34
Turning Up the Heat The Demoralization Effect of Competition in Contests Dawei Fang 1 Thomas Noe 2 Philipp Strack 3 1 Department of Economics, Gothenburg 2 Said Business School, Oxford 3 Department of Economics, UC Berkeley 1 Motivation

Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

  • Upload
    others

  • View
    0

  • Download
    0

Embed Size (px)

Citation preview

Page 1: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Turning Up the HeatThe Demoralization Effect of Competition in Contests

Dawei Fang1 Thomas Noe2 Philipp Strack3

1Department of Economics, Gothenburg

2Said Business School, Oxford

3Department of Economics, UC Berkeley

1

Motivation

Page 2: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Motivation

• Contests are ubiquitous• sales contests• promotion contests• college admissions• exam grades• R&D contests

• But many of these competitions are “soft”.• Salary differences among managers receiving different performance

ratings are small (Medoff and Abraham, 1980)• Corporate sales contests often set multiple prizes (Backes-Gellner and

Pull, 2013)• Some firms employ large promotion rates (Ariga et al., 1999)• Some schools use easy grading curves: “As, Bs, and tuition fees”

2

Basic insight

• The classic tradeoff: increasing competitiveness• increases efficiency, but also• increases inequality

• This paper shows: In a very standard framework for contestcompetitions, all-pay auctions, that there is no tradeoff.

• When the marginal cost of effort is increasing, making contest morecompetitive leads to contestant demoralization, and thereby• reduces the cost efficiency of individual contestant effort• leads to second order stochastically dominated effort distributions• increases risk-taking under conventional statistical measures if cost

convexity effect is non-decreasing in the level of effort.

3

Page 3: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Is demoralization just a theoretical oddmint?

• Demoralization is consistent with property interpreted data from fieldand lab experiments• We develop simple statistical measures to identify demoralization effects

in contests• Using these measures, we identify demoralization effects in laboratory and

field experiments• Develop predictions for the interaction between competitiveness and

reward levels required for valid empirical tests of contest models

• But why are some contests so competitive?• Derive general conditions under which, competitive contest designs are

optimal despite demoralization

4

Testable implications: A sample

• Less competitive bonus schemes lead to higher worker output

• Small classes lead to higher student effort

• Competitive prizes improve the outcome in R&D contests when costconvexity is limited

5

Page 4: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

The model

The model

6

Page 5: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

The all-pay contest

• n≥ 2 homogeneous agents compete in an all-pay contest.

• Each agent i simultaneously chooses an effort (outcome) xi and incurs aneffort cost c(xi).

• c is differentiable, strictly increasing, and weakly convex, with c(0) = 0.

• There is an ordered vector of prizes v ∈ Rn+: v1 ≤ ·· · ≤ vn and

0 = v1 < vn.

• Agent with kth highest effort wins prize vk. Ties are broken randomly.

• Agents are risk neutral.

• An agent’s payoff equals his prize less his effort cost.

• Solution concept: symmetric Nash equilibrium

8

Equilibrium

• The effort distribution in any symmetric equilibrium must satisfy (Barutand Kovenock, 1998):• No point mass• No rents• Interval support [0, x̄], where c(x̄) = vn

LEMMA 1. EQUILIBRIUM

There exists a unique symmetric equilibrium. In this equilibrium, each agent’seffort distribution Fv is given implicitly by

exp. contest reward︷ ︸︸ ︷n

∑i=1

vi

(n−1i−1

)Fv(x)i−1 (1−Fv(x))

n−i =

effort cost︷︸︸︷c(x) , x ∈ [0, x̄].

10

Page 6: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

What is increased competition?

• Three notions of a more competitive environment• Greater prize inequality: Rewards to super-stars increase relative to

rewards to stars• Increase contest size: More students take exam under the same grading

curve• Entry by new contestants without an increase in contest rewards: New

firms enter a stagnant market.

• Basic result: Under all three notions of increased competitiveness:

Increased competitiveness =⇒ Demoralization

12

Competition: inequality

Page 7: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Prize inequality effects:Inequity and inefficiency

13

The tale of two contests

Contest Type Performance Prizes

Worst Median Best

Last-place elimination (E) 0 1 1Winner take all (wta) 0 0 2

15

Page 8: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Equilibrium in Contest E

• Three contestants: A, B, and C

• Each contestant chooses a random performance level with distribution

• Effort cost: c(x) = x2

• Equilibrium in Contest E

FE(x) = 1−√

1− x2,x ∈ [0,1]

• Expected contestant effort =

µ(XE) =∫ 1

0x fE(x)dx =

∫ 1

0(1−FE(x))dx =

π

4≈ 0.785

16

Why is this an equilibrium?

• Consider contestant A:

• Contestant A wins a prize of 1 if and only if she is not the worstperformer.

• Contestant A is the worst performer if both B and C outperform her.

• So if contestant A chooses effort x, the probability that A is the worstperformer is

P{XEB > x & XE

C > x}= P{XEB > x}P{XE

C > x}= (1−FE(x))2

• So, the expected contest reward to A from effort x equals

prize︷︸︸︷1 ×

Prob. Winning︷ ︸︸ ︷(1− (1−FE(x))2)

17

Page 9: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Expected contest rewards

Exp. Contest Reward

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

1.5

2.0

x -Effort

18

Reward = cost: Equilibrium

Exp. Contest Reward

Cost

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

1.5

2.0

x -Effort

19

Page 10: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Switch from E to WTA

• Suppose the contest design is changed to WTA

• and contestants B and C follow their E-contest strategies

• What is the expected contest reward for A?• Contestant A wins with effort x in the WTA contest if and only if A tops

both B and C:• Probability:

P{XEB < x & XE

C < x}= P{XEB < x}P{XE

C < x}= (FE(x))2

• Winner’s prize = 2• Expected contest reward:

Prize︷︸︸︷2 ×

Prob. of Winning︷ ︸︸ ︷(FE(x))2

20

Reward 6= Cost: Disequilibrium

Exp. Contest Reward

Cost

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

1.5

2.0

x -Effort

21

Page 11: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

A’s response

• A maximizes the difference between expected contest rewards and effortcosts• So A’s best response, holding the responses of B and C fixed, is to

increase effort to x = 1• So, it seems that switching to WTA increases A’s effort• But, unfortunately for A, B and C will also change their strategies• They would also want to choose effort level x = 1.• But if A, B and C choose effort level x = 1. A, B, and C will tie and split

equally the winners’ prize of 2.• Their contest reward equals

contest reward =13×2 < c(x = 1) = 1 = effort cost.

so, not an equilibrium

22

Equilibrium in WTA contest

• Under WTA, the highest effort level must exceed the highest effort levelin E. Why?• The highest effort level earns the winner’s prize with probability 1• The winners prize is 2• In equilibrium, cost equals expected reward.• So, at the highest level of effort under WTA, x̄wta,

c(x̄wta) = 2 =⇒ (x̄wta)2 = 2 =⇒ x̄wta =√

2

• Thus, effort spreads out

• Spreading out effort is inefficient

23

Page 12: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Effort and efficiency

• Cost efficiency and effort:

Cost efficiency =Effort

Effort Cost=

xc(x)

=1x↓ in x

Effort Cost×Cost efficiency = Effort

• In both the WTA contest and E contest, expected effort costs equalexpected contest rewards

• Expected contest rewards are the same in both contests, so expectedeffort costs are the same.

• In the WTA, effort costs are spent less efficiently, reaching for occasionalvery high performance levels

24

Equilibrium in the WTA

• Each contest chooses a random performance level with distribution

Fwta(x) =x√2, x ∈ [0,

√2]

• Expected contestant effort =

µ(Xwta) =∫ √2

0(1−Fwta(x))dx =

1√2≈ 0.707

25

Page 13: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Reward = cost: Equilibrium

Exp. Contest Reward

Cost

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.40.0

0.5

1.0

1.5

2.0

x -Effort

26

Effort in the E contest SSD dominates effort in the WTA contest

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.40.0

0.2

0.4

0.6

0.8

1.0

x -Effort

1 - F

1 - FE

1 - Fwta

27

Page 14: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Pop quiz: Field Experiment 1 (Lim et al., 2009)

• 15 salespeople compete

• In Treatment 1, the winner gets 300$

v = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,300)

• In Treatment 2, the 5 best performing salespeople get 60$

v = (0,0,0,0,0,0,0,0,0,0,60,60,60,60,60)

• Question: which reward scheme better motivates the sales staff?

28

Prize inequality

DEFINITION OF PRIZE INEQUALITY

Let w ∈Pn and v ∈Pn be two prize vectors of equal length and sum ofprizes, ∑

ni=1 wi = ∑

ni=1 vi. w is more unequal than v if the Lorenz curve of w is

constantly lower than the Lorenz curve of v, i.e.,

k

∑i=1

wi ≤k

∑i=1

vi for all k ∈ {1, . . . ,n}.

30

Page 15: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Changes in prize inequality

A prize vector becomes more equal when

• Pigou-Dalton transfer: prizes are shifted from better to worse performingagents, e.g.,

(0,1,3)→ (0,2,2)

• Averaging: replace a subset of prizes by the subset average, e.g.,

(0,1,2,3,4)→ (0,2,2,2,4)

• Interpretation: more equal prizes reduce competition

31

THEOREM 1. DEMORALIZATION EFFECT OF PRIZE INEQUALITY

Suppose w ∈Pn is more unequal than v ∈Pn, then

• µ(Xw)≤ µ(Xv) and

• Xw �SSD Xv.

32

Page 16: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Example: Motivating effort through bonuses

2 4 6 8 10 12 14 16 18

0.10

0.12

0.14

0.16

0.18

0.20

0.22

Number of Prizes k

Av

era

ge

Ou

tpu

t → less competitive →

0.0 0.1 0.2 0.3 0.4 0.5

0.0

0.2

0.4

0.6

0.8

1.0

Output

CD

F

FIGURE 1: On the left: the average output in a situation with n = 20 agents,c(x) = x2, a total budget of one and k identical nonzero prizesv = (0, . . . ,0,1/k, . . . ,1/k). On the right: the CDF for 4 nonzero prizes (red) and 8nonzero prizes (blue).

33

Field experiment 1 revisited

(v11,v12,v13,v14,v15) Average Sales Std. Deviation

Treatment 1 (0,0,0,0,300) 897.07 $ 778.31 $Treatment 2 (60,60,60,60,60) 1391.00 $ 1096.97 $

TABLE 1: Average sales in a field experiment performed by Lim et al (2009) withn = 15 salespeople in each contest where the lowest 10 prizes equaled zerov1 = v2 = . . .= v10 = 0. The difference in average sales is statistically significant.

Return to performance riskiness

34

Page 17: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Another field experiment: Field Experiment 2 (Lim et al., 2009)

(v10,v11,v12,v13,v14,v15) Average Sales Std. Deviation

Treatment 3 (0,40,40,40,40,40) 358.33 $ 181.98 $Treatment 4 (7,13,23,36,51,70) 333.33 $ 211.05 $

TABLE 2: Average sales in another field experiment performed by Lim et al (2009)with n = 15 salespeople in each contest where the lowest 9 prizes equaled zerov1 = v2 = . . .= v9 = 0 and the prize vector in Treatment 4 marginally failed to bemore unequal than the prize vector in Treatment 3. The difference in average sales isnot statistically significant.

35

Competition: Crowding

Page 18: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Contest size effects:Bigger is not better

36

This contest is getting pretty crowded

• Consider the simple 2-contestant contest with prize vector (0,1), andeffort cost, c(x) = x2

• Equilibrium effort distribution: F2(x) = x2.• Expected effort: 2/3

• Scale up the contest ten-fold, increasing the number of prizes andnumber of contestants

• Results in a 20-contestant contest with prize vector

(

×10︷ ︸︸ ︷0,0, . . . ,0,

×10︷ ︸︸ ︷1,1, . . . ,1)

• prize inequality is unaffected by the 10 fold scaling• the size of rewards is not affected• the reward per contestant is not affected

• So why should scaling matter?

38

Page 19: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Effect of scaling on incentives

• Under the 2-contestant equilibrium strategy, F2, scaling makes thedistribution of the sample median more peeked around the median.

• Beating sample median effort results in winning the prize

• So, the ten-fold scaling greatly increases the marginal rewards for effortlevels close to the median of F2

• Holding the strategies of the other contestants fixed, each contestantwould increase expected effort to x = 0.825 which “almost” ensureswinning the prize.

39

Reward 6= effort: Disequilibrium

Exp. Contest Reward

Cost

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

x -Effort

40

Page 20: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Equilibrium in the scaled contest

• But, if all contestants shifted to x = 0.825, all contestants would earn lessthan their cost of effort.

• Equilibrium in the scaled contest:• F20(x) = I−1

x2 (10,10), (I−1 is Inverse Beta Distribution.)• Effort shifts away from the median toward extreme effort levels• Expected effort≈ 0.575

41

Effort in scaled contest is SSD dominated

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

x -Effort

1 - F

1 - F2

1 - F20

42

Page 21: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Pop quiz: Effect changing class size

• Consider a simple grading scheme that gives a Pass to better 50% ofstudents and a Fail to the rest.

• Suppose you have one class of 20, and one of 40 students.

• Question: which class of students exert higher effort on average?

43

Increasing contest size

DEFINITION OF CONTEST SCALING

Prize vector w ∈Pn·s is a scaling of v ∈Pn if

w = (v1,v1, . . . ,v1︸ ︷︷ ︸s times

,v2,v2, . . . ,v2︸ ︷︷ ︸s times

, . . .)

Fixing average reward and prize inequality, contest size is increased when

• contest scaling: e.g., increasing # agents when grading on a curve isadopted

• consolidation: consolidating identical contests into a single large contest

45

Page 22: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

THEOREM 2. DEMORALIZATION EFFECT OF CONTEST SIZE

If w ∈Pn·s is a scaling of v, then

• µ(Xw)≤ µ(Xv) and

• Xw �SSD Xv.

46

Example continued

5 10 15 20Contest Scale

0.58

0.60

0.62

0.64

0.66

Average Effort

FIGURE 2: Average effort in an s-scaling the 2-contestant contest example,1≤ s≤ 20. 50% of the contestants receive a prize of 1, the rest receive a prize of 0.

47

Page 23: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Competition: Entry

Entry:Mixing level and size effects

Making demoralized contestants

48

Page 24: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Entry

• Generally, the effect of mixed prize level and contest size changes isindeterminate

• The exception: Adding entrants without adding rewards• Size effect: more contestants• Prize level effect: Lower prize rewards per capita.

DEFINITION OF ENTRY

Prize vector w ∈Pn+1 is an entrant transformation of v ∈Pn if

w = (0,v1,v2, . . . ,vn).

49

Entry discourages individual, but increases total effort

THEOREM 3. ENTRANT EFFECT ON EFFORT

If w ∈Pn+1 is an entrant transformation of v, then individual effort is smaller

µ(Xw)< µ(Xv) and Xw �FSD Xv

while total effort is larger

E

[n+1

∑i=1

Xwi

]≥ E

[ n

∑i=1

Xvi

]

50

Page 25: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Risk taking

Competition and risk-taking:Variances, like first impressions,

can be deceiving

51

Page 26: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

The competitiveness/riskiness relation

• If we define risk in the Rothchild and Stiglitz (1971) sense of “increasingrisk,” i.e., SSD, then increasing competitiveness always increasesriskiness.

• If riskiness and risk-taking are defined by conventional statisticalmeasures, e.g., variance, standard deviation, coefficient of variation, willdemoralization increase risk-taking?

52

Sufficient condition

PROPOSITION 1. INCREASING VARIANCE, STANDARD DEVIATION, AND

COEFFICIENT OF VARIATION

• If c is subquadratic (i.e., less convex than the square function x ↪→ x2),increasing prize inequality or contest scale increases variance, standarddeviation, and coefficient of variation of contest outcomes.

• So the competitiveness/risk-taking relation holds under conventionalmeasures of riskiness if the cost function is either quadratic or linear.

• Useful for lab experiments.

53

Page 27: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

When cost is superquadratic, variance and standard deviation maynot increase with competitiveness

• Suppose c(x) = x3. Consider the two prize vectors:

v = (0, . . . ,0︸ ︷︷ ︸101

,0.01, . . . ,0.01︸ ︷︷ ︸100

) w = (0, . . . ,0︸ ︷︷ ︸200

,1)

• Numerical calculations of contest statistics:

Prize vector µ(X) σ(X) σ(X)/µ(X)v 0.116 0.101 0.869w 0.015 0.085 5.752

• Driving force: (a) Variance and standard deviation are not scale-invariantand (b) demoralization reduces the scale of effort.

54

Sufficient condition for inequality to increase risk under scale-invariantrisk measures

DEFINITION OF value-preserving convexificationw ∈Pn is a value-preserving convexification of v ∈Pn if ∑

ni=1 wi = ∑

ni=1 vi

and

(wj+1−wj)(vi+1− vi)≥ (wi+1−wi)(vj+1− vj), 1≤ i < j≤ n−1

PROPOSITION 2.Suppose c is geometrically convex, i.e.,

x ↪→ c′(x)c(x)/x

is nondecreasing.

If w�VPC v, then Xw/µ(Xw) is a mean-preserving spread of Xv/µ(Xv).

55

Page 28: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

• The geometric convexity condition is much weaker than the subquadraticcondition: simply requires that the effect of convexity not decrease aseffort increases.

• Almost all “standard” convex cost functions are geometrically convex,e.g., power law, exponential, and polynomials with positive coefficients.

• The VPC condition simply requires that the inequality is increasedtransfers that disproportionately increase the largest prizes. Excludesinequality increasing transfers from the bottom to the middle of the prizeschedule.

• Field Experiment 1 (Lim et al., 2009) involves a VPC. Although standarddeviation is lower in WTA, coefficient of variation is higher:(778.31/897.07≈ 0.87 > 0.79≈ 1096.97/1391).

Results in Lim et al.

56

Sufficient condition for contest scaling to increase riskiness underscale-invariant measures

• Simple contests are contests with only two distinct prize values (e.g.,promotion contests, college admissions, winner-take-all contests).

PROPOSITION 3.Suppose c is geometrically convex. If v is a simple contest and w is a scalingof v, then Xw/µ(Xw) is a mean-preserving spread of Xv/µ(Xv).

57

Page 29: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Importance of controlling for total prize

PROPOSITION 3. REWARD LEVEL EFFECT ON PERFORMANCE

RISKINESS

If w = λv, λ > 1, performance is less risky (according to all scale-invariantmeasures of riskiness) under w than under v if c is geometrically convex. If cis a power law (which is geometrically linear), then performance riskinessremains the same.

• Controlling for reward-level effects essential for specify valid empiricaltests of the contest model

• Proposition 3 can be used to generate variance bounds on the effect ofreward level changes on effort variance.

58

Best shot

Page 30: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Competition in best shot contests:If you don’t care about averageeffort, you might not care about

demoralization

59

Designer welfare in development contests

• Let Xmax denote the best outcome.

• In development contests, the contest designer maximizes

E[u(Xmax)], where u is increasing and convex.

• E.g., u(Xmax) = Xmax or u(Xmax) = max[Xmax−K,0], where K dependson development costs, market conditions, etc.

• u is convex in Xmax if development timing is flexible (Dixit and Pindyck,1994).

60

Page 31: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Results

PROPOSITION 4. THE EFFECT OF COMPETITION ON BEST SHOT

If c is logconcave (i.e., less convex than the exponential function), increasingcompetitiveness (prize inequality or contest consolidation) increases designerwelfare in development contests.

• Standard convex cost functions, such as power law, exponential, andquadratic, are both logconcave and geometrically convex.

• Intuition: Increasing competitiveness produces two counteracting effects• positive side: an increase in upside performance dispersion• negative side: a decrease in average performance

61

• Proposition 4 is consistent with the stylized fact that most developmentcontests have a winner-take-all prize schedule.

• If effort costs are very convex, it can be beneficial to use multiple prizesin development contests.

62

Page 32: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Conclusion

Conclusion

63

Page 33: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Conclusion

• We find a demoralization effect of competition under an all-pay contestframework.

• Increasing competition reduces average effort and increases performanceriskiness in the sense of SSD.

• Demoralization leads to qualifications for increased competition toincrease outcome riskiness under conventional measures of riskiness.

• Demoralization also leads to qualifications for increased competition tobenefit designer welfare in development contests.

64

Page 34: Turning Up the Heat - Philipp StrackE(x)= 1 p 1 x2;x 2[0;1] Expected contestant effort = m(XE)= Z 1 0 xf E(x)dx = Z 1 0 (1 F E(x))dx = p 4 ˇ0:785 16 Why is this an equilibrium? Consider

Thank you for your time!

65