MATH 5010–001 SUMMER 2003
SOLUTIONS TO ASSIGNMENT 8
Problems from pp. 290–295:
1. (a) X = the maximum, and Y = the sum of the dots on a roll of two independently1
for 1 ≤ j ≤ 36, whereas
thrown dice. Let p(x, y) = P {X = x, Y = y}. Then p(j, 2j) = 36
2
p(i, j) = 36 if 1 ≤ i ≤ 6 and i < j < 2i.
1. (b) X = value of the first die, and Y = maximum of the two values. Let p(x, y) =
j
1
P {X = x, Y = y}. Then, p(i, j) = 36
if 1 ≤ i < j ≤ 6, and p(j, j) = 36
if 1 ≤ j ≤ 6.
[E.g., p(3, 3) is the probability of tossing a three for the first die and a three or less for the
second.]
1. (c) X = the maximum, and Y = the minimum. Let p(x, y) = P {X = x, Y = y}. Then,
2
1
if i > j and 1 ≤ i, j ≤ 6. On the other hand, p(i, i) = 36
for all 1 ≤ i ≤ 6.
p(i, j) = 36
9. (a)
f (x, y) =
6 2 xy x +
7
2
0 < x < 1, 0 < y < 2.
R1R2
Because f (x, y) ≥ 0, you need to check only that 0 0 f (x, y) dy dx = 1. First, integrate
the inside integral:
Z 2 6 2 xy 6 2
x +
dy =
2x + x .
(1)
2
7
0 7
The integral of this (dx) from x = 0 to x = 1 is one.
9. (b) Integrate out y:
Z
fX (x) =
2
0
6 2
6 2 xy x +
dy =
2x + x ,
7
2
7
0 < x < 1,
by equation (1) above.
9. (c) Plot the region to find that
Z
P {X > Y } =
1
0
Z
x
0
6 2 xy 15
x +
dy dx =
.
7
2
56
9. (d) By part (b),
P
1
X<
2
Z
1/2
=
0
6 2
5
2x + x dx =
.
7
28
1
Also,
P
1
1
Y > , X<
2
2
Therefore,
P
Z
1/2
Z
2
=
0
1/2
1
1
Y >
X<
2
2
6 2 xy 69
x +
dy dx =
.
7
2
448
=
69/448
69
=
.
5/28
80
9. (e) By part (b),
Z
E[X] =
0
1
6
xfX (x) dx =
7
Z
1
0
5
2x3 + x2 dx = .
7
9. (f ) First, integrate out the dx to get fY (y):
6
fY (y) =
7
Z
0
1
h
xy i
6 1 y
2
x +
dx =
+
,
2
7 3 4
Therefore,
6
E[Y ] =
7
Z
2
0
y y2
+
3
4
dy =
0 < y < 2.
8
.
7
R∞
49. (a) For any j = 1, . . . , 5, and for all a > 0, P {Xj > a} = a λe−λx dx = e−λa . Now
min(X1 , . . . , X5 ) > a means that all of the Xj ’s are at least a. Therefore,
P {min (X1 , . . . , X5 ) ≤ a} = 1 − P {X1 > a, . . . , X5 > a}
= 1 − [P {X1 > a} × · · · × P {X5 > a}] = 1 − e−5λa .
49. (b) Similarly, max(X1 , . . . , X5 ) ≤ a means that all Xj ’s are at most a. Therefore,
P {max (X1 , . . . , X5 ) ≤ a} = P {X1 ≤ a} × · · · × P {X5 ≤ a} = 1 − e−λa
2
5
.