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Consider the Ramsey-Cass-Koopmans model in discrete time. Specifically, consider the following phase diagram.

enter image description here

For simplicity, let's consider the trajectory marked A which is in the northwest corner of this diagram.

In this location, supposedly $K_{t+1}<K_t$ and $C_{t+1}>C_t$. But how does one show this? Up to this point, all I have seen is that "supposedly" this trajectory veers off, but I don't see how it necessarily does so.

Let $$C_t = K_t^\alpha + (1-\delta)K_t - K_{t+1}$$

My Quesiton

Can someone explain why A veers off towards the NW direction? Specifically, why must $K_{t+1}<K_t$ and $C_{t+1}>C_t$?

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  • $\begingroup$ Actually, this is Romer (2006, p. 61), right? How come you just switched to discrete time? As far as I can tell phase diagramms are constructed from differential equation. Anyway, you may explictly state the differential equations for consumption $\dot c$ and capital $\dot k$. Differentiate them with respect to $c$ and $k$ and the dynamics will become clear. $\endgroup$ – clueless Apr 12 '16 at 20:16
  • $\begingroup$ Regarding the title of your post, it seems you are rather interested in understanding the dynamics, but convergence properties of trajectories. Actually only the saddle path is converging towards the fixed point. $\endgroup$ – clueless Apr 12 '16 at 20:44
  • $\begingroup$ Is this about something more about than the standard method to deduce the dynamic tendencies of a phase diagram? If not, this may be prove useful to you, alecospapadopoulos.wordpress.com/2015/11/13/… $\endgroup$ – Alecos Papadopoulos Apr 12 '16 at 23:29
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See Romer (2006, Sec 2.3) for details of notation. Dynamics are given by \begin{align} \dot k &= f(k) - (n+g)k - c\\ \dot c &= \frac{c}{\theta}(f'(k) - \rho - \theta g) \end{align}

Presume you are located somewhere on the $\dot k = 0$ locus. Now if we increase $c$ one gets \begin{align} \frac{\partial \dot k}{\partial c} = -1 < 0 \end{align}

That is, if we are above $\dot k = 0$ capital decreaes over time. Hence at point $A$ we move to the left.

Now presume you are located somewhere on the $\dot c = 0$ locus. Now if we increase $k$ one gets \begin{align} \frac{\partial \dot c}{\partial k} = \frac{c}{\theta} f''(k) < 0 \end{align}

That is, if we are right of $\dot c = 0$ consumption decreaes over time. Hence at point $A$ we move up in addition (left of $\dot c = 0$).

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