Motivational Goal Bracketing - School of Economics and ...

Appendix
A Proofs
A.1 Derivation of the incentive constraint (5)
Written out, the incentive constraint for self 1 is:
β {p (¯y − a) − (1 − p) µ a} − c ≥ −β µ a.
Working hard, the goal is surpassed with probability p (resulting in a gain ¯y − a); and
with probability 1 − p the task outcome falls short of the goal (resulting in a loss µ a).
Shirking results in a substandard outcome with associated loss µ a. Rewritten in terms of
the respective probabilities of obtaining a gain or a loss (Pr(gain|ē) = p, Pr(loss|ē) = 1 − p,
Pr(gain|e) = 0, and Pr(loss|e) = 1), the incentive constraint becomes:
β {Pr(gain|ē) (¯y − a) − Pr(loss|ē) µ a} − c ≥ β {Pr(gain|e) (¯y − a) − Pr(loss|e) µ a}.
Rearranging:
β a [{Pr(gain|ē) − Pr(gain|e)} + µ {Pr(loss|e) − Pr(loss|ē)}] ≥ c − β p ¯y.
Using Pr(gain|e) = 1 − Pr(loss|e) yields (5).
A.2 Proof of Lemma 1
With narrow goals, the utility is separable across tasks. So we can consider each task on its
own, using the single-task incentive constraint (5).
(i) To motivate self 1 (self 2), a narrow goal of at least â is required. And this goal has to
be feasible, i.e., â ≤ ¯y ⇔ 1
µ βsc ≤ β (using the definition of βsc in Assumption 1).
(ii) By Assumption 1, for the utility component of self 0 related to task i = 1, 2 (U0i(ani; e))
we have U0i(0; ē)−U0i(0; e) > 0. Moreover, U0i(¯y; ē)−U0i(0; e) = β [−(1 − p) µ ¯y − c] <
0. Note that U0i(a; ē) − U0i(0; e) is a continuous and strictly decreasing function in a,
because µ > 1:
∂
∂ a U0i(a; ē) = −β [µ − p (µ − 1)] < 0.
Applying the intermediate value theorem, there exists a unique value ã ∈ (0, ¯y) such
that U0i(ã; ē) − U0i(0; e) = 0. This is the maximum narrow goal level that self 0 would
ever choose to get a future self to exert high effort in a task.
As tasks are symmetric, consider without loss of generality the effort choice of self 1
in task 1. The next argument shows that there exists a βn ∈ (β, βsc) such that the
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