class: center, middle, inverse, title-slide # Ramsey Model: Consumer Behavior ## Macroeconomics (2017Q4) ### Kenji Sato ### 2018-1-11 --- class: center, middle # Consumer Behavior --- ## Consumer Optimization Each individual decides when and how much to consume by solving the utility maximization: `$$\max \int_0^\infty e^{-\rho t} u(c(t))dt$$` subject to `$$c(t) + \dot{\! s}(t) = r(t) s(t) + w(t)$$` From the perspective of the consumer, `\(r(t)\)` and `\(w(t)\)` are given. --- ## Trade-off between consumption and saving The Ramsey model is a problem of intertemporal resource allocation. **You need to sacrifice future consumption to consume today**. `$$c(t) + \dot{\! s}(t) = r(t) s(t) + w(t)$$` It is possible to increase the amount of consumption, `\(c(t)\)`, at any point in time by reducing saving, `\(\dot s(t)\)`, at the same moment. Since `\(\dot s(t) < 0\)` is allowable (withdrawal), `\(c(t)\)` can be arbitrarily large, at least in the short run. --- ## Trade-off between consumption and saving (cont'd) If `\(\dot s(t)\)` is large and negative for a long time, however, the amount of future assest will also be large and negative. This results in the reduced consumption in the future because the representative agent cannot continue to roll over debt forever. (NPG condition.) The solution to the dynamic optimization problem is typically expressed as a differential equation. --- class: center, middle # Hamiltonian Approach --- ## General Setting Consider the following problem (somewhat more abstract than the Ramsey problem.) `$$\begin{aligned} &\max_{x, y}\int_{0}^{\infty}F(t, x(t), y(t))dt \\ &\text{subject to} \\ &\qquad\dot{x}=G(t, x(t), y(t)), \\ &\qquad x(0):\ \text{given,} \end{aligned}$$` * `\(x\)` is the stock (state) variable * `\(y\)` is the flow (control) variable --- ## Hamiltonian `$$\max_{x, y}\int_{0}^{\infty}F(t, x(t), y(t))dt \quad \text{subj. to}\quad \dot{x}=G(t, x(t), y(t))$$` The Hamiltonian (or less ambiguously, present-value Hamiltonian) is the following function. `$$\mathcal{H}(t, x, y, \lambda) = F(t, x, y) + \lambda G(t, x, y),$$` where `\(\lambda\)` is a multiplier. --- ## Hamiltonian (cont'd) Hamiltonian can be rewritten as `$$\mathcal{H}(t, x, y, \lambda) = F(t, x, y) + \lambda \dot{\!x}.$$` The first term is the flow return, like utility from consumption, etc. at some point in time. The second term, on the other hand, is the value gained from increased stock. `\(\lambda\)` can be interpreted as the price of the stock of the state variable `\(x\)`. The Hamiltonian expresses the intertemporal substitution between now and later. --- ## Canonical Equations The solution of the problem must satisfy (necessary condition) the following equations: `$$\begin{aligned} \frac{\partial\mathcal{H}}{\partial y} &= 0 \\ \dot{\lambda} &=-\frac{\partial\mathcal{H}}{\partial x},\\ \dot{x} &= \frac{\partial\mathcal{H}}{\partial\lambda}. \end{aligned}$$` We do not follow the proof here. Interested students can consult Acemoglu (2009). --- ## Canonical Equation 3 The third condition `$$\dot{x} = \frac{\partial\mathcal{H}}{\partial\lambda}$$` is obvious becuase this is equivalent to the constraint. --- ## Canonical Equation 1 The first condition `$$\frac{\partial\mathcal{H}}{\partial y} = 0$$` is less obvious but it states that control `\(y\)` must be chosen so that the Hamiltonian is maximized at any point in time. --- ## Canonical Equation 2 An increase in `\(x\)` changes the current flow return plus the value of stock by `$$\frac{\partial \mathcal H}{\partial x} = \frac{\partial F}{\partial x} + \lambda \frac{\partial G}{\partial x}$$` Increased stock will be depreciated by the amount `$$- \dot{\! \lambda}$$` In the optimum, the depreciation must be fully covered by the increased value. `$$\dot{\lambda} = -\frac{\partial\mathcal{H}}{\partial x}$$` --- class: center, middle # Solving the Ramsey problem --- ## Ramsey problem Compare `$$\max_{s,c} \int_0^\infty e^{-\rho t} u(c(t))dt \quad \text{subj. to}\quad c(t) + \dot{\! s}(t) = r(t) s(t) + w(t)$$` with `$$\max_{x, y}\int_{0}^{\infty}F(t, x(t), y(t))dt \quad \text{subj. to}\quad \dot{x}=G(t, x(t), y(t))$$` What are the equivalents for `\(F\)`, `\(G\)`, `\(x\)`, and `\(y\)` in the Ramsey problem? --- ## Ramsey problem (cont'd) Stock and flow `$$x \longleftrightarrow s, \qquad y \longleftrightarrow c$$` Objective `$$F(t, x, y) \longleftrightarrow F(t, s, c) = e^{-\rho t} u(c)$$` Constraint `$$G(t, x, y) \longleftrightarrow G(t, s, c) = rs + w - c$$` --- ## Hamiltonian and canonical equations `$$\mathcal H = e^{-\rho t} u(c) + \lambda (rs + w - c)$$` #### Deriv. w.r.t flow is zero `$$\frac{\partial H}{\partial c} = e^{-\rho t} u'(c) - \lambda = 0.$$` #### Depreciation of stock = deriv. w.r.t. stock `$$\dot \lambda = - \frac{\partial H}{\partial s} = - \lambda r$$` #### Constraint `$$\dot{\! s}(t) = r(t) s(t) + w(t)$$` --- ## Euler condition From `$$\begin{aligned} \lambda = e^{-\rho t} u'(c), \end{aligned}$$` we obtain `$$\begin{aligned} \frac{\dot{\! \lambda}}{\lambda} = -\rho + \frac{\dot{\! c} u''(c)}{u'(c)} \end{aligned}$$` Since `\(\dot{\lambda}/\lambda = -r\)` by another canonical equation, we have `$$\begin{aligned} \frac{\dot{\! c} u''(c)}{u'(c)} = \rho - r \end{aligned}$$` --- ## Euler condition By rearranging, `$$\begin{aligned} \dot{\! c} = \frac{u'(c) (r - \rho)}{- u''(c)} \end{aligned}$$` This equation, the Euler equation, determines the plan of consumption. --- ## Relative risk aversion Let the coefficient for the relative risk aversion be denoted as `$$\Theta(c) = - \frac{c u''(c)}{u'(c)}$$` The Euler equation can be rewritten as `$$\begin{aligned} \frac{\dot{\! c}}{c} = \frac{r - \rho}{\Theta(c)} \end{aligned}$$` The larger `\(\Theta (c)\)` is, the larger the curverture of the utility function gets. A consumer with large `\(\Theta\)` has stronger incentive to smooth theier consumption stream than those with small `\(\Theta\)`. --- ## Constant relative risk aversion In practice, we use CRRA utility function very often. `$$u(c) = \frac{c^{1 - \theta} - 1}{1 - \theta},$$` where `\(\theta\)` is a positive parameter. * Show that `\(\Theta(c) = \theta\)` for all `\(c\)`. * Show that when `\(\theta = 1\)`, `\(u(c) = \ln c\)`.