Math 5080–1
Solutions to homework 2
May 29, 2012
# 7, p. 166. We need
1
P
x1
P
x2
f (x1 , x2 ) to be one. That sum is equal to
=
f (0 , 0) + f (0 , 1) + f (0 , 2) + f (1 , 0) + f (1 , 1) + f (1 , 2) + f (2 , 0) + f (2 , 1) + f (2 , 2)
=
0 + c + 2c + c + 2c + 3c + 2c + 3c + 4c
=
18c
⇒
c = 1/18.
# 8, p. 166.
P P
(a) We want
1=c
x
y
f (x , y) = 1; that is,
∞
∞ X
X
2x+y
x=0 y=0
x!y!
=c
∞
∞
X
2x X 2y
x=0
x!
y=0
y!
= ce2 e2 = ce4
⇒
c = e−4 .
(b) The marginal PMF of X is: If x is not a positive integer then fX (x) = 0;
else
∞
∞
X
2x X 2y
e2 2 x
2x
f (x , y) = e−4
fX (x) =
= e−4 e2 =
.
x! y=0 y!
x!
x!
y=0
That is, X ∼ Poisson(2). Similarly, Y ∼ Poisson(2).
(c) Yes. They are independent since the marginal PMF factorizes as a function
of x only times a function of y only.
# 14, p. 167.
R∞
R∞
(a) f1 (x) = −∞ f (x , y) dy = 0 e−(x+y) dy · I{x > 0} = e−x I{x > 0}.
Similarly, f2 (y) = e−y I{y > 0}. This shows that X ∼ Exp(1), Y ∼
Exp(1), and X and Y are independent [by factorization].
(b) F (x , y) = P {X ≤ x , Y ≤ y}. This probability is zero unless both x and
y are ≥ 0. In the case that x ≥ 0 and y ≥ 0,
Z Z
Z x
Z y
−a
F (x , y) =
f (a , b) da db =
e da
e−b dy = 1 − e−x 1 − e−y .
a≤x
b≤y
0
0
1
R∞
R∞
e−x dx = e−2 .
RR
R∞
R∞
R∞
(d) P {X < Y } = 0<x<y e−x e−y dx dy = 0 e−x ( x e−y dy) dx = 0 e−2x dx =
1/2.
(c) P {X > 2} =
f1 (x) dx =
2
2
(e) Perhaps it is easier to first compute P {X + Y ≤ 2}, since it is
Z 2 Z 2−x
Z 2
Z
−y
e dy e−x dx =
P {X+Y ≤ 2} =
1 − ex−2 e−x dx =
0
0
0
2
0
which is 1 − 3e−2 . Therefore, P {X + Y > 2} = 3e−2 .
(f) Yes, by factorization.
# 16, p. 167. The roots of this quadratic equation are
√
p
−2X ± 4X 2 − 4Y
t=
= −X ± X 2 − Y .
2
The roots are real if and only if X 2 ≥ Y , which has probability
!
Z 1 Z x2
Z 1 2
1
x
2
P {X ≥ Y } =
fY (y) dy fX (x) dx =
dx = .
3
−1
0
−1 2
# 19, p. 168.
(a) We want
Z
1
Z
1
1=k
0
Z
(x + y) dy dx = k
x
0
The integral is equal to
Z
0
1
1
1
x(1 − x) + (1 − x2 ) dx.
2
1
1
3x2
+
dx = .
x−
2
2
2
Therefore, k = 2.
R1
R1
(b) f1 (x) = x f (x , y) dy · I{0 ≤ x ≤ 1} = 2 x (x + y) dy · I{0 ≤ x ≤ 1}, which
is
f1 (x) = 2x + 1 − 3x2 · I{0 ≤ x ≤ 1}.
Ry
Similarly, f2 (y) = 2 0 (x + y) dx · I{0 ≤ y ≤ 1} = 3y 2 · I{0 ≤ y ≤ 1}.
(c) There are several cases; you need to draw the region of integration in each
case in order to follow the computations correctly:
• If 0 ≤ x ≤ y ≤ 1, then
Z x Z y
Z x Z y
F (x , y) =
f (a , b) da db = 2
(a + b) da db
0
b
0
b
Z x 2
y − b2
2
=2
+ b(y − b) db = y x + yx2 − x3 .
2
0
2
Z
e−x dx−
0
2
e−2 dx,
• If 0 ≤ y ≤ x, then
Z y Z y
Z y Z y
(a + b) da db
f (a , b) da db = 2
F (x , y) =
b
0
b
0
Z y 2
y − b2
=2
+ b(y − b) db = y 3 .
2
0
• If 0 ≤ x ≤ 1 and y ≥ 1, then
Z x Z 1
Z x Z 1
(a + b) db da
f (a , b) db da = 2
F (x , y) =
a
0
a
0
Z x
1 − a2
=2
a(1 − a) +
da = x2 + x − x3 .
2
0
• If x, y ≥ 1 [both], then F (x , y) = 1.
• In all other cases, F (x , y) = 0.
(d) Note that if 0 ≤ x ≤ 1, then
Z
1
(x + y) dy = 2x(1 − x) + 1 − x2 = 2x + 1 − 3x2 .
fX (x) = 2
x
Therefore, by definition [as a function of y],
fY |X (y | x) =
2(x + y)
I{x ≤ y ≤ 1}.
2x + 1 − 3x2
(e) Note that if 0 ≤ y ≤ 1, then
y
Z
(x + y) dx = 3y 2 .
fY (y) = 2
0
Therefore, by definition [as a function of x],
fX|Y (x | y) =
2(x + y)
I{0 ≤ x ≤ y}.
3y 2
3