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

  • $\begingroup$ Which utility function are you using? $\endgroup$ – caverac Nov 26 '17 at 10:23
  • $\begingroup$ this one: $U=\sum_{t}^{\infty} \beta^{t} \bigg( \frac{c_{t}^{1-\gamma}}{1-\gamma} -1 \bigg)$ $\endgroup$ – user68863 Nov 26 '17 at 12:12
  • $\begingroup$ and calibration: $\gamma=2$, $\beta=0.9964$, $\alpha = 0.36$, $\delta = 0.025$ $\endgroup$ – user68863 Nov 26 '17 at 12:15

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 ${\bf y} = 0$

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  • $\begingroup$ 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? $\endgroup$ – user68863 Nov 26 '17 at 15:59
  • $\begingroup$ @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 ${\bf x}=0$ $\endgroup$ – caverac Nov 26 '17 at 16:34
  • $\begingroup$ 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). $\endgroup$ – user68863 Nov 26 '17 at 16:48
  • $\begingroup$ @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 $\endgroup$ – caverac Nov 26 '17 at 16:51
  • $\begingroup$ 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: ch.mathworks.com/company/newsletters/articles/… 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? $\endgroup$ – user68863 Nov 26 '17 at 17:16

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