Chapter 16: Reference Dependence, Losses, Ownership, and Regret. The Encyclopedia of Economic Principals, interactive companion

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

The loss-adjusted selling price

How large a gap between selling and buying prices does loss aversion create?

Without loss aversion the owner and the buyer share one threshold, the use value. Loss aversion lifts only the owner's threshold, opening a band of prices at which neither trades.

$Equation, written in LaTeX: \mathrm{WTA}=1.5(\$6)=\$9,$

$Equation, written in LaTeX: \mathrm{WTP}=\$6.$

$Equation, written in LaTeX: \$9-\$6=\$3.$

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Dana owns the mug and Eli does not; both value using it at $6. Giving up the mug is coded as a loss weighted by the loss coefficient lambda, so Dana's minimum selling price WTA is lambda times $6. Eli's maximum buying price WTP stays at $6.

Predict first. With lambda = 1, does Dana sell at the posted price of $7.50?

Your prediction Yes, she sells No, she keeps the mug

Loss coefficient lambda

Posted price ($)

WTA (Dana, owner): $9.00
WTP (Eli, buyer): $6.00
Wedge WTA - WTP: $3.00
Dana at the posted price: Keeps the mug
Eli at the posted price: Declines

In dollars, WTA = 1.5 x 6 = 9.00 and WTP = $6.00, a wedge of $9.00 - $6.00 = $3.00. At the posted price, $7.50 is below her $9.00 threshold, so Dana keeps the mug; $7.50 is above his $6.00 use value, so Eli declines.

Worked steps

WTA = 1.5 x $6 = $9.00
WTP = use value = $6.00
Wedge = $9.00 - $6.00 = $3.00
Dana: $7.50 < $9.00, so she keeps the mug
Eli: $7.50 > $6.00, so he declines

Use the idea

When owners hold out for much more than buyers will pay, check whether the gap is a loss-weighted threshold rather than a difference in use value.

Where the conclusion applies

A riskless exchange with negligible wealth effects and transaction costs, a single loss coefficient and equal use values. Ownership alone does not prove loss aversion.

Check your understanding: With lambda = 2, what is the wedge?

WTA = 2 x $6 = $12, so the wedge is $12 - $6 = $6.

Chapter 16 source: section "Loss aversion".

Demonstration 2 of 4

Regret beats a better gamble

Can anticipated regret make a chooser reject a gamble with a higher expected value?

Squaring the payoff difference makes the large $50 regret in the bad state outweigh many small rejoicings in the good state. The expected payoffs never change; the comparison with the rejected option does.

$Equation, written in LaTeX: E[G]=0.9(\$65)+0.1(\$0)=\$58.50.$

Equation, written in LaTeX: D(G,S)=0.9(225)+0.1(-2{,}500)=-47.5.

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S is the sure payment and G pays $65 with probability p, otherwise $0. Both outcomes are revealed. d is the gamble's payoff minus the sure payoff in each state, and Q(d) = d squared for d at or above 0 and minus d squared for d below 0. D(G,S) is the probability-weighted sum of Q.

Predict first. At p = 0.95 with the sure $50, does the chooser still take the sure payment?

Your prediction Yes, the sure payment No, the gamble

Sure payment ($)

Chance of the good state

Figure: Regret beats a better gamble. Left: expected payoffs, sure $50.00 against the gamble's $58.50. Right: the regret terms 202.50 and -250.00 sum to D(G,S) = -47.50, so the chooser takes the sure payment. — Sure payment ($): $50, Chance of the good state: 0.9
Constructed example: the chapter's hypothetical choice (sure $50 against $65 with probability 0.9); sure payments $40 and $45 and probabilities 0.95 and 0.99 are added for comparison.

E[S]: $50.00
E[G]: $58.50
Q(good): 225
Q(bad): -2,500
D(G,S): -47.50
Choice: Sure payment

E[G] = 0.90 x 65 = $58.50, above the sure $50.00. With both outcomes revealed, the good state gives d = 65 - 50 = 15 and Q = 225; the bad state gives d = 0 - 50 = -50 and Q = -2,500. D(G,S) = 0.90 x 225 + 0.10 x (-2,500) = 202.50 + (-250.00) = -47.50, negative, so the chooser takes the sure payment despite the gamble's higher expected value.

Worked steps

E[G] = 0.90 x 65 + 0.10 x 0 = $58.50
Good state: d = 65 - 50 = 15, Q = 15 x 15 = 225
Bad state: d = 0 - 50 = -50, Q = -(50 x 50) = -2,500
D(G,S) = 0.90 x 225 + 0.10 x (-2,500) = 202.50 + (-250.00) = -47.50
D(G,S) < 0, so the chooser takes the sure payment

Use the idea

When people turn down a favorable gamble whose result they will see next to the safe option, ask whether hiding the foregone result would change the choice.

Where the conclusion applies

Linear outcome utility, a single illustrative odd comparison rule Q and full revelation of both outcomes. A different Q or no feedback can reverse the choice.

Check your understanding: With S = 40 and p = 0.9, what is D(G,S)?

d good = 25, Q = 625; d bad = -40, Q = -1,600; D = 0.9 x 625 + 0.1 x (-1,600) = 562.5 - 160 = 402.5 > 0, so the chooser takes the gamble.

Chapter 16 source: section "Regret theory".

Demonstration 3 of 4

When history sneaks into the decision

Can a larger, irrecoverable past spend flip a forward-looking decision?

Only the forward value matters for the decision, and it is the same in every state. The sunk-cost term adds history to the score, and a large enough past spend pushes it above zero.

$Equation, written in LaTeX: B-C=\$45{,}000-\$60{,}000=-\$15{,}000,$

$Equation, written in LaTeX: -\$15{,}000+0.10(\$20{,}000)=-\$13{,}000,$

$Equation, written in LaTeX: -\$15{,}000+0.10(\$200{,}000)=\$5{,}000,$

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B is expected receipts from finishing the documentary and C the remaining cost. Past spending is sunk and conveys no information. The committee's score adds alpha times the disclosed past spending.

Predict first. With alpha = 0.05 and $200,000 sunk, does the committee continue?

Your prediction Yes, it continues No, it stops

Disclosed past spending ($)

Sunk-cost weight alpha

Figure: When history sneaks into the decision. Bars: forward value -$15,000, sunk-cost term $20,000 and continuation score $5,000; the committee continues. — Disclosed past spending ($): $200,000, Sunk-cost weight alpha: 0.1
Constructed example: the chapter's hypothetical documentary (receipts $45,000, cost $60,000, alpha 0.10, sunk $20,000 or $200,000); sunk $100,000 and weights 0, 0.05 and 0.20 are added for comparison.

Forward value B - C: -$15,000
Sunk-cost term: $20,000
Continuation score: $5,000
Committee decision: Continue
Rational decision: Stop

B - C = $45,000 - $60,000 = -$15,000 whatever was spent before. The committee adds 0.10 x 200,000 = 20,000 dollars, so its score is -15,000 + 20,000 = 5,000 and it continues, taking on an expected loss of $15,000.

Worked steps

B - C = 45,000 - 60,000 = -15,000
Sunk-cost term = 0.10 x 200,000 = 20,000
Score = -15,000 + 20,000 = 5,000
Score > 0, so the committee continues

Use the idea

Before continuing a project, write down the remaining cost and remaining benefit alone; if the decision changes once past spending is mentioned, the past spending is doing the work.

Where the conclusion applies

No resale value, past spending that carries no information about receipts, and a linear sunk-cost weight alpha.

Check your understanding: What past spending makes the alpha = 0.10 committee indifferent?

0.10 x S = 15,000, so S = $150,000.

Chapter 16 source: section "Sunk-cost effect".

Demonstration 4 of 4

Six bets as one portfolio

How does looking at a bundle of favorable bets change the chance of losing money?

Seen one at a time, each bet has a 50 percent chance of a loss. Pooled, gains in some bets offset losses in others, so six bets lose with probability 34.375 percent. The fall is not smooth: because payoffs come in whole heads, 12 bets at $120 lose slightly more often (38.721 percent), while at $150 the chance drops to 19.385 percent. The expected total grows with every bet.

$Equation, written in LaTeX: 0.5(\$120)-0.5(\$100)=\$10,$

$Equation, written in LaTeX: 6(\$10)=\$60.$

Equation, written in LaTeX: 120h-100(6-h)=220h-600.

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Each bet pays the heads payoff or loses $100 on a fair coin. h is the number of heads among n independent bets. The loss probability is the chance the bundle's total is below zero.

Predict first. With 12 bets instead of 6, does the loss probability rise or fall?

Your prediction It rises It falls

Bets in the bundle

Payoff after heads ($)

Figure: Six bets as one portfolio. Distribution of the total payoff of 6 bets paying 120 or minus 100: losing totals in red carry probability 34.375%; the expected total is $60. — Bets in the bundle: 6, Payoff after heads ($): $120
Constructed example: the chapter's hypothetical six bets (+$120 or -$100 on a fair coin); bundles of 1 and 12 bets and heads payoffs of $110 and $150 are added for comparison.

Expected value per bet: $10.00
Expected bundle total: $60.00
Loss probability: 34.375%
Worst case: -$600
Losing outcomes: 2 or fewer heads

Each bet is worth 0.5 x 120 - 0.5 x 100 = 10.00 dollars, so 6 bets are worth 6 x 10.00 = 60.00. With h heads the total is 120h - 100(6 - h) = 220h - 600, which is negative when h is 2 or fewer. That happens in 22 of 2^6 = 64 equally likely sequences, a loss probability of 34.375%; the worst case is -$600.

Worked steps

Per bet: 0.5 x 120 - 0.5 x 100 = 10.00
Bundle: 6 x 10.00 = 60.00
Total with h heads: 220h - 600
Loss when h <= 2 (at least 4 tails)
P(loss) = 22 / 64 = 34.375%

Use the idea

Judge a repeated favorable risk by the distribution of the total, not by the chance of losing on any single round.

Where the conclusion applies

Independent fair coins, identical bets and enough liquidity to bear the worst case.

Check your understanding: With one bet paying $120 after heads, what is the loss probability?

A single tail loses $100, so the loss probability is 1/2 = 50%.

Chapter 16 source: section "Narrow bracketing".

The loss-adjusted selling price

Choose an example

Calculated values

Worked steps

Use the idea

Where the conclusion applies

Regret beats a better gamble

Choose an example

Calculated values

Worked steps

Use the idea

Where the conclusion applies

When history sneaks into the decision

Choose an example

Calculated values

Worked steps

Use the idea

Where the conclusion applies

Six bets as one portfolio

Choose an example

Calculated values

Worked steps

Use the idea

Where the conclusion applies