<p>I guess you already went trough the algebra below, but just for context, the problem you're trying to solve is </p>

<p>$$
\max_{c}\sum_{t=0}^{+\infty}\beta^t u(c_t) \\
\text{s.t.}~~ f(k_t) + (1- \delta)k_t = c_t + k_{t+1} \tag{1}
$$</p>

<p>where $f(k_t) = k_t^\alpha$ and</p>

<p>$$
u(c_t) = \frac{c_t^{1-\gamma}}{1-\gamma} - 1 \tag{2}
$$</p>

<p>The problem in (1) can be cast into the two coupled equations</p>

<p>\begin{eqnarray}
u'(c_t) &amp;=&amp; \beta[1 + f'(k_{t+1}) - \delta]u'(c_{t+1}) \\
k_{t+1} &amp;=&amp; f(k_t) + (1-\delta)k_t - c_t \tag{3}
\end{eqnarray}</p>

<p>where $u'(x) = x^{-\gamma}$, and $f'(x) = \alpha x^{\alpha-1}$. These first of Eqns. (3) can be inverted to obtain an expression for $c_{t+1}$ in terms of $(k_t,c_t)$, leading to</p>

<p>\begin{eqnarray}
c_{t+1} &amp;=&amp; \beta^{1/\gamma}c_t [1 + \alpha[k_t^\alpha + (1-\delta)k_t - c_t]^{\alpha-1} - \delta]^{1/\gamma} \\
k_{t+1} &amp;=&amp; f(k_t) + (1-\delta)k_t - c_t \tag{4}
\end{eqnarray}</p>

<p>Which can be expressed as</p>

<p>$$
{\bf x}_{t+1} = {\bf F}({\bf x}_{t})~~~\mbox{with}~~~ {\bf x}_t = \left(\begin{array}{c}c_{t}\\k_{t}\end{array}\right) \tag{5}
$$</p>

<p>A fixed point ${\bf x}^*$ of the map ${\bf F}$ is such that</p>

<p>$$
{\bf x}^* = {\bf F}({\bf x}^*) \tag{6}
$$</p>

<p>that is, a point for which the system does not evolve. If you use $\gamma=2$, $\beta=0.9964$, $\alpha=0.36$, $\delta=0.025$ this point is (found by solving Eq. (6)),</p>

<p>$$
{\bf x}^* = \left(\begin{array}{c}c^*\\k^*\end{array}\right) = \left(\begin{array}{c}2.84829\\52.2808\end{array}\right)  \tag{7}
$$</p>

<p>which is clearly different from zero! You can linearize ${\bf F}$ around ${\bf x}^*$ and write the result as </p>

<p>$$
{\bf y}_{t+1} = {\bf J}{\bf y}_t ~~~\mbox{where}~~~ {\bf y}_t = {\bf x}_{t} - {\bf x}^*, ~~~ {\bf J} = \left.\frac{\partial{\bf J}}{\partial {\bf x}}\right|_{{\bf x} = {\bf x}^*} \tag{8}
$$</p>

<p>Is this last system the one that has a saddle point at ${\bf y} = 0$</p>

CC BY-SA 3.0
2017-11-26T15:59:33.117Z
thank you @caverac! I did the exact same thing and the saddle path really goes through the steady state point, but it just doesn't go through the origin, which i don't know why. saddle path should go through the origin right?
user68863
CC BY-SA 3.0
2017-11-26T16:34:44.953Z
@user68863 It goes through the origin of the linearized system ${\bf y}_{t+1} = J {\bf y}_t$ (Eq. (8)), but does not need to go through ${\bf x}=0$
caverac
CC BY-SA 3.0
2017-11-26T16:48:48.643Z
thank you!, so what i did was right. but can we transform this linearized policy function into something concave? my teacher said the saddle path should look like the locus of capital (the blue line).
user68863
CC BY-SA 3.0
2017-11-26T16:51:55.450Z
@user68863 The properties of the fixed point ${\bf x}^*$ are determined by the eigenvalues of $J$, and these are set by the problem. Maybe if you change the cost function, or the constraint you could make the problem a convex one
caverac
CC BY-SA 3.0
2017-11-26T17:16:28.460Z
thank you! the only problem now is that the area below the saddle path should have the diverging paths heading to the right, but when i try some starting points very close to saddle path (but still in the area below it), i got the diverging paths going up toward the north-west. i used bakward integration as described in this link: https://ch.mathworks.com/company/newsletters/articles/simulating-the-ramsey-cass-koopmans-model-using-matlab-and-simulink.html

I know ode45 is for continuous time and in my case it's discrete time. but even so there shouldn't be problem with the diverging path?
user68863
CC BY-SA 3.0
2017-11-26T15:07:12.737Z
2017-11-26T15:07:12.737Z
caverac
<p>In my case, the utility function is CEIS and discrete, the production fuction is $f(k_{t})=k_{t}^\alpha$, the budget constraint is $f(k_{t})+(1-\delta)k_{t}=c_{t} + k_{t+1}$. I use Jacobian matrix and Schur factorization to get the linearized policy function for consumption, therefore i can plot saddle path and unstable arms. In the end they look like below. However I read that saddle path must go through the origin, which is not right in my plot. </p>

<p>So my question is: does saddle path always go through the origin?</p>

<p><a href="https://i.sstatic.net/Ru8Wt.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Ru8Wt.png" alt="enter image description here"></a></p>

CC BY-SA 3.0
2017-11-26T10:23:04.963Z
Which utility function are you using?
caverac
CC BY-SA 3.0
2017-11-26T12:12:46.163Z
this one: $U=\sum_{t}^{\infty} \beta^{t} \bigg( \frac{c_{t}^{1-\gamma}}{1-\gamma} -1 \bigg)$
user68863
CC BY-SA 3.0
2017-11-26T12:15:06.200Z
and calibration: $\gamma=2$, $\beta=0.9964$, $\alpha = 0.36$, $\delta = 0.025$
user68863
<economic-growth><rbc><balanced-growth>
Can saddle path not go through the origin in Ramsey model?
user68863
# Can saddle path not go through the origin in Ramsey model?

(Asked by: username_0 on 2017-11-25)

In my case, the utility function is CEIS and discrete, the production fuction is $f(k\_{t})=k\_{t}^\alpha$, the budget constraint is $f(k\_{t})+(1-\delta)k\_{t}=c\_{t} + k\_{t+1}$. I use Jacobian matrix and Schur factorization to get the linearized policy function for consumption, therefore i can plot saddle path and unstable arms. In the end they look like below. However I read that saddle path must go through the origin, which is not right in my plot.

So my question is: does saddle path always go through the origin?

[![enter image description here](https://i.sstatic.net/Ru8Wt.png)](https://i.sstatic.net/Ru8Wt.png)

**Tags:** economic-growth, rbc, balanced-growth

> **Comments:**
> **username_1**: Which utility function are you using?  
> **username_0**: this one: $U=\sum_{t}^{\infty} \beta^{t} \bigg( \frac{c_{t}^{1-\gamma}}{1-\gamma} -1 \bigg)$  
> **username_0**: and calibration: $\gamma=2$, $\beta=0.9964$, $\alpha = 0.36$, $\delta = 0.025$  

---

## 1 answers

### Answer by username_1 on 2017-11-26. Score: 1 | Accepted answer

I guess you already went trough the algebra below, but just for context, the problem you're trying to solve is

$$
\max\_{c}\sum\_{t=0}^{+\infty}\beta^t u(c\_t) \\
\text{s.t.}~~ f(k\_t) + (1- \delta)k\_t = c\_t + k\_{t+1} \tag{1}
$$

where $f(k\_t) = k\_t^\alpha$ and

$$
u(c\_t) = \frac{c\_t^{1-\gamma}}{1-\gamma} - 1 \tag{2}
$$

The problem in (1) can be cast into the two coupled equations

\begin{eqnarray}
u'(c\_t) &=& \beta[1 + f'(k\_{t+1}) - \delta]u'(c\_{t+1}) \\
k\_{t+1} &=& f(k\_t) + (1-\delta)k\_t - c\_t \tag{3}
\end{eqnarray}

where $u'(x) = x^{-\gamma}$, and $f'(x) = \alpha x^{\alpha-1}$. These first of Eqns. (3) can be inverted to obtain an expression for $c\_{t+1}$ in terms of $(k\_t,c\_t)$, leading to

\begin{eqnarray}
c\_{t+1} &=& \beta^{1/\gamma}c\_t [1 + \alpha[k\_t^\alpha + (1-\delta)k\_t - c\_t]^{\alpha-1} - \delta]^{1/\gamma} \\
k\_{t+1} &=& f(k\_t) + (1-\delta)k\_t - c\_t \tag{4}
\end{eqnarray}

Which can be expressed as

$$
{\bf x}\_{t+1} = {\bf F}({\bf x}\_{t})~~~\mbox{with}~~~ {\bf x}\_t = \left(\begin{array}{c}c\_{t}\\k\_{t}\end{array}\right) \tag{5}
$$

A fixed point ${\bf x}^\*$ of the map ${\bf F}$ is such that

$$
{\bf x}^\* = {\bf F}({\bf x}^\*) \tag{6}
$$

that is, a point for which the system does not evolve. If you use $\gamma=2$, $\beta=0.9964$, $\alpha=0.36$, $\delta=0.025$ this point is (found by solving Eq. (6)),

$$
{\bf x}^\* = \left(\begin{array}{c}c^\*\\k^\*\end{array}\right) = \left(\begin{array}{c}2.84829\\52.2808\end{array}\right) \tag{7}
$$

which is clearly different from zero! You can linearize ${\bf F}$ around ${\bf x}^\*$ and write the result as

$$
{\bf y}\_{t+1} = {\bf J}{\bf y}\_t ~~~\mbox{where}~~~ {\bf y}\_t = {\bf x}\_{t} - {\bf x}^\*, ~~~ {\bf J} = \left.\frac{\partial{\bf J}}{\partial {\bf x}}\right|\_{{\bf x} = {\bf x}^\*} \tag{8}
$$

Is this last system the one that has a saddle point at ${\bf y} = 0$

> **Comments:**
> **username_0**: thank you @username_1! I did the exact same thing and the saddle path really goes through the steady state point, but it just doesn't go through the origin, which i don't know why. saddle path should go through the origin right?  
> **username_1**: @username_0 It goes through the origin of the linearized system ${\bf y}_{t+1} = J {\bf y}_t$ (Eq. (8)), but does not need to go through ${\bf x}=0$  
> **username_0**: thank you!, so what i did was right. but can we transform this linearized policy function into something concave? my teacher said the saddle path should look like the locus of capital (the blue line).  
> **username_1**: @username_0 The properties of the fixed point ${\bf x}^*$ are determined by the eigenvalues of $J$, and these are set by the problem. Maybe if you change the cost function, or the constraint you could make the problem a convex one  
> **username_0**: thank you! the only problem now is that the area below the saddle path should have the diverging paths heading to the right, but when i try some starting points very close to saddle path (but still in the area below it), i got the diverging paths going up toward the north-west. i used bakward integration as described in this link: https://ch.mathworks.com/company/newsletters/articles/simulating-the-ramsey-cass-koopmans-model-using-matlab-and-simulink.html  I know ode45 is for continuous time and in my case it's discrete time. but even so there shouldn't be problem with the diverging path?  

---

