Ma (1996) studied the random order mechanism, a matching mechanism suggested by Roth and Vande Vate (1990) for marriage markets. By means of an example he showed that the random order mechanism does not always reach all stable matchings. Although Ma's (1996) result is true, we show that the probability distribution he presented - and therefore the proof of his Claim 2 - is not correct. The mistake in the calculations by Ma (1996) is due to the fact that even though the example looks very symmetric, some of the calculations are not as ''symmetric.''

Bettina Klaus

Flip Klijn

English

Research Papers in Economics

Corrigendum to \On Randomized Matching Mechanisms"
[Economic Theory 8(1996)377-381]¤
Bettina Klausy Flip Klijnz
January 2006
Summary: Ma (1996) studied the random order mechanism, a matching mechanism suggested
by Roth and Vande Vate (1990) for marriage markets. By means of an example he showed that
the random order mechanism does not always reach all stable matchings. Although Ma's (1996)
result is true, we show that the probability distribution he presented { and therefore the proof
of his Claim 2 { is not correct. The mistake in the calculations by Ma (1996) is due to the
fact that even though the example looks very symmetric, some of the calculations are not as
\symmetric." JEL classi¯cation: C78
For a description of the marriage model we refer to Roth and Vande Vate (1990). A marriage
market is denoted by (M;W;P) where M = fm1;:::;mag is a set of \men," W = fw1;:::;wag
is a set of \women," and P is a preference pro¯le. The set of stable matchings for (M;W;P) is
denoted by S(P). We now recall the random order mechanism.
Random Order (RO) Mechanism
Input: A marriage market (M;W;P).
Set R0 := ;, ¹0 such that for all i 2 N, ¹0(i) = i, and t := 1.
Step t: Choose an agent it from NnRt¡1 at random. Set Rt := Rt¡1 [ fitg.
Suppose it = w 2 W. (Otherwise replace w by m in Step t.)
Stable Room Procedure
Case (i) There exists no blocking pair (m;w) for ¹t¡1 with m 2 Rt:
Stop if t = n and de¯ne RO(P) := ¹t¡1. Otherwise set ¹t = ¹t¡1 and go to Step t := t + 1.
Case (ii) There exists a blocking pair (m;w) for ¹t¡1 with m 2 Rt:
Choose the blocking pair (m¤;w) for ¹t¡1 with m¤ 2 Rt that w prefers most.
If ¹t¡1(m¤) = m¤, then de¯ne ¹t such that ¹t(w) := m¤, ¹t(m¤) := w, and for all i 2 Nnfw;m¤g,
¹t(i) := ¹t¡1(i). Stop if t = n and de¯ne RO(P) := ¹t. Otherwise go to Step t := t + 1.
If ¹t¡1(m¤) = w0 2 W, then rede¯ne ¹t¡1(w) := m¤, ¹t¡1(m¤) := w, ¹t¡1(w0) := w0, and for all
i 2 Nnfw;m¤;w0g, ¹t¡1(i) := ¹t¡1(i). Set w := w0, and repeat the Stable Room Procedure.
¤We thank two anonymous referees for their helpful comments. B. Klaus's and F. Klijn's research was supported
by Ram¶ on y Cajal contracts of the Spanish Ministerio de Ciencia y Tecnolog¶ ³a. The work of the authors was also
partially supported by Research Grant BEC2002-02130 from the Spanish Ministerio de Ciencia y Tecnolog¶ ³a and
by the Barcelona Economics Program of CREA.
yCorresponding author. Department of Economics, Maastricht University, P.O. Box 616, 6200 MD Maastricht,
The Netherlands; e-mail: B.Klaus@algec.unimaas.nl
zInstitut d'Anµ alisi Econµ omica (CSIC), Campus UAB, 08193 Bellaterra (Barcelona), Spain; e-mail:
Flip.Klijn@uab.es
1The algorithm ends in exactly n := 2a steps and its outcome is a random stable matching
RO(P), generated by a sequence of agents (i1;:::;in). The set of possible sequences of agents
equals the set of permutations of all agents denoted by Q. Hence, jQj = n!. Moreover, for any
¹ 2 S(P), let Q¹ µ Q be the (possibly empty) set of sequences that lead to ¹. Denote q¹ = jQ¹j.
The random order mechanism induces in a natural way a probability distribution P over the set
of stable matchings: for any ¹ 2 S(P), the probability that RO(P) = ¹ equals p¹ =
q¹
n!.
By using the following example, Ma (1996) showed that the random order mechanism may
not reach all stable matchings. Although Ma's (1996) theorem is true, we show that the proba-
bility distribution he presented { and therefore the proof of his Claim 2 { is not correct.
Knuth's (1976) Example. Let (M;W;P) with a = 4 and P given below.
Preferences Stable Matchings
P(m1) = w1 w2 w3 w4 m1 ¹1 = m1 m2 m3 m4
P(m2) = w2 w1 w4 w3 m2 ¹2 = m2 m1 m3 m4
P(m3) = w3 w4 w1 w2 m3 ¹3 = m1 m2 m4 m3
P(m4) = w4 w3 w2 w1 m4 ¹4 = m2 m1 m4 m3
P(w1) = m4 m3 m2 m1 w1 ¹5 = m3 m1 m4 m2
P(w2) = m3 m4 m1 m2 w2 ¹6 = m2 m4 m1 m3
P(w3) = m2 m1 m4 m3 w3 ¹7 = m3 m4 m1 m2
P(w4) = m1 m2 m3 m4 w4 ¹8 = m4 m3 m1 m2
¹9 = m3 m4 m2 m1
¹10 = m4 m3 m2 m1
Ma (1996) claimed that (p¹1;p¹2;p¹3;p¹4;p¹5;p¹6;p¹7;p¹8;p¹9;p¹10) = (1
4; 1
8; 1
8;0;0;0;0; 1
8; 1
8; 1
4).
Below we show that this is not true by calculating the correct probability distribution:
( 9600
40320; 5280
40320; 5280
40320;0;0;0;0; 5280
40320; 5280
40320; 9600
40320). The mistake in the calculations is due to the
fact that even though the example looks very symmetric, some of the calculations are not as
\symmetric," in other words Ma's (1996) statement on page 380 that \the proofs for the re-
maining cases are similar." is not correct. In spite of the mistake, our computation is still based
on Ma's (1996) idea of analyzing sequences of agents backwards, i.e., considering the last agent
that enters, subsequently the last but one agent that enters, etc.. The di®erence is that we
provide more detailed discussions and justify the (restricted) use of \symmetry." ¦
Proof: We show that (p¹1;p¹2;p¹3;p¹4;p¹5;p¹6;p¹7;p¹8;p¹9;p¹10) = ( 9600
40320; 5280
40320; 5280
40320;
0;0;0;0; 5280
40320; 5280
40320; 9600
40320) by checking which stable matchings the random order mechanism
induces for various sequences (i1;:::;i8). Whenever we refer to a unique stable matching ob-
tained for a marriage market not containing all agents, we calculated the man-optimal and the
woman-optimal matching for the \submarket" using the deferred acceptance algorithm and de-
tected that they coincide (this calculation is not included in the proof). Furthermore, whenever
we \satisfy" a blocking pair, the (unique) proposing agent does not propose to agents that are
better than his/her previous match (all these proposals would be rejected).
Case a: m1 enters last; i.e., the sequence of agents is (i1;:::;m1). There are only two stable
matchings ¹0 and ¹00 when the set of agents consists of all women W and the remaining three
men fm2;m3;m4g:
w1 w2 w3 w4
¹0 : j j j j
w1 m2 m3 m4
w1 w2 w3 w4
¹00 : j j j j
w1 m2 m4 m3
2When m1 enters last, he proposes to the single woman w1, who accepts. So, matching ¹0 implies
matching ¹1 and matching ¹00 implies ¹3.
Case a.1: m2 enters before m1; i.e., the sequence is (i1;:::;m2;m1).
Case a.1.1: m3 enters before m2 and m1; i.e., the sequence is (i1;:::;m3;m2;m1). The
unique stable matching before m3, m2, and m1 enter matches m4 to w4 and everybody else to
themselves. Next, when m3 enters he proposes to w3, who accepts. Similarly, when m2 enters
he proposes to w2, who accepts. Thus, w1 is single and the resulting matching is ¹0. Hence, all
5! sequences induce ¹1.
Case a.1.2: m4 enters before m2 and m1; i.e., the sequence is (i1;:::;m4;m2;m1). Similarly
as in Case a.1.1, all 5! sequences induce ¹1.
Case a.1.3: w1 enters before m2 and m1; i.e., the sequence is (i1;:::;w1;m2;m1). There are
only two stable matchings ~ ¹0 and ~ ¹00 before w1, m2, and m1 enter:
w2 w3 w4
~ ¹0 : j j j
w2 m3 m4
w2 w3 w4
~ ¹00 : j j j
w2 m4 m3
It is easy to check that half of the partial sequences (i1;:::;i5) with fi1;:::;i5g\fw1;m2;m1g =
; result in ~ ¹0, the other half in ~ ¹00: if [i5 2 fm3;m4g] then (i1;:::;i5) results in ~ ¹0, if [i5 2
fw3;w4g] then (i1;:::;i5) results in ~ ¹00, if [i4 2 fm3;m4g and i5 = w2] then (i1;:::;i5) results
in ~ ¹0, and if [i4 2 fw3;w4g and i5 = w2] then (i1;:::;i5) results in ~ ¹00. After agents w1, m2,
and m1 enter, ~ ¹0 induces ¹0. Similarly, ~ ¹00 induces ¹00. Hence, 5!
2 sequences induce ¹1 and 5!
2
sequences induce ¹3.
Case a.1.4: w2 enters before m2 and m1; i.e., the sequence is (i1;:::;w2;m2;m1). Similarly as
in Case a.1.3, 5!
2 sequences induce ¹3 and 5!
2 sequences induce ¹1.
Case a.1.5: w3 enters before m2 and m1; i.e., the sequence is (i1;:::;w3;m2;m1). The unique
matching before agents w3, m2, and m1 enter matches m3 to w4, m4 to w2, and w1 to herself.
When w3 enters she proposes to m4, who accepts. Now w2 is single. Next, when m2 enters he
proposes to w2, who accepts. Thus, w1 is single and the resulting matching is ¹00. Hence, all 5!
sequences induce ¹3.
Case a.1.6: w4 enters before m2 and m1; i.e., the sequence is (i1;:::;w4;m2;m1). Similarly as
in Case a.1.5, all 5! sequences induce ¹3.
Summary Case a.1: 360 sequences (i1;:::;m2;m1) induce ¹1 and 360 sequences
(i1;:::;m2;m1) induce ¹3.
Summary Cases a.2-a.7: In a similar way as in Case a.1 we can calculate the number of
sequences that induce ¹1 and ¹3, respectively, in case the last but one position is occupied by
an agent di®erent from m2. We summarize the results in the table below.
Summary Case a: By summing up the boldface numbers in the table below we see that 2400
sequences (i1;:::;m1) induce ¹1 and 2640 sequences (i1;:::;m1) induce ¹3.
3Case Sequences Inducing ¹1 Inducing ¹3
a.1 (i1;:::;m2;m1) 360 360
a.2 (i1;:::;m3;m1) 720 {
a.3 (i1;:::;m4;m1) 720 {
a.4.1 (i1;:::;m2;w1;m1) 60 60
a.4.2 (i1;:::;m3;w1;m1) 120 {
a.4.3 (i1;:::;m4;w1;m1) 120 {
a.4.4 (i1;:::;w2;w1;m1) { 120
a.4.5 (i1;:::;w3;w1;m1) { 120
a.4.6 (i1;:::;w4;w1;m1) { 120
a.4 (i1;:::;w1;m1) 300 420
a.5.1 (i1;:::;m2;w2;m1) 60 60
a.5.2 (i1;:::;m3;w2;m1) 120 {
a.5.3 (i1;:::;m4;w2;m1) 120 {
a.5.4 (i1;:::;w1;w2;m1) { 120
a.5.5 (i1;:::;w3;w2;m1) { 120
a.5.6 (i1;:::;w4;w2;m1) { 120
a.5 (i1;:::;w2;m1) 300 420
a.6 (i1;:::;w3;m1) { 720
a.7 (i1;:::;w4;m1) { 720
Case b: m2 enters last; i.e., the sequence is (i1;:::;m2).
Because of the symmetry of the preferences, by changing the roles of agents [m1 and m2],
[w1 and w2], [m3 and m4], and [w3 and w4] in the proof of Case a we can show that in Case b
2400 sequences (i1;:::;m2) induce ¹1 and 2640 sequences (i1;:::;m2) induce ¹3.
Case c: m3 enters last; i.e., the sequence is (i1;:::;m3).
There are only two stable matchings ^ ¹0 and ^ ¹00 when the set of agents consists of all women
W and the remaining three men fm1;m2;m4g:
w1 w2 w3 w4
^ ¹0 : j j j j
m1 m2 w3 m4
w1 w2 w3 w4
^ ¹00 : j j j j
m2 m1 w3 m4
When m3 enters last, he proposes to the single woman w3, who accepts. So, matching ^ ¹0 implies
matching ¹1 and matching ^ ¹00 implies ¹2.
In order to determine which sequences induce matchings ¹1 and ¹2, we change the roles of
agents [m1 and m3], [w1 and w3], [m2 and m4], and [w2 and w4] in the proof of Case a. Note
that after this change, matching ^ ¹0 corresponds to ¹0 in the proof of Case a and matching ^ ¹00
corresponds to ¹00 in the proof of Case a. Similarly, matching ¹1 corresponds to ¹1 in the proof
of Case a and ¹2 corresponds to ¹3 in the proof of Case a.
Thus, changing the roles of the agents as speci¯ed above in the proof of Case a implies that
in Case c 2400 sequences (i1;:::;m3) induce ¹1 and 2640 sequences (i1;:::;m3) induce ¹2.
Case d: m4 enters last; i.e., the sequence is (i1;:::;m4).
Because of the symmetry of the preferences, by changing the roles of agents [m3 and m4],
[w3 and w4], [m1 and m2], and [w1 and w2] in the proof of Case c we can show that in Case d
2400 sequences (i1;:::;m4) induce ¹1 and 2640 sequences (i1;:::;m4) induce ¹2.
4Summary Cases a to d: Let m 2 M. Then, 9600 sequences (i1;:::;m) induce ¹1, 5280
sequences (i1;:::;m) induce ¹2, and 5280 sequences (i1;:::;m) induce ¹3.
Let w 2 W. Similarly to Cases a to d, 9600 sequences (i1;:::;w) induce ¹10, 5280 sequences
(i1;:::;w) induce ¹9, and 5280 sequences (i1;:::;w) induce ¹8.
Finally, the probability distribution induced by the random order mechanism equals
(p¹1;p¹2;p¹3;p¹4;p¹5;p¹6;p¹7;p¹8;p¹9;p¹10) = ( 9600
40320; 5280
40320; 5280
40320;0;0;0;0; 5280
40320; 5280
40320; 9600
40320) 6=
(1
4; 1
8; 1
8;0;0;0;0; 1
8; 1
8; 1
4).1 2
References
Knuth, D.E. (1976) Marriages Stables. Montreal: Les Presses de l'Universit¶ e Montreal.
Ma, J. (1996) \On Randomized Matching Mechanisms," Economic Theory 8, 377-381.
Roth, A.E. and Vande Vate, J.H. (1990) \Random Paths to Stability in Two-Sided Matching,"
Econometrica 58, 1475-1480.
1For a version of the proof discussing all cases in detail please contact any of the authors.
5

Corrigendum to ''On Randomized Matching Mechanisms'' [Economic Theory 8(1996)377-381]

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