Let us roll a die twice. The number from the first throw is denoted by X1 and the number from the second throw is denoted by X2. Let us define a new variable Y by Y(x1+x2-5)^2. * Make a table that shows all cases of Y with their probabilities. * What is the expectation value of Y, i.e. E[Y]? * Obtain the variance of Y, V[Y]. Solution Let us roll a die twice. The number from the first throw is denoted by X1 and the number from the second throw is denoted by X2. Let us define a new variable Y by Y(x1+x2-5)^2. * Make a table that shows all cases of Y with their probabilities. * What is the expectation value of Y, i.e. E[Y]? * Obtain the variance of Y, V[Y]. We can use the following: Sum of two die Probability 2 1/36 3 2/36 = 1/18 4 3/36 = 1/12 5 4/36 = 1/9 6 5/36 7 6/36 = 1/6 8 5/36 9 4/36 = 1/9 10 3/36 = 1/12 11 2/36 = 1/18 12 1/36 Total 36/36 sum of dice y p y^2 y*p y^2*p 2 9 1/36 81 0.25 2.25 3 4 1/18 16 0.222222 0.888889 4 1 1/12 1 0.083333 0.083333 5 0 1/9 0 0 0 6 1 5/36 1 0.138889 0.138889 7 4 1/6 16 0.666667 2.666667 8 9 5/36 81 1.25 11.25 9 16 1/9 256 1.777778 28.44444 10 25 1/12 625 2.083333 52.08333 11 36 1/18 1296 2 72 12 49 1/36 2401 1.361111 66.69444 sums: 1 9.833333 236.5 To get the y column, subtract 5 from the sum column and then square. The second column and the third column of this table show the possible values of y and their probabilities. E(Y) and V(Y) are calculated using the sums in the last row of this table. E(Y) = 9.8333 Sum of two die Probability 2 1/36 3 2/36 = 1/18 4 3/36 = 1/12 5 4/36 = 1/9 6 5/36 7 6/36 = 1/6 8 5/36 9 4/36 = 1/9 10 3/36 = 1/12 11 2/36 = 1/18 12 1/36 Total 36/36.

1. Let us roll a die twice. The number from the first throw is denoted by X1 and the number from
the second throw is denoted by X2. Let us define a new variable Y by Y(x1+x2-5)^2.
* Make a table that shows all cases of Y with their probabilities.
* What is the expectation value of Y, i.e. E[Y]?
* Obtain the variance of Y, V[Y].
Solution
Let us roll a die twice. The number from the first throw is denoted by X1 and the number from
the second throw is denoted by X2. Let us define a new variable Y by Y(x1+x2-5)^2.
* Make a table that shows all cases of Y with their probabilities.
* What is the expectation value of Y, i.e. E[Y]?
* Obtain the variance of Y, V[Y].
We can use the following:
Sum of two die
Probability
2
1/36
3
2/36 = 1/18
4
3/36 = 1/12
5
4/36 = 1/9
6
5/36
7
6/36 = 1/6
8
5/36
9
4/36 = 1/9
10
3/36 = 1/12
11

2. 2/36 = 1/18
12
1/36
Total
36/36
sum of dice
y
p
y^2
y*p
y^2*p
2
9
1/36
81
0.25
2.25
3
4
1/18
16
0.222222
0.888889
4
1
1/12
1
0.083333
0.083333
5
0
1/9
0
0
0
6

3. 1
5/36
1
0.138889
0.138889
7
4
1/6
16
0.666667
2.666667
8
9
5/36
81
1.25
11.25
9
16
1/9
256
1.777778
28.44444
10
25
1/12
625
2.083333
52.08333
11
36
1/18
1296
2
72
12

4. 49
1/36
2401
1.361111
66.69444
sums:
1
9.833333
236.5
To get the y column, subtract 5 from the sum column and then square.
The second column and the third column of this table show the possible values of y and their
probabilities.
E(Y) and V(Y) are calculated using the sums in the last row of this table.
E(Y) = 9.8333
Sum of two die
Probability
2
1/36
3
2/36 = 1/18
4
3/36 = 1/12
5
4/36 = 1/9
6
5/36
7
6/36 = 1/6
8
5/36
9
4/36 = 1/9
10
3/36 = 1/12
11
2/36 = 1/18

5. 12
1/36
Total
36/36