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I ran into this function reading a macroeconomic paper" $$ \psi (n)=G(1-n)-\beta \frac{G'(1-n)F(n)}{F'(n)}.$$ The paper claims this function is monotonic decreasing and the central result depends on this claim. But I find this is may not be true. The first order of this function seems to change sign when $n$ varies between $0$ and $1$. Both $F$ and $G$ are concave functions.$n$ is in the close interval of $0$ and $1$ and $\beta$ is in the open interval of $0$ and $1$

Can anyone corroborate on this?

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There must be some restrictions on $\beta$. For $F(x)=G(x)=-x^2$ and $\beta=-1$, $\psi'(n) = 1$. – user26872 Jun 28 '12 at 19:13
$\beta$ has to be between $0$ and $1$.I should add that. But the functions in the example you give are not concave functions – Yan Song Jun 28 '12 at 19:28
There must be some more restrictions on $F$ and $G$; for example, $G$ decreasing in its argument implies $G$ increasing in $n$ and the second term on the r.h.s. can be driven arbitrarily close to zero by making $\beta$ smaller, implying that somewhere $\psi(n)$ can be made increasing. – jbowman Jun 28 '12 at 21:17
@YanSong: $F(x) = -x^2$ is not concave? – user26872 Jun 28 '12 at 22:43
I agree with your argument. But if $\beta$ is somewhere close to one, is it still possible to come up with a counter example? It's an economic paper and the $\beta$ is interpreted as a discount factor, meaning it should be smaller than one, but should not be not arbitrarily close to one – Yan Song Jun 28 '12 at 22:46

Here are two simple counterexamples to the claim.

(I) Let $F(x) = x$ and $G(x) = -x$. These functions are concave (and convex). But $\psi'(n) = 1+\beta > 0$.

(II) Let $F(x) = G(x) = -x^2$. If $n < \frac{2+\beta}{2(1+\beta)}$, then $\psi'(n) > 0$. For example, if $\beta = 1/2$, then $\psi'(n) > 0$ for $n<5/6$.

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