<p>so to the best of my knowledge, it is possible only with log utility and <span class="math-container">$\delta = 1$</span>, namely the stock of capital fully depreciates, so yes, we are often restricted to implicit answers (or you linearize and you solve the system of difference equations by means of an eigenvalue-eigenvector decomposition, problem is in some setups linearizing gives you a terrible approximation of the true solution, namely if you have outside options or stuff like borrowing constraints).</p>
<p>Let's find the solution for <span class="math-container">$U(c_t) = \ln c_t$</span>, <span class="math-container">$\delta = 1$</span> and <span class="math-container">$F(k_t) = k_t^\alpha$</span> with <span class="math-container">$\alpha \in (0,1)$</span></p>
<p>From the F.O.C.s, we have a system of two equations:</p>
<p><span class="math-container">\begin{equation}
\frac{1}{c_t} = \beta \frac{1}{c_{t+1}}\alpha k_{t+1}^{\alpha - 1}  \hspace{20pt} (1) 
\end{equation}</span>
<span class="math-container">\begin{equation}
\hspace{1pt} c_t = k^{\alpha}_t - k_{t+1} \hspace{29pt} (2) 
\end{equation}</span></p>
<p>Substituting <span class="math-container">$(2)$</span> into <span class="math-container">$(1)$</span>, we get the infamous second order non-linear difference equation:</p>
<p><span class="math-container">\begin{equation}
\frac{1}{k^{\alpha}_t - k_{t+1} } = \alpha\beta \frac{ k_{t+1}^{\alpha - 1}}{k^{\alpha}_{t+1} - k_{t+2}}  \hspace{20pt} (3) 
\end{equation}</span></p>
<p>For <span class="math-container">$t = T-2$</span>, <span class="math-container">$(3)$</span> boils down to:</p>
<p><span class="math-container">\begin{equation}
\frac{1}{k^{\alpha}_{T-2} - k_{T-1} } = \alpha\beta \frac{ k_{T-1}^{\alpha - 1}}{k^{\alpha}_{T-1} - k_{T}}  \hspace{20pt} (4) 
\end{equation}</span></p>
<p>The limit of <span class="math-container">$(4)$</span> as <span class="math-container">$T \rightarrow +\infty$</span> is (since <span class="math-container">$k_T \rightarrow 0$</span> as <span class="math-container">$T \rightarrow +\infty$</span>)</p>
<p><span class="math-container">\begin{equation}
\frac{1}{k^{\alpha}_{T-2} - k_{T-1} } = \alpha\beta \frac{ 1}{k_{T-1}}  \hspace{20pt} (5) 
\end{equation}</span></p>
<p>Solving <span class="math-container">$(5)$</span> for the control variable in <span class="math-container">$T-2$</span>, <span class="math-container">$k_{T-1}$</span>, we get:</p>
<p><span class="math-container">\begin{equation}
k_{T-1} = \frac{\alpha\beta}{1+\alpha\beta}k_{T-2}^{\alpha}\hspace{20pt} (6) 
\end{equation}</span></p>
<p>Now, we iterate this process by substituting <span class="math-container">$t= T-3$</span> into equation <span class="math-container">$(3)$</span> we get</p>
<p><span class="math-container">\begin{equation}
\frac{1}{k^{\alpha}_{T-3} - k_{T-2} } = \alpha\beta \frac{ k_{T-2}^{\alpha - 1}}{k^{\alpha}_{T-2} - k_{T-1}}  \hspace{20pt} (7) 
\end{equation}</span></p>
<p>Plugging <span class="math-container">$(6)$</span> into <span class="math-container">$(7)$</span> and solving for <span class="math-container">$k_{T-2}$</span> yields:</p>
<p><span class="math-container">\begin{equation}
k_{T-2} = \frac{\alpha\beta + (\alpha\beta)^2}{1 + \alpha\beta + (\alpha\beta)^2}k^{\alpha}_{T-3} \hspace{20pt} (8) 
\end{equation}</span></p>
<p>Iterating <span class="math-container">$(8)$</span>, we get:</p>
<p><span class="math-container">\begin{equation}
k_{t+1} = \frac{\sum^{T-(t+1)}_{j=1} (\alpha\beta)^j}{\sum^{T-(t+1)}_{k=0} (\alpha\beta)^k} k_{t}^{\alpha} \hspace{20pt} (9) 
\end{equation}</span></p>
<p><span class="math-container">$(9)$</span> boils down to:</p>
<p><span class="math-container">\begin{equation}
k_{t+1} = \alpha\beta\frac{1-(\alpha\beta)^{T-(t+1)}}{1-(\alpha\beta)^{T-t}}k_{t}^{\alpha}  \hspace{20pt} (10)
\end{equation}</span></p>
<p>The limit of <span class="math-container">$(10)$</span> as <span class="math-container">$T\rightarrow + \infty$</span> is simply
<span class="math-container">\begin{equation}
k_{t+1} = \alpha\beta k_{t}^{\alpha}  \hspace{20pt} (*)
\end{equation}</span></p>

CC BY-SA 4.0
2024-01-23T09:56:51.273Z
Note that you can't do the trick of rearranging the Euler equation if there is uncertainty, due to the Jensen's inequality. Namely, a necessary condition for having a closed form solution is that households are homogenous and know in t=0 every future income realization...
Matteo Bulgarelli
CC BY-SA 4.0
2024-01-13T13:06:50.713Z
2024-01-13T13:06:50.713Z
Matteo Bulgarelli
<p>Suppose we have a discrete time neoclassical growth model with CRRA utility where we solve <span class="math-container">$$\max_{\{c,k\}_{t=0}^{\infty}} \sum_{t=0}^{\infty}\beta^t \frac{c_t^{1-\sigma}-1}{1-\sigma}$$</span> <span class="math-container">$$\text{st. } \begin{cases} k_{t+1}=AF(k_t)-\delta k_t -c_t \\
c_t\geq 0, k_t\geq 0,k_0=\hat{k}_0\end{cases}$$</span>
Where <span class="math-container">$F$</span> is a production function satisfying the Inada conditions.</p>
<p>Optimality conditions from this problem are the Euler equation for consumption, the budget constraint, and two boundary conditions. <span class="math-container">$$\begin{align*} \frac{c_t^{-\sigma}}{\beta c_{t+1}^{-\sigma}}&amp;=F'(k_{t+1})+1-\delta \\
k_{t+1}&amp;=AF(k_t)-\delta k_t -c_t \\
k_0&amp;=\hat{k}_0 \\
\lim_{\tau \to \infty}\beta^{\tau}c_{\tau}^{-\sigma}k_{\tau+1}&amp;=0 \end{align*}$$</span></p>
<p>Given some initial level of consumption and capital, we can use the Euler equation and the budget constraint to iterate forward the optimal path, and the bottom two conditions provide boundary values which pin down our optimal <span class="math-container">$c_0$</span> and <span class="math-container">$k_0$</span>.</p>
<p>Whilst I can program a shooting algorithm to approximate the initial consumption level using the TVC condition, I was wondering if there was an analytic solution for the initial consumption level in this basic model. I tried deriving an answer by iterating the Euler equation and budget constraint forward, but it got messy quickly and I couldn't get an answer out of it.</p>
<p>Does this mean we are restricted to implicit answers for the initial consumption levels in these models? Or is there a simple way to derive an analytic answer that I'm missing? I realise the initial consumption level isn't the focus of the model, since it is mainly about transition dynamics and behaviour around the steady state, but thought it would be nice if there was an exact solution for this toy example.</p>

<macroeconomics><economic-growth>
From the Transversality Condition to Initial Consumption
Joseph Basford
# From the Transversality Condition to Initial Consumption

(Asked by: username_0 on 2024-01-07)

Suppose we have a discrete time neoclassical growth model with CRRA utility where we solve $$\max\_{\{c,k\}\_{t=0}^{\infty}} \sum\_{t=0}^{\infty}\beta^t \frac{c\_t^{1-\sigma}-1}{1-\sigma}$$ $$\text{st. } \begin{cases} k\_{t+1}=AF(k\_t)-\delta k\_t -c\_t \\
c\_t\geq 0, k\_t\geq 0,k\_0=\hat{k}\_0\end{cases}$$
Where $F$ is a production function satisfying the Inada conditions.

Optimality conditions from this problem are the Euler equation for consumption, the budget constraint, and two boundary conditions. $$\begin{align\*} \frac{c\_t^{-\sigma}}{\beta c\_{t+1}^{-\sigma}}&=F'(k\_{t+1})+1-\delta \\
k\_{t+1}&=AF(k\_t)-\delta k\_t -c\_t \\
k\_0&=\hat{k}\_0 \\
\lim\_{\tau \to \infty}\beta^{\tau}c\_{\tau}^{-\sigma}k\_{\tau+1}&=0 \end{align\*}$$

Given some initial level of consumption and capital, we can use the Euler equation and the budget constraint to iterate forward the optimal path, and the bottom two conditions provide boundary values which pin down our optimal $c\_0$ and $k\_0$.

Whilst I can program a shooting algorithm to approximate the initial consumption level using the TVC condition, I was wondering if there was an analytic solution for the initial consumption level in this basic model. I tried deriving an answer by iterating the Euler equation and budget constraint forward, but it got messy quickly and I couldn't get an answer out of it.

Does this mean we are restricted to implicit answers for the initial consumption levels in these models? Or is there a simple way to derive an analytic answer that I'm missing? I realise the initial consumption level isn't the focus of the model, since it is mainly about transition dynamics and behaviour around the steady state, but thought it would be nice if there was an exact solution for this toy example.

**Tags:** macroeconomics, economic-growth


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## 1 answers

### Answer by username_1 on 2024-01-13. Score: 1 | Accepted answer

so to the best of my knowledge, it is possible only with log utility and $\delta = 1$, namely the stock of capital fully depreciates, so yes, we are often restricted to implicit answers (or you linearize and you solve the system of difference equations by means of an eigenvalue-eigenvector decomposition, problem is in some setups linearizing gives you a terrible approximation of the true solution, namely if you have outside options or stuff like borrowing constraints).

Let's find the solution for $U(c\_t) = \ln c\_t$, $\delta = 1$ and $F(k\_t) = k\_t^\alpha$ with $\alpha \in (0,1)$

From the F.O.C.s, we have a system of two equations:

\begin{equation}
\frac{1}{c\_t} = \beta \frac{1}{c\_{t+1}}\alpha k\_{t+1}^{\alpha - 1} \hspace{20pt} (1)
\end{equation}
\begin{equation}
\hspace{1pt} c\_t = k^{\alpha}\_t - k\_{t+1} \hspace{29pt} (2)
\end{equation}

Substituting $(2)$ into $(1)$, we get the infamous second order non-linear difference equation:

\begin{equation}
\frac{1}{k^{\alpha}\_t - k\_{t+1} } = \alpha\beta \frac{ k\_{t+1}^{\alpha - 1}}{k^{\alpha}\_{t+1} - k\_{t+2}} \hspace{20pt} (3)
\end{equation}

For $t = T-2$, $(3)$ boils down to:

\begin{equation}
\frac{1}{k^{\alpha}\_{T-2} - k\_{T-1} } = \alpha\beta \frac{ k\_{T-1}^{\alpha - 1}}{k^{\alpha}\_{T-1} - k\_{T}} \hspace{20pt} (4)
\end{equation}

The limit of $(4)$ as $T \rightarrow +\infty$ is (since $k\_T \rightarrow 0$ as $T \rightarrow +\infty$)

\begin{equation}
\frac{1}{k^{\alpha}\_{T-2} - k\_{T-1} } = \alpha\beta \frac{ 1}{k\_{T-1}} \hspace{20pt} (5)
\end{equation}

Solving $(5)$ for the control variable in $T-2$, $k\_{T-1}$, we get:

\begin{equation}
k\_{T-1} = \frac{\alpha\beta}{1+\alpha\beta}k\_{T-2}^{\alpha}\hspace{20pt} (6)
\end{equation}

Now, we iterate this process by substituting $t= T-3$ into equation $(3)$ we get

\begin{equation}
\frac{1}{k^{\alpha}\_{T-3} - k\_{T-2} } = \alpha\beta \frac{ k\_{T-2}^{\alpha - 1}}{k^{\alpha}\_{T-2} - k\_{T-1}} \hspace{20pt} (7)
\end{equation}

Plugging $(6)$ into $(7)$ and solving for $k\_{T-2}$ yields:

\begin{equation}
k\_{T-2} = \frac{\alpha\beta + (\alpha\beta)^2}{1 + \alpha\beta + (\alpha\beta)^2}k^{\alpha}\_{T-3} \hspace{20pt} (8)
\end{equation}

Iterating $(8)$, we get:

\begin{equation}
k\_{t+1} = \frac{\sum^{T-(t+1)}\_{j=1} (\alpha\beta)^j}{\sum^{T-(t+1)}\_{k=0} (\alpha\beta)^k} k\_{t}^{\alpha} \hspace{20pt} (9)
\end{equation}

$(9)$ boils down to:

\begin{equation}
k\_{t+1} = \alpha\beta\frac{1-(\alpha\beta)^{T-(t+1)}}{1-(\alpha\beta)^{T-t}}k\_{t}^{\alpha} \hspace{20pt} (10)
\end{equation}

The limit of $(10)$ as $T\rightarrow + \infty$ is simply
\begin{equation}
k\_{t+1} = \alpha\beta k\_{t}^{\alpha} \hspace{20pt} (\*)
\end{equation}

> **Comments:**
> **username_1**: Note that you can't do the trick of rearranging the Euler equation if there is uncertainty, due to the Jensen's inequality. Namely, a necessary condition for having a closed form solution is that households are homogenous and know in t=0 every future income realization...  

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