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Wikipedia states GHH preferences to have the general form

$$V(c, l) = U(c - G(l))$$

with $U$ increasing concave, and $G$ increasing convex.

In interior solution w.r.t. $l$ requires

$$ U'(c - G(l))(\frac{dc}{dl} - G'(l)) = 0$$

As wikipedia says, this is given by

$$ \frac{dc}{dl} = G'(l)$$

However, what about the peculiar case where $c = G(l)$? Could someone elaborate a bit on the intuition

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  • $\begingroup$ If $c=G(l)$, the first-order condition is not satisfied, is it? Since $U'(0)>0$. $\endgroup$
    – Oliv
    Aug 25 '15 at 10:32
  • $\begingroup$ @Oliv With for example $U(x) = x^{1-\sigma}$ it would be - a standard case of an increasing concave function. $\endgroup$
    – FooBar
    Aug 25 '15 at 11:03
  • $\begingroup$ in that case, $U'(x)=(1-\sigma)x^{-\sigma} \rightarrow + \infty$ when $x \rightarrow 0$ (since $\sigma>0$). More generally, since $U$ is concave, $U'(0)=0$ would imply that $U'(x) \leq 0$ for $x>0$, and this contradicts the assumption that $U$ is increasing. It is probably implicitly assumed that $U'>0$ on the relevant interval. $\endgroup$
    – Oliv
    Aug 25 '15 at 11:25
  • $\begingroup$ Thanks, I think the two comments together suffice as an answer. $\endgroup$
    – FooBar
    Aug 25 '15 at 11:28
  • $\begingroup$ You're welcome, I'm glad it answered your concern. $\endgroup$
    – Oliv
    Aug 25 '15 at 11:30
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The first-order condition would be satisfied under the assumption $c=G(l)$ if $U'(0)=0$. But this latter requirement is incompatible with the assumptions: indeed, $U'$ is nonincreasing (by the concavity of $U$) which yields $U'(x) \leq 0$ for any $x \geq 0$, which contradicts the fact that $U$ is increasing.

If $U'(x)>0$ for all $x$, the only possibility to meet the first-order condition is indeed $\dfrac{dc}{dl}=G'(l)$.

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