A Study of the Suprenewal Process

Abstract: The classical coupon collector’s problem is concerned with the number of purchases in order to have a complete collection, assuming that on each purchase a consumer can obtain a randomly chosen coupon. For most real situations, a consumer may not just get exactly one coupon on each purchase. Motivated by the classical coupon collector’s problem, in this work, we study the so-called suprenewal process. Let {Xi, i ≥ 1} be a sequence of independent and identically distributed random variables, Sn = ∑n i=1 Xi, n ≥ 1, S0 = 0. For every t ≥ 0, define Qt = inf{n | n ≥ 0, Sn ≥ t}. For the classical coupon collector’s problem, Qt denotes the minimal number of purchases, such that the total number of coupons that the consumer has owned is greater than or equal to t, t ≥ 0. First the process {Qt, t ≥ 0} and the renewal process {Nt, t ≥ 0}, where Nt = sup{n|n ≥ 0, Sn ≤ t}, generated by the same sequence {Xi, i ≥ 1} are compared. Next some fundamental and interesting properties of {Qt, t ≥ 0} are provided. Finally limiting and some other related results are obtained for the process {Qt, t ≥ 0}.


Introduction
Starting from the end of April 2005, collecting Hello Kitty magnets became an immensely popular hobby in Taiwan.President Chain Stores Corp., which runs Taiwanese largest convenience store chain, 7-Eleven, was giving away one of a series of commemorative Hello Kitty magnets for each NTD77 a consumer spends at 7-Eleven store.There are 41 different patterns of Hello Kitty magnets in total.Because the cover of each package of magnet is the same, it is reasonable to assume that the magnets are given randomly.
We now review the classical coupon collector's problem.Assume there are N distinct coupons in a collection, and a series of random draws is made with replacement from these.Let T denote the number of draws necessary for all N coupons to have been drawn at least once.Properties of T had been studied by many authors, see e.g.Goodwin (1949) and Feller (1968).Among others, expectation and variance of T can be obtained as follows.For every k ≥ 1, let C k ∈ {1, 2, • • • , N } be the type of coupon obtained at the k-th draw.The k-th draw is called a success, if C k has not been obtained before the k-th draw.For 1 ≤ i ≤ N , let T i denote the number of draws after the (i − 1)-th success, till the i-th success.Then T = ∑ N i=1 T i .Obviously, T 1 , T 2 , • • • , T N are independent, and T i has a geometric distribution with parameter p i = (N − i + 1)/N , then E(T i ) = N/(N − i + 1), and Var(T i ) = (1 − (N − i + 1)/N )/((N − i + 1)/N ) 2 , 1 ≤ i ≤ N .Thus where for N ≥ 1, H N = ∑ N i=1 1/i is the N -th Harmonic number, and (1.2) The above coupon collector's problem can be generalized.Assume the i-th coupon has probability p i of being drawn, where 0 < p i < 1, 1 ≤ i ≤ N , such that ∑ N i=1 p i = 1, the p i 's are allowed to be unequal.This was studied by von Schelling (1954).Some limiting results were derived by Baum and Billingsley (1965) and Hoslt (1971), and others.Related problems had also been discussed, such as the collector's brotherhood problem.As an example, Foata et al. (2001) and Foata and Zeilberger (2003) considered the situation that the collector shares his harvest with his brothers.They answered the question that when the collection of the collector is completed, the number of coupons each brother still lacks.
For our present problem, the expected number of magnets needed for collecting a complete set of 41 magnets is 41 ∑ 41 i=1 1/i .= 176.42.For a particular consumer, in each purchase, if his spending is less than NTD77, then he gets 0 magnet, if his spending is at least NTD77 and less than NTD154 (= 2 × 77), than he gets 1 magnet, if his spending is at least NTD154 and less than NTD231 (= 3 × 77), then he gets 2 magnets on that purchase, and so on.Now what is the number of purchases needed in order to get magnets greater than or equal to 176.42?To solve this problem, first we introduce a new process and study some of its properties.
Let {X i , i ≥ 1} be a sequence of independent and identically distributed (i.i.d.) random variables, For the magnets problem, X i can be viewed as the number of magnets received on the i-th purchase, i ≥ 1, and Q t can be viewed as the minimal number of purchases, such that the total number of magnets is greater than or equal to t, t ≥ 0.
Recall that the renewal process {N t , t ≥ 0} generated by the same sequence {X i , i ≥ 1}, where for t ≥ 0, N t = sup{n | n ≥ 0, S n ≤ t}, N t denotes the number of renewals in [0, t].Q t can be referred to as the minimal number of renewals in [t, ∞), and we call {Q t , t ≥ 0} the suprenewal process.In Section 2, we compare {Q t , t ≥ 0} with {N t , t ≥ 0}.In Section 3, some fundamental and interesting properties of {Q t , t ≥ 0} are studied.Those tedious proofs will be given in the Appendix.Also some limiting results are presented in Section 4. Finally, in Section 5, we give an example to provide a partial answer of the Hello Kitty magnets problem.

Comparisons of {Q
. random variables with the same distribution as X, where X, a nonnegative random variable, has the distribution function F with F (0−) = 0 and F (0 Let {Q t , t ≥ 0} and {N t , t ≥ 0} be the suprenewal process and renewal process generated by Hence for every t > 0 and integer n ≥ 1, (2.1) where F n is the n-fold convolution of F with itself, n ≥ 1, and F 0 (t) = 1, t ≥ 0. If F is continuous, then S n is a continuous random variable for every integer n ≥ 0. Consequently, We now compare the two processes {Q t , t ≥ 0} and {N t , t ≥ 0}.First instead of having right continuous sample paths for {N t , t ≥ 0}, {Q t , t ≥ 0} has left continuous sample paths.Next instead of (2.1) and (2.3), whether F is continuous or not, On the other hand, {Q t , t ≥ 0} and {N t , t ≥ 0} have the same jump times.Denote the sequence of jump times by 0 and where and Y τ i and N τ i −1 are also independent.If F is continuous, then F (0) = 0, and (2.8) As an example, let F (x) = 1 − e −λx , λ > 0, x > 0. Then it is well known that N t has a P(λt) distribution, t > 0. By (2.8), Q t − 1 is also P(λt) distributed for almost all t on [0, ∞).Note that except (2.5) and (2.6), we also have the following relationship (2.9) Although Q t may be less than N t (if F (0) > 0), from the definitions of Q t and N t , we have (2.10) In particular Recall that S Nt+1 − t, t − S Nt , and X Nt+1 = S Nt+1 − S Nt are called residual life at time t, current life at time t, and total life at time t, respectively, for the renewal process {N t , t ≥ 0}.It is known that P (X Nt+1 > x) ≥ P (X 1 > x), x ≥ 0, and E(X Nt+1 ) ≥ E(X 1 ), t ≥ 0. This is the so-called inspection paradox.Similarly, it can be shown (2.13) Furthermore, using the fact that a renewal process probabilistically starts over when a renewal occurs, for every increasing function g, the following inequality is immediate: Similarly, we have In particular We give a typical sample paths of {Q t , t ≥ 0} and {N t , t ≥ 0}, respectively, to illustrate the relationships (2.5), (2.6) and (2.9).Assume X

Some Fundamental Properties of {Q t , t ≥ 0}
There are many investigations for properties of renewal process in the literatures.In this section, we explore some basic properties of the process {Q t , t ≥ 0}, especially for the case that X takes on nonnegative integer values.Throughout this section, let P (X < ∞) = 1, and First we introduce some notation which will be used often in this work.Let t and t denote the ceiling function and the floor function, respectively, namely t = the least integer greater than or equal to t, and t = the greatest integer less than or equal to t.For example, 3.7 = 4, 3.7 = 3, and 6 = 6 = 6.For integers a, b, c, with a ≤ b ≤ c, and nonnegative integers and x i = a, and and x i = a, and , and B a,a,N a = C a .We give three simple examples in the following.
Remark 1.As a comparison, for the {X i , i ≥ 1} defined in Example 3, we have S Nt = t , and This also can be seen by ( 2.5), (2.6) and Example 3. Now Although it is rather cumbersome, the distribution of Q t , t > 0, can be obtained.We present this in the following.
Theorem 1.For every integer n ≥ 1 and t > 0, , p 0 = 0, By using Theorem 1, the Laplace transform φ t (s) of Q t , s ≥ 0, t ≥ 0, that is and the moments of Q t can be obtained immediately.We summarize the results in the following corollary.
The proofs of Theorem 1 and Corollary 1 will be given in the Appendix.

Example 3.(Continued)
We use Theorem 1 and (i) of Corollary 1, respectively, to demonstrate (3.7) Hence where ) is defined to be 0.This shows Q t ∼ N B( t , p 1 ), t > 0. Next from (i) of Corollary 1 and (24), we have which is exactly the Laplace transform of a N B( t , p 1 ) distributed random variable.
Note that for integers a, b, with a ≤ b, , n ≥ 1, k ≥ 0, denote the number of k-combinations with repetition of n distinct things.Before giving Example 4, we need the following lemma.
Lemma 1.For integers n ≥ 1, and k ≥ 0, we have The proof of Lemma 1 will also be given in the Appendix.
Example 4. Assume p 0 = 0 and and Proof.For t < n, the result is obvious.We now prove the case for t ≥ n.By letting p 0 = 0 and p k = p(1 − p) k−1 , k ≥ 1, in Theorem 1, and from Lemma 1, we obtain (3.9) That the last equality of (3.9) holds is because Remark 2. As a comparison, for the p k , k ≥ 0, defined in Example 4, we have and P (N t = n) = 0, for t < n.

Limiting and Some Other Related Results
For the process {Q t , t ≥ 0}, properties about moments and limiting behaviors are similar to the renewal process {N t , t ≥ 0}, which can be obtained immediately by using (2.5), (2.6) and (2.9).
The central limit theorem also holds for {Q t , t ≥ 0}.
For the process {N t , t ≥ 0}, it is well known that Although N t is not a stopping time, Q t nevertheless is a stopping time.Hence by the Wald equality and (i) of Theorem 2, we have We use Example 3 to illustrate (4.3).

An Example
We now give an example to provide a partial answer of the Hello Kitty magnets problem mentioned in the Introduction.Note that if one magnet is given at each purchase, the expected number of purchases to collect a complete set of 41 magnets is t = 176.42.
Example 5. Let Y 1 , Y 2 , • • • be i.i.d.random variables with the same distribution as Y , where for every i ≥ 1, Y i denotes the amount that a consumer spends at the i-th purchase at 7-Eleven.Assume that Y ∼ Uniform{1, 2, • • • , 250}.Then Thus by routine computations, we obtain and 466,and Var(Q 176.42 ) .= 101.888.Hence the expected number of purchases to get magnets greater than or equal to 176.42 is about 152.466.Furthermore from (4.3), we have  Finally, we give the curve of the probability density function of Q 176.42 in Figure 2, and plot the probability density functions of Z 176.42 and N (0, 1) in Figure 3, where is the normalization of Q 176.42 .As expected, due to Theorem 3, the normal approximation to the probability density function of Z 176.42 is very accurate.

Appendix
Proof of Theorem 1. Obviously we only need to prove (3.1) holds for positive integer t.We prove this by induction.(i) First we prove the case p 0 = 0 and N < ∞.That (3.1) holds for t = 1 can be seen as following.From the assumptions, we have P (S n < 1) = 0, n ≥ 1, P (S 0 < 1) = 1.Thus where the last equality holds is due to for every n ≥ 1, B n,n,0 is a null set, hence g n,1 = 0.This together with g 0,1 = 1 implies g 0,1 − g 1,1 = 1, and g n−1,1 − g n,1 = 0, n ≥ 2. Now suppose (3.1) is true for t = r ≥ 1, i.e. we have This proves (3.1) holds for t = r + 1.By the induction argument this completes the proof for the case p 0 = 0 and N < ∞.The proof of (3.1) for the case p 0 = 0 and N = ∞ is similar, hence is omitted.
(ii) Next we prove the case 0 < p 0 < 1 and N < ∞.The proof of (3.1) for t = 1 is as following.
suppose (3.1) is true for t = r ≥ 1, i.e. we have Then and That the second equality of (A.3) holds is because if Substituting (A.1), (A.3) and (A.4) into (A.2), it yields where the third equality holds is because if The assertion follows by substituting A and B in (A.12) (A.13) into (A.11), and noting that g 0,t = 1.The proof for the case p 0 = 0 is given below.(iii).As in (ii), we obtain as required.
Proof of Lemma 1.For n ≥ 1, let y 1 , y 2 , • • • , y n be any n integers.For every i ≥ 1, let z i = ∑ n j=1 I {y j =i} , where as desired.