Math 361 X1
Homework 6 Solutions
Spring 2003
Graded problems:
2(a);4(a)(b);5(b);6(iii); each worth 3 pts., maximal score is 12 pts.
Problem 1.
[3.1:4] Let
X
1
and
X
2
be the numbers obtained on two rolls of a fair die. Let
Y
1
= max(
X
1
,X
2
)
and
Y
2
= min(
X
1
,X
2
). Find the joint distribution of (a) (
X
1
,X
2
), (b) (
Y
1
,Y
2
).
Solution.
(a)
Values:
The values of
X
1
and
X
2
are 1
,... ,
6, so the joint distribution is a 6
×
6 matrix with
entries indexed by 1
,
2
,... ,
6.
Computation of probabilities
P
(
k,l
) =
P
(
X
1
=
k,X
2
=
l
)
:
This is easy:
P
(1
,
1) =
P
(1 appears in ﬁrst roll and 1 appears in second roll) =
±
1
6
²
2
=
1
36
,
and the same calculation gives
P
(
k,l
) =
1
36
(
k,l
= 1
,
2
,... ,
6)
Check:
Since there are 36 entries and each is 1
/
36, the sum of all entries is 1.
(b)
Values:
The possible values of
Y
1
and
Y
2
are again 1
,... ,
6, so the joint distribution is a 6
×
6
matrix with entries indexed by 1
,
2
,... ,
6.
Computation of probabilities
P
(
k,l
) =
P
(
Y
1
=
k,Y
2
=
l
)
:
P
(1
,
1) =
P
(
Y
1
= 1
,Y
2
= 1) =
P
(max(
X
1
,X
2
) = 1
,
min(
X
1
,X
2
) = 1)
=
P
(
X
1
= 1
,X
2
= 1) =
1
36
,
P
(2
,
1) =
P
(
Y
1
= 2
,Y
2
= 1) =
P
(max(
X
1
,X
2
) = 2
,
min(
X
1
,X
2
) = 1)
=
P
(
X
1
= 1
,X
2
= 2) +
P
(
X
1
= 2
,X
2
= 1) =
1
36
+
1
36
=
1
18
,
P
(1
,
2) =
P
(
Y
1
= 1
,Y
2
= 2) =
P
(max(
X
1
,X
2
) = 1
,
min(
X
1
,X
2
) = 2) = 0
,
and in general,
P
(
k,l
) =
P
(
Y
1
=
k,Y
2
=
l
) =
1
/
36 if
l
=
k
1
/
18 if
l < k
0
if
l > k
Check:
The sum of these entries is 6
·
(1
/
36) + 15
·
(1
/
18) = 1.
1