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Motivational Goal Bracketing - School of Economics and ...

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Similarly, the incentive constraint <strong>of</strong> self 2 after he observes that self 1 shirked is:<br />

ab β (µ − 1) [Pr(loss|e2, e1) − Pr(loss|ē2, e<br />

1)] ≥ c − β p ¯y. (15)<br />

<br />

1−(1−p)=p<br />

Again this coincides with the incentive constraint for self 1, if self 1 expects self<br />

2 to shirk. Note also that the incentive constraint (15) coincides with the single-<br />

task/narrow-goals incentive constraint (5), so it binds at ab = ân. As ân < âb, self<br />

t ∈ {1, 2} will work hard under âb, even if the self responsible for the other task does<br />

not.<br />

ii) Part (i) <strong>of</strong> Lemma 2.<br />

The upper bound on an “easy” broad goal, ab ≤ ¯y, implies that âb is feasible if <strong>and</strong><br />

only if<br />

iii) Part (ii) <strong>of</strong> Lemma 2.<br />

c ≤ β p ¯y [(1 − p) µ + p] ⇔<br />

1<br />

(1 − p) µ + p βsc ≤ β.<br />

The pro<strong>of</strong> <strong>of</strong> Part (ii) is analogous to that <strong>of</strong> Lemma 1 <strong>and</strong> therefore omitted. Again,<br />

the precise formula for the cut<strong>of</strong>f value is <strong>of</strong> little importance, but we state it here for<br />

completeness:<br />

βb =<br />

A.4 Pro<strong>of</strong> <strong>of</strong> Lemma 3<br />

See Section 4.2.<br />

A.5 Pro<strong>of</strong> <strong>of</strong> Lemma 4<br />

Comparing the narrow goal ân =<br />

forward that âb = ân/(1 − p).<br />

A.6 Pro<strong>of</strong> <strong>of</strong> Proposition 1<br />

[µ (1 − p) 2 + p (2 − p)] c<br />

p [µ ((1 − p 2 ) µ + p 2 ) ¯y − 2 (µ − 1)(1 − p) c] .<br />

c−β p ¯y<br />

β p (µ−1) with the broad goal âb =<br />

For narrow goals the (normalized) expected utility <strong>of</strong> self 0 is:<br />

For the broad goal it is:<br />

U0(ân, ân; ē1, ē2)/β = 2 (p ¯y − c) − ân [2 p + 2 (1 − p) µ].<br />

U0(âb; ē1, ē2)/β = 2 (p ¯y − c) − âb [2 p − p 2 + (1 − p) 2 µ].<br />

22<br />

c−β p ¯y<br />

, it is straight-<br />

β p (1−p) (µ−1)

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