••tí
6. Particular consideración del problema de la identidad del pago y de la calidad de la cosa debida. Delimitación del objeto
Basic Exercises 5.1 Answers will vary.
5.2
a) Yes, X is a random variable because it is a real-valued function whose domain is the sample space of a random experiment, namely, the random experiment of tossing a balanced die twice and observing the two faces showing. Yes, X is a discrete random variable because it is a random variable with a finite (and, hence, countable) range, namely, the set {2, 3, . . . , 12}.
b)–f) We have the following table:
Part Event The sum of the two faces showing . . . Subset of the sample space
b) {X = 7} is seven. {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
c) {X > 10} exceeds 10 (i.e., is either 11 or 12). {(5, 6), (6, 5), (6, 6)}
d) {X = 2 or 12} is either 2 or 12. {(1, 1), (6, 6)}
e) {4 ≤ X ≤ 6} is four, five, or six. {(1, 3), (2, 2), (3, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
f) {X ∈ A} is an even integer. {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}
5.3
a) Yes, X is a random variable because it is a real-valued function whose domain is the sample space of a random experiment, namely, the random experiment of observing which of five components of a unit are working and which have failed. Yes, X is a discrete random variable because it is a random variable with a finite (and, hence, countable) range, namely, the set {0, 1, 2, 3, 4, 5}.
b)–e) We have the following table:
Part Event In words . . . Subset of the sample space
b) {X ≥ 4} At least four (i.e., four or five) of the {(s, s, s, s, f ), (s, s, s, f, s), (s, s, f, s, s), components are working. (s, f, s, s, s), (f, s, s, s, s), (s, s, s, s, s)} c) {X = 0} None of the components are working. {(f, f, f, f, f )}
d) {X = 5} All of the components are working. {(s, s, s, s, s)}
e) {X ≥ 1} At least one of the components is working. {(f, f, f, f, f )}c 5-1
f) {X = 2} g) {X ≥ 2} h) {X ≤ 2} i) {2 ≤ X ≤ 4}
5.4
a) Let m denote male and f denote female. Then we can represent each possible outcome as an ordered triple, where each entry of the triple is either m or f . For instance, (m, f, f ) represents the outcome that the first born is male, the second born is female, and the third born is female. Therefore, a sample space for this random experiment is ! = { (x1, x2, x3): xj ∈ {m, f } , 1 ≤ j ≤ 3 }.
b) We have the following table:
ω (m, m, m) (m, m, f ) (m, f, m) (m, f, f ) (f, m, m) (f, m, f ) (f, f, m) (f, f, f )
Y (ω) 0 1 1 2 1 2 2 3
c) Yes, Y is a random variable because it is a real-valued function whose domain is the sample space of a random experiment. Yes, Y is a discrete random variable because it is a random variable with a finite (and, hence, countable) range, namely, the set {0, 1, 2, 3}.
5.5 Take ! = { (x, y) : −3 ≤ x ≤ 3, −3 ≤ y ≤ 3 } and, for brevity, set d(x, y) =!
x2+ y2. a) We have
S(x, y)=" 10, if d(x, y) ≤ 1;
5, if 1 < d(x, y) ≤ 2;
0, otherwise.
b) Yes, S is a discrete random variable because it is a random variable with a finite (and, hence, countable) range, namely, the set {0, 5, 10}.
c)–h) We have the following table:
Part Event The archer’s score is . . . Subset of the sample space c) {S = 5} 5 points; that is, she hits the ring with inner
radius 1 foot and outer radius 2 feet centered at the origin.
{ (x, y) ∈ ! : 1 < d(x, y) ≤ 2 }
d) {S > 0} positive; that is, she hits the disk of radius
2 feet centered at the origin. { (x, y) ∈ ! : d(x, y) ≤ 2 } e) {S ≤ 7} at most 7 points; that is, she does not hit the
bull’s eye. { (x, y) ∈ ! : d(x, y) > 1 }
f) {5 < S ≤ 15} more than 5 points and at most 15 points;
that is, she hits the bull’s eye. { (x, y) ∈ ! : d(x, y) ≤ 1 } g) {S < 15} less than 15 points, which is certain. !
h) {S < 0} negative, which is impossible. ∅
5.6
a) Let n denote nondefective and d denote defective. Then we can represent each possible outcome as an ordered five-tuple, where each entry of the five-tuple is either n or d. For instance, (n, d, n, n, d) represents the outcome that the first, third, and fourth parts selected are nondefective and the second and fifth parts selected are defective. Thus, a sample space for this random experiment is
!=#
(x1, x2, x3, x4, x5): xj ∈ {n, d} , 1 ≤ j ≤ 5$ . b) We have Y = 1 if X ≤ 1, and Y = 0 if X > 1.
5.7
a) Let r denote red and w denote white. Then we can represent each possible outcome as an ordered triple, where the first entry of the triple is the number of the urn chosen, and the second and third entries are the colors of the first and second balls selected, respectively. For instance (II, r, w) represents the outcome that Urn II is chosen, the first ball selected is red, and the second ball selected is white. Hence, a possible sample space is ! = {(I, r, r), (I, r, w), (I, w, r), (II, r, w), (II, w, r), (II, w, w)}.
b) We have the following table:
ω (I, r, r) (I, r, w) (I, w, r) (II, r, w) (II, w, r) (II, w, w)
X (ω) 2 1 1 1 1 0
5.8
a) As the interview pool consists of three people, the number of women in the interview pool must be between zero and three, inclusive. Hence, the possible values of the random variable X are 0, 1, 2, and 3.
b) The number of ways that the interview pool can contain exactly x women (x = 0, 1, 2, 3) is the number of ways that x women can be chosen from the five women and 3 − x men can be chosen from the six men. The number of ways that x women can be chosen from the five women is%5
x
&
, and the number of ways that 3 − x men can be chosen from the six men is% 6
3−x
&
. Consequently, by the BCR, the number of ways that the interview pool can contain exactly x women equals%5
x
&% 6
3−x
&
. We, therefore, have the following table:
x Number of ways {X = x} can occur
0 '5
0 ('6
3 (
= 1 · 20 = 20
1 '5
1 ('6
2 (
= 5 · 15 = 75
2 '5
2 ('6
1 (
= 10 · 6 = 60
3 '5
3 ('6
0 (
= 10 · 1 = 10
c) {X ≤ 1} is the event that at most one (i.e., 0 or 1) woman is in the interview pool. From the table in part (b), we see that the number of ways that event can occur is 20 + 75 = 95.
5.9
a) We have the following table:
ω HHH HHT HTH HTT THH THT TTH TTT
Y (ω) 3 1 1 −1 1 −1 −1 −3
b) The event {Y = 0} is, in words, that the number of heads equals the number of tails, which is impos-sible. Hence, as a subset of the sample space, we have {Y = 0} = ∅.
5.10 Denote the two salads by s1 and s2, the three entrees by e1, e2, and e3, and the two desserts by d1 and d2. Also, let s0, e0, and d0 represent a choice of no salad, entree, and dessert, respectively. Thus, each possible choice of a meal is of the form siejdk, where i, k ∈ {0, 1, 2} and j ∈ {0, 1, 2, 3}. We note that the sample space consists of 3 · 4 · 3 = 36 possible outcomes.
a) We have the following table:
ω X (ω) ω X (ω) ω X (ω) ω X (ω)
s0e0d0 0.00 s0e3d0 3.50 s1e2d0 4.00 s2e1d0 4.00 s0e0d1 1.50 s0e3d1 5.00 s1e2d1 5.50 s2e1d1 5.50 s0e0d2 1.50 s0e3d2 5.00 s1e2d2 5.50 s2e1d2 5.50 s0e1d0 2.50 s1e0d0 1.00 s1e3d0 4.50 s2e2d0 4.50 s0e1d1 4.00 s1e0d1 2.50 s1e3d1 6.00 s2e2d1 6.00 s0e1d2 4.00 s1e0d2 2.50 s1e3d2 6.00 s2e2d2 6.00 s0e2d0 3.00 s1e1d0 3.50 s2e0d0 1.50 s2e3d0 5.00 s0e2d1 4.50 s1e1d1 5.00 s2e0d1 3.00 s2e3d1 6.50 s0e2d2 4.50 s1e1d2 5.00 s2e0d2 3.00 s2e3d2 6.50
b) The event {X ≤ 3} is, in words, that the meal costs $3.00 or less. As as subset of the sample space, we have that
{X ≤ 3} = {s0e0d0, s0e0d1, s0e0d2, s0e1d0, s0e2d0, s1e0d0, s1e0d1, s1e0d2, s2e0d0, s2e0d1, s2e0d2} . 5.11 We can take the sample space to be ! = {w, ℓw, ℓℓw, . . .}, where
ℓ . . . ℓ ) *+ ,
n−1 times
w
represents the outcome that you don’t win a prize the first n − 1 weeks and you do win a prize the nth week; that is, it takes n weeks before you win a prize.
a) Yes, W is a random variable because it is a real-valued function whose domain is the sample space !.
We have
W ( ℓ . . . ℓ) *+ ,
n−1 times
w )= n, n∈ N .
b) Yes, W is a discrete random variable because, as we see from part (a), the range of W is N .
c) {W > 1} is the event that it takes you more than 1 week to win a prize or, equivalently, you don’t win a prize the first week.
d) {W ≤ 10} is the event that you win a prize by the 10th week.
e) {15 ≤ W < 20} is the event that you first win a prize between weeks 15 and 19, inclusive; that is, it takes you at least 15 weeks but less than 20 weeks to win a prize.
5.12
a) Define g(x) = ⌊m + (n − m + 1)x⌋. For 0 < x < 1, we have
m− 1 < m + (n − m + 1)x − 1 < g(x) ≤ m + (n − m + 1)x < n + 1.
Because g is integer valued, it follows that g% (0, 1)&
⊂ {m, m + 1, . . . , n}. However, if k is an integer with m + 1 ≤ k ≤ n, then
0 < k− m
n− m + 1 <1 and g
' k− m n− m + 1
(
= k.
Furthermore,
g
' 1
2(n − m + 1) (
= ⌊m + 1/2⌋ = m.
Thus, g% (0, 1)&
= {m, m + 1, . . . , n} and, therefore, the possible values of Y are m, m + 1, . . . , n.
b) Yes, Y is a discrete random variable because, from part (a), its range is finite and, hence, countable.
c) For u ∈ R, we have ⌊u⌋ ≤ u < ⌊u⌋ + 1. Hence,
Y = y if and only if y ≤ m + (n − m + 1)X < y + 1.
Therefore,
{Y = y} =
- y− m
n− m + 1 ≤ X < y+ 1 − m n− m + 1 .
.
Theory Exercises
5.13 Let X be a random variable with countable range, say, R. Then we have {X ∈ R} = !. Hence, P (X ∈ R) = P (!) = 1,
so that Definition 5.2 holds with K = R.
5.14
a) Suppose to the contrary that there is an n ∈ N such that { x ∈ R : P (X = x) > 1/n } contains more than n − 1 elements. Then there is an m ∈ N with m ≥ n and real numbers x1, . . . , xm such that P (X = xj) >1/n for j = 1, . . . , m. Therefore,
P%
X ∈ {x1, . . . , xm}&
= /m j=1
P (X= xj) > m/n≥ 1.
Thus, P%
X ∈ {x1, . . . , xm}&
>1, which is impossible.
b) We have
{ x ∈ R : P (X = x) ̸= 0 } = 0∞ n=1
{ x ∈ R : P (X = x) > 1/n }.
From part (b), each set in the union is finite and, hence, countable. Therefore, the set on the left of the preceding display is countable, being a countable union of countable sets.
c) For any random variable X, we have P (X = x) = 0 except for countably many real numbers, x.
Advanced Exercises
5.15 We provide two examples of which the first uses the hint. Let ! = [0, 1] and let P be length in the extended sense. Note that P is a probability measure on !. Define X(ω) = ω if ω ∈ C, and X(ω) = 2 otherwise, where C is a set as described in the hint. The range of X is C ∪ {2}, which is uncountable because C is uncountable. Let K = {2}. Then K is countable and
P (X ∈ K) = P (X = 2) = P%
[0, 1] \ C&
= P% [0, 1]&
− P (C) = |[0, 1]| − |C| = 1 − 0 = 1.
Hence, X is a discrete random variable with an uncountable range.
An a second example, let ! = [0, 1] and define P on the events of ! by P (E)=
-1, if 1 ∈ E;
0, if 1 /∈ E.
It is easy to see that P is a probability measure on !. Now, define X(ω) = ω for all ω ∈ !. The range of X is [0, 1], which is uncountable. Let K = {1}. Then K is countable and
P (X∈ K) = P (X = 1) = P ({1}) = 1.
Hence, X is a discrete random variable with an uncountable range.
5.16 From Exercise 5.14, we know that A = { x ∈ R : P (X = x) ̸= 0 } is countable. Let x1, x2, . . . be an enumeration of A. As probabilities are always between 0 and 1, inclusive, we have 0 ≤ P (X ∈ A) ≤ 1, However, by the additivity axiom,
/
x
P (X= x) =/
x∈A
P (X= x) +/
x /∈A
P (X= x) = /∞ n=1
P (X = xn)+ 0
= /∞ n=1
P%
X∈ {xn}&
= P 1
X∈ 0∞ n=1
{xn} 2
= P (X ∈ A).
Hence,
0 ≤/
x
P (X= x) ≤ 1.
5.17 Suppose X is a discrete random variable. Then there is a countable set K such that P (X ∈ K) = 1.
Applying the additivity axiom, we get
1 = P (X ∈ K) = /
x∈K
P (X = x) ≤/
x
P (X = x) ≤ 1, where the last inequality follows from Exercise 5.16. Hence,3
xP (X = x) = 1.
Conversely, suppose3
xP (X= x) = 1. From Exercise 5.14, the set A = {x ∈ R : P (X = x) ̸= 0}
is countable. Applying the additivity axiom, we get 1 =/
x
P (X = x) =/
x∈A
P (X = x) = P (X ∈ A).
Hence, there is a countable set A such that P (X ∈ A) = 1; that is, X is a discrete random variable.
5.18 Suppose that X is a discrete random variable. Then, from Definition 5.2, there is a countable set K ⊂ R such that P (X ∈ K) = 1. From the complementation rule,
P (X ∈ Kc)= 1 − P (X ∈ K) = 1 − 1 = 0, and, in particular then, P (X = x) = 0 if x /∈ K. Let x0 ∈ Kcand define
X0(ω)=
-X(ω), if X(ω) ∈ K;
x0, if X(ω) /∈ K.
The range of X0 is a subset of (perhaps equal to) K ∪ {x0}, which is countable, being the union of two countable sets. Therefore, the range of X0is countable, being a subset of a countable set. We note that
{X0= x} =" {X = x}, if x ∈ K;
{X ∈ Kc} , if x = x0;
∅, otherwise.
Hence,
P (X0 = x) =
"P (X = x), if x ∈ K;
P (X ∈ Kc), if x = x0;
0, otherwise.
=
-P (X = x), if x ∈ K;
0, if x /∈ K.
= P (X = x).
Thus, X0is a random variable with a countable range such that P (X0 = x) = P (X = x) for all x ∈ R.
Conversely, suppose that there is a random variable X0with a countable range such that P (X0 = x) = P (X = x), x ∈ R.
Let R denote the range of X0. Applying the additivity axiom twice, we conclude that P (X∈ R) =/
x∈R
P (X= x) =/
x∈R
P (X0 = x) = P (X0 ∈ R) = P (!) = 1.
Consequently, there is a countable set R ∈ R such that P (X ∈ R) = 1, meaning that X is a discrete random variable.
5.19 The sample space for this random experiment consists of all real numbers between 0 and 1, ex-clusive: ! = (0, 1) = { x : 0 < x < 1 }. Because a number is being selected at random, a geometric probability model is appropriate. Thus, for each event E,
P (E)= |E|
|!| = |E|
1 = |E|, (∗)
where |E| denotes the length (in the extended sense) of the set E.
a) Yes, X is a random variable because it is a real-valued function whose domain is !. In fact, X is defined on ! by X(x) = x; that is, X is the identity function on !.
b) No, X is not a discrete random variable. Indeed, as we will presently show, P (X = x) = 0 for all x ∈ R. This result, in turn, implies that3
xP (X = x) = 0 ̸= 1. Therefore, from Exercise 5.17, X is not a discrete random variable.
To show that P (X = x) = 0 for all x ∈ R, we can proceed as follows. Clearly, the range of X is the interval (0, 1) and, consequently, we have P (X = x) = 0 if x ̸∈ (0, 1). So, from now on, we assume that x ∈ (0, 1). From Equation (∗),
P (X = x) = |{X = x}| = |{x}| = 0.
Alternatively, for n ∈ N sufficiently large, we have [x, x + 1/n] ⊂ (0, 1) = !. Let An= [x, x + 1/n].
Then {X = x} ⊂ {X ∈ An} for all n ∈ N and, hence, from Equation (∗), P (X = x) ≤ P (X ∈ An)= |X ∈ An| = |An| = 1
n, for n sufficiently large. Thus, P (X = x) = 0.
5.20 A sample space for this random experiment is the unit disk: ! = { (x, y) : x2+ y2<1 }. Because the dish is smeared with a uniform suspension of bacteria, use of a geometric probability model is reasonable. Specifically, we can think of the location of the center of the first spot (visible bacteria colony) as a point selected at random from the unit disk. Consequently, for each event E,
P (E)= |E|
|!| = |E|
π , (∗∗)
where |E| denotes the area (in the extended sense) of the set E.
a) Yes, Z is a random variable because it is a real-valued function whose domain is !. In fact, Z is defined on ! by Z(x, y) =!
x2+ y2.
b) No, Z is not a discrete random variable. Indeed, as we will presently show, P (Z = z) = 0 for all z ∈ R. This result, in turn, implies that3
zP (Z= z) = 0 ̸= 1. Therefore, from Exercise 5.17, Z is not a discrete random variable.
To show that P (Z = z) = 0 for all z ∈ R, we can proceed as follows. Clearly, the range of Z is the interval [0, 1) and, hence, P (Z = z) = 0 if z ̸∈ [0, 1). So, from now on, we assume that z ∈ [0, 1). The event {Z = z} is that the center of the first spot to appear is z units from the center of the petri dish or,
equivalently, that the center of the first spot to appear lies on the circle C =#
(x, y): x2+ y2 = z2$ . Hence, from Equation (∗∗),
P (Z = z) = P (C) = |C|
π = 0,
where the last equality follows from the fact that the area of a circle (boundary of a disk) is 0. Alterna-tively, let
An=
-(x, y): z ≤4
x2+ y2 ≤ z + 1/n .
, n∈ N ,
and note that An⊂ ! for n sufficiently large. As {Z = z} ⊂ {z ≤ Z ≤ z + 1/n} = Anfor all n ∈ N , we have, in view of Equation (∗∗), that
P (Z = z) ≤ P (An)= |An|
π = π(z+ 1/n)2− πz2
π = 2z
n + 1 n2, for sufficiently large n. Hence, P (Z = z) = 0.
5.21 We can take the sample space for this random experiment to be all positive real numbers; that is, ! = (0, ∞) = { x ∈ R : x > 0 }. The outcome of the random experiment is x means that the duration of the total solar eclipse is x minutes.
a) Note that X is a random variable because it is a real-valued function whose domain is !. In fact, X is defined on ! by X(x) = x; that is, X is the identity function on !. However, we claim that X is not a discrete random variable. Indeed, as we will presently show, P (X = x) = 0 for all x ∈ R. This result, in turn, implies that 3
xP (X= x) = 0 ̸= 1. Hence, from Exercise 5.17, X is not a discrete random variable.
To show that P (X = x) = 0 for all x ∈ R, we can proceed as follows. Clearly, the range of X is (a subset of) the interval (0, ∞) and, thus, we have P (X = x) = 0 if x ̸∈ (0, ∞). So, from now on, we assume that x ∈ (0, ∞). Intuitively speaking, it is impossible that we would ever observe any specific duration, x, more than once. Hence, in n independent repetitions of the random experiment (observing the duration of a solar eclipse), we would have n%
{X = x}&
≤ 1. Consequently, from the frequentist interpretation of probability, for large n,
0 ≤ P (X = x) ≈ n%
{X = x}&
n ≤ 1
n. Because 1/n → 0 as n → ∞, we conclude that P (X = x) = 0.
b) Note that Y is a random variable because it is a real-valued function whose domain is !. In fact, Y is defined on ! by Y (x) = ⌊x⌋. Clearly, the range of Y is (a subset of) the nonnegative integers and, hence, is countable. Therefore, Y is a discrete random variable.
5.22 Note: Referring to the argument presented in Exercise 5.21(a), we see that, generally speaking, a random variable, X, denoting the measurement of a quantity such as time, weight, height, or temperature, has the property that P (X = x) = 0 for all x ∈ R. Therefore, in view of Exercise 5.17, such a random variable is not discrete.
a) From the note, as T6denotes a (random) time, it is not a discrete random variable.
b) The number, X3, of patients who arrive during the first 3 hours is a discrete random variable because its range is (a subset of) the nonnegative integers, which is countable.
c) We have S = ⌊T6⌋. Hence, S is a discrete random variable because its range is (a subset of) the nonnegative integers, which is countable.
d) From the note, as the elapsed time between the arrivals of the sixth and eighth patients denotes a (random) time, it is not a discrete random variable.
e) From the note, as the elapsed time between the arrival of the last patient to arrive before 3:00A.M.and the arrival of the first patient to arrive after 3:00A.M.denotes a (random) time, it is not a discrete random variable.
f) The event {X3 ≥ 6} occurs if and only if at least six patients arrive during the first 3 hours, which happens if and only if the sixth patient arrives within the first 3 hours, that is, if and only if the event {T6 ≤ 3}
occurs. Hence, {X3≥ 6} = {T6 ≤ 3}.
5.23
a) We note that a value of X is always a ratio of two integers. Hence, the range of X is countable, being a subset of the set of rational numbers (which is a countable set).
b) First we show that the range of X, which we denote R, consists precisely of the rational numbers in the interval (0, 1], which we denote K. As a value of X is always a ratio of two positive integers of which the numerator does not exceed the denominator, we have R ⊂ K. Conversely, suppose that x ∈ K.
Then we can write x = m/n where m and n are positive integers with m ≤ n. Now, from the domination principle and the general multiplication rule,
P (X= x) = P (X = m/n) ≥ P (N = n, M = m) = P (N = n)P (M = m | N = n)
= p(1 − p)n−1·1 n >0.
Thus, x ∈ R. We have now shown that R = K; that is, the range of X consists of all rational numbers in (0, 1]. The required result now follows from the fact that between any two rational numbers is another rational number.