  B  PBsince A and a are complements.
Therefore, Pa  B = PB - PA  B =0.80-0.28=0.52
(e) Pa / B = Pa  B / PB =0.52/0.8=0.65
(f) PB / a = Pa  B / Pa =0.52/0.7=0.74
U5 - Problem 8
Of great interest to the electric power industry today is the detection of the "hot spot" temperature in power
transformers. This is the temperature of the hottest part of the winding and is a function of many variables
including loading and ambient temperature. A simplified approach assumes that if the hot spot temperature
exceeds a certain level, the transformer will fail in one hour; otherwise it will not fail. The hot spot
temperature may be approximated if both the oil temperature and the loading current are known.
We desire to predict transformer failure for a particular, rather low, loading current I. Let's denote the event
"transformer failure" to be A. Let's also denote the event "oil temperature test indicates eminent failure" to
be B. It is known that at the loading level I, PA =0.003 and that the oil temperature test has 2% false
positives and 1% false negatives.
What is Pa ? (a is the complement of A)
What is PB / A ?
What is PB / a ?
What is PB  A ?
What is PB  a ?
What is PB ?
If, at this loading current I, the oil temperature test indicates the transformer will fail,
(i) What is the probability that the transformer will fail ?
(ii) What is the probability that the transformer will not fail ?
(h) Consider that the event A is "having cancer" and the event B is "positive test for cancer". What does
the above result suggest about screening tests for diseases with low prevalence?
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Solution to problem 8
I provide two answers below, depending on whether you though a "positive" test indicated "good"
transformer or failed transformer. I will accept either answer.
Solution 1 : If you assumed that a "positive" test indicates that the transformer is "good" and a "negative"
test indicates that the transformer is failed, then
(a) Pa =1-0.003=0.997
(b) PB / A =1-0.01=0.99
(c) PB / a =0.02
(d) PB  A = PA PB / A =0.003*0.98=0.00294
(e) PB  a = Pa PB / a =0.997*0.02=0.01994
(f) PB = PB  A + PB  a =0.02291
(g) Given B, we use
PA / B = PB  A / PB =0.00297/0.02291=0.13
Pa / B = PB  a / PB =0.01994/0.02291=0.87=1- PA / B
Note that each of the above two probabilities could have been computed using Bayes Theorem, e.g.,
PA / B = PB  A /  PB / Ai PAi  where in this case, A1 =A and A2 =a
2
i 1
(h) it suggests that if I am in a population where cancer occurs infrequently (e.g., person under 40), and if
a reliable, but imperfect test indicates I have cancer, the probability that I actually do have cancer
(13%) is still in my favor (i.e., there is 87% chance I do not have cancer ).
Solution 2 : If you assumed that a "positive" test indicates that the transformer is "good" and a "negative"
test indicates that the transformer is failed, then
(i) Pa =1-0.003=0.997
(j) PB / A =1-0.02=0.98
(k) PB / a =0.01
(l) PB  A = PA PB / A =0.003*0.98=0.00294
(m) PB  a = Pa PB / a =0.997*0.01=0.00997
(n) PB = PB  A + PB  a =0.01291
(o) Given B, we use
PA / B = PB  A / PB =0.00294/0.01291=0.2277
Pa / B = PB  a / PB =0.0997/0.01291=0.7723=1- PA / B
Note that each of the above two probabilities could have been computed using Bayes Theorem, e.g.,
PA / B = PB  A /  PB / Ai PAi  where in this case, A1 =A and A2 =a
2
i 1
(h) it suggests that if I am in a population where cancer occurs infrequently (e.g., person under 40), and if
a reliable, but imperfect test indicates I have cancer, the probability that I actually do have cancer
(22.77%) is still in my favor (i.e., there is 77.23% chance I do not have cancer ).
U6 - Problem 1
An engineer at a large industrial is looking for an energy seller to supply its electric energy needs. He has
the names of five energy sellers, A, B, C, D, and E, and email addresses. He wants to know which ones
have load following capability, so he sends them a message.
A and B are the only ones that have load following capability. If we assume that the order of reply is
random, then we can define the number of replies necessary before a load following supplier is identified a
the random variable Y, e.g., y=1 indicates that the first supplier to reply is a load following generator.
(a) Find and plot the probability mass function for Y. Hint: Evaluate fY  y , y  1... 5 one at a time. The
multiplication rule may help.
(b) Find and plot the cumulative distribution function for Y.
Solution to problem 1
Note that Y is the RV associated with the first “positive” request.
(a)

fY 1  prob A or B

 2
first    0.4
 5





 2
fY 2  prob C , D, or E 1st  AND A or B 2nd    0.4
5






X
X
 P X 1  X 2   P X 1   P X 2 / X 1 
1
2
3
 3  2 6
    

 0.3
5
4
20
10
   






fY 3  prob C , D, or E 1st or 2nd  AND A or B 3rd 








X
X
 P  X 3  X 4   P  X 3   P X 4 / X 3 
3
4
 3  2  2 2
     
 0.2
 5   4   3  10


fY 4   probC , D, E all in first 3


 3  2 1 1
     
 0.1
 5   4   3  10
(b)
U7 - Problem 1
Suppose that the temperature T of a certain conductor has a uniform pdf in the range 40 to 60 degrees C.
Then compute:
(a) PT  50
(b) P45  T  55
(c) P55  T  70
Solution to problem 1
PDF
 20  k  1
1
k
 0.05
20
(a) PT  50  10 0.05  0.5
(b) P45  T  55  10 0.05  0.5
(c) P55  T  70  5 0.05  0.25
U7 - Problem 2
Let x denote the vibratory stress (psi) on a wind turbine blade for a particular wind speed y. Assume that
the pdf for X is given by
 x2 





x
 e 2 y 2  , x  0
f X  x    y 2 

 0,
o.w.
Note that y is just a constant.
(a) Verify that f X x  is a legitimate pdf
(b) Suppose y=100. What is the probability that x is
(i) less than 200 ?
(ii) less than or equal to 200 ?
(iii) Greater than 200 ?
(c) What is the probability that x is between 100 and 200 ?
(d) Give an expression for PX  x
Solution to problem 2
(a) For f(x) to be a legitimate pdf, we must have
(i) f x   0
(ii)



0
 f x dx  1.0  
let u  
 x2
x 2 y2
e dx
y2
x2
 2x  x
 du 
 2
2y2
2y2
y


x2 
 x2 



 2 y2 
  x    2 y 2 
    2 e 
dx   e  
y 
0
0
 e
(b) (i)
 2

 2 y2





e
PX  200  
 02

 2 y2

200




 0  1  1
f x dx 

200

0

x   x
e
y2
2


2 y2 
dx
  200 22 



 e
  e  2 100    e 0 




0
 0.1353  1  0.8647
 x2



2 y2 

200
(ii) PX  200  0.8647
(iii) PX  200  1  PX  200  0.1353
(c) P100  x  200  
200

f x dx 
100
 e
  x2



2 y2 

200
 e
 200 2

 2 100 2






x   x
 y2 e
100
200
e
 100 2

 2 100 2





2


2 y2 
dx
 0.1353  0.6065   0.4712
100
(d) for x  0, F(x)=0
(e) for
x  0, F  x    f  d   e
x

 2



2 y2 

x
 e
 x2
2 y2
e
 2
2 y2
 1 e
 x2

U8 – Problem 9
1.
Two random variable, X and Y, are described by the following bi-variate probability density function:
fXY(x,y) =6x2y 0.8736<x<1, 1<y<2
=0
otherwise
a. Prove that this function is a valid pdf.
b. Find both marginal distributions.
c. Find the conditional pdf, fX(x|y).
d. Compute Prob[X<0.9, Y< 1.3].
e. Prove that X and Y are dependent, or prove that they are independent.
f. Compute the mean value of X, X.
Solution to problem 9
(a)
(i)
f X ,Y  x, y   0
x, y
2 y2
(ii)
 
1 
2 
 
  f x, y dxdy    f x, y dxdy    f x, y dxdy    f x, y dxdy
X ,Y
X ,Y
  
X ,Y
  
X ,Y
1 
2 
2 

 f x, y dxdy
X ,Y
1 
2 0.8736
1



    f X ,Y  x, y dxdy   f X ,Y  x, y dxdy   f X ,Y  x, y dxdy 
1  
0.8736
1

2

1
2
1
 f x, y dxdy    6 x
X ,Y
1  0.8736
1  0.8736
2
y 2
  0.6666 ydy  0.3333 y 2
y 1
2
2

ydxdy   2 x 3 y

x 1
x  0.8736
1
 0.33334  1  1
1
So, f X ,Y  x, y  is a valid pdf.
(b)
f X x  

1
f X ,Y x, y dy 


2


2
2

1
2
f X ,Y  x, y dy   f X ,Y  x, y dy   f X ,Y x, y dy
f X x    f X ,Y x, y dy   6 x 2 ydy  3x 2 y 2
1
9 x
f X x   
 0
0 .8736  x  1


Otherwise
f X ,Y x, y dx 
0.8736

fY  y 
 9x 2
y 1
1
2
fY  y 
y 2

f X ,Y x, y dx 

1
1

0.8736
1
 f x, y dx   6 x
X ,Y
0.8736

f X ,Y x, y dx   f X ,Y x, y dx
2
ydx  2 x 3 y 2
x 1
x  0.8736
1
 0.6666 y
0.8736
0.6666 y 1  y  2
fY  y  
Otherwise
 0
(c)
(d)
f X x / y  
f X ,Y  x, y 
fY  y
P X  0.9, Y  1.3 
 6x 2 y

 9x 2
  0.6666 y

0
1.3 0.9
  f x, y dxdy
X ,Y
  
0.8736  x  1
1 y  2
Otherwise
dy
P X  0.9, Y  1.3 
1 0. 9
1. 3 0. 9
  
1 

 0dxdy 
  f x, y dxdy
X ,Y
1. 3 0. 9

  f x, y dxdy
X ,Y
1 
1.3 0.8736
0.9


    0dx   f X ,Y  x, y dx dy
1  
0.8736

1.3 0.9

 
f X ,Y x, y dxdy 
1.3 0.9
  6x
1 0.8736
ydxdy
1 0.8736
1. 3
  2x y
3
1.3
x  0.9
x  0.8736
dy   0.1246 ydy  0.0623 y 2
1
y 1.3
y 1
1
PX  0.9, Y  1.3  0.04298
(e)
2
According to the results obtained in (b) and (c):
9 x 2
0.8736  x  1
1 y  2
f X x   
Otherwise
0
9 x 2
f X x / y   
0
0.8736  x  1
Otherwise
1 y  2
f X x  f X x / y   X and Y are independent
(f)
 X  EX mY n 
X 
 
m  1, n  0
2
1
  xf x, y dxdy    x6 x
X ,Y
  
2
2
2
ydxdy   1.5 x 4 y
1 0.8736
  0.6263 ydy  0.3132 y 2
y 2
y 1
x 1
x  08736
dy
1
 0.9395
1
U9 - Problem 3
George wants to buy electric energy. He can buy from company A, B, C, or D, but he must choose only
one. George obtains past data on these four suppliers, and he then determines that the prices for some of
them have been quite volatile, influenced by the state of the economy. George assesses this data and then
forms the table below.
Costs ($/MW-hr)
Energy seller
Recession
Stagnation
Inflation
A
30
60
90
B
10
40
70
C
20
20
50
D
45
45
45
Probability of economic state

0.1
0.4
0.5
a)
Assuming that George simply wants to minimize his average costs over the long run,
determine the expected costs for all energy suppliers. Select the supplier based on minimum
expected cost.
b) Assuming that George cannot afford any surprises and therefore wants to minimize his
uncertainty, determine the standard deviation for each supplier. Select the supplier based on
minimum variance (same as minimum standard deviation).
Solution to problem 3
Energy seller
Recession
Stagnation
Inflation
Ex (cost)
x
A
B
C
D
Probability of economic state
30
10
20
45
0.1
60
40
20
45
0.4
90
70
50
45
0.5
72
52
35
45
19.9
19.9
15
0
Sample calculation for Ex(cost)
Seller A: (0.1)(30) + (0.4)(60)+(0.5)(90)=72
Sample calculation for VAR(cost)
Seller A: 0.130  72   0.460  72   0.590  72   396
2
  x  396  19.9
(a) Pick Seller C
(b) Pick Seller D
2
2