Two-Period Brock–Mirman Model
Two-Period Brock–Mirman Model
We consider a two-period Brock–Mirman model with general period utility \(u(c)\), rather than imposing a specific functional form.
Environment
Time is indexed by \(t=0,1\). The initial capital stock \(k_0>0\) is given. Output is produced according to the stochastic production function
\[ y_t = z_t k_t^\alpha, \qquad 0<\alpha<1, \]
where \(z_t\) is an exogenous productivity shock.
We assume full depreciation of capital, so the resource constraint in each period is
\[ c_t + k_{t+1} = z_t k_t^\alpha. \]
Planner’s Problem
Using no bequest terminal condition, the agent consumes all remaining resources in period 1, so
\[ k_2 = 0. \]
Hence the two-period problem is
\[ \max_{\{c_0,k_1,c_1\}} \; \mathbb{E}_0 \left[ u(c_0) + \beta u(c_1) \right] \]
subject to
\[ c_0 + k_1 = z_0 k_0^\alpha, \]
\[ c_1 = z_1 k_1^\alpha, \]
with feasibility conditions
\[ c_0 > 0, \qquad c_1 > 0, \qquad k_1 \ge 0, \]
and given initial condition
\[ k_0 \text{ is given}. \]
Reduced Form
Substituting the constraints into the objective function, the problem can be written as
\[ \max_{k_1 \ge 0} \; \mathbb{E}_0 \left[ u\!\left(z_0 k_0^\alpha - k_1\right) + \beta \, u\!\left(z_1 k_1^\alpha\right) \right]. \]
This gives the complete formulation of the two-period Brock–Mirman model under terminal condition A while keeping the utility function in the general form \(u(\cdot)\).
Solution
We solve the model by backward substitution.
From the period-0 resource constraint, \[ c_0 = z_0 k_0^\alpha - k_1. \]
From the period-1 resource constraint and terminal condition \(k_2=0\), \[ c_1 = z_1 k_1^\alpha. \]
Substituting these two expressions into the objective function, the planner’s problem becomes \[ \max_{k_1 \geq 0} \; \mathbb{E}_0 \left[ u\!\left(z_0 k_0^\alpha - k_1\right) + \beta u\!\left(z_1 k_1^\alpha\right) \right]. \]
Therefore, the only choice variable is \(k_1\).
First-Order Condition
Assuming an interior solution, the first-order condition with respect to \(k_1\) is \[ -u'\!\left(z_0 k_0^\alpha - k_1\right) + \beta \, \mathbb{E}_0 \left[ u'\!\left(z_1 k_1^\alpha\right) z_1 \alpha k_1^{\alpha-1} \right] = 0. \]
Rearranging, we obtain \[ u'(c_0) = \beta \, \mathbb{E}_0 \left[ u'(c_1) \, \alpha z_1 k_1^{\alpha-1} \right]. \]
This is the Euler equation for the two-period problem.
Interpretation
The left-hand side, \(u'(c_0)\), is the marginal utility loss from reducing current consumption by one unit in order to increase saving.
The right-hand side is the discounted expected marginal utility gain in period 1. Saving one more unit in period 0 increases \(k_1\) by one unit, which raises next-period output and consumption by \[ \alpha z_1 k_1^{\alpha-1}. \]
Hence the optimal choice of \(k_1\) equates the marginal cost of saving today with the discounted expected marginal benefit tomorrow.
Characterization of the Optimal Allocation
Once the optimal \(k_1^*\) is determined from \[ u'\!\left(z_0 k_0^\alpha - k_1\right) = \beta \, \mathbb{E}_0 \left[ u'\!\left(z_1 k_1^\alpha\right) \alpha z_1 k_1^{\alpha-1} \right], \]
the optimal consumptions are given by \[ c_0^* = z_0 k_0^\alpha - k_1^*, \]
and \[ c_1^* = z_1 (k_1^*)^\alpha. \]
Therefore, solving the model amounts to solving the first-order condition for \(k_1^*\) and then recovering optimal consumption from the resource constraints.
Special Case: Log Utility
If we further assume \[ u(c) = \ln c, \] then \[ u'(c) = \frac{1}{c}. \]
The Euler equation becomes \[ \frac{1}{z_0 k_0^\alpha - k_1} = \beta \, \mathbb{E}_0 \left[ \frac{1}{z_1 k_1^\alpha} \alpha z_1 k_1^{\alpha-1} \right]. \]
Inside the expectation, the term simplifies as \[ \frac{1}{z_1 k_1^\alpha} \alpha z_1 k_1^{\alpha-1} = \frac{\alpha}{k_1}. \]
Hence \[ \frac{1}{z_0 k_0^\alpha - k_1} = \beta \frac{\alpha}{k_1}. \]
Solving for \(k_1\), we obtain \[ k_1^* = \frac{\alpha \beta}{1+\alpha \beta} z_0 k_0^\alpha. \]
Then optimal period-0 consumption is \[ c_0^* = z_0 k_0^\alpha - k_1^* = \frac{1}{1+\alpha \beta} z_0 k_0^\alpha, \]
and optimal period-1 consumption is \[ c_1^* = z_1 \left( \frac{\alpha \beta}{1+\alpha \beta} z_0 k_0^\alpha \right)^\alpha. \]
Conclusion
To solve the two-period Brock–Mirman model with no bequest terminal condition, we proceed in three steps:
- Use the resource constraints to substitute out \(c_0\) and \(c_1\).
- Rewrite the problem as a maximization over \(k_1\) only.
- Derive the first-order condition and solve for \(k_1^*\).
Under general utility \(u(\cdot)\), the solution is characterized implicitly by the Euler equation. Under log utility, the solution can be obtained in closed form.