# Consumer theory with subproblem

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.

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