Question

In: Statistics and Probability

For Problems 8-10, throw a die twice and let X, Y be the results (each from...

For Problems 8-10, throw a die twice and let X, Y be the results (each from 1 to 6).

Problem 8. Find the expectation and variance of X.

Problem 9. Find the variance of X - 2Y.

Problem 10. Find the expected value of X^2Y.

Solutions

Expert Solution

Problem 8.

X : Results of throwing a single die

P(X) : Probability of X

Expectation :

Variance : Var(X) = E(X2) - E(X)2

X P(X) P(X) XP(X) X2P(X)
1 1/6 0.1667 0.1667 0.1667
2 1/6 0.1667 0.3333 0.6667
3 1/6 0.1667 0.5000 1.5000
4 1/6 0.1667 0.6667 2.6667
5 1/6 0.1667 0.8333 4.1667
6 1/6 0.1667 1.0000 6.0000
Total 1.0000 3.5000 15.1667

Expectation

Expectation = 3.5

Variance : Var(X) = E(X2) - E(X)2 = 15.1667 - 3.52 = 2.9167

Problem 9

Variance of X-2Y

The following table provides, All possibles of X,Y and their probability.

X Y P(X,Y)
1 1 1/36
1 2 1/36
1 3 1/36
1 4 1/36
1 5 1/36
1 6 1/36
2 1 1/36
2 2 1/36
2 3 1/36
2 4 1/36
2 5 1/36
2 6 1/36
3 1 1/36
3 2 1/36
3 3 1/36
3 4 1/36
3 5 1/36
3 6 1/36
4 1 1/36
4 2 1/36
4 3 1/36
4 4 1/36
4 5 1/36
4 6 1/36
5 1 1/36
5 2 1/36
5 3 1/36
5 4 1/36
5 5 1/36
5 6 1/36
6 1 1/36
6 2 1/36
6 3 1/36
6 4 1/36
6 5 1/36
6 6 1/36

The following table extends the above by calculating X-2Y

X Y X-2Y
1 1 -1
1 2 -3
1 3 -5
1 4 -7
1 5 -9
1 6 -11
2 1 0
2 2 -2
2 3 -4
2 4 -6
2 5 -8
2 6 -10
3 1 1
3 2 -1
3 3 -3
3 4 -5
3 5 -7
3 6 -9
4 1 2
4 2 0
4 3 -2
4 4 -4
4 5 -6
4 6 -8
5 1 3
5 2 1
5 3 -1
5 4 -3
5 5 -5
5 6 -7
6 1 4
6 2 2
6 3 0
6 4 -2
6 5 -4
6 6 -6

The above table is sorted on X-2Y, so that same value of X-2Y are together

X Y X-2Y
1 6 -11
2 6 -10
1 5 -9
3 6 -9
2 5 -8
4 6 -8
1 4 -7
3 5 -7
5 6 -7
2 4 -6
4 5 -6
6 6 -6
1 3 -5
3 4 -5
5 5 -5
2 3 -4
4 4 -4
6 5 -4
1 2 -3
3 3 -3
5 4 -3
2 2 -2
4 3 -2
6 4 -2
1 1 -1
3 2 -1
5 3 -1
2 1 0
4 2 0
6 3 0
3 1 1
5 2 1
4 1 2
6 2 2
5 1 3
6 1 4
X Y X-2Y X-2Y P(X-2Y)
1 6 -11 -11 1/36
2 6 -10 -10 1/36
1 5 -9 -9 2/36
3 6 -9
2 5 -8 -8 2/36
4 6 -8
1 4 -7 -7 3/36
3 5 -7
5 6 -7
2 4 -6 -6 3/36
4 5 -6
6 6 -6
1 3 -5 -5 3/36
3 4 -5
5 5 -5
2 3 -4 -4 3/36
4 4 -4
6 5 -4
1 2 -3 -3 3/36
3 3 -3
5 4 -3
2 2 -2 -2 3/36
4 3 -2
6 4 -2
1 1 -1 -1 3/36
3 2 -1
5 3 -1
2 1 0 0 3/36
4 2 0
6 3 0
3 1 1 1 2/36
5 2 1
4 1 2 2 2/36
6 2 2
5 1 3 3 1/36
6 1 4 4 1/36
Z:(X-2Y) P(Z)=P(X-2Y) ZP(Z) Z2P(Z)
-11 0.0278 -0.3056 3.3611
-10 0.0278 -0.2778 2.7778
-9 0.0556 -0.5000 4.5000
-8 0.0556 -0.4444 3.5556
-7 0.0833 -0.5833 4.0833
-6 0.0833 -0.5000 3.0000
-5 0.0833 -0.4167 2.0833
-4 0.0833 -0.3333 1.3333
-3 0.0833 -0.2500 0.7500
-2 0.0833 -0.1667 0.3333
-1 0.0833 -0.0833 0.0833
0 0.0833 0.0000 0.0000
1 0.0556 0.0556 0.0556
2 0.0556 0.1111 0.2222
3 0.0278 0.0833 0.2500
4 0.0278 0.1111 0.4444
1.0000 -3.5000 26.8333

Let Z= X-2Y

Variance of X-2T i.e Var(Z)

Var(Z) = E(Z2) - E(Z)2

E(Z) = -3.5

E(Z2) = 26.8333

Var(Z) = E(Z2) - E(Z)2 = 26.8333 - (-3.5)2 = 14.5833

Variance of X-2Y = 14.5833

Problem 10:

Following the above procedure,

W=X2Y

W

P(W)

WP(W)

1

0.0278

0.0278

2

0.0278

0.0556

3

0.0278

0.0833

4

0.0556

0.2222

5

0.0278

0.1389

6

0.0278

0.1667

8

0.0278

0.2222

9

0.0278

0.2500

12

0.0278

0.3333

16

0.0556

0.8889

18

0.0278

0.5000

20

0.0278

0.5556

24

0.0278

0.6667

25

0.0278

0.6944

27

0.0278

0.7500

32

0.0278

0.8889

36

0.0556

2.0000

45

0.0278

1.2500

48

0.0278

1.3333

50

0.0278

1.3889

54

0.0278

1.5000

64

0.0278

1.7778

72

0.0278

2.0000

75

0.0278

2.0833

80

0.0278

2.2222

96

0.0278

2.6667

100

0.0278

2.7778

108

0.0278

3.0000

125

0.0278

3.4722

144

0.0278

4.0000

150

0.0278

4.1667

180

0.0278

5.0000

216

0.0278

6.0000

1.0000

53.0833


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