class: center, middle, inverse, title-slide # Ramsey Model: Analysis of the dynamics ## Macroeconomics (2017Q2) ### Kenji Sato ### 2017-07-13 --- $$\let\oldhat\hat \renewcommand{\hat}[1]{\oldhat{\hspace{0pt} #1}}$$ <div style="margin-top: -2em"></div> ## Homework / Exercises Visit [kjst.jp/ma17q2pr](http://kjst.jp/ma17q2pr) now to check your submission status. Make sure all your submissions got purple icons. ### Midterm Report Get invitation for `mid`. Visit [kjst.jp/ma17q2hw](kjst.jp/ma17q2hw) > Due 2017-07-14 18:00. > Hand in by Pull Request. --- ## Additional assumption Recall the objective function: `$$\max \int_0^\infty e^{-(\rho - n) t} u(c(t)) dt,$$` where `$$u(c) = \frac{c^{1 - \theta} - 1}{1 - \theta}.$$` Obviously, we have to assume `\(\rho - n > 0\)`. In the following analysis we assume a slightly stronger condition: `$$\rho - n > \max \{ 0, (1-\theta)g\}$$` --- ## Equilibrium: First-order Dynamics The dynamics of the Ramsey model is determined by the following two differential equations. Capital accumulation equation: `$$\dot{\hat{k}} = f\left(\hat{k}\right) - \hat{c} - (\delta + g + n) \hat{k}$$` Euler equation: `$$\frac{\dot{\hat{c}}}{\hat{c}} = \frac{f'\left(\hat{k}\right) - \delta - \rho - \theta g}{\theta}$$` --- ## Locus of `\(\dot{\hat{k}} = 0\)` Let's derive the condition under which `\(\dot{\hat{k}} = 0\)`. From the capital accumulation equation, `$$\dot{\hat{k}} = f\left(\hat{k}\right) - \hat{c} - (\delta + g + n) \hat{k},$$` we obtain `$$\begin{aligned} \dot{\hat{k}} = 0 \Leftrightarrow \hat{c} = f\left(\hat{k}\right) - (\delta + g + n) \hat{k} \end{aligned}$$` --- ## Regions for `\(\dot{\hat{k}} > 0\)` and `\(\dot{\hat{k}} < 0\)` Moreover, `$$\begin{aligned} \dot{\hat{k}} > 0 \Leftrightarrow \hat{c} < f\left(\hat{k}\right) - (\delta + g + n) \hat{k} \end{aligned}$$` and `$$\begin{aligned} \dot{\hat{k}} < 0 \Leftrightarrow \hat{c} > f\left(\hat{k}\right) - (\delta + g + n) \hat{k} \end{aligned}$$` `\(\dot{\hat{k}}\)` is increasing below the `\(\dot{\hat{k}} = 0\)` locus and decreasing above the locus. --- ## Regions for `\(\dot{\hat{k}} > 0\)` and `\(\dot{\hat{k}} < 0\)` (cont'd) You can rewrite these conditions as below: `$$\begin{aligned} \dot{\hat{k}} \gtreqless 0 \Longleftrightarrow f\left(\hat{k}\right) - \hat{c} \gtreqless (\delta + g + n) \hat{k} \end{aligned}$$` Capital increases if and only if the investment, `\(f(\hat{k}) - \hat{c}\)`, is greater than the break-even level, `\((\delta + g + n)\hat{k}\)`. This condition is exactly what you have already encountered in the analysis of the Solow model. --- ## Phase diagram <img src="ramsey/ramsey.001.jpeg" width="800px" style="display: block; margin: auto;" /> --- ## Golden Rule Notice that the curve has a unique maximum, `\(\hat{k}_G\)`, where `$$f'\left(\hat{k}_G \right) = \delta + g + n$$` must hold. In the Solow model, the steady state that satisfies this condition is called the Golden Rule steady state, at which the consumption level is maximized over all steady states. --- ## Golden Rule (cont'd) <img src="ramsey/ramsey.002.jpeg" width="800px" style="display: block; margin: auto;" /> --- ## Locus of `\(\dot{\hat{c}} = 0\)` From `$$\frac{\dot{\hat{c}}}{\hat{c}} = \frac{f'\left(\hat{k}\right) - \delta - \rho - \theta g}{\theta}$$` we can derive the locus on which `\(\dot{\hat{c}} = 0\)`. `$$\dot{\hat{c}} = 0 \Longleftrightarrow f'\left( \hat{k} \right) = \delta + \rho + \theta g$$` --- ## Regions for `\(\dot{\hat{c}} > 0\)` and `\(\dot{\hat{c}} < 0\)` Obviously, `$$\dot{\hat{c}} \gtreqless 0 \Longleftrightarrow f'\left( \hat{k} \right) \gtreqless \delta + \rho + \theta g$$` Since `\(f'\)` is a decreasing function of `\(\hat{k}\)`, `$$\dot{\hat{c}} \gtreqless 0 \Longleftrightarrow \hat{k} \lesseqgtr (f')^{-1}\left(\delta + \rho + \theta g\right)$$` Let's define `$$\hat{k}^* = (f')^{-1}\left(\delta + \rho + \theta g\right)$$` --- ## Regions for `\(\dot{\hat{c}} > 0\)` and `\(\dot{\hat{c}} < 0\)` (cont'd) Recall `\(f'(\hat{k}) - \delta = r\)`. The first-order condition for the consumption is easy to grasp when `\(g = 0\)`. `$$\dot{\hat{c}} \gtreqless 0 \Longleftrightarrow r \gtreqless \rho + \theta\cdot 0$$` Consumption is growing when benefit from waiting `\(\rho\)` is greater than utility cost of waiting `\(\rho\)`. When `\(g \neq 0\)`, an additional term is needed to adjust for technology change. --- ## Regions for `\(\dot{\hat{c}} > 0\)` and `\(\dot{\hat{c}} < 0\)` (cont'd) <img src="ramsey/ramsey.003.jpeg" width="800px" style="display: block; margin: auto;" /> --- ## `\(\hat{k}^*\)` and `\(\hat{k}_G\)` Notice how I arranged the locus, `\(\hat{k} = \hat{k}^*\)`, and the Golden Rule level, `\(\hat{k} = \hat{k}_G\)`. By the assumption that `\(\rho - n > \max \{ 0, (1-\theta)g\}\)`, `$$f'\left(\hat{k}^* \right) = \delta + \rho + \theta g > \delta + n + g = f'\left(\hat{k}_G\right)$$` which implies $$\hat{k}^* < \hat{k}_G $$ --- ## `\(\hat{k}^*\)` and `\(\hat{k}_G\)` The intersection, `\(\left(\hat{k}^*, \hat{c}^*\right)\)`, of the loci for `\(\dot{\hat{k}} = 0\)` and `\(\dot{\hat{c}} = 0\)` corresponds to the steady state level of `\(\left(\hat{k}, \hat{c}\right)\)`. In the steady state, the level of capital stock per unit of effective labor is necessarily smaller than the golden rule level. `\(\hat{k}^*\)` is sometimes called, modified golden rule. --- ## Phase Diagram <img src="ramsey/ramsey.004.jpeg" width="800px" style="display: block; margin: auto;" /> --- ## A trajectory from `\(\hat{k}(0) < k^*\)`: 1 <img src="ramsey/ramsey.005.jpeg" width="800px" style="display: block; margin: auto;" /> --- ## A trajectory from `\(\hat{k}(0) < k^*\)`: 2 <img src="ramsey/ramsey.006.jpeg" width="800px" style="display: block; margin: auto;" /> --- ## A trajectory from `\(\hat{k}(0) < k^*\)`: 3 <img src="ramsey/ramsey.007.jpeg" width="800px" style="display: block; margin: auto;" /> --- ## Optimal trajectory The problem yet to be solved is the determination of `\(\hat{c}(0)\)`. Given `\(\hat{k}(0)\)`, depending on the level of candidates for `\(\hat{c}(0)\)`, the trajectories for the differential equations exibit either of the following. 1. `\(\hat{c}(t) \to 0\)` 2. `\(\hat{k}(t) \to 0\)` 3. `\(\hat{k}(t) \to \hat{k}^*\)` and `\(\hat{c}(t) \to \hat{c}^*\)` --- ## Optimal trajectory (cont'd) 1. `\(\hat{c}(t) \to 0\)` 2. `\(\hat{k}(t) \to 0\)` The first trajectory doesn't solve the maximization problem because by raising `\(\hat{c}(0)\)` by a small amount we can make `\(\hat{c}(t)\)` larger for any `\(t\)`. The second trajectory will eventually violate the feasibility condition. --- ## Optimal trajectory (cont'd) 3. `\(\hat{k}(t) \to \hat{k}^*\)` and `\(\hat{c}(t) \to \hat{c}^*\)` The third trajectory (which lies in the middlle of 1 and 2) converges to the steady state, `\(\left(\hat{k}^*, \hat{c}^*\right)\)`, where `\(\dot{\hat{k}} = 0\)` and `\(\dot{\hat{c}} = 0\)` hold simultaneously. This trajectory is the optimal path. To obtain the convergence, the trajectory must start from an appropriate `\(\hat{c}(0)\)`. The steady state is said to be **saddle-point stable**. --- ## Optimal Consumption For a given `\(\hat{k}(0)\)`, optimal `\(\hat{c}(0)\)` is uniquely determined by the converging path to the steady state. This mapping gives the optimal consumption function. (blue curve) <img src="ramsey/ramsey.008.jpeg" width="500px" style="display: block; margin: auto;" /> --- ## Balanced Growth In the long-run, the Ramsey model converges to the balanced growth path, in which all variables grow at constat rates. In Problem Set K, you are ask to obtain the growth rates for `$$y, k, c, Y, K, C$$`