HONR 399
October 25, 2010
Chapter 24 Answers
1. Consider the joint density of X and Y from Example 22.1 of Chapter 22, namely,
1
fX,Y (x, y) = (1 − x2 )(3 − y)
8
for −1 ≤ x ≤ 1 and −1 ≤ y ≤ 1,
and fX,Y (x, y) = 0 otherwise.
Find E[X].
The density of X is, for −1 ≤ x ≤ 1,
Z 1
1
1
1
3
fX (x) =
(1 − x2 )(3 − y) dy = (1 − x2 ) (3y − y 2/2)y=−1 = (1 − x2 )
8
4
−1 8
and fX (x) = 0 otherwise. So the expected value of X is
Z
Z 1
1
3 2
3
3 1
2
(x − x3 ) dx =
(x)
(1 − x ) dx =
x /2 − x4/4)x=−1 = 0.
E[X] =
4
4 −1
4
−1
Find E[Y ].
The density of Y is, for −1 ≤ y ≤ 1,
Z 1
1
1
1
1
fY (y) =
(1 − x2 )(3 − y) dx = (x − x3/3)(3 − y)x=−1 = (3 − y)
8
6
−1 8
and fY (y) = 0 otherwise. So the expected value of Y is
Z 1
Z
1
1
1 1
1
(3y − y 2 ) dy = (3y 2/2 − y 3/3)y=−1 = −1/9.
E[Y ] =
(y)
(3 − y) dy =
6
6 −1
6
−1
1
2. Consider the joint density of X and Y from Example 22.3 of Chapter 22, namely,
3
for 0 ≤ x and 0 ≤ y and x + y ≤ 2,
fX,Y (x, y) = xy
2
and fX,Y (x, y) = 0 otherwise.
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Find E[Y ].
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Figure 1: The joint density is defined as fX,Y (x, y) = 32 xy for x, y in this triangular region.
For each y with 0 ≤ y ≤ 2, the joint density fX,Y (x, y) is defined as fX,Y (x, y) = 32 xy
when 0 ≤ x ≤ 2 − y. This should be clear from the picture below.
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0x:;l
f,-t o+
i
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X
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Figure 2: If 0 ≤ y ≤ 2, then the joint density is defined as fX,Y (x, y) = 32 xy for 0 ≤ x ≤ 2−y.
Thus, the density of Y is, for 0 ≤ y ≤ 2,
2−y
Z 2−y
3
3 x2
3
fY (y) =
xy dx =
(y)
= (2 − y)2 (y)
2
2 2
4
0
x=0
and fY (y) = 0 otherwise. So the expected value of Y is
Z 2 Z
2
3 2 2
3
3
2
E[Y ] =
(y)
(2−y) (y)dy =
(4y −4y 3 +y 4 )dy = (4y 3/3 − 4y 4/4 + y 5/5)y=0 = 4/5.
4
4 0
4
0
2
3. Consider the joint density of X and Y from Example 23.1 of Chapter 23, namely:
A bird lands in a grassy region described as follows: 0 ≤ x, and 0 ≤ y, and x + y ≤ 10.
Let X and Y be the coordinates of the bird’s landing. Assume that X and Y have the joint
density
fX,Y (x, y) = 1/50
for 0 ≤ x and 0 ≤ y and x + y ≤ 10,
and fX,Y (x, y) = 0 otherwise.
Find E[Y ].
[Hint: The answer is not “E[Y ] = 5”.]
$
$
lo
lo
Figure 3: The joint density is fX,Y (x, y) = 1/50 for x, y in this triangle.
u4,
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*+
?>
For each y with 0 ≤ y ≤ 10, we have fX,Y (x, y) = 1/50 for 0 ≤ x ≤ 10 − y. So we get
X
u4,
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X
Figure 4: If 0 ≤ y ≤ 10, the joint density is fX,Y (x, y) = 1/50 for 0 ≤ x ≤ 10 − y.
Thus, the density of Y is, for 0 ≤ y ≤ 10,
Z 10−y
fY (y) =
1/50 dx = (10 − y)/50
0
and fY (y) = 0 otherwise. So the expected value of Y is
Z 10
Z 10
10
1
1
E[Y ] =
(y)(10 − y)/50 dy =
(10y − y 2 ) dy =
(10y 2/2 − y 3/3)y=0 = 10/3.
50 0
50
0
3
4. Consider two points that are independently placed on a line of length 10, at locations X
and Y . Thus the joint density of X and Y is
fX,Y (x, y) = 1/100
for 0 ≤ x ≤ 10 and 0 ≤ y ≤ 10,
and fX,Y (x, y) = 0 otherwise.
First, try to show that, if Z denotes the distance between X and Y (so 0 ≤ Z ≤ 10),
then the cumulative distribution function FZ (z) of Z is
1 2
1
z
for 0 ≤ z ≤ 10
FZ (z) = z −
5
100
and FZ (z) = 0 for z ≤ 0 and FZ (z) = 1 for z ≥ 10. [Hint: Use the complement; try to find
P (Z > z), i.e., the probability that the distance between X and Y is more than z. A picture
should help.]
Since X and Y are always between 0 and 10, then the distance Z between X and Y is
always between 0 and 10 too. So FZ (z) = 0 for z ≤ 10 and FZ (z) = 1 for z ≥ 10. All we
need to consider now is 0 ≤ z ≤ 10. For any such z, we see that X and Y are more than z
apart if X and Y are found in either of the shaded triangles below. The area of each triangle
14
0-
b-7
lo
-?
\0
= [
I o- Z
Figure 5: The regions where X and Y are more than z apart.
is (10 − z)2 /2, and the joint density is constant, so we do not even need to integrate. We
get:
FZ (z) = P (Z ≤ z) = 1 − P (Z > z) = 1 −
4
(10 − z)2 /2 (10 − z)2 /2
z
z2
−
= −
.
100
100
5 100
Whether or not you can accomplish the part above, just find fZ (z), by differentiating FZ (z).
We know fZ (z) =
d
F (z).
dz Z
d
fZ (z) =
dz
Thus
1
1 2
z−
z
5
100
=
1
z
−
5 50
for 0 ≤ z ≤ 10,
and fZ (z) = 0 otherwise.
Use the density fZ (z) to find E[Z], i.e., the expected distance between X and Y .
We compute
Z
E[Z] =
10
(z)
0
1
z
−
5 50
Z
dz =
0
10
z
z2
−
5 50
dz =
10
z2
z 3 −
= 10/3.
10 150 z=0
5. Review question:
Consider the baseball scenario discussed in Example 22.4 of Chapter 22, namely:
Each time a pitcher delivers a fast ball, the speed is distributed between 90 and 100 miles
per hour, with density 1/10. Assume that the speeds of pitches are independent.
On Monday, the pitcher throws fast balls repeatedly, until the very first fast ball that
exceeds 98.7 miles per hour. How many fast balls does he expect to throw, to accomplish
this task?
We see that the probability an individual pitch exceeds 98.7 miles per hour
R 100is 1.3/10 = .13.
[For those who love to integrate instead, we can get the .13 by integrating 98.7 1/10 dt = .13,
just temporarily using T as the speed of an isolated fast ball].
The number of fast balls that he expects to throw until one exceeds 98.7 miles per hour
is a geometric random variable with p = .13, so the expected number that he expects to
throw until one exceeds 98.7 miles per hour is 1/p = 1/.13 = 100/13 ≈ 7.69230.
On Tuesday, the pitcher throws exactly 15 fast balls. How many of them does he expect
to exceed 98.7 miles per hour?
The number of fast balls that exceed 98.7 miles per hour—out of 15 fast balls total—is a
binomial random variable with n = 15 and p = .13, so the expected number that he expects
to exceed 98.7 miles per hour is np = (15)(.13) = 1.95.
5