Given the simplest form of a Lucas model, i.e., a Bellman equation given by \begin{align} J(x_t) & = \max_{c_t, x_{t+1}} \{ u(c_t) + \beta E_{\pi} [ J(x_{t+1})] \} \\ & \textrm{ s.t. } p_{t}x_{t+1} = (p_t + d_t)x_t - c_t, \nonumber \end{align} with a FOC given by \begin{align} u'(c_t)p_t = \beta E_{\pi} [u'(c_{t+1}) (p_{t+1}+d_{t+1})], \end{align} is there any way to calculate the value function $J(x_t)$ explicitly(i.e. without numerical solution algorithms)?

More generally, when is it possible to calculate the value function from a Bellman set-up explicitly, and what are the standard approaches to do this?

Thanks a lot in advance!

Just to explain the above variables:

$c_t$ : consumption, $x_t$ : risky asset holdings, $d_t$ : dividends, $\pi$ : probability law for $d_{t+1}$, $u(\cdot)$ : standard utility function


1 Answer 1


As far as I know, the only method that works is to guess and verify: You guess the functional form of the value function $J(x)$ and verify that it indeed satisfies the Bellman equation.

Of course, depending on how lucky you are, guessing might be a rather time consuming approach.

  • In some cases, you can make a more educated guess by solving several finite horizon versions of the problem, i.e. solve, \begin{align*} J_T(x) = \max \sum_{t = 0}^T u(c_t) \text{ s.t. } &p_t x_{t+1} = (p_t + d_t)x_t-c_t.\\ &x_0 \text{ given } \end{align*} for values $T = 1,2,\dots$. If you are able to get a closed form solution for $J_T(x)$, you could try a similar functional form for the general problem $J(x)$.

  • Sometimes, it is also possible to get shape restrictions on the function $J(x)$. For example, you might be able to show that $J$ is monotone, or concave, or homogeneous. This sometimes restricts the class of possible solutions.

  • Finally, you could solve the problem numerically and then try to fit a particular shape (e.g. linear in $\ln(x)$). This can guide you to a more particular functional form.

  • $\begingroup$ Thanks a lot for your answer - I'll try out the finite horizon approach soon. Do you already know if it is possible to solve for the value function in the above example model? $\endgroup$
    – Eddie
    Aug 28, 2023 at 11:11

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