Discrete Random Variables and Their Distributions
5.7 Other Important Discrete Random Variables
p1(1 − p)(z−1)−1= (z − 1)p(1 − p)z−2. Hence, for z = 2, 3, . . . ,
pX+Y(z)= P (X + Y = z) = P (A ∩ B) = P (A)P (B)
= (z − 1)p(1 − p)z−2· p = (z − 1)p2(1 − p)z−2. This result agrees with the one found in part (b).
5.7 Other Important Discrete Random Variables
Basic Exercises
5.110 We note that the only possible values of I{X=1}and X are 0 and 1. Hence, to prove that those two random variables are equal, it suffices to show that X = 1 if and only if I{X=1} = 1. However,
X(ω)= 1 ⇔ ω ∈ {X = 1} ⇔ I{X=1}(ω)= 1.
5.111
a) We note that both IE∩F and IE· IF are 1 if E ∩ F occurs and both are 0 otherwise. Consequently, we have IE∩F = IE· IF.
b) Suppose that E ∩ F = ∅. The only possible values of IE∪F are 0 and 1 and, because E ∩ F = ∅, the same is true for IE+ IF. However,
IE∪F(ω)= 0 ⇔ ω /∈ E ∪ F ⇔ ω /∈ E and ω /∈ F ⇔ IE(ω)= IF(ω)= 0 ⇔ (IE+ IF)(ω)= 0.
Consequently, IE∪F = IE+ IF. Conversely, suppose that E ∩ F ̸= ∅. Let ω0 ∈ E ∩ F . Then IE∪F(ω0)= 1 ̸= 2 = 1 + 1 = IE(ω0)+ IF(ω0)= (IE + IF)(ω0).
Consequently, IE∪F ̸= IE+ IF. We have now established that a necessary and sufficient condition for IE∪F = IE+ IF is that E ∩ F = ∅, that is, that E and F are mutually exclusive.
c) We claim that IE∪F = IE+ IF − IE∩F, that is, that
IE∪F(ω)= IE(ω)+ IF(ω)− IE∩F(ω), ω∈ !.
To establish the preceding equation, we consider four cases.
Case 1: ω ∈ (E ∪ F )c
Then ω /∈ E ∪ F , ω /∈ E, ω /∈ F , and ω /∈ E ∩ F . Hence,
IE∪F(ω)= 0 = 0 + 0 − 0 = IE(ω)+ IF(ω)− IE∩F(ω).
Case 2: ω ∈ E ∩ F
Then ω ∈ E ∪ F , ω ∈ E, and ω ∈ F . Hence,
IE∪F(ω)= 1 = 1 + 1 − 1 = IE(ω)+ IF(ω)− IE∩F(ω).
Case 3: ω ∈ E ∩ Fc
Then ω ∈ E ∪ F , ω ∈ E, ω /∈ F , and ω /∈ E ∩ F . Hence,
IE∪F(ω)= 1 = 1 + 0 − 0 = IE(ω)+ IF(ω)− IE∩F(ω).
Case 4: ω ∈ Ec∩ F
Then ω ∈ E ∪ F , ω /∈ E, ω ∈ F , and ω /∈ E ∩ F . Hence,
IE∪F(ω)= 1 = 0 + 1 − 0 = IE(ω)+ IF(ω)− IE∩F(ω).
5.112 The only possible values of I?nAn are 0 and 1 and, because A1, A2, . . . are mutually exclusive, the same is true for3
nIAn. Hence, to prove that those two random variables are equal, it suffices to show that I?
nAn = 0 if and only if3
nIAn = 0. However, I?nAn(ω)= 0 ⇔ ω /∈0
n
An⇔ ω /∈ Anfor all n ⇔ IAn(ω)= 0 for all n ⇔'/
n
IAn
(
(ω)= 0.
5.113
a) For 1 ≤ k ≤ n, let Ekdenote the event that the kth trial results in success. Then IEk = 0 if the kth trial results in failure and IEk = 1 if the kth trial results in success. Hence,3n
k=1IEk gives the total number of successes in the n trials; that is, X =3n
k=1IEk.
b) For 1 ≤ k ≤ n, let Xk be 1 or 0 depending on whether the kth trial results in success or failure, respectively. Then each Xk is a Bernoulli random variable with parameter p and, moreover,3n
k=1Xk
gives the total number of successes in the n trials; that is, X =3n k=1Xk.
5.114 Because we have a classical probability model, we know that P ({ω}) = 1/N for all ω ∈ !. Thus, for k = 1, 2, . . . , N, we have
P (X= k) = P ({ ω : X(ω) = k }) = P ({ωk}) = 1/N.
Hence, pX(x)= 1/N if x ∈ {1, 2, . . . , N}, and pX(x)= 0 otherwise. In other words, X has the discrete uniform distribution on the set {1, 2, . . . , N}.
5.115 We have
Hence, by the law of total probability, for y ∈ S, P (Y = y) =/
Thus, Y has the discrete uniform distribution on S.
b) By symmetry, the value of Y is equally likely to be any of the 10 digits. Therefore,
pY(y)=
-1/10, if y ∈ S;
0, otherwise.
Thus, Y has the discrete uniform distribution on S.
5.116
a) From Proposition 5.13 on page 241, we see that X ∼ N B(3, 0.67). Hence, pX(x)=
From the complementation rule, the FPF, and part (a),
P (X ≥ 4) = 1 − P (X < 4) = 1 −/
From the FPF and part (a),
P (X≤ 4) =/
5.117 Let X denote the number of at-bats up to and including the second hit. Then X ∼ N B(2, 0.26), so that
pX(x)=
'x− 1 1
(
(0.26)2(0.74)x−2= (x − 1)(0.26)2(0.74)x−2, x= 2, 3, . . . , and pX(x)= 0 otherwise.
a) We have
P (X= 5) = pX(5) = (5 − 1)(0.26)2(0.74)5−2 = 0.110.
b) From the complementation rule and the FPF,
P (X >5) = 1 − P (X ≤ 5) = 1 −/
x≤5
pX(x)= 1 − /5 k=2
pX(k)
= 1 − /5 k=2
(k− 1)(0.26)2(0.74)k−2= 1 − (0.26)2 /5 k=2
(k− 1)(0.74)k−2
= 0.612.
Alternatively, because X > 5 if and only if the number of hits in the first five at-bats is at most 1, we have
P (X >5) = /1 k=0
'5 k (
(0.26)k(0.74)5−k = (0.74)5+ 5(0.26)(0.74)4 = 0.612.
c) From the FPF,
P (3 ≤ X ≤ 10) = /
3≤x≤10
pX(x)= /10 k=3
pX(k)= (0.26)2 /10 k=3
(k− 1)(0.74)k−2 = 0.710.
5.118 For each r ∈ N , we know that X is a positive-integer valued random variable. From Proposi-tion 5.11 on page 234, the only positive-integer valued random variables that have the lack-of-memory property are geometric random variables. A negative binomial random variable is a geometric random variable if and only if r = 1. Hence, X has the lack-of-memory property if and only if r = 1.
5.119 Let E denote the event of a 7 or 11 when two balanced dice are rolled. Then
P (E)= 6 + 2 36 =
2 9.
Let X denote the number of rolls required for the fourth occurrence of E. Then X ∼ N B(4, 2/9). Hence, by the complementation rule and the FPF,
P (X >6) = 1 − P (X ≤ 6) = 1 −/
x≤6
pX(x)= 1 − /6 k=4
pX(k)
= 1 − /6 k=4
'k− 1 4 − 1 (
(2/9)4(7/9)k−4= 1 − (2/9)4 /6 k=4
'k− 1 3
(
(7/9)k−4
= 0.975.
5.120
a) Consider Bernoulli trials with success probability p. Let X denote the number of trials until the rth suc-cess and let Y denote the number of sucsuc-cesses in the first n trials. Then X ∼ N B(r, p) and Y ∼ B(n, p).
Event {X > n} occurs if and only if the rth success occurs after the nth trial, which happens if and only if the number of successes in the first n trials is less than r, that is, event {Y < r} occurs. There-fore, {X > n} = {Y < r} and, in particular, P (X > n) = P (Y < r).
Applying part (a) and the FPF, we get /∞ c) From the complementation rule and the FPF,
P (X > n)= 1 − P (X ≤ n) = 1 −/
Hence, with this computation, we must evaluate n − r + 1 terms of the PMF of X.
d) From the FPF,
Hence, with this computation, we must evaluate r terms of the PMF of Y .
e) From parts (c) and (d), in computing P (X > n), the difference between the numbers of terms that must be evaluated using the PMF of X and the PMF of Y is
(n− r + 1) − r = n − 2r + 1 = n
Thus, when n is large relative to r, it is necessary to evaluate roughly n more terms using the PMF of X than using the PMF of Y . In other words, when n is large relative to r, there is considerable computational savings in using the PMF of Y rather than the PMF of X to evaluate P (X > n).
5.121 For x = r, r + 1, . . . ,
5.122 Let X and Z denote the calls on which the second and fifth sales are made, respectively. Also, and, arguing as in Example 5.26(c), we find that
P (Z = k | X = j) = P (Y = k − j) = From the FPF and the preceding display, it follows immediately that
P (Z≤ 15 | X = j) =
a) Applying the law of total probability, we now get that
P (X ≤ 5, Z = 15) =
b) Again applying the law of total probability, we get
P (X ≤ 5, Z ≤ 15) =
Moreover, because the only two possible values of IE are 0 and 1, we have, from the complementation rule, that
pIE(0) = P (IE = 0) = 1 − P (IE = 1) = 1 − pIE(1) = 1 − P (E).
Hence,
pIE(x)=" 1 − P(E), if x = 0;
P (E), if x = 1;
0, otherwise.
5.124 The possible values of X are r, r + 1, . . . . For such a positive integer, the event {X = k} occurs if and only if the rth success is obtained on the kth trial. That happens if and only if (1) among the first k − 1 trials, exactly r − 1 are successes, and (2) the kth trial is a success. Denote the former event by A and the latter event by B. We have {X = k} = A ∩ B and, because the trials are independent, event A and event B are independent. Hence,
P (X = k) = P (A ∩ B) = P (A)P (B).
The number of successes in the first k − 1 trials has the binomial distribution with parameters k − 1 and p. Thus, by Proposition 5.3 on page 201,
P (A)=
'k− 1 r− 1 (
pr−1(1 − p)(k−1)−(r−1)=
'k− 1 r− 1 (
pr−1(1 − p)k−r. Also, the probability that the kth trial is a success is p, so P (B) = p. Therefore,
P (X= k) =
'k− 1 r− 1 (
pr−1(1 − p)k−r· p =
'k− 1 r − 1 (
pr(1 − p)k−r. Consequently, the PMF of the random variable X is
pX(x)=
'x− 1 r− 1 (
pr(1 − p)x−r, x = r, r + 1, . . . , and pX(x)= 0 otherwise.
5.125 We must verify that pXsatisfies the three properties stated in Proposition 5.1 on page 187. Clearly, we have pX(x)≥ 0 for all x. Moreover, { x ∈ R : pX(x)̸= 0 } = {r, r + 1, . . .}, which is countable, being a subset of the countable set N . Applying Exercise 5.121 and the binomial series [Equation (5.45) on page 241], we get
/
x
pX(x)= /∞ k=r
'k− 1 r− 1 (
pr(1 − p)k−r = /∞ k=r
' −r k− r
(
pr(p− 1)k−r
= pr /∞ j=0
'−r j
(
(p− 1)j = pr%
1 + (p − 1)&−r
= prp−r = 1.
5.126
a) In Bernoulli trials, the trials are independent and, furthermore, the success probability is positive.
Therefore, from Proposition 4.6, in repeated Bernoulli trials, the probability is 1 that a success will eventually occur. Because of independence in Bernoulli trials, once a success occurs, the subsequent process of Bernoulli trials is independent of the prior trials and behaves the same probabilistically as the process of Bernoulli trials starting from the beginning. Hence, we conclude that the probability is 1 that a second success will eventually occur. Continuing in this way, we deduce that, for each positive integer r, the probability is 1 that an rth success will eventually occur.
b) If we let X denote the number of trials up to and including the rth success, then X ∼ N B(r, p). The probability that an rth success will eventually occur is
P (X <∞) = /
x<∞
pX(x)=/
x
pX(x)= 1, where the last equality is a consequence of Exercise 5.125.
5.127 We must verify that the function defined in Equation (5.50) satisfies the three properties stated in Proposition 5.1 on page 187. We note that p(0) =%−α union of the countable sets {0} and N . Applying now the binomial series, Equation (5.45) on page 241, we get
a) Let k be an integer between 0 and N, inclusive. Let Rk denote the event that the smoker finds the matchbox in his right pocket empty at the moment when there are k matches in the matchbox in his left pocket; and let Lk denote the event that the smoker finds the matchbox in his left pocket empty at the moment when there are k matches in the matchbox in his right pocket. Events Rk and Lk are mutually exclusive, have the same probability (by symmetry), and have union {Y = k}. To find P (Rk), we consider a success to be that the smoker goes to his right pocket for a match. Event Rk occurs if and only if the (N + 1)st success occurs on trial (N + 1) + (N − k) = 2N + 1 − k. Applying the negative binomial PMF with parameters N + 1 and 1/2, we see that the probability of that happening is
P (Rk)=
b) Referring to part (a), we obtain the following table for N = 5:
y 0 1 2 3 4 5
pY(y ) 0.246094 0.246094 0.218750 0.164063 0.093750 0.031250 And, for N = 10, we have
y pY(y ) y pY(y ) 0 0.176197 6 0.061096 1 0.176197 7 0.034912 2 0.166924 8 0.016113 3 0.148376 9 0.005371 4 0.122192 10 0.000977 5 0.091644
c) We proceed as in part (a) until we apply the negative binomial PDF. At that point, we use the one with parameters N + 1 and p. Hence, we find that
P (Rk)='2N − k N
(
pN+1(1 − p)N−k and P (Lk)='2N − k N
(
(1 − p)N+1pN−k. Therefore,
P (Y = k) = P (Rk)+ P (Lk)='2N − k N
( 8
pN+1(1 − p)N−k+ (1 − p)N+1pN−k 9
. Consequently,
pY(y)='2N − y N
( 8
pN+1(1 − p)N−y + (1 − p)N+1pN−y9
, y = 0, 1, . . . N, and pY(y)= 0 otherwise.
d) Referring to part (c), we obtain the following table for N = 5:
y 0 1 2 3 4 5
pY(y ) 0.200658 0.217380 0.216760 0.187730 0.126720 0.050752 And, for N = 10, we have
y pY(y ) y pY(y ) 0 0.117142 6 0.098410 1 0.126903 7 0.068997 2 0.134867 8 0.039308 3 0.138435 9 0.016240 4 0.134671 10 0.003670 5 0.121357
e) The answers here are the same as those in part (d) because, if we replace p by 1 − p, the probabilities for Y do not change. We can see this fact by interchanging the roles of the left and right pockets or, more directly, by replacing p by 1 − p (and, therefore, 1 − p by p) in the formula for the PMF of Y determined in part (c).
5.129
a) Consider a success a win of one point by Jane on any particular trial and let X denote the number of trials until the 21st success. Then X ∼ N B(21, p). Jane wins if and only if she gets 21 points before Joan does, that is, if and only if the 21st success occurs on or before trial 41. Hence, by the FPF, the required probability is
b) Suppose that p = 1/2, that is, that Jane and Joan are evenly matched. Then, by symmetry, the probability is 1/2 that Jane wins the game. Hence, in view of part (a), we have
1
a) For each nonnegative integer y,
pYr(y)= Moreover, from the hint given in Exercise 5.86(a),
rlim→∞
c) Suppose that X ∼ N B(r, p), where r is large and p is close to 1. Let Y = X − r. Then the PDF of Y is given by Equation (5.49). Now,
pX(x)= P (X = x) = P (Y + r = x) = P (Y = x − r) = pY(x− r).
Hence, in view of parts (a) and (b), pX(x)≈ e−r(1−p)
%r(1 − p)&x−r
(x− r)! , x = r, r + 1, . . . .
In other words, for large r and p close to 1, the PMF of a N B(r, p) random variable can be approximated by the right side of the previous display.