Klaus, Bettina

Klijn, Flip

English

Research Papers in Economics

Economic Theory (2007) 32: 411–416
DOI 10.1007/s00199-006-0117-3
EXPOSITA NOTE
Bettina Klaus · Flip Klijn
Corrigendum to
“On randomized matching mechanisms”
[Economic Theory 8(1996)377–381]∗
Received: 29 June 2005 / Accepted: 10 April 2006 / Published online: 18 May 2006
© Springer-Verlag 2006
Abstract Ma (in Econ. Theory 8, 377–381, 1996) studied the random order
mechanism,amatchingmechanismsuggestedbyRothandVandeVate(Econome-
trica 58, 1475–1480, 1990) for marriage markets. By means of an example he
showed that the random order mechanism does not always reach all stable match-
ings. 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.”
Keywords Random order mechanism · Stability · Two-sided matching
JEL Classiﬁcation Number 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 ={ m1,...,ma} is a set
of “men,” W ={ w1,...,w b} is a set of “women” (possibly a  = b), and P is a
∗Wethanktwoanonymousrefereesfortheirhelpfulcomments.B.Klaus’sandF.Klijn’sresearch
was supported by Ramón y Cajal contracts of the Spanish Ministerio de Ciencia y Tecnología.
The work of the authors was also partially supported through the Spanish Plan Nacional I+D+I
(BEC2002-02130 and SEJ2005-01690) and the Generalitat de Catalunya (SGR2005-00626 and
the Barcelona Economics Program of CREA).
B. Klaus (B)
Department of Economics, Maastricht University,
P.O. Box 616, 6200 MD Maastricht, The Netherlands
E-mail: B.Klaus@algec.unimaas.nl
F. Klijn
Institut d’Anàlisi Econòmica (CSIC), Campus UAB,
08193 Bellaterra (Barcelona), Spain
E-mail: ﬂip.klijn@uab.es412 B. Klaus and F. Klijn
preference proﬁle. 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 := ∅, µ0 such that for all i ∈ M ∪ W, µ0(i) = i (all agents are matched to
themselves), and t := 1.
Step t: Choose an agent it from (M ∪ W) \ Rt−1 at random and independently of
previous choices. Set Rt := Rt−1 ∪{ it}. Suppose it = w ∈ W (otherwise replace
w by m in Step t) and enter the following procedure.
Stable room procedure
Case (i) There exists no blocking pair (m,w)for µt−1 with m ∈ Rt:
Stop if t = a +b and deﬁne RO(P) := µt−1. Otherwise set µt = µt−1 a n dg ot o
Step t := t + 1.
Case (ii) There exists a blocking pair (m,w)for µt−1 with m ∈ Rt:
Choose the blocking pair (m∗,w)for µt−1 with m∗ ∈ Rt that w prefers most.
If µt−1(m∗) = m∗ (man m∗ is matched to himself), then deﬁne µt such that
µt(w) := m∗ (womanw ismatchedtomanm∗),µt(m∗) := w (manm∗ ismatched
towomanw),andforalli ∈ (M ∪ W)\{w,m∗},µt(i) := µt−1(i)(allotheragents
keeptheirpreviousmatches).Stopift = a+banddeﬁne RO(P) := µt.Otherwise
go to Step t := t + 1.
If µt−1(m∗) = w  ∈ W (man m∗ is matched to woman w ), then redeﬁne
µt−1(w) := m∗ (woman w is matched to man m∗), µt−1(m∗) := w (man m∗
is matched to woman w), µt−1(w ) := w  (woman w  is matched to herself), and
for all i ∈ (M ∪ W)\{w,m∗,w }, µt−1(i) := µt−1(i) (all other agents keep their
previous matches). Set w := w , and repeat the Stable Room Procedure.1
The algorithm ends in exactly a + b steps and its outcome is a random stable
matching RO(P), generated by a sequence of agents (i1,...,ia+b).T h es e to f
possible sequences of agents equals the set of permutations of all agents denoted
by Q. Hence, |Q|=(a + b)! Moreover, for any µ ∈ S(P),l e tQµ ⊆ Q be the
(possibly empty) set of sequences that lead to µ. Denote qµ =| Qµ|. The random
order mechanism induces in a natural way a probability distribution P over the set
of stable matchings: for any µ ∈ S(P), the probability that RO(P) = µ equals
pµ = qµ/((a + b)!).
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 probability distribution he presented – and therefore the
proof of his Claim 2 – is not correct.
Knuth’s (1976) Example Let (M,W, P) with a = b = 4a n dP given in the left
part of the table below. The right part of the table lists the set of stable matchings
S(P) ={ µ1,...,µ 10},w h e r ew1, w2, w3,a n dw4 are matched to the men that
1 Note that our presentation of the random order mechanism is not exactly the same as Ma’s
(1996). In Ma (1996), each random order of agents is withdrawn with equal probability and then
the order is ﬁxed. We determine the random order of agents sequentially by withdrawing at the
beginning of each step with equal probability one of the agents from the remaining population.Corrigendum to “On randomized matching mechanisms” 413
are listed for each of the matchings µ1, ..., µ10, e.g., µ1 = m1 m2 m3 m4 means
that w1 is matched to m1, w2 is matched to m2, w3 is matched to m3,a n dw4 is
matched to m4.
Ma (1996) claimed that

pµ1, pµ2, pµ3, pµ4, pµ5, pµ6, pµ7, pµ8, pµ9, pµ10

= 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
correctprobabilitydistribution:

9600
40320, 5280
40320, 5280
40320,0,0,0,0, 5280
40320, 5280
40320, 9600
40320

.
Themistakeinthecalculationsisduetothefactthateventhoughtheexamplelooks
very symmetric, some of the calculations are not as “symmetric;” in other words
Ma’s (1996) statement on p. 380 that “the proofs for the remaining cases are simi-
lar.” 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 differ-
ence is that we provide more detailed discussions and justify the (restricted) use
of “symmetry.”  
Preferences Stable matchings
P(m1) = w1 w2 w3 w4 m1 µ1 = m1 m2 m3 m4
P(m2) = w2 w1 w4 w3 m2 µ2 = m2 m1 m3 m4
P(m3) = w3 w4 w1 w2 m3 µ3 = m1 m2 m4 m3
P(m4) = w4 w3 w2 w1 m4 µ4 = m2 m1 m4 m3
P(w1) = m4 m3 m2 m1 w1 µ5 = m3 m1 m4 m2
P(w2) = m3 m4 m1 m2 w2 µ6 = m2 m4 m1 m3
P(w3) = m2 m1 m4 m3 w3 µ7 = m3 m4 m1 m2
P(w4) = m1 m2 m3 m4 w4 µ8 = m4 m3 m1 m2
µ9 = m3 m4 m2 m1
µ10 = m4 m3 m2 m1
Proof We show that

pµ1, pµ2, pµ3, pµ4, pµ5, pµ6, pµ7, pµ8, pµ9, pµ10

=

9600
40320, 5280
40320, 5280
40320,0,0,0,0, 5280
40320, 5280
40320, 9600
40320

by checking which stable
matchingstherandomordermechanisminducesforvarioussequences(i1,...,i8).
Whenever we refer to a unique stable matching obtained for a marriage market
not containing all agents, we calculated the man-optimal and the woman-optimal
matchingforthe“submarket”usingthedeferredacceptancealgorithmanddetected
thattheycoincide(thiscalculationisnotincludedintheproof).Furthermore,when-
ever 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 µ  and µ   when the set of agents consists of all women W
and the remaining three men {m2,m3,m4}:
w1 w2 w3 w4
µ  : ||||
w1 m2 m3 m4
w1 w2 w3 w4
µ   :| | | |
w1 m2 m4 m3
.
Whenm1 enterslast,heproposestothesinglewomanw1,whoaccepts.So,match-
ing µ  implies matching µ1 and matching µ   implies µ3.414 B. Klaus and F. Klijn
Case a.1 m2 enters before m1; i.e., the sequence is (i1,...,m2,m1).
Case a.1.1 m3entersbeforem2 andm1;i.e.,thesequenceis(i1,...,m3,m2,m1).
The unique stable matching before m3, m2,a n dm1 enter matches m4 to w4 and
everybody else to themselves. Next, when m3 enters he proposes to w3, who ac-
cepts.Similarly,whenm2 entersheproposestow2,whoaccepts.Thus,w1 issingle
and the resulting matching is µ . Hence, all 5! sequences induce µ1.
Case a.1.2 m4 entersbeforem2 andm1;i.e.,thesequenceis(i1,...,m4,m2,m1).
Similarly as in Case a.1.1,a l l5 ! sequences induce µ1.
Case a.1.3 w1 entersbeforem2 andm1;i.e.,thesequenceis(i1,...,w 1,m2,m1).
There are only two stable matchings ˜ µ  and ˜ µ   before w1, m2,a n dm1 enter:
w2 w3 w4
˜ µ  :| | |
w2 m3 m4
w2 w3 w4
˜ µ   :| | |
w2 m4 m3
.
It is easy to check that half of the partial sequences (i1,...,i5) with {i1,...,i5}∩
{w1,m2,m1}=∅result in ˜ µ , the other half in ˜ µ  :i f[ i5 ∈{ m3,m4}]t h e n
(i1,...,i5) results in ˜ µ ,i f[ i5 ∈{ w3,w 4}]t h e n(i1,...,i5) results in ˜ µ  ,i f
[i4 ∈{ m3,m4} and i5 = w2]t h e n(i1,...,i5) results in ˜ µ ,a n di f[ i4 ∈{ w3,w 4}
and i5 = w2]t h e n(i1,...,i5) results in ˜ µ  . After agents w1, m2,a n dm1 enter, ˜ µ 
induces µ . Similarly, ˜ µ   induces µ  . Hence, 5!/2 sequences induce µ1 and 5!/2
sequences induce µ3.
Case a.1.4 w2 entersbeforem2 andm1;i.e.,thesequenceis(i1,...,w 2,m2,m1).
Similarly as in Case a.1.3,5 !/2 sequences induce µ3 and 5!/2 sequences induce
µ1.
Case a.1.5 w3 entersbeforem2 andm1;i.e.,thesequenceis(i1,...,w 3,m2,m1).
The unique matching before agents w3, m2,a n dm1 enter matches m3 to w4, m4
to w2,a n dw1 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 µ  . Hence, all 5! sequences induce µ3.
Case a.1.6 w4 entersbeforem2 andm1;i.e.,thesequenceis(i1,...,w 4,m2,m1).
Similarly as in Case a.1.5,a l l5 ! sequences induce µ3.
Summary Case a.1 360 sequences (i1,...,m2,m1) induce µ1 and 360 sequences
(i1,...,m2,m1) induce µ3.
Summary Cases a.2–a.7 In a similar way as in Case a.1 we can calculate the
number of sequences that induce µ1 and µ3, respectively, in case the last but one
position is occupied by an agent different from m2. We summarize the results in
the table below.Corrigendum to “On randomized matching mechanisms” 415
Case Sequences Inducing Inducing
µ1 µ3
a.1 (i1,...,m2,m1) 360 360
a.2 (i1,...,m3,m1) 720 –
a.3 (i1,...,m4,m1) 720 –
a.4.1 (i1,...,m2,w 1,m1) 60 60
a.4.2 (i1,...,m3,w 1,m1) 120 –
a.4.3 (i1,...,m4,w 1,m1) 120 –
a.4.4 (i1,...,w 2,w 1,m1) – 120
a.4.5 (i1,...,w 3,w 1,m1) – 120
a.4.6 (i1,...,w 4,w 1,m1) – 120
a.4 (i1,...,w1,m1) 300 420
a.5.1 (i1,...,m2,w 2,m1) 60 60
a.5.2 (i1,...,m3,w 2,m1) 120 –
a.5.3 (i1,...,m4,w 2,m1) 120 –
a.5.4 (i1,...,w 1,w 2,m1) – 120
a.5.5 (i1,...,w 3,w 2,m1) – 120
a.5.6 (i1,...,w 4,w 2,m1) – 120
a.5 (i1,...,w2,m1) 300 420
a.6 (i1,...,w3,m1) – 720
a.7 (i1,...,w4,m1) – 720
Summary Case a: By summing up the boldface numbers in the table below we
seethat2,400sequences(i1,...,m1)induceµ1 and2,640sequences(i1,...,m1)
induce µ3.
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 2,400 sequences (i1,...,m2) induce µ1 and 2,640
sequences (i1,...,m2) induce µ3.
Case c m3 enters last; i.e., the sequence is (i1,...,m3).
There are only two stable matchings ˆ µ  and ˆ µ   when the set of agents consists
of all women W and the remaining three men {m1,m2,m4}:
w1 w2 w3 w4
ˆ µ  : ||||
m1 m2 w3 m4
w1 w2 w3 w4
ˆ µ   :| | | |
m2 m1 w3 m4
.
Whenm3 enterslast,heproposestothesinglewomanw3,whoaccepts.So,match-
ing ˆ µ  implies matching µ1 and matching ˆ µ   implies µ2.
Inordertodeterminewhichsequencesinducematchingsµ1 andµ2,wechange
the roles of agents [m1 and m3], [w1 and w3], [m2 and m4], and [w2 and w4]i nt h e
proof of Case a. Note that after this change, matching ˆ µ  corresponds to µ  in the
proof of Case a and matching ˆ µ   corresponds to µ   in the proof of Case a.S i m -
ilarly, matching µ1 corresponds to µ1 in the proof of Case a and µ2 corresponds
to µ3 in the proof of Case a.416 B. Klaus and F. Klijn
Thus, changing the roles of the agents as speciﬁed above in the proof of Case a
impliesthatinCasec2,400sequences(i1,...,m3)induceµ1 and2,640sequences
(i1,...,m3) induce µ2.
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 2,400 sequences (i1,...,m4) induce µ1 and 2,640
sequences (i1,...,m4) induce µ2.
SummaryCasesatod: Letm ∈ M.Then,9,600sequences(i1,...,m)induceµ1,
5,280 sequences (i1,...,m) induce µ2, and 5,280 sequences (i1,...,m) induce
µ3.
Let w ∈ W. Similarly to Cases a to d, 9,600 sequences (i1,...,w)induce µ10,
5,280 sequences (i1,...,w)induce µ9, and 5,280 sequences (i1,...,w)induce
µ8.
Finally,theprobabilitydistributioninducedbytherandomordermechanismequals

pµ1, pµ2, pµ3, pµ4, pµ5, pµ6, pµ7, pµ8, pµ9, pµ10

=

9600
40320, 5280
40320, 5280
40320,
0,0,0,0, 5280
40320, 5280
40320, 9600
40320

 =
1
4, 1
8, 1
8,0,0,0,0, 1
8, 1
8, 1
4

.2    
References
Knuth, D.E.: Marriages Stables. Montréal: Les Presses de l’Université Montréal (1978)
Ma, J.: On randomized matching mechanisms. Econ. Theory 8, 377–381 (1996)
Roth, A.E., Vande Vate, J.H.: Random paths to stability in two-sided matching. Econometrica
58, 1475–1480 (1990)
2 For a version of the proof discussing all cases please contact any of the authors.

For a version of the proof discussing all cases please contact any of the authors.

Marriages Stables. Montréal: Les Presses de l’Université Montréal

Corrigendum to “On randomized matching mechanisms" [Economic Theory 8(1996)377–381].

http://arno.unimaas.nl/show.cgi?fid=16364

Corrigendum to “On randomized matching mechanisms" [Economic Theory 8(1996)377–381].

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