From the Transversality Condition to Initial Consumption

Question

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.

Answer 1

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}