Consider a household which solves the following problem: $$v(k,r,w)=\underset{c,l\in B{(k,r,ω)}}{\ {max}} \{u(c,l)\}$$

where $u : R_+^2 \rightarrow R$ is a strictly concave, twice continuously differentiable, strictly increasing function in its two arguments: consumption, $c$, and leisure, $l$. The constraints the household must obey in selecting $c, l$ are summarized by $B$: $$B(k, r, w) = {\{c, l : 0 ≤ c ≤ rk + w(1 − l), 0 ≤ l ≤ 1\}}$$

Here, $k, r, w > 0$ are numbers over which the household has no control. Prove that $v$ is concave in $k$ and that the derivative of $v$ with respect to $k$ exists for ‘interior $k$’. Display a formula for the derivative of $v$.

What I was thinking for solving this is by following Benveniste & Scheinkman theorem on differentiability $ω : D → R$ defined on the neighborhood $D$ of $x_0$, i.e. $D ⊂ X$ and $x_0 ∈ int(D)$ such that: $ω(x) ≤ v(x)$ and $ω(x_0) = v(x_0)$.

And $ω(.)$ is concave and differentiable, then $v(.)$ is differentiable at $x_0$ and $v'(x_0) =ω'(x_0)$.

I'm guessing we have to substitute c with $rk + w(1 − l)$ in the utility function, but I'm confused a bit with leisure. Because after that, I think we should just get the envelope theorem or not?

  • $\begingroup$ Note that you could write $v(kr,w)$ instead of $v(k,r,w)$, this simplifies the expression of the derivatives of $v$ wrt $k$ and $r$. $\endgroup$ – Bertrand Dec 7 '20 at 11:38
  • $\begingroup$ @Bertrand I'm sorry, I'm not following. Could you please elaborate more on that? $\endgroup$ – Maybeline Lee Dec 7 '20 at 12:03
  • $\begingroup$ If you define $z = rk$ and replace it in your problem, you end up with $v(z,w)$. This establishes a relationship between the marginal utility of $k$ and of $r$. $\endgroup$ – Bertrand Dec 7 '20 at 12:27
  • $\begingroup$ You should probably use the Inverse Function Theorem to establish that solution exists and is differentiable. $\endgroup$ – Jesper Hybel Jan 9 at 17:18

Sharing my answer out there, correct me if I'm wrong. $u(c,l)=u(rk+w(1-l), l)$

$U$ is strictly concave and differentiable. Let $max$ u attained at $(c^*, l^*)$ i.e $(k^*, l^*)$.

Then $v(k^*)=u(k^*)$ and $u(k) ≤ v(k)$

Then by Theorem 4.10 (Benveniste & Scheinkman), $v$ is differentiable at $k^*$.

$V_k^*=U_k(rk^* + w(1-l^*, l*) = U_c(rk^* + w(1-l^*, l*)$

  • $\begingroup$ Just few comments: 1). It is not (fully) clear to me how this is related to Benveniste and Scheinkman (who, in their 1979 paper, propose results that apply to dynamic problems). 2). Why do you write $k^*$? Is it optimally chosen? This fully change the statement of you question in which $k$ is exogenous. 3). You should avoid using the same notations for different functions $u(c,l)$ and then $u(k)$ $\endgroup$ – Bertrand Dec 7 '20 at 18:52

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