Problem Set 4 With Solutions
:
ECO 207: 3 and 4
8 October 2019
1.
What is the expected value for a game in which you roll two fair dice and get paid the
amount in the lower number if the two numbers, or the number if they are equal.
Ans: If X is the random variable which is the number displayed when one fair dice is thrown,
then the expected value of X is
E(X) = (1 + 2+ 3+ 4+ 5+ 6)/6
= 21/6
= 7/2 or 3.5
If 2 fair dice are thrown, then the possible outcomes are:
(1,1)
....
(1,6)
.
.
(6,1) ... (6,6)
If Y is the random variable which is equal to the lower number, then
E(Y) = [(11*1) + (9*2) + (7*3) + (5*4) + (3*5) + (1*6)]/36
= (11 + 18+ 21+ 20 + 15 + 6)/36
= 91/36
= approx 2.5278
2.
The joint distribution of X and Y is as follows. X takes 3 values: 0,1,2 and Y (1,2, …6)
Y
X
0
1
2
1
0.025
0.05
0.025
2
0.025
0.05
0.025
3
0.025
0.05
0.025
4
0.025
0.05
0.025
5
0.05
0.1
0.05
6
0.1
0.2
0.1
1

Find the marginal PMFs of X and Y. Also find E(X), E (Y) and E[min(X, Y, 1)]. What is E (X|Y =6)
Answer: The reason they are called marginal distributions is that they are written into the
margins of the table. So just make a column for the total of Y and a row for the total of X and
add across rows and down columns. Also add the marginal totals to make sure they are both 1.
I’ll do Y, you do X. I’ll write the Y totals as a row, but they should be in a column. For Y = 1, 2, …,
6 you should get 0.1, 0.1, 0.1, 0.1, 0.2 and 0.4. And the sum
is
1.

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