I want to prove that a function ψ(a1,a0), which is additively decomposed in two quasiconcave functions, is also quasiconcave. My numerical simulations seem to suggest it is possible, but I cannot get around it. I have tried using the definition of quasiconcavity (i.e. plugging ψ(λa + (1-λ)b) >...etc), but I can not prove the inequality. I also tried using the border Hessian, but i cannot sign the determinant. Any suggestions??(adsbygoogle = window.adsbygoogle || []).push({});

Here are the details of the problem:

ψ(a1,a0)=R(a1)+R(a0)

Where a1 [itex]\in[0,1][/itex] and a0 [itex]\in[0,1][/itex], and

R(aj)=F(Bl(aj)) ∫[itex]^{1}_{aj}[/itex] t*P(t)dt + F(Bh(aj)) ∫[itex]^{1}_{aj}[/itex] t*(1-P(t))dt for j[itex]\in{1,0}[/itex]

and the functions F(.), Bl(.), Bh(.) and P(.) have the following properties:

F',Bl',Bh',P'>0

Bl''>0

Bh''<0

F'' and P'' can be signed as required for the proof

Thanks a lot for any hints!

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# Specific proof of quasiconcavity for the sum of quasiconcave functions

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