Math 5010-1, Spring 2005
Assignment 8
Problems
#2, p. 290.
(a) Write p(x, y) = P {X1 = x, X2 = y}, and Ri and Wi for the events that the ith
draw is respectively red or white. Note that
8
7
×
12 13
8
5
p(0, 1) = P (R1 )P (W2 | R1 ) =
× ,
13 12
5
8
× ,
p(1, 0) = P (W1 )P (R2 | W1 ) =
13 12
5
4
p(1, 1) = P (W1 )P (W2 | W1 ) =
× .
13 12
(b) Next, write p(a, b, c) = P {X1 = a, X2 = b, X3 = c} to find that
p(0, 0) = P (R2 | R1 )P (R1 ) =
8
7
6
×
× ,
13 12 11
8
7
5
p(0, 0, 1) =
×
× ,
13 12 11
5
7
8
×
× ,
p(0, 1, 0) =
13 12 11
5
8
7
p(1, 0, 0) =
×
× ,
13 12 11
5
8
4
p(1, 0, 1) =
×
× ,
13 12 11
4
9
5
×
× ,
p(1, 1, 0) =
13 12 11
8
5
4
p(0, 1, 1) =
×
× ,
13 12 11
5
4
3
p(1, 1, 1) =
×
× .
13 12 11
[These are it, because of the principle of counting.]
p(0, 0, 0) =
#8, p. 290.
(a) Want
RR
f (x, y) dx dy = 1. In this case, dx dy is easier to compute. So,
Z ∞Z ∞
Z ∞Z y
f (x, y) dx dy = c
(y 2 − x2 )e−y dx dy
−∞ −∞
0
−y
Z
2 3 −y
∞
3
=c 0
2y − y e dy
3
Z ∞
4c
4c
=
y 3 e−y dy = Γ(4).
3 0
3
1
Γ(4) = 3! = 6. Therefore, c = 1/8.
(b) First fX :
Z
1 ∞ 2
y − x2 e−y dy
(−∞ < x < ∞)
fX (x) =
8 |x|
!
Z ∞
Z ∞
1
e−y dy
=
y 2 e−y dy − x2
8
|x|
|x|
!
Z ∞
1
=
y 2 e−y dy − x2 e−|x| .
8
|x|
The remaining integral is computed by two applications of integration-by-parts:
Z
Z
2 −y
2 −y
y e dy = −y e + 2 ye−y dy
Z
2 −y
−y
−y
= −y e + 2 −ye + e dy
= −y 2 e−y − 2ye−y − 2e−y .
Therefore,
i
1 h 2 −|x|
−|x|
−|x|
fX (x) =
x e
+ 2|x|e
+ 2e
8
(−∞ < x < ∞).
The fY -computation is easier:
Z
1 y 2
y − x2 e−y dx
fY (y) =
8 −y
1
1
= y 3 e−y − y 3 e−y
(y > 0).
4
12
(c) E[X] =
R∞
−∞
xfX (x) dx, so E[X] = 0 by symmetry.
#14, p. 291. Let X and Y denote the distance to 0 of the accident and the ambulance, respectively. We know that X and Y are independent, both uniformly distributed on [0, 1].
Therefore,
1/L, if 0 < x < L
1/L, if 0 < y < L,
fX (x) =
, and fY (y) =
0,
otherwise
0,
otherwise.
Thanks to independence,
f (x, y) =
1/L2 , if 0 < x < L and 0 < y < L,
0,
otherwise.
Therefore, for all 0 < a < L,
Z Z
P {|X − Y | ≤ a} =
f (x, y) dx dy,
R
2
where R is the region between y = x + a and y = x − a in the square of side L. The
only values of interest for a are 0 ≤ a ≤ L. So,
Z LZ L
Z a Z x+a
−2
−2
dy dx
dy dx + L
P {|X − Y | ≤ a} = L
0
0
= L−2
a
Z
(x + a) dx + L−2
0
=
Z
a
L
x−a
(L − x + a) dx
a
2a2 + L2
.
2L2
Therefore, the pdf of |X − Y | is
f|X−Y | (a) =
2a/L2 , if 0 ≤ a ≤ L,
0,
otherwise.
In particular, the ambulance is expected to travel
Z L
2
2L
E |X − Y | = 2
a2 da =
units.
L 0
3
#15, p. 291.
(a) Integrate.
(b) In this case, f (x, y) = 1/4 when −2 < x < 2 and −2 < y < 2 (and zero,
otherwise). Compute the marginal s:
Z 2
1
1
fX (x) =
dy =
(−2 < x < 2)
2
−2 4
Z 2
1
1
fY (y) =
dx =
(−2 < y < 2).
2
−2 4
Therefore, in all cases, f (x, y) = fX (x)fY (y).
(c) In the general uniform case P {(X, Y ) ∈ A} = area(A)/area(R). Here, area(R) =
4, A = π, so P {X 2 + Y 2 ≤ 1} = π/4.
#17, p. 292. Implict here are the assumptions that the Xi ’s are uniform on [0, L], and independent
from one another. Note that whenever i, j, and k are all different P {Xi < Xj <
Xk } = P {Xj < Xk < Xi } = P {Xk < Xi < Xj } = · · ·. Since there are 3! = 6 such
possibilities, and because they are all equi-probable, they must add up to one; i.e.,
P {X1 < X2 < X3 } = 1/6.
#20, p. 292. In the second case, the answer is a resounding, “no.” This is because, there, P {X <
Y } = 1. (Work out the details in terms of the pdf’s.) In the first case, compute
marginals (do it for the second case, too):
Z ∞
−x
fX (x) = xe
e−y dy = xe−x (x > 0)
0
Z ∞
fY (y) = e−y
xe−x dx = e−y Γ(2) = e−y (y > 0).
0
3
Because fX (x)fY (y) = f (x, y), X and Y are independent.
Theoretical Exercises
#14, p. 297. This was worked out during lecture.
4