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- Zb |E[z2|z3, z1 = 0]|dFz3|z1=0 > 0 Therefore, applying the same argument as Kim and Pollard (1990) pp. 214-215 yields ∂ ∂b E[z2(sgn(z0 3b)|z1 = 0]|b=β0 = 0 (A.4) and − ∂2 E[z2(sgn(z0 3b) − sgn(z0 3β0))|z1 = 0] ∂b∂b0 = Z 1[z0 3β0 = 0] ∂ ∂z3 E[z2|z3, z1 = 0] 0 β0z3z0 3fz3|z1=0(z3)dβ0 (A.5) where β0 is the surface measure on the boundary of {z3 : z0 3β0 ≥ 0}.
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