The Encyclopedia of Economic Principals

Chapter 89

Norms, Identity, Social Interaction, and Multipliers

How other people's choices amplify, stall and reverse our own.

Four of the chapter's hypothetical cases, made interactive: a peer multiplier, a stalled threshold cascade, a tipping neighborhood, and a buyer who values exclusivity.

Every example here is a constructed teaching example: it uses the hypothetical numbers of the chapter's worked examples, plus a few values added for comparison and labelled as such in each panel. Nothing here measures a real market, firm or household.

Demonstration 1 of 4

How peers multiply a subsidy

How much does peer feedback add to a subsidy's direct effect on adoption?

Each round of adoption triggers a gamma-sized response in the next, so the rounds form a geometric series summing to beta / (1 - gamma). Positive feedback amplifies, negative feedback dampens.

Equation, written in LaTeX: \frac{d\bar y}{dx}=\frac{\beta}{1-\gamma}, M_s=\frac{1}{1-\gamma}.

Equation, written in LaTeX: \frac{6}{1-0.4}=10

Equation, written in LaTeX: \frac{6}{1.2}=5

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beta is the direct effect of the subsidy with coworker behavior fixed (points of adoption); gamma is the extra adoption induced by each one-point rise in peer adoption. M_s is the social multiplier.

Predict first. With crowded bicycle storage, gamma = -0.2. Is the total above or below the direct 6 points?

Your prediction

Choose an example

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Figure: How peers multiply a subsidy. Bars of each adjustment round starting at 6 points, a running total line, and the equilibrium total of 10.00 points.
Direct effect (points): 6, Peer feedback: 0.4
Constructed example: the book's hypothetical workplace has beta = 6 and gamma = 0.4, 0 and -0.2; beta of 3 and 9 and gamma of 0.6 are added.

Calculated values

Equilibrium change
10.00 points
Direct effect
6 points
Feedback
4.00 points
Multiplier 1 / (1 - gamma)
1.67

A direct effect of 6 points with peer feedback 0.40 settles at 6 / (1 - 0.40) = 10.00 points: peers amplify the subsidy: 4.00 points come through feedback.

Worked steps

  1. Rounds: 6, 6 x 0.40 = 2.40, 2.40 x 0.40 = 0.96, ...
  2. Total = 6 / (1 - 0.40) = 6 / 0.60 = 10.00
  3. Feedback = 10.00 - 6 = 4.00

Use the idea

Separate a program's direct effect from peer feedback before scaling it; the multiplier is local and fails as gamma nears one.

Where the conclusion applies

Linear, stable feedback with gamma below one; the chapter warns that capacity limits can shrink or reverse the peer response.

Check your understanding: With beta = 9 and gamma = 0.6, what is the equilibrium change?
9 / (1 - 0.6) = 22.5 points.

Chapter 89 source: section "Social multiplier".

Demonstration 2 of 4

One threshold stops a cascade

How can a one-unit change in one worker's threshold stop a cascade, and can a seed restart it?

A cascade continues only while each next threshold is within reach of the current count. One gap stalls it; a seed that fills the gap restarts it.

Equation, written in LaTeX: (0,1,2,3,4,5,6,7,8,9).

Equation, written in LaTeX: (0,1,2,4,4,5,6,7,8,9).

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Each worker joins once the visible count of active workers (plus any temporary, publicly observed seeds) reaches the worker's threshold. Joining is irreversible.

Predict first. With thresholds (0, 1, 2, 4, 4, 5, ..., 9), does one visible seed revive the stalled cascade?

Your prediction

Choose an example

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Figure: One threshold stops a cascade. Ten workers plotted by threshold; 3 are filled as joined. Thresholds 0, 1, 2, 4, 4, 5, 6, 7, 8, 9 with 0 seeds.
Fourth worker's threshold: 4, Temporary visible seeds: 0
Constructed example: the book's hypothetical thresholds (0 to 9), the change from 3 to 4 and one seed; a fourth threshold of 5 and two seeds are added.

Calculated values

Final participation
3 of 10
Visible seeds
0
Thresholds
0, 1, 2, 4, 4, 5, 6, 7, 8, 9

With the fourth threshold at 4 and 0 visible seeds, the cascade stalls at 3 of 10: the next threshold is 4, above the visible count of 3 + 0 = 3. The final visible count is 3 + 0 = 3.

Worked steps

  1. Thresholds: (0, 1, 2, 4, 4, 5, 6, 7, 8, 9); seeds = 0
  2. Active counts by round: 0, 1, 2, 3
  3. Result: the cascade stalls at 3 of 10: the next threshold is 4, above the visible count of 3 + 0 = 3

Use the idea

Look for the next unmet threshold, and compare the cost of a seed with the cost of lowering that one boundary.

Where the conclusion applies

Workers observe the true count, joining is irreversible, and seeds count as visible participants.

Check your understanding: With the fourth threshold at 5 and one seed, how many join?
Count 3 + 1 = 4 brings in the threshold-4 worker, then 4 + 1 = 5 brings both 5s, and all 10 join.

Chapter 89 source: section "Threshold model of collective behavior".

Demonstration 3 of 4

Five exits that tip a neighborhood

Can a few unrelated departures tip a mixed neighborhood into complete A exit?

Each departure lowers the share seen by those who remain, and the denominator shrinks with it. Once one group's boundary is crossed, its exit can push the share below the next group's.

Equation, written in LaTeX: s_A=\frac{47}{95}\approx49.5\%.

Equation, written in LaTeX: \frac{27}{75}=36\%,

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52 A and 48 B households occupy 100 homes. Twenty A households stay only while the A share is at least the sensitive threshold; the other 32 accept shares down to 40 percent. External exits come from the less sensitive group.

Predict first. If moving costs lower the sensitive group's threshold to 45 percent, does the cascade start?

Your prediction

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Figure: Five exits that tip a neighborhood. A share of occupied homes by stage, starting at 52% and falling to 0.0%, with threshold lines at 50% and 40%.
External A exits: 5, Sensitive group's threshold: 50%
Constructed example: the book's hypothetical neighborhood has 52 A, 48 B, groups of 20 and 32 with thresholds 50 and 40 percent, 5 exits and a 45 percent alternative; 3 and 7 exits are added.

Calculated values

A share after exits
49.5%
Final A households
0
Vacancies
52

5 external exits leave 47 A households among 95 occupied homes, a share of 47 / 95 = 49.5% against the sensitive group's 50% boundary, so the neighborhood tips: every A household leaves, leaving 48 B households and 52 vacancies.

Worked steps

  1. After 5 exits: 47 / (47 + 48) = 47 / 95 = 49.5%
  2. Wave 1: 27 / (27 + 48) = 36.0%
  3. Wave 2: 0 / (0 + 48) = 0.0%
  4. Result: the neighborhood tips: every A household leaves, leaving 48 B households and 52 vacancies

Use the idea

Track the share each household observes and its threshold; small changes in mobility or entry timing can stop a chain.

Where the conclusion applies

No entry into vacancies, groups move together and B households stay; the chapter treats this as an amplification mechanism, not the cause of any real city's pattern.

Check your understanding: With 7 exits and the 45 percent threshold, what happens?
45 / 93 = 48.4%, above 45% and 40%, so everyone else stays: 45 A households remain.

Chapter 89 source: section "Schelling segregation tipping".

Demonstration 4 of 4

Exclusivity and willingness to pay

How does expected ownership change a status buyer's decision at a given price?

For a snob buyer, each extra owner lowers willingness to pay, so wider distribution alone can lose the sale. A lower price can still win her back, and a bandwagon buyer moves the opposite way.

Equation, written in LaTeX: W=300-2N,

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W is Lena's willingness to pay for one watch and N the number of other visible owners she expects in her reference group. She buys when W exceeds the price.

Predict first. Ownership rises to 70 at price $180. Does Lena still buy?

Your prediction

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Figure: Exclusivity and willingness to pay. Willingness to pay falls with expected owners for the snob buyer and rises for the bandwagon buyer; at N = 40 the snob buyer's W is $220 against a price line at $180.
Expected other owners: 40, Price ($): 180
Constructed example: the book's hypothetical buyer has W = 300 - 2N, N of 40 and 70 and prices of 150, 180 and 210; N = 55 is added.

Calculated values

Willingness to pay W
$220
Surplus W - price
$40
Buys
Yes
Bandwagon buyer's W
$220

With 40 other expected owners, Lena's willingness to pay is 300 - 2 x 40 = 220 dollars against a price of 180 dollars, so Lena buys with surplus $40. A bandwagon buyer values the same watch at 140 + 2 x 40 = 220 dollars.

Worked steps

  1. W = 300 - 2 x 40 = 220
  2. Surplus = 220 - 180 = 40
  3. Bandwagon buyer: 140 + 2 x 40 = 220
  4. Result: Lena buys with surplus $40

Use the idea

Estimate the price response holding prevalence fixed and the prevalence response holding price fixed before choosing an edition size.

Where the conclusion applies

A linear willingness-to-pay rule with fixed quality, income and resale value, as in the chapter's hypothetical case.

Check your understanding: At N = 55 and price $210, does she buy?
W = 300 - 110 = 190 < 210, so no.

Chapter 89 source: section "Snob effect".