Proof of Lemma 1

In order to prove Lemma 1, we introduce the following notation. For every integer k≥1, let X(Q,h,k) be the set of students that will have proposed to hospital h by step k under DA(Q), i.e., in some step l∈{1,…,k} of DA(Q), and let m(Q,h,k) be the set of students tentatively matched to h in some step l∈{1,…,k} of DA(Q). Let X(Q,h) be the set of students that will have proposed to h by the last step of DA(Q), i.e., X(Q,h)=∪kX(Q,h,k).

We prove Lemma 1 by proving a series of claims. Let Qh′ list all students in μ(h′) in the same relative order as in Ph′ and report every other student as unacceptable. Let Q=(Qh′,Q˜−h′). Let Qh′′ lists all students in μ(h′)∪{s′} in the same relative order as in Ph′ and report every other student as unacceptable. Let Q′=(Qh′′,Q˜−h′) and let μ′=DA(Q′).

Claim 1. DA(Q)=μ.

Proof. Clearly, μ is individually rational at Q. Moreover, if a pair (s,h) blocks μ at Q, then (s,h) would also block μ at P. Hence, μ is stable at Q.

Now we show that DA(Q) is stable at Q˜. Clearly, DA(Q) is individually rational at Q˜. Assume that s∈S and h∈H∖{h′} block DA(Q) at Q˜. Since Qs=Q˜s and Qh=Q˜h, then (s,h) would also block DA(Q) at Q which is a contradiction to the stability of DA. So, we conclude that no s∈S and h∈H∖{h′} block DA(Q) at Q˜.

Now we show that no s∈S and h′ block DA(Q) at Q˜. Since both μ and DA(Q) are stable at Q, by Roth (1984a), |DA(Q)(h′)|=|μ(h′)|. Moreover, since only the students in μ(h′) are acceptable to h′ under Qh′,DA(Q)(h′)=μ(h′). Assume by contradiction that s∈S and h′ block DA(Q) at Q˜. Then, [|μ(h′)|\ltqh′ or sQ˜h′s∗ for some s∗∈μ(h′)] and h′PsDA(Q)(s). Since DA(Q) and μ are stable at Q and DA(Q) is the student optimal stable matching at Q,DA(Q)(s)Rsμ(s). Hence, h′Psμ(s). Therefore, s and h′ also block μ=DA(Q˜) at Q˜, but this contradicts the stability of DA. Thus, DA(Q) is stable at Q˜.

Since μ is stable at Q, DA(Q) is weakly preferred by all students to μ. Moreover, since DA(Q) is stable at Q˜,μ=DA(Q˜) is weakly preferred by all students to DA(Q). Therefore, μ=DA(Q) as desired. ▴

Claim 2. (A) holds, i.e., μ(h′)≠μ′(h′).

Proof. Student s′ proposes to h′ at the same step under DA(Q) and under DA(Q′). Furthermore, s′ is tentatively accepted by h′ at DA(Q′). Since s′ and h′ block μ at P,|μ(h′)|\ltqh′ or s′Ph′s∗ for some s∗∈μ(h′). Since Qh′′ lists students in μ(h′)∪{s′} according to Ph′ and lists any other student as unacceptable, s′ is not rejected in any latter step of DA(Q′). Thus, s′∈μ′(h′), and hence μ′(h′)≠μ(h′). ▴

Claim 3. For each hospital h and each step k,X(Q′,h,k)⊆X(Q,h,k).

Proof. For k=1 the inclusion is in fact an equality since at step 1 of DA(Q′) and DA(Q) each student proposes to exactly the same hospital.

Assume that the inclusion holds for k. We will show that the inclusion also holds for k+1. Let s∈X(Q′,h,k+1). If s∈X(Q′,h,k), then by induction, s∈X(Q,h,k) and hence s∈X(Q,h,k)⊆X(Q,h,k+1). So, assume s∈X(Q′,h,k+1)∖X(Q′,h,k). Then, in DA(Q′), student s proposed to h at step k+1 but not at step k. So, s was rejected by some hospital hˉ≠h at step k of DA(Q′). By the induction hypothesis, s∈X(Q′,hˉ,k)⊆X(Q,hˉ,k). If hˉ≠h′, then hˉ will also have rejected s by step k of DA(Q) since Qhˉ′=Qhˉ. Assume hˉ=h′. We consider two cases.

1. Case 1.s=s′ or s∉μ(h′). Then, ∅Qh′s. Thus, hˉ will also have rejected s by step k of DA(Q).

2. Case 2.s≠s′ and s∈μ(h′). By definition of Qh′′,sQh′′∅. Since s is rejected by h′ at step k of DA(Q′), (i) |m(Q′,h′,k)|=qh′ and (ii) for each student s∗∈m(Q′,h′,k),s∗Qh′′s. By responsiveness and the fact that Qh′′ is a dropping strategy obtained from Ph′,m(Q′,h′,k)Ph′μ(h′). Moreover, by the definition of DA,μ′(h′)Rh′m(Q′,h′,k). Hence, μ′(h′)Ph′μ(h′). This implies that Qh′′ is a profitable deviation for h′ contradicting that Q is an equilibrium. Therefore, s is not rejected by h′ at step k of DA(Q′). This contradicts the fact that s is rejected by h′ at step k of DA(Q′). Thus, Case 2 is impossible.

Since s is rejected by hˉ at step k of DA(Q) and he makes his proposals in the same order in DA(Q) and DA(Q′), he will have proposed to h by step k+1 of DA(Q). Hence, s∈X(Q,h,k+1). ▴

We complete the proof. (B) By Claim 3 and the fact that μ′≠μ (Claim 2), μ′ is weakly preferred by all students to μ. Then, the first part of (B) follows from (i) and (iii) of Lemma 1 in Erdil and Ergin (2008), page 684. The second part of (B) follows from (i) and (ii) of Lemma 1 in Erdil and Ergin (2008), page 684. (C) If h≠h′ and μ(h)≠μ′(h), then by (B) there is r∈μ(h)∖μ′(h) and s∈μ′(h)∖μ(h) such that r,s∈S. Moreover, since Qh=Qh′,r and s are acceptable under both Qh and Qh′. Assume by contradiction that |μ′(h)|\ltqh. By Claim 3, s∈μ′(h)⊆X(Q′,h)⊆X(Q,h). Therefore, s∈μ(h), but this is a contradiction. Hence, |μ′(h)|=qh. Using (B), we conclude |μ(h)|=|μ′(h)|=qh. (D) follows from (C) and the fact that μ(h)≠μ′(h). (E) Since μ′ Pareto dominates μ for students and s∈μ′(h)∖μ(h),hPsμ(s). Assume by contradiction that sQ˜hr∈μ(h). Then, s and h block μ at Q˜ contradicting the stability of DA. ▪

Proof of Proposition 1

Assume that PH has a cycle of length l≥2, given by

We define a preference profile for the students as follows. For each i=1,…,l let Psi:hi,hi−1,∅, where h0≡hl. For each s,s′∈S∖{s1,…,sl} let Ps=Ps′ and ∅Psh for each h∈{h1,…hl}. That is, all other students have the same preferences and any hospital h1,…,hl is unacceptable to them. The preferences of hospitals {h1,…,hl} and students {s1,…,sl} are depicted below, vertical dots mean that preferences can be arbitrary, while horizontal dots mean that the pattern of the given sequence continues until it reaches its final element.

Let qh=1 for all h∈H∖{h1} and qh1=2. Consider the dropping strategy for h1,Ph1′:s2,∅ and the profile P′=(Ph1′,P−h0). Note that at P′ all hospitals play a dropping strategy and all students report their true preferences. Let μ′=DA(P′).

We show that for each i=1,…,l,μ′(hi)={si+1}, where sl+1≡s1. At DA(P′) no student s∈S∖{s1,…,sl} proposes to a hospital in {h1…hl}. Furthermore, each student si,i=1,…,l, proposes to hi at the first step of DA(P′). Each student si,i=2,…,l, is tentatively accepted. However, s1 is rejected by h1. Therefore, s1 proposes to hl at step 2, hl tentatively accepts s1 and rejects sl,sl proposes to hl−1 at step 3, hl−1 tentatively accepts sl and rejects sl−1. This process continues until s2 proposes to h1,h1 tentatively accepts s2. At this point no si,i=1,…,l, makes more proposals so the the acceptances become final.

We show that P′ is a Nash equilibrium. Each hi, i=2,…sl fills its capacity with the best student among the students who find hi acceptable. Therefore, no hi, i=2,…sl has a profitable deviation. To show that h1 has no profitable deviations either, it is enough to show that h1 cannot improve by listing s1 as acceptable. Let Ph′′′ be such that s1Ph1′′∅ and P′′=(Ph1′,P−h0). At the first step of DA(P′′) each student si,i=1,…,l, proposes to hi. Furthermore, each students si,i=1,…,l, is tentatively accepted. Therefore, no si,i=1,…,l, makes more proposals and the acceptances become final. Since DA(P′′)(h1)={s1} and s2Ph1s1, P′′ is not a profitable deviation. Hence P′ is an equilibrium.

To complete the proof note that since (i) h1Ps1μ′(s1)=hl, (ii) |μ′(h1)|=1\ltqh1=2 and (iii) s1 is acceptable to h1,h1 and s1 block μ′ under the true preferences. Thus, μ′ is not stable with respect to P. ▪