In addition to accepting or rejecting a candidate arrive at time r, we may consider purchasing an option at a cost cx to recall the candidate at time r+x, but this privilege may be invoked only once. For large sample size, using the best-choice criterion and deducting option costs, the optimal strategy and return are obtained

Rose, John S.

English

University of Richmond

University of Richmond UR Scholarship Repository Robins School of Business White Paper Series, 1980-2011 Robins School of Business 1983 The Secretary Problem with a Call Option John S. Rose University of Richmond Follow this and additional works at: https://scholarship.richmond.edu/robins-white-papers  Part of the Business Commons Recommended Citation Rose, John S. 1983. "The Secretary Problem with a Call Option." E.C.R.S.B. 83-8. Robins School of Business White Paper Series. University of Richmond, Richmond, Virginia. This White Paper is brought to you for free and open access by the Robins School of Business at UR Scholarship Repository. It has been accepted for inclusion in Robins School of Business White Paper Series, 1980-2011 by an authorized administrator of UR Scholarship Repository. For more information, please contact scholarshiprepository@richmond.edu. 1983-8 THE SECRETARY PROBLEM WITH A CALL OPTION* 1 Johns. Rose University of Richmond In addition to accepting or rejecting a candidate arriving at timer, we may consider purchasing an option at a cost ex to recall the candidate at time r+x, but this privilege may be invoked only once. For large sample size, using the best-choice criterion and deducting option c:osts, the optimal strategy and return are obtained. Secretary problem, optimal choice, recall 1. Johns. Rose, Department of Management Systems, Robins School of Business, University of Richmond, Richmond, VA, 23173. * Partially supported by grants from the DuPont Company and the Committee on Faculty Research, University of Richmond. 1. INTRODUCTION AND SUMMARY The classical model of the secretary problem can be found in Gilbert and Mosteller [3, Section 2a] or De Groot [2, pp. 325-331], and its solution is reviewed briefly in the next section. A restrictive assumption of the model is the impossibility of re-calling previously rejected applicants. This restriction was relaxed by Yang [7].: at any stage of the process we may attempt to recall a previously rejected applicant, whose availability is uncertain and (maybe) stochastically decreasing in time. Others have subsequently extended Yang's work; see Corbin [1] or Petruc-celli [5]. A different approach to the recall issue is proposed here. Let n be the population size and suppose that, at some stage m, we observe an applicant which is preferable to the m-1 previously sampled, and rejected, applicants. Then, this applicant is a candidate (for the best). If we select (reject) the candidate, then we risk (not) finding a better one among the n-m applicants yet unsampled. In order to mitigate this risk, we may purchase an option granting us the privilege to recall the candidate at stage m+k, k=l, ••. , n-m. The cost to acquire such an option is bk, b~O. If the option is acquired, we say that we are holding the candidate fork periods; the interval {m+l, .•• , m+k} is called the holding period. The essential operational difference between our model and Yang's is that we must decide at the time of observation whether or not we shall later have the option to recall the candidate. 1 In Rose [6], the call options were renewable, so it was suf-ficient to consider only the one-stage holding period, k=l. Here, we expressly prohibit renewal so that the duration, k, of the holding period becomes a crucial decision variable •. Consequently, if no better applicant is observed during the holding period, the option will be exercised and the candidate selected - clearly, in this context, it is suboptimal to purchase an option and then let it expire. Should a better applicant appear during the holding period, then we might take out another option to recall this new candidate, giving rise to the rather awkward situation in which two, or even more, applicants are being held simultaneously. This phenomenon is under investigation and will be reported on else-where; to avoid that complication here, assume that we are permit-ted to hold only one candidate. Then, a better applicant appearing during the holding period will be accepted or rejected according to the classical procedure. Our goal is to determine at what stages a candidate should be held, and for how long, in order to maximize the probability of selecting the best applicant less the holding costs. In the next section, letting n+<x> and b+O such that nb+c, the infinite model is derived from the finite problem introduced above. The solution to the infinite model is summarized in the following theorem, whose proof is relegated to Section 3. The function U(•) defined on (0,1] denotes the optimal return function; see (1) and (8). The dependence of U on c is suppressed. 2 THEOREM I. If c~l, then the classical procedure is optimal and -1 U(r)=max{e , r}. II. For OSc<l, there exist s 1 (c) and s 2 (c), with O<s1ss2<1, such that, for a candidate arriving at timer, it is optimal to: a. reject , rss 1 ; b. hold until . time r/c, s 1<rss 2; c. hold until time 1, s 2<r<l, provided that no candidate has been held previously. Furthermore, d. e. f. g. -1 -1 there exists c 1~.32136 such that c 1e <s1<e s 1 (c)=s 2 (c) if and only if csc1 ; s 1 and s 2 are increasing inc. III. For OSc<l, there exist t 1 (c), t 2 (c), t 3 (c), t 4 (c), with O<t1st2st3st4<1, such that a. b. c. d. e. f. U(r)=U(tl), rStl; . -1 U(r)=cr+e +rlog r/c, t 1<rst2; -1 U(r)=cr+e +r-c, t 2<rst3 ; U(r)=cr-rlog c, t 3<rst4 ; U(r)=cr-rlog r+r-c, t 4<rSl; U(O)Se- 1+e- 312 . -1 Note that, from (II.a,e), applicants arriving prior to c 1e , or approximately 12% of the population, will be rejected automa-tically, even if the option cost, c, vanishes. Holding a candid-ate too soon, we may encounter a better applicant before time 3 -1 e , in which case nothing has been gained - the new applicant will be rejected according to the classical procedure. There would be no such lower bound on s 1 (c) if several applicants could be held simultaneously. Even when c=O, the maximal return is only e-1+e- 312 , the upper bound in (III.f), which is a relative -1 improvement of about 60% over the classical return, e • Again, if options could be taken on several applicants, then we should get U(O)=l for c=O. Also, consider (II.b,d,f). If c is not too small, then there are times, r, for which it is optimal to hold for a period r/c-r, which is only part of the time remaining, 1-r. Contrast this result with those of [1 i Special Case C-Additively Decreasing Recall, p. 21), in which it is optimal to observe the entire population before attempting recall, or of [6], in which it is optimal to renew the option until the end of the process and only then to exercise it. According to (II.d,e), we should hold a -1 candidate appearing somewhat earlier thane until some time -1 later thane , which is the critical cut-off time for the classical problem (see (7)). Finally, it is implied in (II.a,b,c) that a candidate is accepted immediately upon its arrival only if the call option alternative was used to hold an earlier applicant. Numerical solutions of the finite problem are displayed in Table 1. Fortunately, these results are fairly well approximated by the solution of the infinite model as given in the Theorem. 4 2. FINITE AND INFINITE MODELS The optimality equations follow from the customary backward induction argument of dynamic programming. Keep in mind that our (net} return is 1 or O minus option price. Define two events: A ={candidate appears at stage m}, and B ={first m-1 applicants m m rejected}, m=l, ••• , n. Let U (m} denote the maximum expected n return given A and B. Also conditioning on A, let V (m,k} be m m m n the probability that the best applicant is chosen, given that the candidate is held fork periods. Finally, let v (m} denote the n maximum expected return, given B 1 • Then, m+ U (m} = n max k=l, ••• ,n-m {m/n, V (m), V (m,k)-bk}, n n ( 1} where the three bracketed terms in (1) are the returns expected from accepting, rejecting, and holding (fork periods) the candidate, respectively. The second and third terms are computed recursively: v (m) = [U (m+l) +mv (m+l)] / (m+l), (v (n) =O) n n n n ( 2) V (m,k)=[U0 (m+l)+mV (m+l,k-1)]/(m+l), (V (m,O)=m/n) (3) n n n n k=l, ••• ,n-m, where u0 is the optimal return function under the n classical model (see (6) below). The interpretation of (3) should be clear. With probability 1/(m+l) a better applicant appears at stage m+l and we apply the classical procedure. Otherwise, we still hold the candidate, but for only k~l periods beyond stage m+l. The condition V (m,O)=m/n means merely that an expired n option should be exercised. 5 From the recursions (2) and (3) we solve for v and V (•,k) n n in terms of U and u0 respectively, obtaining n n V (m) = n V (m,k) n n m E i=m+l U (i) /i (i-1), n m+k = m E u0 (i)/i(i-l) + m/n. i=m+l n (4) (5) Now, we need to review briefly the solution to the classical problem. There exists a positive integer, r*, such that the first n r*-1 applicants are rejected and the first candidate, if any, n observed thereafter is selected. The optimal return function is u0 (m) = n m/n, m~r* n n-1 (r*-1)/n E 1/i n i=r*-1 n , m<r* • n -1 o + o + In the limit, r*/n+e · and U ([rn] )+U (r) as n+oo, where [p] n n (6) denotes the smallest positive integer equal to or greater than p, and 1 -1 r r>e u0 (r) ' - -1 -1 e , r$e , 0$r$l. (7) Follow the method given in Mucci [4] to obtain the infinite model. To standardize and extend the return functions on the unit interval, + V ([rn] . , n write f (r)=U ([rn]+), h (r)=v ([rn]+), and n n n n . g (r,x)= n + [xn] ) . Then (1) is rewritten as f (r)= max n + O<x$l-[rn] /n + + {[rn] /n, h (r), g (r,x)-b[xn] }, and the expressions (4) and (5) n n can be appropriately rewritten, too. Now, take limits as n+oo. The option alternative is dominated unless b+Oi to keep the 6 holding cost linear, suppose that nb+c~O as n+~. h +v, and g +V where n n U(r) = max {r, v(r), V(r,x)-cx}, O~x~l-r V (r) 1 2 = r fr U(y)dy/y V(r,x) Thus, f +U, n (8) (9) (10) OSrSl. From the continuity of V (r, •), use of "max" in (8) is permissible; closing the interval with x=O is a trivial matter, since V(r,O)=r. The next section is devoted to solving (8)-(10). 3. PROOF OF THEOREM Some abbreviated notation will be helpful. Let w(r,x)=V(r,x) -ex, let x*=x*(r) maximize w(r,x) on [0,1-r], and let W(r)=w(r,x*). LEMMA 3.1 PROOF -1 If U(r)=W(r) _for r<e o -1 Obviously, U(r)~U (r)=e • -1 then r+x*~e • -1 Suppose r+x*<e • -1 -1 Substituting (7) into (10) yields V(r,x*)<V(r,e -r)=e , so -1 W(r)<e . D LEMMA 3.2 Let c<l and suppose U(r)=W(r). Then x*=l-r on r>c and x*=r/c-r on rsc. 0 If c~l, then U(r)=U (r) and (I) holds. PROOF -1 Using (7) and (10) and holdin•g r+x~e , we obtain w (r,x)=r/(x+r)-c. For r>c, w (r~x)>O and x*=l-r; and for rsc, X X the maximum is achieved at x*=r/c-r. Now, suppose c~l and that o -1 U(r)>U (r) for some r. Then U(r)=W(r), so r+x*~e • However, 7 . -1 w (r,x)~O on r+x~e • X Either r~e-l and x*=O, in which case W(r)= o -1 V(r,O)=r=U (r), or r<e and -1 x*=e -r, in which case -1 -1 o W(r)<V(r,e -r)=e =U (r). D Henceforth, assume c<l. Also, as the next lemma shows, we may eliminate the "r" term from (8) and just compare the reject and hold decisions. LEMMA 3. 3 PROOF For r<l, U(r)>r. -1 o -1 If r<e , then U(r)~U (r)=e • Suppose -1 r~e • Then, U(r)~w(r,x)=r+rlog(l+x/r)-cx, from (7) and (10). For x suffi-ciently small, iog(l+x/r)>cx/r, so U(r)>r. D Finally, we state a lemma which will be proved several times for different values of rand c. LEMMA 3.4 If W(r)>v(r), then W(s)>v(s), s>r. The constant c 1 appearing in (II.e) is the solution of t = exp(-3/2(t+l)). (11) The role of this equation becomes apparent subsequently. It is -1 convenient now to consider separately the three cases: c~e , 3.1 -1 THE CASE c~e . First, hold r>c. x*=l-r, so W(r)=w(r,1-r)=cr-rlogr+r-c. 8 By Lemma 3.2, PROOF of Lemma 3.4: Let D(r)=W(r)-v(r), where vis given in (9). 2 2 Then, d/dr[D(r)/r]=-1/r+c/r +U(r)/r )0, since U(r)~r. It follows that D(r) is increasing. 0 Now, let r be such that W(r)>v(r). By Lemma 3.4, we may substi-tute W(•) for U(•) in (9), obtaining v(r)=W(r)-r-crlogr+(r/2)lolr. 2 It is easy to show that 1/2log r-clogr-1<0, so v(r)<W(r). Thus, we have established (II.c) and (III.e), with s 2=t4=c. -1 Next, keep e <rSc and follow the approach used in the pre-ceding paragraph. Here, x*=r/c-r and W(r)=cr-rlogc. The proof of Lemma 3.4 is even simpler: d/dr[D(r)/r]=U(rYr2>0. Assume W(r)> v(r). On (e-1 ,c] substitute W(•) for U(•), while on (c,l] use U(•) given in (III.e), thereby obtaining from (9), v(r)=W(r)-r/2• 2 2 log c+rlogclogr-crlogr-r. Let f(c)=(l+l/2log c)/(logc-c). Then, v(r)<W(r) provided logr>f(c). Because r>e-1 , it suffices to show f(c)<-1, which is readily obtained. Hence, (III.d) is proved with -1 t 3=e • -1 Finally, let rSe and repeat the same sort of argument. The optimal holding period is still x*=r/c-r and, in evaluating -1 V(r,x*), use Lemma 3.1 to get W(r)=cr+e +rlogr/c. Also by Lemma -1 3.1, if r<ce , then v(r)>W(r). To verify Lemma 3.4, we get d/dr[D(r)/r]=-e-1/r2+1/r+U(r)/r2>0, since U(r)~e-1 • Again, assume W(r)>v(r). In the computation of v(r), U(•) is given by W(•) on -1 -1 (r,e ] , by (III.d) on (e ,cl, and by (III.e) on (c,1]. We get 2 v(r)=W(r)-rg(r), where g(r) is quadratic in logr: g(r)=l/2log r+ 2 (l+c-logc)logr+l/2log c+3/2. Hence, v(r)<W(r) if and only if 9 g(r)>O. Let r 1 denote the exp(-(l+c)+logc+fci), where larger root of g(r)=O, so r 1=r1 (c)= . 2 d=d(c)=(l+c) -2(1+c)logc-3. It is . · -1 -1 fairly straightforward to show that ce <r1<e and that g(r) -1 is strictly increasing once Sr<r1 • Hence, we have now verified -1 (II.a,b,d,e,f,g) and (III.b), where s 1=t1=r1 and t 2=e • Because t 2=t3 , (III.c) is trivial. To verify (III.a), note that U(r)=v(r) on rsr1 . Then, substitute the appropriate values for U(•), as calculated above, into (9) and solve the resulting differential equation for v(r), given v(r1 )=cr1+e-1+r1logr1/c. -2 -1 Finally, note that v(r1) is decreasing inc, so U(O)=v(r1)se +e , and (III.f) holds. The remaining two cases are handled similarly. For the sake of parsimony, we omit most of the argument and concentrate on the rough spots. Once the sand t values are known, the results can be verified directly by substitution into (8)-(10), anyway. 3.2 -1 THE CASE cl<c<e • -1 -1 Here, t 4=t3=e and, on r>e , U(r)= W(r) is given by (III.e), as in the previous section. -1 For c<rSe , -1 x*=l-r, W(r)=cr+e +r-c, and Lemma 3.4 holds. If U(s)=W(s) on s>r, then v(r)=W(r)-r(3/2+(c+l)logr). if r>h(c), where h(c)=exp(-3/2(c+l)). -1 -1 increasing, with h(O)>O and h(e )<e • Thus, v(r)<W(r) if and only -1 On (O,e ) , his concave Because c>c1 , as given by (11), it follows that h(c)<c. Therefore, r>h(c) and (II.c) and (III.c) are established with s 2=t2=c. Finally, if rsc, then (II.a,b,f,g) and (III.a,b,f) are readily obtained as before, with -1 Also, ce ~r1<c, so (II.d,e) hold. 10 3.3 THE CASE csc1 • The previous results are applicable here, -1 -1 for r>e For c<rSe , we still get v(r)<W(r) if and only if r>h(c). However, since csc1 , h(~)~c. Thus (II.c) and (III.c) hold now with s 2=t27h(c). Furthermore, U(r)=v(r)>W(r), c<r<h(c), and it remains to show that U(r)=v(r) on rsc. To this end, sub-stitute (III.c,e) for U(•) into (9) and solve the differential equation for v, obtaining v(r)=v(s 2)=W(s 2)=cs 2+e-1+s 2-c, a -1 constant. For rsc, W(r)=cr+e +rlogr/c<W(s2), and the result is established. That also proves (II.a,d,g) and (III.a), with s 1=t1=h(c), and finally establishes the "if" part of (II.f). -3/2 -1 Finally,s1=s 2=h(c)>h(O)=e >c1e , so (II.e) and (III.f) are verified. 11 REFERENCES [1] R.M. Corbin, "The secretary problem as a model of choice," J. Math. Psych. 21, 1-29 (1980) [2] M.H. De Groot, Optimal Statistical Decisions, McGraw-Hill, New York (1970) [31 J.P. Gilbert and F. Mosteller, "Recognizing the maximum of a sequence," J. Am. Stat. Assoc. 61, 35-73 (1966) -, [4] A.G. Mucci, "Differential equations and optimal choice problems," Annals Stat._!, 104-113 (1973) [51 J. D. Petruccelli, "Best-choice problems involving uncertainty of selection and recall of observations," J. Appl. Probab. !§, 415-425 (1981) [61 J.S. Rose, "Optimal sequential selection based on relative ranks with renewable call options," J. Am. Stat. Assoc. (to appear) [7] M.C.K. Yang, "Recognizing the maximum of a random sequence based on relative rank with backward solicitation," J. Appl. Probab. 11, 504-512 (1974) Cost (c) o.o • 2 .35 .5 .9 Table 1 Optimal Procedure (s1/n,s 2/n) and Maximal Return (Un(O)) for b=c/n(l) 25 . 24,. 24,. 6121 .32,.32,.5322 .36,.36,.4823 .40,.48,.4402 .40,.88,.3842 Population Size (n) 100 .23,.23,.5962 .29,.29,.5167 • 34,. 35,. 4671 .36,.50,.4266 .38, .90, .3733 500 .224,.224,.5920 .288,.288,.5127 .330,.350,.4633 .354,.500,.4229 .370,.900,.3705 00 .2231,.2231,.5910 .2865,.2865,.5117 ,3287,.3500,.4623 .3534,.5000,.4220 .3678,.9000,.3698 (1) For a candidate arriving at stage m, the optimal procedure prescribes rejecting if m<s 1 , holding for a period k<n-m if s 1~m<s2 , and holding for a period n-m if m~s2 • 

The Secretary Problem with a Call Option

University of Richmond UR Scholarship Repository Robins School of Business White Paper Series, 1980-2011 Robins School of Business 1982 The Secretary Problem with a Call Option John S. Rose University of Richmond Follow this and additional works at: https://scholarship.richmond.edu/robins-white-papers  Part of the Business Commons Recommended Citation Rose, John S. 1982. "The Secretary Problem with a Call Option." E.C.R.S.B. 82. Robins School of Business White Paper Series. University of Richmond, Richmond, Virginia. This White Paper is brought to you for free and open access by the Robins School of Business at UR Scholarship Repository. It has been accepted for inclusion in Robins School of Business White Paper Series, 1980-2011 by an authorized administrator of UR Scholarship Repository. For more information, please contact scholarshiprepository@richmond.edu. THE SECRETARY PROBLEM WITH A CALL OPTION* 1 Johns. Rose University of Richmond In addition to accepting or rejecting a candidate arriving at timer, we may consider purchasing an option at a cost ex to recall the candidate at time r+x, but this privilege may be invoked only once. For large sample size, using the best - choice criterion and deducting option costs, the optimal strategy and return are obtained. Secretary problem; optimal choice, recall 1. Johns. Rose, Department of Management Systems , Robins Schoo l of Business, _University of Richmond, Richmond, VA, 23173. * Partially supported by grants from the DuPont Company and the Committee on Faculty Research, University of Richmond. 1. INTRODUCTION AND SUMMARY The classical model of the secretary problem can be found in Gilbert and Mosteller (3, Section 2a] or De Groot [2, pp. 325-331], and its solution is reviewed briefly in the next section. A restrictive assumption of the model is the impossibility of re-' calling previously rejected applicants. This restriction was relaxed by Yang [7] .: at any stage of the process we may attempt to recall a previously _rejected applicant, whose availability is uncertain and (maybe) stochastically decreasing in time. Others have subsequently extended Yang's work; see Corbin [1] or Petruc-celli [SJ. A different approach to the recall issue is proposed here. Let n be the population siz~ and suppose that, at some stage m, we observe an applicant which is preferable to the m-1 previously sampled, and rejected, applicants. Then, this applicant is a candidate (for the best). If we select (reject) the candidate, then we risk (not) finding a better one among the n-m applicants yet unsampled. In order to mitigate this risk, we may purchase an option granting us the privilege to recall the candidate at stage m+k, k=l, ••• , n-m. The cost to acquire such an option is bk, b~O. If the option is acquired, we say that we are holding the candidate fork periods; the interval {m+l, ••• , m+k} is called the holding period. The essential operational difference between our model and Yang's is that we must decide at the time of observation whether or not we shall later have the option to recall the candidate. 1 In Rose ["6), the call options were renewable, so it was suf-ficient to consider only the one-stage holding period, k=l. Here, we expressly prohibit renewal so that the duration, k, of th ,e holding period becomes a crucial decision variable •. Conseque~tly, if no better applicant is observed during the holding period, the option will be exercised and the candidate selected - clearly, in this context, it is suboptimal to purchase an option and then let it expire. Should a better applicant appear during the holding period, then we might take out another option to recall this new candidate, giving rise to the rather awkward situation . in which two, or even more, applicants are being held simultaneously. This phenomenon is under investigation and will be reported on else- . where1 to avoid that complication here, assume that we are permit-ted to hold only one candidate. Then, a better applicant appearing during the holding period will be accepted or rejected according to the classical procedure. Our qoal is to determine at what stages a candidate should be held, and for how long, in order to maximize the probability of selecting the best applicant less the holding costs. In the next section, letting n-+<o and _~O s~ch that nb+c, the infinite model is derived from the finite problem introduced above. The solution to the infinite model is summarized in the following theorem, whose proof is reiegated to Section 3. The function U(•) defined on [O,l) denotes the optimal return functioni see (1) and (8). The dependence of U on c is suppressed. 2 rHEOREM I~ If c~l, then the cl~ssical procedure is optimal and -1 U(r)=max{e , r}. II. For OSc<l, there exist s 1 (c) and ·s 2 (c), with O<s1ss 2<1, such that, for a candidate arriving at timer, it is optimal to: a. reject , rss 1; b. hold until time r/c, s 1<rss 2t c. hold until time 1, s 2 <r<l, provided that no candidate has been held previously. Furthermore, d. -1 e <r/c<l, for s 1<rss 2 t e. f. g. s 1 and s 2 are increasing inc. III. For OSc<l, there .exist t 1 (c), t 2 (c), t 3 (c), t 4 (c), with O<t 1st 2st 3st 4<1, such that a. b. c. d. e. f. U(r)=U(tl), rstl; . -1 U(r)=cr+e +rlog r/c, t 1<rst 2; -1 U(r)=cr+e +r-c, t 2<rst 3 ; U(r)=cr-rlog c, t 3<rst 4 ; . U(r)=cr-rlog r+r-c, t 4<rSl; U(O)Se- 1+e-J/Z_ -1 Note that, from (II.a,e), applicants arriving prior to c 1e , or approximately 12\ of the population, will be rejected automa-tically, even if the option cost, c, vanishes. Holding a candid-ate too soon, we may encounter a better applicant before time 3 -1 e , in which case nothing has been gained - the new applicant will be rejected according to the classical procedure • . There would be no such lower bound on s 1 (c) if several applicants could be held simultaneously. Even when c=O, the maximal return is only -1 -3/2 , e +e , the upper bound in (III.f), which is a relative -1 improvement of about 60% over the classical return, e • Again, if options could be taken on several applicants, then we should get U(O)=l for c=O. Also, consider (II.b,d,f). If c is not too small, then there .are times, r, for which it is optimal to hold for a period r/c-r, which is only part of the time remaining, 1-r. Contrast this result with those of (1; Special Case C-Additively Decreasing Recall, p. 211, in which it is optimal to observe the entire population before attempting recall, or of (61, in which it is optimal to renew the option until the end of the process an~ only then to exercise it. According to (II.d,e), we should hold a candidate appearing somewhat earlier than e-l until some time later than e- 1, which is the critical cut-off time for the classical problem (see (7)). Finally, it is implied in (II.a,b,c) that a candidate is accepted immediately upon its arrival only if the call option alternative was used to hold an earlier applicant. Numerical solutions of the finite problem are displayed in Table 1. Fortunately, these results are fairly well approximated by the solution of the infinite model as given in the Theorem. 4 2. FINITE AND INFINITE MODELS The optimality equations follow from the customary backward induction argument of dynamic programming. Keep in mind that our (net) return isl .or O minus option price. Define two events: A ={candidate appears at stage m}, and B ={first rn-1 applicants rn m . rejected}, rn=l, ••• , n. Let U (m) denote the maximum expected n return · given A and B. Also conditioning on A, let v (m,k) be rn rn rn n the probability that the best applicant is chosen, given that the candidate is held fork periods. Finally, let v (m) denote the n maximum expected return, given B 1• Then, rn+ U (m) = max {m/n, v (m), V (m,k)-bk}, n n n k=l, ••• ,n-m (1) where the three bracketed terms in (1) are the returns expected from accepting, rejecting, and holding (fork periods) the candidate, respectively. The second and third terms are computed recursively: v (rn) = [U (m+l)+mv (m+l))/(m+l), (v (n)=O) n n n n ( 2) V (rn,k)=[U 0 (m+l)+mV (m+l,k-1) )/(m+l), (V (rn,O)=m/n) (3) n . n Q . n k=l, ••• ,n-rn, where u0 is the optimal return function under the n classical model (see (6) below). The interpretation of (3) should . be clear. With probability 1/(rn+l) a better applicant appears at · stage rn+l and we apply the classical procedure. Otherwise, we still hold the candidate, but for only k-1 periods beyond stage rn+l. The condition V (m,O)=rn/n means merely that an expired n option should be exercised. 5 From the recursions (2) and (3) we solve for v and V (•,k) n n in terms of U and u0 respectively, obtaining n n n V (m) n = m I: i=m+l U (i) /i (i-1), n V (m,k) n m+k = m I: u0 (i)/i(i-1) + m/n. i=m+l n (4) (5) Now, we need to review briefly the solution to the classical problem. There exists a positive inte ger, r*, such that the first n r*-1 applicants are rejected and the first candidate, if any, n observed thereafter is selected. The optimal return function is u0 (m) = n m/n, ~r• n n-1 (r*-1)/n I: 1/i n i=r*-1 n , m<r*. n -1 0 . . + 0 + In the limit, r*/n+e and U ((rn] )+U (r) as n+«>, where [p) n n (6) denotes the smallest positive integer equal to or greater than p, and l . -1 r r>e u0 (r) ' - -1 -1 e , r:S:e , O~r~a. (7) Follow the method given in Muc·ci {4) to obtain the infinite model. To standardize and extend the return functions on the unit interval, + V ({rn) . , n + + write f (r)=U ([rn] ), h (r)~v ([rn] ), and g (r,x)= n n n n . n + [xn] ). Then (1) is rewritten as fn(r)= max + O<xSl-[rn] /n + + l[rn] /n, h (r), g (r,x)-b[xn) }, and the expressions (4) and (5) n n can be appropriately rewritten, too. Now, take limits as n+co. The option alternative is dominated unless b+O; to keep the 6 holding cost linear, suppose that nb+c~O as n+co, h +v, and g +V where n n U(r) = max {r, v(r), V(r,x)-cx}, OSxSl-r 1 2 v(r) = r fr U(y)dy/y ,. V(r,x) = r r;+rUo(y)dy/y 2 + r, Thus, f +U, n (8) (9) (10) OSrS:l, From the continuity of V(r,•), use of "max" in (8) is permis 'sible1 closing the interval with x=O is a trivial matter, since V(r,O)=r. The next section is devoted to solving (8)-(10). 3. PROOF OF THEOREM Some abbreviated notation will be helpful. Let w (r ,x)=V(r ,x) -ex, let x*=x*(r) maximize w(r,x) on [0,1-r], and let W(r)=w(r,x*). LEMMA 3.1 PROOF -1 If U(r)=W(r) _for r<e Obviously, ·u (r) ~u0 (r) =e -l. -1 then r+x*~e • -1 Suppose r+x*<e • · -1 -1 Substituting (7) into (10) yields V(r,x*)<V(r,e -r)=e , so W(r)<e- 1 • D LEMMA 3.2 Let c<l and suppose U(r)=W(r). Then x*=l-r on r>c and x*=r/c-r on rsc. If c~l, then U(r)=U 0 (r) and (I) holds. PROOF . -1 Using (7) and (10) and holding r+x~e , we obtain w (r,x)=r/(x+r)-c. For r>c, w (r,x)>O and x*=l-r1 and for rS:c, X X the maximum is achieved at x*•r/c-r. Now, suppose c~l and that o -1 U(r)>U (r) for some r. Then U(r)=W(r), so r+x*~e • However, 7 w (r,x)SO on r+x~e- 1• Either r~e- 1 and x*=O, in which case W(r)= X o -1 -1 V(r,O)=r=U (r), or r<e and x*=e -r, in which case -1 -1 o W(r)<V(r,e -r)=e =U (r). D Henceforth, assume c<l. Also, as the next lemma shows, we may eliminate the "r" term from (8) and just compare the reject and hold decisions. LEMMA 3.3 PROOF For r<l, U(r)>r. -1 o -1 -1 If r<e , then U(r)~U (r)=e • Suppose r~e • Then, U(r)~w(r,x)=r+rlog(l+x/r)-cx, from (7) and (10). For x suffi-ciently small, iog(l+x/r)>cx/r, so U(r)>r. 0 Finally, we state a lemma which will be proved several times for different values of rand c. LEMMA 3.4 If W(r)>v(r), then W(s)>v(s), s>r. The constant c 1 appearing in (II.e) is the solution of t = exp(-3/2(t+l)). (11) The role of this equation becomes apparent subsequently. It is convenient now to consider separately the three cases: c~e- 1 ; 3.1 -1 THE CASE c~e • First, hold r>c. x*=l-r, so W(r)=w(r,1-r)=cr-rlogr+r-c. 8 By Lenna 3.2, PROOF of Lemma 3.4: Let D(r)=W(r)-v(r), where vis given in (9). 2 2 Then, d/dr[D(r)/r]=-1/r+c/r +U(r)/r )0, since U(r)~r. It follows that D(r) is increasing. D Now, let r be such that W(r)>v(r). By Lerrana 3.4, we may substi-tute W(•) for U(•) in (9), obtaining v(r)=W(r)-r-crlogr+(r/2)log 2r. 2 It is easy to show that 1/2109 r-clogr-1<0, so v(r)<W(r). Thus, we have established (II.c) and (III.e), with s 2=t 4=c • . -1 Next, keep e <rSc and follow the approach used in the pre-ceding paragraph. Here, x*=r/c-r and W(r)=cr-rlogc. The proof of Lemma 3.4 is even simpler: d/dr[D(r)/r)=U(r~ 2>o. Assume W(r)> V (r). -1 On (e ,cl substitute W(•) for U(•), while on (c,l] use U( •) given in (III.e), thereby obtaining from (9), v(r)=W(r)-r/2• 2 2 log c+rlogclogr-crlogr-r. Let f(c)=(l+l /2 log c)/(logc-c). Then, v(r)<W(r) provided logr>f(c). -1 Because r>e , it suffices to show f(c)<-1, which is readily obtained. Hence, (III.d) is proved with -1 t 3=e • -1 Finally, let rSe and repeat the same sort of argument. The optimal holding period is still x*=r/c-r and, in evaluating . -1 V(r,x*), use Lemma 3.1 to get W(r)=cr+e +rlogr/c. Also by Lemma -1 3.1, if r<ce , then v(r)>W(r). To verify .Lemma 3.4, we get -1 2 2 1 d/dr[D(r)/r} =-e /r +1/r+U(r)/r >0, since U(r)~e-. Again, assume W(r)>v(r). In t .he computation of v(r), U(•) is given by W(•) on -1 -1 (r,e ] , by (III.d) on (e ,c), and by (Ill.e) on (c,1}. We get 2 v(r)=W(r)-rg(r), where g(r) is quadratic in logr: g(r)=l/2log r+ 2 (l+c-logc)logr+l/2log c+3/2. Hence, v(r)<W(r) if and only if 9 g(r)>0. Let r 1 denote the exp(-(l+c)+logc+fci), where larger root of g(r)=0, so r 1=r 1 (c}= . 2 d=d(c)=(l+c) -.2(1+c)logc-3 • . It is · -1 -1 fairly straightforward to show that ce <r 1<e and that g(r) -1 is strictly increasing once Sr<r 1 • Hence, we have now verified -1 (II.a,b,d,e,f,g) and (III.b), where s 1=t 1=r 1 and t 2=e • Because t 2=t 3 , (III.c) is trivial. To verify (III.a), note that U(r)=v(r) on rsr 1 • Then, substitute the appropriate values for U(•), as calculated above, into (9) and solve the resulting differential equation for v(r), given v(r 1)=cr 1+e-1+r 1logr 1/c. -2 -1 Fi~ally, note that v(r 1) is decreasing inc, so U(0)=v(r 1)se +e , and (III.f) holds. The remaining two cases are handled similarly. For the sake of parsimony, we omit most of the argument and concentrate on the rough spots. Once the sand t values are known, the results can be verified directly by substitution into (8)-(10), anyway. 3.2 -1 THE CASE cl<c<e • -1 -1 Here, t 4=t 3=e and, on r>e , U(r)= W(r) is given by (III.e), as in the previous section. -1 For c<rSe , -1 x*=l-r, W(r)=cr+e +r-c, and Lemma 3.4 holds. If U(s)EW(s) on s>r, then v(r)=W(r)-r(3/2+(c+l)logr). if r>h(c), where h(c)=exp(-3/2(c+l)). -1 -1 increasing, with h(0)>O and h(e )<e • Thus, v(r)<W(r) if and only -1 On (0,e . ), his concave Because c>c 1 , as given by (11), it follows that h(c)<c. Therefore, r>h(c) and (11.c) and (III.c) are established with s 2=t 2=c • . Finally, if rsc, then (II.a,b,f,g) and (III.a,b,f) are readily obtained as before, . with -1 Also, ce ~r 1<c, so (II.d,e) hold. 10 3.3 THE CASE c~c 1• The previous results are applicable here, -1 -1 for r>e For c<rSe , we still get v(r)<W(r) if and only if r>h(c). However, since csc 1 , h(c)~c. Thus (II.c) and (III.c) hold now with s 2=t 27h(c). Furthermore, U(r)=v(r)>W(r), c<r<h(c), and it remains to show that U(r)=v(r) on rsc. To this end, sub-stitute (III.c,e) for U(•) into (9) and solve the differential equation for v, obtaining v(r)=v(s 2)=W(s 2)=cs 2+e-1+s 2-c, a -1 constant. For rsc, W(r)=cr+e . +rlogr/c<W(s 2), and the result is established. That also proves (II.a,d,g) and (III.a), with s 1=t 1=h(c), and finally establishes the "if" part of (II.f). -3/2 ..;1 Finally,s 1=s 2=h(c)>h(O)=e >c1e , so (II.e) and (III.f) are verified . 11 REFERENCES [ll R.M. Corbin, "The secretary problem as a model of choice," J. Math. Psych • .!.!_, 1-29 (1980) (2) M.H. De Groot, Optimal Statistical Decisions, McGraw-Hill, New York (1970) [31 J.P. Gilbert and F. Mosteller, "Recognizing the maximum of a sequence," J. Am. Stat ·. Assoc. 61, 35-73 (1966) - , (4) A.G. Mucci, "Differential equations and optimal choice problems," Annals Stat.!, 104-113 (1973) [SJ J. D. Petruccelli, "Best-choice problems involving uncertainty of selection and recall of observations," J. Appl. Probab. 18, 415-425 (1981) [6] J.S. Rose, "Optimal sequential selection based on relative ranks with renewable call options," J. Am. Stat. Assoc. (to appear) [7] M.C.K, Yang, "Recognizing the maximum of a random sequence based on relative rank with backward solicitation," J. Appl. Probab. 11, 504-512 (1974) Cost (c) 0.0 • 2 .35 .5 .9 Table 1 Optimal Procedure (s 1/n,s 2/n) and Maximal Return (Un(0)) for b=c/n(l) 25 . 24,.24,.6121 .32,.32,.5322 .36, .36, .4823 .40, .48, .4402 .40, .88,. 3842 Population Size (n) 100 .23,.23,.5962 .29,.29,.5167 .34, .35, .4671 .36,.50,.4266 .38,.90,.3733 500 .224,.224,.5920 .288,.288,.5127 .330,.350,.4633 .354,.500,.4229 .370,.900,.3705 ... .2231,.2231,.5910 .2865~.2865,.5117 .3287,.3500,.4623 .3534,.5000,.4220 .3678,.9000,.3698 (1) For a candidate arriving at stage m, the optimal procedure prescribes rejecting if m<s1, holding for a period k<n-m if s 1Sm<s2 , and holding for a period n-m if m~s2• 

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The Secretary Problem with a Call Option

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