Appendix C: Derivation of Equation 16
We first re-formulate equation (15) in the main text for convenience.
(C1)
∂L∖
***
∂s I s=o,t ,T ,g ,z
∂ ls ( w )
∂s
dF ( w )
+ μ1 - z - [1 - F(W)] - μ3 - λs (W) - z
Differentiation of the Lagrangean in (14) with respect to z yields the following first-
order condition:
(C2)
∂L
****
∂z s =0,t ,T ,g ,z
w
ʃ[^ ' [Vs ( w )] - λs ( w )]dF ( w ) - μ1 -1 -
W
∂ls(w)
W--
∂z
dF( w )
+ (μ - μ2) - [1 - F(W)] - μ3 - λs( W) = 0.
Substituting for the term μ3 - λs(W) from (C2) into (C1) yields:
∂L∣
***
∂s I s=0,t ,t ,g ,z
W
=μ1 - z - [1 - F(W)] - z -

'[Vs(W)] -λs(W)]dF(W)
W
(C3)
W
fΓ ∂ls ( W )
I w--—
J|_ ∂s
W
dF ( W ) + μ1 -1 - z

∂ls(W )’
∂z
W
dF( w )
W
+ z - ∫[∏ ' '[Vs (w)] - λs (w)]dF(w) - (μ1 - μ2 ) - z - [1 - F(τ^)].
W
Re-formulating the budget constraint faced by an individual who engages in signaling
yields:
(C4) J (l, c, t, T, s, z) ≡ (1 -1) - w - (1 -1) + T - c - (1 + s) - z = 0 .
Differentiation of the expression in (C4) yields:
,r.. ∂J, ∂J,
(C5) z-- s =0-- s=0 = 0.
∂z ∂s
Thus, for any ability level W, the following holds:
27
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