Consumer theory with subproblem

Question

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.

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: $$ \theta_i v_{\theta_i} = h_i^\ast v_H. $$