Say the agent's problem is $$\max_{c,\{h\}, N}\{U(c, v(\boldsymbol{h} ; \boldsymbol{\theta}))+\lambda(w N-c)\}$$, subject to $\sum_{i=1}^{I} h_{i}+N \leq 1, \quad N \in \mathcal{N}$.

Assume $U(c, v(\boldsymbol{h} ; \boldsymbol{\theta}))$ is weak separability, $v\left(\boldsymbol{h} ; \boldsymbol{\theta}\right)=\sum_{i=1}^{I} \frac{\left(\theta_{i} h_{i}\right)^{1-\left(1 / \eta_{i}\right)}}{1-\left(1 / \eta_{i}\right)}$, and the analysis is done for a fixed $\lambda$.

FOC: $U_{c}=\lambda$; $U_{v} v_{i}=\omega \quad \text { for } i=1, \ldots, I $, where $v_{i}=\partial v / \partial h_{i}$, and denote $\hat{\omega} \equiv \omega / U_{v}$.

Weak separability allows us to consider the subproblem: $$\mathrm{v}(H ; \boldsymbol{\theta}) \equiv \max _{\left\{h_{i}\right\}} v\left(h_{1}, \ldots, h_{I} ; \boldsymbol{\theta}\right), \quad \text { subject to } \sum_{i} h_{i} \leq H$$ with $\mathrm{v}_{H}=\hat{\omega}$.

At an optimum, we have $U(c, v(\boldsymbol{h} ; \boldsymbol{\theta}))=U(c, \mathrm{v}(H ; \boldsymbol{\theta}))$, and $$U_{v} \mathrm{v}_{H}=\omega$$.

Differentiating the last equation: $$\begin{aligned} \frac{\partial \ln \omega}{\partial \ln \theta_{i}} &=\left(\frac{U_{v v}-U_{c v}^{2} / U_{c c}}{U_{v}}\right) \mathrm{v}_{\theta_{i}} \theta_{i}+\frac{\partial \ln \mathrm{v}_{H}}{\partial \ln \theta_{i}} \end{aligned}$$.

My question is that how should we derive the $\mathrm{v}_{\theta_i}$. If we don't consider any effect of $\theta_i$ on $H$, $\mathrm{v}_{\theta_i} = h_i/ \theta_i$ following the Envelop theorem. However, I feel that then the derivation is entirely conditional on $H$, while we should somehow also take into account of how $\theta_i$ affects $H$. Or maybe we should do the analysis at a given amount of leisure $H$? Either way, the paper I am reading says it should be something like $h_{i} \mathrm{v}_{H}=\mathrm{v}_{\theta_{i}} \theta_{i}$ but I have no idea where the $\mathrm{v}_{H}$ comes from.


1 Answer 1


The subproblem gives: $$ v(H,\theta) = \max_{h_i} \sum_{i = 1}^I \frac{(\theta_i h_i)^{1 - 1/\eta_i}}{1 - 1/\eta_i} \text{ s.t. } \sum_{i = 1}^I h_i \le H. $$ The first order condition for $h_i$ gives, $$ \theta_i (\theta_i h_i^\ast)^{-1/\eta_i} = \lambda^\ast, $$ where $\lambda$ is the Lagrange multiplier. From the Envelope theorem, we get: $$ v_H = \lambda^\ast, \tag{1} $$ and $$ v_{\theta_i} = h_i^\ast (\theta_i h_i^\ast)^{-1/\eta_i} = \frac{h_i^\ast}{\theta_i} \lambda^\ast. \tag{2} $$ where the latter equality uses the first order condition.

Subsituting out $(1)$ into $(2)$ gives: $$ \theta_i v_{\theta_i} = h_i^\ast v_H. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.