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Problems on Mathematical Expectation

Module by: Paul E Pfeiffer. E-mail the author

Exercise 1

(See Exercise 1 from "Problems on Distribution and Density Functions", m-file npr07_01.m). The class {Cj:1≤j≤10}{Cj:1≤j≤10} is a partition. Random variable X has values {1,3,2,3,4,2,1,3,5,2}{1,3,2,3,4,2,1,3,5,2} on C1 through C10, respectively, with probabilities 0.08, 0.13, 0.06, 0.09, 0.14, 0.11, 0.12, 0.07, 0.11, 0.09. Determine E[X]E[X].

Solution

% file npr07_01.m
% Data for Exercise 1 from "Problems on Distribution and Density Functions"
T  = [1 3 2 3 4 2 1 3 5 2];
pc = 0.01*[ 8 13  6  9 14 11 12  7 11  9];
disp('Data are in T and pc')
npr07_01
Data are in T and pc
EX = T*pc'
EX =  2.7000
[X,PX] = csort(T,pc);  % Alternate using X, PX
ex = X*PX'
ex =  2.7000

Exercise 2

(See Exercise 2 from "Problems on Distribution and Density Functions", m-file npr07_02.m ). A store has eight items for sale. The prices are $3.50, $5.00, $3.50, $7.50, $5.00, $5.00, $3.50, and $7.50, respectively. A customer comes in. She purchases one of the items with probabilities 0.10, 0.15, 0.15, 0.20, 0.10 0.05, 0.10 0.15. The random variable expressing the amount of her purchase may be written

X = 3 . 5 I C 1 + 5 . 0 I C 2 + 3 . 5 I C 3 + 7 . 5 I C 4 + 5 . 0 I C 5 + 5 . 0 I C 6 + 3 . 5 I C 7 + 7 . 5 I C 8 X = 3 . 5 I C 1 + 5 . 0 I C 2 + 3 . 5 I C 3 + 7 . 5 I C 4 + 5 . 0 I C 5 + 5 . 0 I C 6 + 3 . 5 I C 7 + 7 . 5 I C 8
(1)

Determine the expection E[X]E[X] of the value of her purchase.

Solution

% file npr07_02.m
% Data for Exercise 2 from "Problems on Distribution and Density Functions"
T = [3.5 5.0 3.5 7.5 5.0 5.0 3.5 7.5];
pc = 0.01*[10 15 15 20 10  5 10 15];
disp('Data are in T, pc')
npr07_02
Data are in T, pc
EX = T*pc'
EX =  5.3500
[X,PX] = csort(T,pc);
ex = X*PX'
ex =  5.3500

Exercise 3

(See Exercise 12 from "Problems on Random Variables and Probabilities", and Exercise 3 from "Problems on Distribution and Density Functions," m-file npr06_12.m). The class {A,B,C,D}{A,B,C,D} has minterm probabilities

p m = 0 . 001 * [ 5 7 6 8 9 14 22 33 21 32 50 75 86 129 201 302 ] p m = 0 . 001 * [ 5 7 6 8 9 14 22 33 21 32 50 75 86 129 201 302 ]
(2)

Determine the mathematical expection for the random variable X=IA+IB+IC+IDX=IA+IB+IC+ID, which counts the number of the events which occur on a trial.

Solution

% file npr06_12.m
% Data for Exercise 12 from "Problems on Random Variables and Probabilities"
pm = 0.001*[5 7 6 8 9 14 22 33 21 32 50 75 86 129 201 302];
c = [1 1 1 1 0];
disp('Minterm probabilities in pm, coefficients in c')
npr06_12
Minterm probabilities in pm, coefficients in c
canonic
 Enter row vector of coefficients  c
 Enter row vector of minterm probabilities  pm
Use row matrices X and PX for calculations
Call for XDBN to view the distribution
EX = X*PX'
EX =  2.9890
T = sum(mintable(4));
[x,px] = csort(T,pm);
ex = x*px
ex =  2.9890

Exercise 4

(See Exercise 5 from "Problems on Distribution and Density Functions"). In a thunderstorm in a national park there are 127 lightning strikes. Experience shows that the probability of of a lightning strike starting a fire is about 0.0083. Determine the expected number of fires.

Solution

X∼X∼ binomial (127, 0.0083). E[X]=127·0.0083=1.0541E[X]=127·0.0083=1.0541

Exercise 5

(See Exercise 8 from "Problems on Distribution and Density Functions"). Two coins are flipped twenty times. Let X be the number of matches (both heads or both tails). Determine E[X]E[X].

Solution

X∼X∼ binomial (20, 1/2). E[X]=20·0.5=10E[X]=20·0.5=10.

Exercise 6

(See Exercise 12 from "Problems on Distribution and Density Functions"). A residential College plans to raise money by selling “chances” on a board. Fifty chances are sold. A player pays $10 to play; he or she wins $30 with probability p=0.2p=0.2. The profit to the College is

X = 50 · 10 - 30 N , where N is the number of winners X = 50 · 10 - 30 N , where N is the number of winners
(3)

Determine the expected profit E[X]E[X].

Solution

N∼N∼ binomial (50, 0.2). E[N]=50·0.2=10E[N]=50·0.2=10. E[X]=500-30E[N]=200E[X]=500-30E[N]=200.

Exercise 7

(See Exercise 19 from "Problems on Distribution and Density Functions"). The number of noise pulses arriving on a power circuit in an hour is a random quantity having Poisson (7) distribution. What is the expected number of pulses in an hour?

Solution

X∼X∼ Poisson (7). E[X]=7E[X]=7.

Exercise 8

(See Exercise 24 and Exercise 25 from "Problems on Distribution and Density Functions"). The total operating time for the units in Exercise 24 is a random variable T∼T∼ gamma (20, 0.0002). What is the expected operating time?

Solution

X∼X∼ gamma (20, 0.0002). E[X]=20/0.0002=100,000E[X]=20/0.0002=100,000.

Exercise 9

(See Exercise 41 from "Problems on Distribution and Density Functions"). Random variable X has density function

f X ( t ) = ( 6 / 5 ) t 2 for 0 ≤ t ≤ 1 ( 6 / 5 ) ( 2 - t ) for 1 < t ≤ 2 = I [ 0 , 1 ] ( t ) 6 5 t 2 + I ( 1 , 2 ] ( t ) 6 5 ( 2 - t ) f X ( t ) = ( 6 / 5 ) t 2 for 0 ≤ t ≤ 1 ( 6 / 5 ) ( 2 - t ) for 1 < t ≤ 2 = I [ 0 , 1 ] ( t ) 6 5 t 2 + I ( 1 , 2 ] ( t ) 6 5 ( 2 - t )
(4)

What is the expected value E[X]E[X]?

Solution

E [ X ] = ∫ t f X ( t ) d t = 6 5 ∫ 0 1 t 3 d t + 6 5 ∫ 1 2 ( 2 t - t 2 ) d t = 11 10 E [ X ] = ∫ t f X ( t ) d t = 6 5 ∫ 0 1 t 3 d t + 6 5 ∫ 1 2 ( 2 t - t 2 ) d t = 11 10
(5)

Exercise 10

Truncated exponential. Suppose X∼X∼ exponential (λ)(λ) and Y=I[0,a](X)X+I(a,∞)(X)aY=I[0,a](X)X+I(a,∞)(X)a.

  1. Use the fact that
    ∫0∞te-λtdt=1λ2and∫a∞te-λtdt=1λ2e-λa(1+λa)∫0∞te-λtdt=1λ2and∫a∞te-λtdt=1λ2e-λa(1+λa)
    (6)
    to determine an expression for E[Y]E[Y].
  2. Use the approximation method, with λ=1/50,a=30λ=1/50,a=30. Approximate the exponential at 10,000 points for 0≤t≤10000≤t≤1000. Compare the approximate result with the theoretical result of part (a).

Solution

E [ Y ] = ∫ g ( t ) f X ( t ) d t = ∫ 0 a t λ e - λ t d t + a P ( X > a ) = E [ Y ] = ∫ g ( t ) f X ( t ) d t = ∫ 0 a t λ e - λ t d t + a P ( X > a ) =
(7)
λ λ 2 [ 1 - e - λ a ( 1 + λ a ) ] + a e - λ a = 1 λ ( 1 - e - λ a ) λ λ 2 [ 1 - e - λ a ( 1 + λ a ) ] + a e - λ a = 1 λ ( 1 - e - λ a )
(8)
tappr
Enter matrix [a b] of x-range endpoints  [0 1000]
Enter number of x approximation points  10000
Enter density as a function of t  (1/50)*exp(-t/50)
Use row matrices X and PX as in the simple case
G = X.*(X<=30) + 30*(X>30);
EZ = G8PX'
EZ = 22.5594
ez = 50*(1 - exp(-30/50))    % Theoretical value
ez = 22.5594

Exercise 11

(See Exercise 1 from "Problems On Random Vectors and Joint Distributions", m-file npr08_01.m). Two cards are selected at random, without replacement, from a standard deck. Let X be the number of aces and Y be the number of spades. Under the usual assumptions, determine the joint distribution. Determine E[X]E[X], E[Y]E[Y], E[X2]E[X2], E[Y2]E[Y2], and E[XY]E[XY].

Solution

npr08_01
Data in Pn, P, X, Y
jcalc
Enter JOINT PROBABILITIES (as on the plane)  P
Enter row matrix of VALUES of X  X
Enter row matrix of VALUES of Y  Y
 Use array operations on matrices X, Y, PX, PY, t, u, and P
EX = X*PX'
EX =  0.1538
 
ex = total(t.*P)             % Alternate
ex =  0.1538
EY = Y*PY'
EY =  0.5000
EX2 = (X.^2)*PX'
EX2 =  0.1629
EY2 = (Y.^2)*PY'
EY2 =  0.6176
EXY = total(t.*u.*P)
EXY =  0.0769

Exercise 12

(See Exercise 2 from "Problems On Random Vectors and Joint Distributions", m-file npr08_02.m ). Two positions for campus jobs are open. Two sophomores, three juniors, and three seniors apply. It is decided to select two at random (each possible pair equally likely). Let X be the number of sophomores and Y be the number of juniors who are selected. Determine the joint distribution for {X,Y}{X,Y} and E[X]E[X], E[Y]E[Y], E[X2]E[X2], E[Y2]E[Y2], and E[XY]E[XY].

Solution

npr08_02
Data are in X, Y,Pn, P
jcalc
- - - - - - - - - - - -
EX = X*PX'
EX =  0.5000
EY = Y*PY'
EY =  0.7500
EX2 = (X.^2)*PX'
EX2 = 0.5714
EY2 = (Y.^2)*PY'
EY2 = 0.9643
EXY = total(t.*u.*P)
EXY = 0.2143

Exercise 13

(See Exercise 3 from "Problems On Random Vectors and Joint Distributions", m-file npr08_03.m ). A die is rolled. Let X be the number of spots that turn up. A coin is flipped X times. Let Y be the number of heads that turn up. Determine the joint distribution for the pair {X,Y}{X,Y}. Assume P(X=k)=1/6P(X=k)=1/6 for 1≤k≤61≤k≤6 and for each k, P(Y=j|X=k)P(Y=j|X=k) has the binomial (k,1/2)(k,1/2) distribution. Arrange the joint matrix as on the plane, with values of Y increasing upward. Determine the expected value E[Y]E[Y].

Solution

npr08_03
Answers are in X, Y, P, PY
jcalc
- - - - - - - - - - - -
EX = X*PX'
EX =   3.5000
EY = Y*PY'
EY =   1.7500
EX2 = (X.^2)*PX'
EX2 = 15.1667
EY2 = (Y.^2)*PY'
EY2 =  4.6667
EXY = total(t.*u.*P)
EXY =  7.5833

Exercise 14

(See Exercise 4 from "Problems On Random Vectors and Joint Distributions", m-file npr08_04.m ). As a variation of Exercise 13, suppose a pair of dice is rolled instead of a single die. Determine the joint distribution for {X,Y}{X,Y} and determine E[Y]E[Y].

Solution

npr08_04
Answers are in X, Y, P
jcalc
- - - - - - - - - - - -
EX = X*PX'
EX =   7
EY = Y*PY'
EY =   3.5000
EX2 = (X.^2)*PX'
EX2 = 54.8333
EY2 = (Y.^2)*PY'
EY2 =  15.4583

Exercise 15

(See Exercise 5 from "Problems On Random Vectors and Joint Distributions", m-file npr08_05.m). Suppose a pair of dice is rolled. Let X be the total number of spots which turn up. Roll the pair an additional X times. Let Y be the number of sevens that are thrown on the X rolls. Determine the joint distribution for {X,Y}{X,Y} and determine E[Y]E[Y].

Solution

npr08_05
Answers are in X, Y, P, PY
jcalc
- - - - - - - - - - - -
EX = X*PX'
EX =  7.0000
EY = Y*PY'
EY =  1.1667

Exercise 16

(See Exercise 6 from "Problems On Random Vectors and Joint Distributions", m-file npr08_06.m). The pair {X,Y}{X,Y} has the joint distribution:

X = [ - 2 . 3 - 0 . 7 1 . 1 3 . 9 5 . 1 ] Y = [ 1 . 3 2 . 5 4 . 1 5 . 3 ] X = [ - 2 . 3 - 0 . 7 1 . 1 3 . 9 5 . 1 ] Y = [ 1 . 3 2 . 5 4 . 1 5 . 3 ]
(9)
P = 0 . 0483 0 . 0357 0 . 0420 0 . 0399 0 . 0441 0 . 0437 0 . 0323 0 . 0380 0 . 0361 0 . 0399 0 . 0713 0 . 0527 0 . 0620 0 . 0609 0 . 0551 0 . 0667 0 . 0493 0 . 0580 0 . 0651 0 . 0589 P = 0 . 0483 0 . 0357 0 . 0420 0 . 0399 0 . 0441 0 . 0437 0 . 0323 0 . 0380 0 . 0361 0 . 0399 0 . 0713 0 . 0527 0 . 0620 0 . 0609 0 . 0551 0 . 0667 0 . 0493 0 . 0580 0 . 0651 0 . 0589
(10)

Determine E[X]E[X], E[Y]E[Y], E[X2]E[X2], E[Y2]E[Y2], and E[XY]E[XY].

Solution

npr08_06
Data are in X, Y, P
jcalc
- - - - - - - - - - - -
EX = X*PX'
EX =   1.3696
EY = Y*PY'
EY =   3.0344
EX2 = (X.^2)*PX'
EX2 =  9.7644
EY2 = (Y.^2)*PY'
EY2 = 11.4839
EXY = total(t.*u.*P)
EXY =  4.1423

Exercise 17

(See Exercise 7 from "Problems On Random Vectors and Joint Distributions", m-file npr08_07.m). The pair {X,Y}{X,Y} has the joint distribution:

P ( X = t , Y = u ) P ( X = t , Y = u )
(11)
Table 1
t = -3.1 -0.5 1.2 2.4 3.7 4.9
u = 7.5 0.0090 0.0396 0.0594 0.0216 0.0440 0.0203
4.1 0.0495 0 0.1089 0.0528 0.0363 0.0231
-2.0 0.0405 0.1320 0.0891 0.0324 0.0297 0.0189
-3.8 0.0510 0.0484 0.0726 0.0132 0 0.0077

Determine E[X]E[X], E[Y]E[Y], E[X2]E[X2], E[Y2]E[Y2], and E[XY]E[XY].

Solution

npr08_07
Data are in X, Y, P
jcalc
- - - - - - - - - - - -
EX = X*PX'
EX =   0.8590
EY = Y*PY'
EY =   1.1455
EX2 = (X.^2)*PX'
EX2 =  5.8495
EY2 = (Y.^2)*PY'
EY2 = 19.6115
EXY = total(t.*u.*P)
EXY =  3.6803

Exercise 18

(See Exercise 8 from "Problems On Random Vectors and Joint Distributions", m-file npr08_08.m). The pair {X,Y}{X,Y} has the joint distribution:

P ( X = t , Y = u ) P ( X = t , Y = u )
(12)
Table 2
t = 1 3 5 7 9 11 13 15 17 19
u = 12 0.0156 0.0191 0.0081 0.0035 0.0091 0.0070 0.0098 0.0056 0.0091 0.0049
10 0.0064 0.0204 0.0108 0.0040 0.0054 0.0080 0.0112 0.0064 0.0104 0.0056
9 0.0196 0.0256 0.0126 0.0060 0.0156 0.0120 0.0168 0.0096 0.0056 0.0084
5 0.0112 0.0182 0.0108 0.0070 0.0182 0.0140 0.0196 0.0012 0.0182 0.0038
3 0.0060 0.0260 0.0162 0.0050 0.0160 0.0200 0.0280 0.0060 0.0160 0.0040
-1 0.0096 0.0056 0.0072 0.0060 0.0256 0.0120 0.0268 0.0096 0.0256 0.0084
-3 0.0044 0.0134 0.0180 0.0140 0.0234 0.0180 0.0252 0.0244 0.0234 0.0126
-5 0.0072 0.0017 0.0063 0.0045 0.0167 0.0090 0.0026 0.0172 0.0217 0.0223

Determine E[X]E[X], E[Y]E[Y], E[X2]E[X2], E[Y2]E[Y2], and E[XY]E[XY].

Solution

npr08_08
Data are in X, Y, P
jcalc
- - - - - - - - - - - - -
EX = X*PX'
EX =   10.1000
EY = Y*PY'
EY =    3.0016
EX2 = (X.^2)*PX'
EX2 = 133.0800
EY2 = (Y.^2)*PY'
EY2 =  41.5564
EXY = total(t.*u.*P)
EXY =  22.2890

Exercise 19

(See Exercise 9 from "Problems On Random Vectors and Joint Distributions", m-file npr08_09.m). Data were kept on the effect of training time on the time to perform a job on a production line. X is the amount of training, in hours, and Y is the time to perform the task, in minutes. The data are as follows:

P ( X = t , Y = u ) P ( X = t , Y = u )
(13)
Table 3
t = 1 1.5 2 2.5 3
u = 5 0.039 0.011 0.005 0.001 0.001
4 0.065 0.070 0.050 0.015 0.010
3 0.031 0.061 0.137 0.051 0.033
2 0.012 0.049 0.163 0.058 0.039
1 0.003 0.009 0.045 0.025 0.017

Determine E[X]E[X], E[Y]E[Y], E[X2]E[X2], E[Y2]E[Y2], and E[XY]E[XY].

Solution

npr08_09
Data are in X, Y, P
jcalc
- - - - - - - - - - - -
EX = X*PX'
EX =   1.9250
EY = Y*PY'
EY =   2.8050
EX2 = (X.^2)*PX'
EX2 =  4.0375
EY2 = (Y.^2)*PY'       EXY = total(t.*u.*P)
EY2 =  8.9850          EXY =  5.1410

For the joint densities in Exercises 20-32 below

  1. Determine analytically E[X]E[X], E[Y]E[Y], E[X2]E[X2], E[Y2]E[Y2], and E[XY]E[XY].
  2. Use a discrete approximation for E[X]E[X], E[Y]E[Y], E[X2]E[X2], E[Y2]E[Y2], and E[XY]E[XY].

Exercise 20

(See Exercise 10 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=1fXY(t,u)=1 for 0≤t≤10≤t≤1, 0≤u≤2(1-t)0≤u≤2(1-t).

f X ( t ) = 2 ( 1 - t ) , 0 ≤ t ≤ 1 , f Y ( u ) = 1 - u / 2 , 0 ≤ u ≤ 2 f X ( t ) = 2 ( 1 - t ) , 0 ≤ t ≤ 1 , f Y ( u ) = 1 - u / 2 , 0 ≤ u ≤ 2
(14)

Solution

E [ X ] = ∫ 0 1 2 t ( 1 - t ) d t = 1 / 3 , E [ Y ] = 2 / 3 , E [ X 2 ] = 1 / 6 , E [ Y 2 ] = 2 / 3 E [ X ] = ∫ 0 1 2 t ( 1 - t ) d t = 1 / 3 , E [ Y ] = 2 / 3 , E [ X 2 ] = 1 / 6 , E [ Y 2 ] = 2 / 3
(15)
E [ X Y ] = ∫ 0 1 ∫ 0 2 ( 1 - t ) t u d u d t = 1 / 6 E [ X Y ] = ∫ 0 1 ∫ 0 2 ( 1 - t ) t u d u d t = 1 / 6
(16)
tuappr:  [0 1] [0 2] 200 400   u<=2*(1-t)
EX = 0.3333   EY = 0.6667  EX2 = 0.1667  EY2 = 0.6667
EXY = 0.1667 (use t, u, P)

Exercise 21

(See Exercise 11 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=1/2fXY(t,u)=1/2 on the square with vertices at (1,0),(2,1),(1,2),(0,1)(1,0),(2,1),(1,2),(0,1).

f X ( t ) = f Y ( t ) = I [ 0 , 1 ] ( t ) t + I ( 1 , 2 ] ( t ) ( 2 - t ) f X ( t ) = f Y ( t ) = I [ 0 , 1 ] ( t ) t + I ( 1 , 2 ] ( t ) ( 2 - t )
(17)

Solution

E [ X ] = E [ Y ] = ∫ 0 1 t 2 d t + ∫ 1 t ( 2 t - t 2 ) d t = 1 , E [ X 2 ] = E [ Y 2 ] = 7 / 6 E [ X ] = E [ Y ] = ∫ 0 1 t 2 d t + ∫ 1 t ( 2 t - t 2 ) d t = 1 , E [ X 2 ] = E [ Y 2 ] = 7 / 6
(18)
E [ X Y ] = ( 1 / 2 ) ∫ 0 1 ∫ 1 - t 1 + t d u d t + ( 1 / 2 ) ∫ 1 2 ∫ t - 1 3 - t d u d t = 1 E [ X Y ] = ( 1 / 2 ) ∫ 0 1 ∫ 1 - t 1 + t d u d t + ( 1 / 2 ) ∫ 1 2 ∫ t - 1 3 - t d u d t = 1
(19)
tuappr: [0 2] [0 2] 200 200  0.5*(u<=min(t+1,3-t))&(u>= max(1-t,t-1))
EX = 1.0000  EY = 1.0002  EX2 = 1.1684  EY2 = 1.1687 EXY = 1.0002

Exercise 22

(See Exercise 12 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=4t(1-u)fXY(t,u)=4t(1-u) for 0≤t≤10≤t≤1, 0≤u≤10≤u≤1.

f X ( t ) = 2 t , 0 ≤ t ≤ 1 , f Y ( u ) = 2 ( 1 - u ) , 0 ≤ u ≤ 1 f X ( t ) = 2 t , 0 ≤ t ≤ 1 , f Y ( u ) = 2 ( 1 - u ) , 0 ≤ u ≤ 1
(20)

Solution

E [ X ] = 2 / 3 , E [ Y ] = 1 / 3 , E [ X 2 ] = 1 / 2 , E [ Y 2 ] = 1 / 6 E [ X Y ] = 2 / 9 E [ X ] = 2 / 3 , E [ Y ] = 1 / 3 , E [ X 2 ] = 1 / 2 , E [ Y 2 ] = 1 / 6 E [ X Y ] = 2 / 9
(21)
tuappr:  [0 1] [0 1]  200 200 4*t.*(1-u)
EX = 0.6667  EY = 0.3333  EX2 = 0.5000  EY2 = 0.1667  EXY = 0.2222

Exercise 23

(See Exercise 13 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=18(t+u)fXY(t,u)=18(t+u) for 0≤t≤20≤t≤2, 0≤u≤20≤u≤2.

f X ( t ) = f Y ( t ) = 1 4 ( t + 1 ) , 0 ≤ t ≤ 2 f X ( t ) = f Y ( t ) = 1 4 ( t + 1 ) , 0 ≤ t ≤ 2
(22)

Solution

E [ X ] = E [ Y ] = 1 4 ∫ 0 2 ( t 2 + t ) d t = 7 6 , E [ X 2 ] = E [ Y 2 ] = 5 / 3 E [ X ] = E [ Y ] = 1 4 ∫ 0 2 ( t 2 + t ) d t = 7 6 , E [ X 2 ] = E [ Y 2 ] = 5 / 3
(23)
E [ X Y ] = 1 8 ∫ 0 2 ∫ 0 2 ( t 2 u + t u 2 ) d u d t = 4 3 E [ X Y ] = 1 8 ∫ 0 2 ∫ 0 2 ( t 2 u + t u 2 ) d u d t = 4 3
(24)
tuappr:  [0 2] [0 2] 200 200  (1/8)*(t+u)
EX = 1.1667  EY = 1.1667  EX2 = 1.6667  EY2 = 1.6667  EXY = 1.3333

Exercise 24

(See Exercise 14 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=4ue-2tfXY(t,u)=4ue-2t for 0≤t,0≤u≤10≤t,0≤u≤1.

f X ( t ) = 2 e - 2 t , 0 ≤ t , f Y ( u ) = 2 u , 0 ≤ u ≤ 1 f X ( t ) = 2 e - 2 t , 0 ≤ t , f Y ( u ) = 2 u , 0 ≤ u ≤ 1
(25)

Solution

E [ X ] = ∫ 0 ∞ 2 t e - 2 t d t = 1 2 , E [ Y ] = 2 3 , E [ X 2 ] = 1 2 , E [ Y 2 ] = 1 2 , E [ X Y ] = 1 3 E [ X ] = ∫ 0 ∞ 2 t e - 2 t d t = 1 2 , E [ Y ] = 2 3 , E [ X 2 ] = 1 2 , E [ Y 2 ] = 1 2 , E [ X Y ] = 1 3
(26)
tuappr: [0 6] [0 1] 600 200  4*u.*exp(-2*t)
EX = 0.5000  EY = 0.6667  EX2 = 0.4998  EY2 = 0.5000  EXY = 0.3333

Exercise 25

(See Exercise 15 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=388(2t+3u2)fXY(t,u)=388(2t+3u2) for 0≤t≤20≤t≤2, 0≤u≤1+t0≤u≤1+t.

f X ( t ) = 3 88 ( 1 + t ) ( 1 + 4 t + t 2 ) = 3 88 ( 1 + 5 t + 5 t 2 + t 3 ) , 0 ≤ t ≤ 2 f X ( t ) = 3 88 ( 1 + t ) ( 1 + 4 t + t 2 ) = 3 88 ( 1 + 5 t + 5 t 2 + t 3 ) , 0 ≤ t ≤ 2
(27)
f Y ( u ) = I [ 0 , 1 ] ( u ) 3 88 ( 6 u 2 + 4 ) + I ( 1 , 3 ] ( u ) 3 88 ( 3 + 2 u + 8 u 2 - 3 u 3 ) f Y ( u ) = I [ 0 , 1 ] ( u ) 3 88 ( 6 u 2 + 4 ) + I ( 1 , 3 ] ( u ) 3 88 ( 3 + 2 u + 8 u 2 - 3 u 3 )
(28)

Solution

E [ X ] = 313 220 , E [ Y ] = 1429 880 , E [ X 2 ] = 49 22 , E [ Y 2 ] = 172 55 , E [ X Y ] = 2153 880 E [ X ] = 313 220 , E [ Y ] = 1429 880 , E [ X 2 ] = 49 22 , E [ Y 2 ] = 172 55 , E [ X Y ] = 2153 880
(29)
tuappr:  [0 2] [0 3] 200 300 (3/88)*(2*t + 3*u.^2).*(u<1+t)
EX = 1.4229  EY = 1.6202  EX2 = 2.2277 EY2 = 3.1141  EXY = 2.4415

Exercise 26

(See Exercise 16 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=12t2ufXY(t,u)=12t2u on the parallelogram with vertices

( - 1 , 0 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 0 , 1 ) ( - 1 , 0 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 0 , 1 )
(30)
f X ( t ) = I [ - 1 , 0 ] ( t ) 6 t 2 ( t + 1 ) 2 + I ( 0 , 1 ] ( t ) 6 t 2 ( 1 - t 2 ) , f Y ( u ) = 12 u 3 - 12 u 2 + 4 u , 0 ≤ u ≤ 1 f X ( t ) = I [ - 1 , 0 ] ( t ) 6 t 2 ( t + 1 ) 2 + I ( 0 , 1 ] ( t ) 6 t 2 ( 1 - t 2 ) , f Y ( u ) = 12 u 3 - 12 u 2 + 4 u , 0 ≤ u ≤ 1
(31)

Solution

E [ X ] = 2 5 , E [ Y ] = 11 15 , E [ X 2 ] = 2 5 , E [ Y 2 ] = 3 5 , E [ X Y ] = 2 5 E [ X ] = 2 5 , E [ Y ] = 11 15 , E [ X 2 ] = 2 5 , E [ Y 2 ] = 3 5 , E [ X Y ] = 2 5
(32)
tuappr: [-1 1] [0 1] 400 200 12*t.^2.*u.*(u>= max(0,t)).*(u<= min(1+t,1))
EX = 0.4035  EY = 0.7342  EX2 = 0.4016  EY2 = 0.6009  EXY = 0.4021

Exercise 27

(See Exercise 17 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=2411tufXY(t,u)=2411tu for 0≤t≤20≤t≤2, 0≤u≤min{1,2-t}0≤u≤min{1,2-t}.

f X ( t ) = I [ 0 , 1 ] ( t ) 12 11 t + I ( 1 , 2 ] ( t ) 12 11 t ( 2 - t ) 2 , f Y ( u ) = 12 11 u ( u - 2 ) 2 , 0 ≤ u ≤ 1 f X ( t ) = I [ 0 , 1 ] ( t ) 12 11 t + I ( 1 , 2 ] ( t ) 12 11 t ( 2 - t ) 2 , f Y ( u ) = 12 11 u ( u - 2 ) 2 , 0 ≤ u ≤ 1
(33)

Solution

E [ X ] = 52 55 , E [ Y ] = 32 55 , E [ X 2 ] = 57 55 , E [ Y 2 ] = 2 5 , E [ X Y ] = 28 55 E [ X ] = 52 55 , E [ Y ] = 32 55 , E [ X 2 ] = 57 55 , E [ Y 2 ] = 2 5 , E [ X Y ] = 28 55
(34)
tuappr:  [0 2] [0 1] 400 200 (24/11)*t.*u.*(u<=min(1,2-t))
EX = 0.9458  EY = 0.5822  EX2 = 1.0368 EY2 = 0.4004   EXY = 0.5098

Exercise 28

(See Exercise 18 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=323(t+2u)fXY(t,u)=323(t+2u) for 0≤t≤20≤t≤2, 0≤u≤max{2-t,t}0≤u≤max{2-t,t}.

f X ( t ) = I [ 0 , 1 ] ( t ) 6 23 ( 2 - t ) + I ( 1 , 2 ] ( t ) 6 23 t 2 , f Y ( u ) = I [ 0 , 1 ] ( u ) 6 23 ( 2 u + 1 ) + I ( 1 , 2 ] ( u ) 3 23 ( 4 + 6 u - 4 u 2 ) f X ( t ) = I [ 0 , 1 ] ( t ) 6 23 ( 2 - t ) + I ( 1 , 2 ] ( t ) 6 23 t 2 , f Y ( u ) = I [ 0 , 1 ] ( u ) 6 23 ( 2 u + 1 ) + I ( 1 , 2 ] ( u ) 3 23 ( 4 + 6 u - 4 u 2 )
(35)

Solution

E [ X ] = 53 46 , E [ Y ] = 22 23 , E [ X 2 ] = 397 230 , E [ Y 2 ] = 261 230 , E [ X Y ] = 251 230 E [ X ] = 53 46 , E [ Y ] = 22 23 , E [ X 2 ] = 397 230 , E [ Y 2 ] = 261 230 , E [ X Y ] = 251 230
(36)
tuappr:  [0 2] [0 2] 200 200 (3/23)*(t + 2*u).*(u<=max(2-t,t))
EX = 1.1518  EY = 0.9596  EX2 = 1.7251  EY2 = 1.1417  EXY = 1.0944

Exercise 29

(See Exercise 19 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=12179(3t2+u)fXY(t,u)=12179(3t2+u), for 0≤t≤20≤t≤2, 0≤u≤min{2,3-t}0≤u≤min{2,3-t}.

f X ( t ) = I [ 0 , 1 ] ( t ) 24 179 ( 3 t 2 + 1 ) + I ( 1 , 2 ] ( t ) 6 179 ( 9 - 6 t + 19 t 2 - 6 t 3 ) f X ( t ) = I [ 0 , 1 ] ( t ) 24 179 ( 3 t 2 + 1 ) + I ( 1 , 2 ] ( t ) 6 179 ( 9 - 6 t + 19 t 2 - 6 t 3 )
(37)
f Y ( u ) = I [ 0 , 1 ] ( u ) 24 179 ( 4 + u ) + I ( 1 , 2 ] ( u ) 12 179 ( 27 - 24 u + 8 u 2 - u 3 ) f Y ( u ) = I [ 0 , 1 ] ( u ) 24 179 ( 4 + u ) + I ( 1 , 2 ] ( u ) 12 179 ( 27 - 24 u + 8 u 2 - u 3 )
(38)

Solution

E [ X ] = 2313 1790 , E [ Y ] = 778 895 , E [ X 2 ] = 1711 895 , E [ Y 2 ] = 916 895 , E [ X Y ] = 1811 1790 E [ X ] = 2313 1790 , E [ Y ] = 778 895 , E [ X 2 ] = 1711 895 , E [ Y 2 ] = 916 895 , E [ X Y ] = 1811 1790
(39)
tuappr: [0 2] [0 2] 400 400 (12/179)*(3*t.^2 + u).*(u<=min(2,3-t))
EX = 1.2923  EY = 0.8695  EX2 = 1.9119  EY2 = 1.0239  EXY = 1.0122

Exercise 30

(See Exercise 20 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=12227(3t+2tu)fXY(t,u)=12227(3t+2tu), for 0≤t≤20≤t≤2, 0≤u≤min{1+t,2}0≤u≤min{1+t,2}.

f X ( t ) = I [ 0 , 1 ] ( t ) 12 227 ( t 3 + 5 t 2 + 4 t ) + I ( 1 , 2 ] ( t ) 120 227 t f X ( t ) = I [ 0 , 1 ] ( t ) 12 227 ( t 3 + 5 t 2 + 4 t ) + I ( 1 , 2 ] ( t ) 120 227 t
(40)
f Y ( u ) = I [ 0 , 1 ] ( u ) 24 227 ( 2 u + 3 ) + I ( 1 , 2 ] ( u ) 6 227 ( 2 u + 3 ) ( 3 + 2 u - u 2 ) f Y ( u ) = I [ 0 , 1 ] ( u ) 24 227 ( 2 u + 3 ) + I ( 1 , 2 ] ( u ) 6 227 ( 2 u + 3 ) ( 3 + 2 u - u 2 )
(41)
= I [ 0 , 1 ] ( u ) 24 227 ( 2 u + 3 ) + I ( 1 , 2 ] ( u ) 6 227 ( 9 + 12 u + u 2 - 2 u 3 ) = I [ 0 , 1 ] ( u ) 24 227 ( 2 u + 3 ) + I ( 1 , 2 ] ( u ) 6 227 ( 9 + 12 u + u 2 - 2 u 3 )
(42)

Solution

E [ X ] = 1567 1135 , E [ Y ] = 2491 2270 , E [ X 2 ] = 476 227 , E [ Y 2 ] = 1716 1135 , E [ X Y ] = 5261 3405 E [ X ] = 1567 1135 , E [ Y ] = 2491 2270 , E [ X 2 ] = 476 227 , E [ Y 2 ] = 1716 1135 , E [ X Y ] = 5261 3405
(43)
tuappr: [0 2] [0 2] 400 400 (12/227)*(3*t + 2*t.*u).*(u<=min(1+t,2))
EX = 1.3805  EY = 1.0974  EX2 = 2.0967  EY2 = 1.5120   EXY = 1.5450

Exercise 31

(See Exercise 21 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=213(t+2u)fXY(t,u)=213(t+2u), for 0≤t≤20≤t≤2, 0≤u≤min{2t,3-t}0≤u≤min{2t,3-t}.

f X ( t ) = I [ 0 , 1 ] ( t ) 12 13 t 2 + I ( 1 , 2 ] ( t ) 6 13 ( 3 - t ) f X ( t ) = I [ 0 , 1 ] ( t ) 12 13 t 2 + I ( 1 , 2 ] ( t ) 6 13 ( 3 - t )
(44)
f Y ( u ) = I [ 0 , 1 ] ( u ) ( 4 13 + 8 13 u - 9 52 u 2 ) + I ( 1 , 2 ] ( u ) ( 9 13 + 6 13 u - 21 52 u 2 ) f Y ( u ) = I [ 0 , 1 ] ( u ) ( 4 13 + 8 13 u - 9 52 u 2 ) + I ( 1 , 2 ] ( u ) ( 9 13 + 6 13 u - 21 52 u 2 )
(45)

Solution

E [ X ] = 16 13 , E [ Y ] = 11 12 , E [ X 2 ] = 219 130 , E [ Y 2 ] = 83 78 , E [ X Y ] = 431 390 E [ X ] = 16 13 , E [ Y ] = 11 12 , E [ X 2 ] = 219 130 , E [ Y 2 ] = 83 78 , E [ X Y ] = 431 390
(46)
tuappr:  [0 2] [0 2] 400 400 (2/13)*(t + 2*u).*(u<=min(2*t,3-t))
EX = 1.2309  EY = 0.9169  EX2 = 1.6849  EY2 = 1.0647 EXY = 1.1056

Exercise 32

(See Exercise 22 from "Problems On Random Vectors and Joint Distributions"). fXY(t,u)=I[0,1](t)38(t2+2u)+I(1,2](t)914t2u2fXY(t,u)=I[0,1](t)38(t2+2u)+I(1,2](t)914t2u2,

                                     for 0≤u≤10≤u≤1.

f X ( t ) = I [ 0 , 1 ] ( t ) 3 8 ( t 2 + 1 ) + I ( 1 , 2 ] ( t ) 3 14 t 2 , f Y ( u ) = 1 8 + 3 4 u + 3 2 u 2 0 ≤ u ≤ 1 f X ( t ) = I [ 0 , 1 ] ( t ) 3 8 ( t 2 + 1 ) + I ( 1 , 2 ] ( t ) 3 14 t 2 , f Y ( u ) = 1 8 + 3 4 u + 3 2 u 2 0 ≤ u ≤ 1
(47)

Solution

E [ X ] = 243 224 , E [ Y ] = 11 16 , E [ X 2 ] = 107 70 , E [ Y 2 ] = 127 240 , E [ X Y ] = 347 448 E [ X ] = 243 224 , E [ Y ] = 11 16 , E [ X 2 ] = 107 70 , E [ Y 2 ] = 127 240 , E [ X Y ] = 347 448
(48)
tuappr  [0 2] [0 1] 400 200   (3/8)*(t.^2+2*u).*(t<=1) + (9/14)*(t.^2.*u.^2).*(t > 1)
EX = 1.0848  EY = 0.6875  EX2 = 1.5286  EY2 = 0.5292  EXY = 0.7745

Exercise 33

The class {X,Y,Z}{X,Y,Z} of random variables is iid (independent, identically distributed) with common distribution

X = [ - 5 - 1 3 4 7 ] P X = 0 . 01 * [ 15 20 30 25 10 ] X = [ - 5 - 1 3 4 7 ] P X = 0 . 01 * [ 15 20 30 25 10 ]
(49)

Let W=3X-4Y+2ZW=3X-4Y+2Z. Determine E[W]E[W]. Do this using icalc, then repeat with icalc3 and compare results.

Solution

Use x and pxpx to prevent renaming.

x = [-5 -1 3 4 7];
px = 0.01*[15 20 30 25 10];
icalc
Enter row matrix of X-values  x
Enter row matrix of Y-values  x
Enter X probabilities  px
Enter Y probabilities  px
 Use array operations on matrices X, Y, PX, PY, t, u, and P
 G = 3*t -4*u;
 [R,PR] = csort(G,P);
 icalc
Enter row matrix of X-values  R
Enter row matrix of Y-values  x
Enter X probabilities  PR
Enter Y probabilities  px
 Use array operations on matrices X, Y, PX, PY, t, u, and P
H = t + 2*u;
EH = total(H.*P)
EH =  1.6500
[W,PW] = csort(H,P);  % Alternate
EW = W*PW'
EW =  1.6500
icalc3                % Solution with icalc3
Enter row matrix of X-values  x
Enter row matrix of Y-values  x
Enter row matrix of Z-values  x
Enter X probabilities  px
Enter Y probabilities  px
Enter Z probabilities  px
Use array operations on matrices X, Y, Z,
PX, PY, PZ, t, u, v, and P
K = 3*t - 4*u + 2*v;
EK = total(K.*P)
EK =  1.6500

Exercise 34

(See Exercise 5 from "Problems on Functions of Random Variables") The cultural committee of a student organization has arranged a special deal for tickets to a concert. The agreement is that the organization will purchase ten tickets at $20 each (regardless of the number of individual buyers). Additional tickets are available according to the following schedule:

11-20, $18 each; 21-30 $16 each; 31-50, $15 each; 51-100, $13 each

If the number of purchasers is a random variable X, the total cost (in dollars) is a random quantity Z=g(X)Z=g(X) described by

g ( X ) = 200 + 18 I M 1 ( X ) ( X - 10 ) + ( 16 - 18 ) I M 2 ( X ) ( X - 20 ) + g ( X ) = 200 + 18 I M 1 ( X ) ( X - 10 ) + ( 16 - 18 ) I M 2 ( X ) ( X - 20 ) +
(50)
( 15 - 16 ) I M 3 ( X ) ( X - 30 ) + ( 13 - 15 ) I M 4 ( X ) ( X - 50 ) ( 15 - 16 ) I M 3 ( X ) ( X - 30 ) + ( 13 - 15 ) I M 4 ( X ) ( X - 50 )
(51)
where M 1 = [ 10 , ∞ ) , M 2 = [ 20 , ∞ ) , M 3 = [ 30 , ∞ ) , M 4 = [ 50 , ∞ ) where M 1 = [ 10 , ∞ ) , M 2 = [ 20 , ∞ ) , M 3 = [ 30 , ∞ ) , M 4 = [ 50 , ∞ )
(52)

Suppose X∼X∼ Poisson (75). Approximate the Poisson distribution by truncating at 150. Determine E[Z]E[Z] and E[Z2]E[Z2].

Solution

X = 0:150;
PX = ipoisson(75,X);
G = 200 + 18*(X - 10).*(X>=10) + (16 - 18)*(X - 20).*(X>=20) + ...
     (15 - 16)*(X- 30).*(X>=30) + (13 - 15)*(X - 50).*(X>=50);
[Z,PZ] = csort(G,PX);
EZ = Z*PZ'
EZ =  1.1650e+03
EZ2 = (Z.^2)*PZ'
EZ2 = 1.3699e+06

Exercise 35

The pair {X,Y}{X,Y} has the joint distribution (in m-file npr08_07.m):

P ( X = t , Y = u ) P ( X = t , Y = u )
(53)
Table 4
t = -3.1 -0.5 1.2 2.4 3.7 4.9
u = 7.5 0.0090 0.0396 0.0594 0.0216 0.0440 0.0203
4.1 0.0495 0 0.1089 0.0528 0.0363 0.0231
-2.0 0.0405 0.1320 0.0891 0.0324 0.0297 0.0189
-3.8 0.0510 0.0484 0.0726 0.0132 0 0.0077

Let Z=g(X,Y)=3X2+2XY-Y2Z=g(X,Y)=3X2+2XY-Y2. Determine E[Z]E[Z] and E[Z2]E[Z2].

Solution

npr08_07
Data are in X, Y, P
jcalc
- - - - - - - - -
G = 3*t.^2 + 2*t.*u - u.^2;
EG = total(G.*P)
EG =  5.2975
ez2 = total(G.^2.*P)
EG2 = 1.0868e+03
[Z,PZ] = csort(G,P);   % Alternate
EZ = Z*PZ'
EZ =  5.2975
EZ2 = (Z.^2)*PZ'
EZ2 = 1.0868e+03

Exercise 36

For the pair {X,Y}{X,Y} in Exercise 35, let

W = g X , Y ) = X for X + Y ≤ 4 2 Y for X + Y > 4 = I M ( X , Y ) X + I M c ( X , Y ) 2 Y W = g X , Y ) = X for X + Y ≤ 4 2 Y for X + Y > 4 = I M ( X , Y ) X + I M c ( X , Y ) 2 Y
(54)

Determine E[W]E[W] and E[W2]E[W2].

Solution

H = t.*(t+u<=4) + 2*u.*(t+u>4);
EH = total(H.*P)
EH =   4.7379
EH2 = total(H.^2.*P)
EH2 = 61.4351
[W,PW] = csort(H,P);  % Alternate
EW = W*PW'
EW =   4.7379
EW2 = (W.^2)*PW'
EW2 = 61.4351

For the distributions in Exercises 37-41 below

  1. Determine analytically E[Z]E[Z] and E[Z2]E[Z2].
  2. Use a discrete approximation to calculate the same quantities.

Exercise 37

fXY(t,u)=388(2t+3u2)fXY(t,u)=388(2t+3u2) for 0≤t≤20≤t≤2, 0≤u≤1+t0≤u≤1+t (see Exercise 25).

Z = I [ 0 , 1 ] ( X ) 4 X + I ( 1 , 2 ] ( X ) ( X + Y ) Z = I [ 0 , 1 ] ( X ) 4 X + I ( 1 , 2 ] ( X ) ( X + Y )
(55)

Solution

E [ Z ] = 3 88 ∫ 0 1 ∫ 0 1 + t 4 t ( 2 t + 3 u 2 ) d u d t + 3 88 ∫ 1 2 ∫ 0 1 + t ( t + u ) ( 2 t + 3 u 2 ) d u d t = 5649 1760 E [ Z ] = 3 88 ∫ 0 1 ∫ 0 1 + t 4 t ( 2 t + 3 u 2 ) d u d t + 3 88 ∫ 1 2 ∫ 0 1 + t ( t + u ) ( 2 t + 3 u 2 ) d u d t = 5649 1760
(56)
E [ Z 2 ] = 3 88 ∫ 0 1 ∫ 0 1 + t ( 4 t ) 2 ( 2 t + 3 u 2 ) d u d t + 3 88 ∫ 1 2 ∫ 0 1 + t ( t + u ) 2 ( 2 t + 3 u 2 ) d u d t = 4881 440 E [ Z 2 ] = 3 88 ∫ 0 1 ∫ 0 1 + t ( 4 t ) 2 ( 2 t + 3 u 2 ) d u d t + 3 88 ∫ 1 2 ∫ 0 1 + t ( t + u ) 2 ( 2 t + 3 u 2 ) d u d t = 4881 440
(57)
tuappr:  [0 2] [0 3] 200 300 (3/88)*(2*t+3*u.^2).*(u<=1+t)
G = 4*t.*(t<=1) + (t + u).*(t>1);
EG = total(G.*P)
EG =   3.2086
EG2 = total(G.^2.*P)
EG2 = 11.0872

Exercise 38

fXY(t,u)=2411tufXY(t,u)=2411tu for 0≤t≤20≤t≤2, 0≤u≤min{1,2-t}0≤u≤min{1,2-t} (see Exercise 27).

Z = I M ( X , Y ) 1 2 X + I M c ( X , Y ) Y 2 , M = { ( t , u ) : u > t } Z = I M ( X , Y ) 1 2 X + I M c ( X , Y ) Y 2 , M = { ( t , u ) : u > t }
(58)

Solution

E [ Z ] = 12 11 ∫ 0 1 ∫ t 1 t 2 u d u d t + 24 11 ∫ 0 1 ∫ 0 t t u 3 d u d t + 24 11 ∫ 1 2 ∫ 0 2 - t t u 3 d u d t = 16 55 E [ Z ] = 12 11 ∫ 0 1 ∫ t 1 t 2 u d u d t + 24 11 ∫ 0 1 ∫ 0 t t u 3 d u d t + 24 11 ∫ 1 2 ∫ 0 2 - t t u 3 d u d t = 16 55
(59)
E [ Z 2 ] = 6 11 ∫ 0 1 ∫ t 1 t 3 u d u d t + 24 11 ∫ 0 1 ∫ 0 t t u 5 d u d t + 24 11 ∫ 1 2 ∫ 0 2 - t t u 5 d u d t = 39 308 E [ Z 2 ] = 6 11 ∫ 0 1 ∫ t 1 t 3 u d u d t + 24 11 ∫ 0 1 ∫ 0 t t u 5 d u d t + 24 11 ∫ 1 2 ∫ 0 2 - t t u 5 d u d t = 39 308
(60)
tuappr:  [0 2] [0 1] 400 200  (24/11)*t.*u.*(u<=min(1,2-t))
G = (1/2)*t.*(u>t) + u.^2.*(u<=t);
EZ = 0.2920  EZ2 = 0.1278

Exercise 39

fXY(t,u)=323(t+2u)fXY(t,u)=323(t+2u) for 0≤t≤20≤t≤2, 0≤u≤max{2-t,t}0≤u≤max{2-t,t} (see Exercise 28).

Z = I M ( X , Y ) ( X + Y ) + I M c ( X , Y ) 2 Y , M = { ( t , u ) : max ( t , u ) ≤ 1 } Z = I M ( X , Y ) ( X + Y ) + I M c ( X , Y ) 2 Y , M = { ( t , u ) : max ( t , u ) ≤ 1 }
(61)

Solution

E [ Z ] = 3 23 ∫ 0 1 ∫ 0 1 ( t + u ) ( t + 2 u ) d u d t + 3 23 ∫ 0 1 ∫ 1 2 - t 2 u ( t + 2 u ) d u d t + 3 23 ∫ 1 2 ∫ 1 t 2 u ( t + 2 u ) d u d t = 175 92 E [ Z ] = 3 23 ∫ 0 1 ∫ 0 1 ( t + u ) ( t + 2 u ) d u d t + 3 23 ∫ 0 1 ∫ 1 2 - t 2 u ( t + 2 u ) d u d t + 3 23 ∫ 1 2 ∫ 1 t 2 u ( t + 2 u ) d u d t = 175 92
(62)
E [ Z 2 ] = 3 23 ∫ 0 1 ∫ 0 1 ( t + u ) 2 ( t + 2 u ) d u d t + 3 23 ∫ 0 1 ∫ 1 2 - t 4 u 2 ( t + 2 u ) d u d t + 3 23 ∫ 1 2 ∫ 1 t 4 u 2 ( t + 2 u ) d u d t = 2063 460 E [ Z 2 ] = 3 23 ∫ 0 1 ∫ 0 1 ( t + u ) 2 ( t + 2 u ) d u d t + 3 23 ∫ 0 1 ∫ 1 2 - t 4 u 2 ( t + 2 u ) d u d t + 3 23 ∫ 1 2 ∫ 1 t 4 u 2 ( t + 2 u ) d u d t = 2063 460
(63)
tuappr:  [0 2] [0 2] 400 400 (3/23)*(t+2*u).*(u<=max(2-t,t))
M = max(t,u)<=1;
G = (t+u).*M + 2*u.*(1-M);
EZ = total(G.*P)
EZ =  1.9048
EZ2 = total(G.^2.*P)
EZ2 = 4.4963

Exercise 40

fXY(t,u)=12179(3t2+u)fXY(t,u)=12179(3t2+u), for 0≤t≤20≤t≤2, 0≤u≤min{2,3-t}0≤u≤min{2,3-t} (see Exercise 29).

Z = I M ( X , Y ) ( X + Y ) + I M c ( X , Y ) 2 Y 2 , M = { ( t , u ) : t ≤ 1 , u ≥ 1 } Z = I M ( X , Y ) ( X + Y ) + I M c ( X , Y ) 2 Y 2 , M = { ( t , u ) : t ≤ 1 , u ≥ 1 }
(64)

Solution

E [ Z ] = 12 179 ∫ 0 1 ∫ 1 2 ( t + u ) ( 3 t 2 + u ) d u d t + 12 179 ∫ 0 1 ∫ 0 1 2 u 2 ( 3 t 2 + u ) d u d t + E [ Z ] = 12 179 ∫ 0 1 ∫ 1 2 ( t + u ) ( 3 t 2 + u ) d u d t + 12 179 ∫ 0 1 ∫ 0 1 2 u 2 ( 3 t 2 + u ) d u d t +
(65)
12 179 ∫ 1 2 ∫ 0 3 - t 2 u 2 ( 3 t 2 + u ) d u d t = 1422 895 12 179 ∫ 1 2 ∫ 0 3 - t 2 u 2 ( 3 t 2 + u ) d u d t = 1422 895
(66)
E [ Z 2 ] = 12 179 ∫ 0 1 ∫ 1 2 ( t + u ) 2 ( 3 t 2 + u ) d u d t + 12 179 ∫ 0 1 ∫ 0 1 4 u 4 ( 3 t 2 + u ) d u d t + E [ Z 2 ] = 12 179 ∫ 0 1 ∫ 1 2 ( t + u ) 2 ( 3 t 2 + u ) d u d t + 12 179 ∫ 0 1 ∫ 0 1 4 u 4 ( 3 t 2 + u ) d u d t +
(67)
12 179 ∫ 1 2 ∫ 0 3 - t 4 u 4 ( 3 t 2 + u ) d u d t = 28296 6265 12 179 ∫ 1 2 ∫ 0 3 - t 4 u 4 ( 3 t 2 + u ) d u d t = 28296 6265
(68)
tuappr:  [0 2] [0 2] 400 400 (12/179)*(3*t.^2 + u).*(u <= min(2,3-t))
M = (t<=1)&(u>=1);
G = (t + u).*M + 2*u.^2.*(1 - M);
EZ = total(G.*P)
EZ =  1.5898
EZ2 = total(G.^2.*P)
EZ2 = 4.5224

Exercise 41

fXY(t,u)=12227(3t+2tu)fXY(t,u)=12227(3t+2tu), for 0≤t≤20≤t≤2, 0≤u≤min{1+t,2}0≤u≤min{1+t,2} (see Exercise 30).

Z = I M ( X , Y ) X + I M c ( X , Y ) X Y , M = { ( t , u ) : u ≤ min ( 1 , 2 - t ) } Z = I M ( X , Y ) X + I M c ( X , Y ) X Y , M = { ( t , u ) : u ≤ min ( 1 , 2 - t ) }
(69)

Solution

E [ Z ] = 12 227 ∫ 0 1 ∫ 0 1 t ( 3 t + 2 t u ) d u d t + 12 227 ∫ 1 2 ∫ 0 2 - t t ( 3 t + 2 t u ) d u d t + E [ Z ] = 12 227 ∫ 0 1 ∫ 0 1 t ( 3 t + 2 t u ) d u d t + 12 227 ∫ 1 2 ∫ 0 2 - t t ( 3 t + 2 t u ) d u d t +
(70)
12 227 ∫ 0 1 ∫ 1 1 + t t u ( 3 t + 2 t u ) d u d t + 12 227 ∫ 1 2 ∫ 2 - t 2 t u ( 3 t + 2 t u ) d u d t = 5774 3405 12 227 ∫ 0 1 ∫ 1 1 + t t u ( 3 t + 2 t u ) d u d t + 12 227 ∫ 1 2 ∫ 2 - t 2 t u ( 3 t + 2 t u ) d u d t = 5774 3405
(71)
E [ Z 2 ] = 56673 15890 E [ Z 2 ] = 56673 15890
(72)
tuappr: [0 2] [0 2] 400 400 (12/227)*(3*t + 2*t.*u).*(u <= min(1+t,2))
M = u <= min(1,2-t);
G = t.*M + t.*u.*(1 - M);
EZ = total(G.*P)
EZ =  1.6955
EZ2 = total(G.^2.*P)
EZ2 = 3.5659

Exercise 42

The class {X,Y,Z}{X,Y,Z} is independent. (See Exercise 16 from "Problems on Functions of Random Variables", m-file npr10_16.m)

          X=-2IA+IB+3ICX=-2IA+IB+3IC. Minterm probabilities are (in the usual order)

0 . 255 0 . 025 0 . 375 0 . 045 0 . 108 0 . 012 0 . 162 0 . 018 0 . 255 0 . 025 0 . 375 0 . 045 0 . 108 0 . 012 0 . 162 0 . 018
(73)

          Y=ID+3IE+IF-3Y=ID+3IE+IF-3. The class {D,E,F}{D,E,F} is independent with

P ( D ) = 0 . 32 P ( E ) = 0 . 56 P ( F ) = 0 . 40 P ( D ) = 0 . 32 P ( E ) = 0 . 56 P ( F ) = 0 . 40
(74)

          Z has distribution

Table 5
Value -1.3 1.2 2.7 3.4 5.8
Probability 0.12 0.24 0.43 0.13 0.08

W=X2+3XY2-3ZW=X2+3XY2-3Z . Determine E[W]E[W] and E[W2]E[W2] .

Solution

npr10_16
Data are in cx, pmx, cy, pmy, Z, PZ
[X,PX] = canonicf(cx,pmx);
[Y,PY] = canonicf(cy,pmy);
icalc3
input: X, Y, Z, PX, PY, PZ
- - - - - - -
Use array operations on matrices X, Y, Z,
PX, PY, PZ, t, u, v, and P
G = t.^2 + 3*t.*u.^2 - 3*v;
[W,PW] = csort(G,P);
EW = W*PW'
EW =   -1.8673
EW2 = (W.^2)*PW'
EW2 = 426.8529

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