Demonstration 1 of 4
The cost of signing before the signal
What does an early binding offer give up when a fit signal is coming?
Early contracts give certainty but destroy the chance to pair on the signal. The information lost is the gap between waiting and early surplus; it is worth paying only when delay costs less.
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Matches are worth 110 when the fit is high and 55 when low; the two fit states are equally likely. w is the total waiting cost across the four participants.
Predict first. Above what waiting cost does early commitment become efficient?
Choose an example
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Constructed example: the chapter's hypothetical firms and graduates (110, 55, waiting cost 70); the waiting cost 40 is added.
Calculated values
- Waiting total
- 220
- Waiting net of cost
- 220
- Early total
- 165
- Information loss
- 55
- More surplus
- waiting
Waiting picks the high-fit pairs: 110 + 110 = 220, net 220 - 0 = 220. Early commitment fixes pairs worth (110 + 55)/2 = 82.5 each: 82.5 + 82.5 = 165. Waiting wins by 55.0. Waiting costs above 220 - 165 = 55 make early commitment efficient.
Worked steps
- Waiting = 110 + 110 = 220
- Net = 220 - 0 = 220
- Early pair = (110 + 55)/2 = 82.5
- Early = 82.5 + 82.5 = 165
- Information loss = 220 - 165 = 55
Use the idea
Coordinate a common offer date, or make early offers reversible, when the information still to come is worth more than the cost of waiting.
Where the conclusion applies
Two equally likely fit states, binding early offers and a known waiting cost.
Check your understanding: With waiting cost 40, which arrangement yields more surplus?
Chapter 31 source: section "Market unraveling".
Demonstration 2 of 4
When to hide your favorite school
When does the Boston mechanism push a student to rank her favorite school second?
Under Boston a school fills in round one, so a first choice that fails wastes a high-priority second choice. Whether to rank honestly depends on beliefs about rivals.
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Nia values school P at 120 and S at 85. Zara, who has priority at P, applies there with probability q. Nia's priority at S guarantees a seat if she lists S first.
Predict first. At what Zara probability is Nia indifferent?
Choose an example
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Constructed example: the chapter's hypothetical district (values 120 and 85, q = 0.40 and 0.20); q = 0.30 and 0.60 are added.
Calculated values
- List P first
- 72
- List S first
- 85
- Best report
- S first (demote favorite)
- Indifference probability
- 0.292
Listing P first pays (1 - 0.40)(120) = 72; listing S first pays 85 for sure. Since 72 < 85, Nia's best report is S first (demote favorite). She is indifferent at q = 1 - 85/120 = 0.292.
Worked steps
- P first = (1 - 0.40)(120) = 72
- S first = 85
- 72 < 85: S first (demote favorite)
- Indifference q = 1 - 85/120 = 0.292
Use the idea
A strategy-proof rule such as deferred acceptance removes the need for this guesswork.
Where the conclusion applies
Known values, a known chance that Zara applies and one seat at each school.
Check your understanding: At q = 0.30, does Nia report truthfully?
Chapter 31 source: section "Boston-mechanism priority loss".
Demonstration 3 of 4
Thick markets reveal the better pairs
How much does bringing everyone to market at once add, and when is it worth the wait?
A thick market lets each side compare all partners at once, so pairs form on fit rather than on who arrived first. The gain must exceed the cost of making everyone wait.
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v is the surplus of an employer and analyst pair. w is the waiting cost per participant; there are four participants.
Predict first. What per-person waiting cost erases the pooling gain?
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Constructed example: the chapter's hypothetical clinic and lab (values 110, 65, 75, 95; cost 18); the cost 10 is added.
Calculated values
- Pooled
- 205
- Pooled net of waiting
- 205
- Fragmented
- 140
- Pooling gain
- 65
- Better market
- pooling
Pooled pairs Clinic with Aria and Lab with Bo: 110 + 95 = 205, net 205 - 4(0) = 205. Fragmented: 65 + 75 = 140. The gain 35 + 30 = 65 is the two option losses. Pooling wins by 65. Pooling stops paying above 65/4 = 16.25 per participant.
Worked steps
- Pooled = 110 + 95 = 205
- Net = 205 - 4(0) = 205
- Fragmented = 65 + 75 = 140
- Gain = 35 + 30 = 65
Use the idea
Synchronize market dates when match values differ a lot across pairs; skip it when pairs are alike.
Where the conclusion applies
Known match values and an equal waiting cost for everyone. With every pair worth 80, as in the chapter's counterfactual, pooling adds nothing.
Check your understanding: At cost 10 per participant, does pooling still win?
Chapter 31 source: section "Market thickness".
Demonstration 4 of 4
Fewer applications, more reviews
When does capping applications help graduates get read?
Congestion arises when messages exceed the capacity to read them. Cutting volume or raising capacity both raise the review rate; neither helps once load is within capacity.
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120 graduates each send a applications spread over 12 firms; m_j is the load per firm and K the number each firm can review. rho is the chance an application is reviewed; each graduate has two suitable firms.
Predict first. From the book case (12 applications, K = 40), does doubling capacity or capping applications at 8 help more?
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Constructed example: the chapter's hypothetical platform (120 graduates, 12 firms, 12 or 4 applications, capacity 40 or 150, two suitable firms); 8 applications and capacity 80 are added.
Calculated values
- Total applications
- 1,440
- Load per firm
- 120
- Review probability
- 0.3333
- Unreviewed applications
- 960
- P(a suitable application reviewed)
- 55.56%
120(12) = 1,440 applications over 12 firms is 120 per firm. rho = 40/120 = 0.3333. Reviews: 480, leaving 960 unread. With two suitable firms, P = 1 - (1 - 40/120)^2 = 1 - (80/120)^2 = 55.56%.
Worked steps
- Total = 120(12) = 1,440
- Load = 1,440/12 = 120
- rho = 40/120 = 0.3333
- Unreviewed = 1,440 - 480 = 960
- P = 1 - (1 - 40/120)^2 = 1 - (80/120)^2 = 55.56%
Use the idea
Diagnose load against processing capacity before restricting applications.
Where the conclusion applies
Applications balanced across firms, random screening and independent review draws.
Check your understanding: With 8 applications and K = 40, what is the chance at least one suitable application is reviewed?
Chapter 31 source: section "Market congestion".