Finger Exercise: Diamond Model

Last edited: 2019-08-17

Brad DeLong: The Finger-Exercise Cobb-Douglas Production Case

1. The Zero-Uncertainty No-Debt Case

Simplify the already-simple near-canonical model. Express all variables in per-worker values. Require that there be neither population growth nor technological progress. Require that there be time-separable log utility.

Assume that the economy is populated by people who live for two periods, working when young, receiving profits from their savings when old, and consuming in both periods. Their utility function is:

(1.1) $ V = C_y + \beta{\ln(C_o)} + P $

where $ C_y $ is consumption when young, $ C_o $ is consumption when old, and $ P $ is the quantity of public goods purchased by the government with the profits from its debt operations. Maximizing utility produces this consumption first-order condition for an agent young ("y") in period t and old ("o") in period t+1:

(1.2) $ 1 = {R^f}_{t+1} \left[ \frac{\beta}{C_{o,t+1}} \right] $

(1.3) $ C_{o,t+1} = \beta R^f_{t+1} $

Now let's add in a Cobb-Douglas production function, where a single worker's productivity is:

(1.4) $ Y_t = A K_{t-1}^\alpha $

The wage rate is:

(1.5) $ w_t = (1 - \alpha) A K_{t-1}^\alpha $

The profit rate is:

(1.6) $ R^f_{t+1} = \alpha A K_{t}^{(\alpha -1)} $

Using:

(1.7) $ C_{y,t} = w_t - K_{t} $

(1.8) $ C_{o,t+1} = R^f_t K_{t} = \alpha A K_{t}^\alpha $

Substitute into the first order condition (1.3):

(1.9) $ A \alpha K_{t}^\alpha = \beta A \alpha K_{}^{(\alpha -1)} $

Then:

(1.10) $ K_{} = \beta^{(1/\alpha)} K_{}^{(1 - 1/\alpha)} $

With the steady-state equilibrium:

(1.11) $ K^* = \beta $

In this steady-state equilibrium:

(1.12) $ R^{f*} = \alpha A \beta^{(\alpha - 1)} $

(1.13) $ C^*_y = (1-\alpha) A \beta^\alpha - \beta $

(1.14) $ C^*_o = \alpha A \beta^\alpha $

And if $ \alpha = \frac{1}{2} $ and $ \beta = \frac{1}{4} $, then: $ K^* = \frac{1}{4}, R^{f*} = A, C^*_o = \frac{A}{4} $, and $ C^*_y = \frac{A - 1}{4} $ is the steady-state equilibrium.

 

2. Adding Debt

The government (a) issues a debt amount $ \delta $ in period t, (b) rolls it over by issuing the same amount $ \delta $ in period t+1, and (c) uses the surplus to provided public goods: $ P_{t+1} = \left( 1 - R^f_{t+1} \right) \delta $ to provide public goods.

(1.7) and (1.8) are then changed:

(2.1) $ C_{y,t} = w_t - K_{t} - \delta $

(2.2) $ C_{o,t+1} = R^f_{t+1} K_{t} + R^f_{t+1} \delta = \alpha A K_{t}^\alpha + \alpha A K_{t}^{(\alpha-1)} \delta $

Substitute into the first order condition to get:

(2.3) $ \alpha A K_{t}^\alpha + \alpha A K_{t}^{(\alpha-1)} \delta = \beta A \alpha K_{t}^{(\alpha -1)} $

(2.4) $ 1 + \frac{\delta}{K_{t}} = \frac{\beta}{K_{t}} $

And so we have full crowding out:

(2.5) $ K^{*\delta} = \beta - \delta $

With:

(2.6) $ R^{f*\delta} = R^{*\delta} = \alpha A K^{(\alpha-1)} = \alpha A \left( \beta - \delta \right)^{(\alpha - 1)} $

(2.7) $ C^{*\delta}_y = (1-\alpha) A K^{\alpha} - \beta = (1-\alpha) A \left( \beta - \delta \right)^\alpha - \beta $

(2.8) $ C^{*\delta}_o = \alpha A \left( \beta - \delta \right)^\alpha + \alpha A \left( \beta - \delta \right)^{(\alpha - 1)}\delta $

(2.9) $ P^{*\delta} = \left( 1 - \alpha A \left( \beta - \delta \right)^{(\alpha - 1)} \right)\delta $

And if $ A = 1, \alpha = \frac{1}{2} $, $ \beta = \frac{1}{4} $, and $ \delta = \frac{1}{20} $, then: $ K^* = 0.23, R^{f*} = 1.0426, C^*_o = 0.2606 $, and $ C^*_y = -0.0102 $ is the steady-state equilibrium.

 

3. Adding Uncertainty (No Debt)

Let's add two-point uncertainty:

(3.1) $ Y_t = A(1 + \epsilon_t)K_{t-1}^\alpha $

where $ \epsilon_t $ has a two-point distribution:

(3.2) $ \epsilon_t = +\sigma $ or $ \epsilon_t = -\sigma $

It is not just production but also the profit rate that becomes risky, so we need to drop the "f" superscript and taker expectations in the first-order condition (1.2), which then becomes:

(3.3) $ 1 = E_t \left[ \frac{\beta R_{t+1}}{C_{o,t+1}} \right] $

Expanding and substituting in the profit rate R:

(3.4) $ 1 = \left( \frac{1}{2} \right) \left[ \frac{\beta \alpha (1+\sigma) A \beta^{(\alpha - 1)}}{C_{o,t+1,(\epsilon=+\sigma)}} \right] + \left( \frac{1}{2} \right) \left[ \frac{\beta \alpha (1-\sigma) A \beta^{(\alpha - 1)}}{C_{o,t+1,{(\epsilon=-\sigma)}}} \right] $

Substituting in consumption when old:

(3.5) $ 1 = \left( \frac{1}{2} \right) \left[ \frac{\beta \alpha (1+\sigma) A \beta^{(\alpha - 1)}}{\alpha (1+\sigma) A \beta^{(\alpha-1)}K_{t}} \right] $ $ + \left( \frac{1}{2} \right) \left[ \frac{\beta \alpha (1-\sigma) A \beta^{(\alpha - 1)}}{\alpha (1-\sigma) A \beta^{(\alpha-1)}K_{t}} \right] $

(3.6) $ 1 = \left( \frac{1}{2} \right) \left[ \frac{\beta}{K_{t}} \right] + \left( \frac{1}{2} \right) \left[ \frac{\beta}{K_{t}} \right] = \frac{\beta}{K_{t}} $

(3.7) $ K_{t} = K^* = \beta $

Because we have log utility, income and substitution effects cancel each other out as they affect the risky interest rate, and so the savings rate is unaffected.

In this steady-state equilibrium with uncertainty, the expected profit rate is still:

(3.8) $ E_t \left[ R^* \right] = \left( \frac{1}{2} \right) \alpha (1+\sigma) A \beta^{(\alpha - 1)} + \left( \frac{1}{2} \right) \alpha (1-\sigma) A \beta^{(\alpha - 1)} = \alpha A \beta^{(\alpha - 1)} $

And expected consumption when young and old are still:

(3.9) $ E_t \left[ C^*_y \right] = \left( \frac{1}{2} \right) \left[ (1-\alpha) (1+\sigma) A \beta^\alpha - \beta \right] + \left( \frac{1}{2} \right) \left[ (1-\alpha) (1-\sigma) A \beta^\alpha - \beta \right] = (1-\alpha) A \beta^\alpha - \beta $

(3.10) $ E_t \left[ C^*_o \right] = \left( \frac{1}{2} \right) \left[ \alpha (1+\sigma) A \beta^\alpha \right] + \left( \frac{1}{2} \right) \left[ \alpha (1-\sigma) A \beta^\alpha \right] = \alpha A \beta^\alpha $

with profit rates, consumption when young, and consumption when old fluctuating depending on whether the productivity shock was positive or negative.

But there is now a risk-free rate of interest different from the expected profit rate—the rate at which risk-free debt would sell if there were any. The first-order condition for its return is:

(3.11) $ 1 = {R^f}_{t+1} E_t \left[ \frac{\beta}{C_{o,t+1}} \right] = R^f_{t+1} \left[ \left( \frac{1}{2} \right) \left[ \frac{\beta}{\alpha (1+\sigma) A \beta^\alpha} \right] + \left( \frac{1}{2} \right) \left[ \frac{\beta}{\alpha (1-\sigma) A \beta^\alpha} \right] \right] = R^f_{t+1}\left[ \frac{\beta}{\alpha (1-\sigma^2) A \beta^\alpha} \right] $

(3.12) $ {R^f}_{t+1} = (1-\sigma^2) \alpha A \beta^{(\alpha-1)} = (1-\sigma^2) E_t \left[ R_{t+1} \right] $

Thus there is a spread of proportional magnitude $ (1-\sigma^2) $ between the risk-free rate of return and the profit rate.

And if $ \alpha = \frac{1}{2} $, $ \beta = \frac{1}{4} $, and $ \sigma = \frac{1}{2} $, then: $ K^* = \frac{1}{4}, E_t R^* = A, R^{f*} = \frac{3A}{4}, E_t C^*_o = \frac{A}{4} $, and $ E_t C^*_y = \frac{A - 1}{4} $ is the steady-state equilibrium.

 

4. Adding Debt (with Uncertainty)

With uncertainty and with debt, the first-order condition for investment in the capital stock becomes:

(4.1) $ 1 = E_t \left[ \frac{\beta R_{t+1}}{C_{o,t+1}} \right] = E_t \left[ \frac {\beta R_{t+1}} {R_{t+1}K_{t} + R^f_{t+1}\delta} \right] $

Substituting in for the profit rate in the two cases—a good shock and a bad shock:

(4.2) $ 1 = \left( \frac{1}{2} \right) \left[ \frac {\beta \alpha (1+\sigma) A \beta^{(\alpha - 1)}} {\alpha (1+\sigma) A \beta^{(\alpha-1)}K_{t} + R^f_{t+1}\delta} \right] + \left( \frac{1}{2} \right) \left[ \frac {\beta \alpha (1-\sigma) A \beta^{(\alpha - 1)}} {\alpha (1-\sigma) A \beta^{(\alpha-1)}K_{t} + R^f_{t+1}\delta} \right] $

Log utility gives us the result that savings are independent of returns, so it is still the case that we have full crowding out and so $ K_t = \beta - \delta $. Therefore:

(4.3) $ E_tR^{*\delta} = \alpha A \left( \beta - \delta \right)^{(\alpha - 1)} $

(4.4) $ E_{t-1}\left[ C^{*\delta}_y \right] = (1-\alpha) A \left( \beta - \delta \right)^\alpha - \beta $

Now what about the risk-free rate, consumption when old, and the profits of the government's debt operations?

Our first-order condition is:

(4.5) $ 1 = {R^f}_{t+1} E_t \left[ \frac{\beta}{C_{o,t+1}} \right] = R^f_{t+1} $ $ \left[ \left( \frac{1}{2} \right) \left[ \frac{\beta}{(1+\sigma) \alpha A (\beta-\delta)^\alpha + R^f_{t+1}\delta} \right] + \left( \frac{1}{2} \right) \left[ \frac{\beta}{(1-\sigma) \alpha A (\beta-\delta)^\alpha + R^f_{t+1}\delta} \right] \right] $

(4.6) $ 1 = \left[ \left( \frac{1}{2} \right) \left[ \frac{\beta}{(1+\sigma) \left( \frac{E_tR_{t+1}K_t}{R^f_{t+1}} \right) + \delta} \right] + \left( \frac{1}{2} \right) \left[ \frac{\beta}{(1-\sigma) \left( \frac{E_tR_{t+1}K_t}{R^f_{t+1}} \right) + \delta} \right] \right] $

Let's call:

(4.7) $ \phi = \frac{E_tR_{t+1}K_t}{R^f_{t+1}} $

(4.8) $ 1 = \left[ \left( \frac{1}{2} \right) \left[ \frac{\beta}{(1+\sigma) \phi + \delta} \right] + \left( \frac{1}{2} \right) \left[ \frac{\beta}{(1-\sigma) \phi + \delta} \right] \right] $

(4.9) $ 1 = \frac {\beta(\phi + \delta)} {(\phi + \delta)^2 - \sigma^2\phi^2}$

(4.10) $ (\phi + \delta)^2 - \sigma^2\phi^2 - \beta(\phi + \delta) = 0 $

(4.11) $ (1 - \sigma^2)\phi^2 + (2\delta - \beta)\phi - \delta(\beta - \delta) = 0 $

(4.12)$ \phi = \frac {\beta - 2\delta ± \sqrt{(\beta - 2\delta)^2 + 4(1 - \sigma^2)\delta(\beta-\delta)} } {2(1 - \sigma^2)} $

(4.13) $ R^f_{t+1} = \frac{E_tR_{t+1}K_t}{\phi} $

(4.14) $ R^f_{t+1} = \frac {(E_tR_{t+1}K_t)(2(1 - \sigma^2))} {\beta - 2\delta + \sqrt{(\beta - 2\delta)^2 + 4(1 - \sigma^2)\delta(\beta-\delta)}} $

(4.15) $ R^{f*\delta}_{t+1} = \frac {2 \alpha A (\beta - \delta)^\alpha(1 - \sigma^2)} {\beta - 2\delta + \sqrt{(\beta - 2\delta)^2 + 4(1 - \sigma^2)\delta(\beta-\delta)}} $

Check: What happens when $ \sigma^2 = 0 $?

(4.16) $ \sigma^2 = 0 $ -> $ R^f_{t+1} = \frac {2 \alpha A (\beta - \delta)^\alpha} {\beta - 2\delta + \sqrt{(\beta - 2\delta)^2 + 4\delta(\beta-\delta)}} = \frac {2 \alpha A (\beta - \delta)^\alpha} {2\beta - 2 \delta} = \alpha A (\beta - \delta)^{(\alpha-1)} = E_tR_{t+1} $

as it should.

Check: What happens when $ \delta = 0 $?

(4.17) $ \delta = 0 $ -> $ R^f_{t+1} = \frac {2 \alpha A \beta^\alpha(1 - \sigma^2)} {\beta + \sqrt{\beta^2 }} = \alpha A \beta^{(\alpha-1)}(1 - \sigma^2) = (1 - \sigma^2)E_tR_{t+1} $

as it should.

(4.18) $ E_tC^{*\delta}_{o,t+1} = \alpha A (\beta-\delta)^\alpha + \delta R^{f*\delta} = \alpha A (\beta-\delta)^\alpha + \delta $ $ \left[ \frac {2 \alpha A (\beta - \delta)^\alpha(1 - \sigma^2)} {\beta - 2\delta + \sqrt{(\beta - 2\delta)^2 + 4(1 - \sigma^2)\delta(\beta-\delta)}} \right] $

(4.19) $ P^{*\delta} = \delta \left[ 1 - \frac {2 \alpha A (\beta - \delta)^\alpha(1 - \sigma^2)} {\beta - 2\delta + \sqrt{(\beta - 2\delta)^2 + 4(1 - \sigma^2)\delta(\beta-\delta)}} \right] $

In order to keep the government's public goods from messing up the model, recall that we have walled their utility effect off from the savings-investment decision:

(1.1) $ V = C_y + \beta{\ln(C_o)} + P $

So that utility will be easy to calculate...

We want to pick a situation in which $ R^{f*\delta}_{t+1} $ is just a hair less than one and $ E_tR^{*\delta}_{t+1} $ is substantially in excess of one, and show that debt is nevertheless good...

(And show that utility as a function of debt is maximized when we push debt to the point where $ R^{f*\delta}_{t+1} = 1 $

 

5. $ \tau $, Derivatives, and such

With respect to the tax rate on capital to pin the profit rate to its baseline value...

(5.?) $ \alpha A \beta^{(\alpha - 1)} = (1 - \tau) \alpha A (\beta - \delta)^{(\alpha - 1)} $

(5.?) $ \tau = 1 - \frac{ \alpha A \beta^{(\alpha - 1)}}{\alpha A (\beta - \delta)^{(\alpha - 1)}} = 1 - \frac{(\beta - \delta)^{(1-\alpha)}}{(\beta)^{(1-\alpha)}} $

The first-order condition for the risk-free rate then becomes:

(5.?) $ 1 = {R^f}_{t+1} E_t \left[ \frac{\beta}{C_{o,t+1}} \right] = R^f_{t+1} $ $ \left[ \left( \frac{1}{2} \right) \left[ \frac{\beta}{(1+\sigma) \alpha A \beta^{(\alpha-1)} (\beta-\delta) + R^f_{t+1}\delta} \right] + \left( \frac{1}{2} \right) \left[ \frac{\beta}{(1-\sigma) \alpha A \beta^{(\alpha-1)} (\beta-\delta) + R^f_{t+1}\delta} \right] \right] $

and the profits of the government become:

(5.?) $ \delta(1-R^f) + (\beta - \delta)( \alpha A (\beta - \delta)^{(\alpha-1)}- \alpha A \beta^{(\alpha-1)}) $

We have:

(4.3) $ E_tR^{*\delta} = \alpha A \left( \beta - \delta \right)^{(\alpha - 1)} $

(4.4) $ E_{t-1}\left[ C^{*\delta}_y \right] = (1-\alpha) A \left( \beta - \delta \right)^\alpha - \beta $

(4.15) $ R^{f*\delta}_{t+1} = \frac {2 \alpha A (\beta - \delta)^\alpha(1 - \sigma^2)} {\beta - 2\delta + \sqrt{(\beta - 2\delta)^2 + 4(1 - \sigma^2)\delta(\beta-\delta)}} $

(4.18) $ E_tC^{*\delta}_{o,t+1} = \alpha A (\beta-\delta)^\alpha + \delta R^{f*\delta} = \alpha A (\beta-\delta)^\alpha + \delta $ $ \left[ \frac {2 \alpha A (\beta - \delta)^\alpha(1 - \sigma^2)} {\beta - 2\delta + \sqrt{(\beta - 2\delta)^2 + 4(1 - \sigma^2)\delta(\beta-\delta)}} \right] $

(4.19) $ P^{*\delta} = \delta \left[ 1 - \frac {2 \alpha A (\beta - \delta)^\alpha(1 - \sigma^2)} {\beta - 2\delta + \sqrt{(\beta - 2\delta)^2 + 4(1 - \sigma^2)\delta(\beta-\delta)}} \right] $

Now take derivatives with respect to $ \delta $:

(5.1) $ \frac{\partial E_t R^{*\delta}}{\partial \delta} = (1- \alpha) \alpha A \left( \beta - \delta \right)^{(\alpha - 2)} $

(5.2) $ \frac{\partial E_{t-1}\left[ C^{*\delta}_y \right]}{\partial \delta} = (1-\alpha) \alpha A \left( \beta - \delta \right)^{(\alpha-1)} $

(5.3) $ \frac{\partial R^{f*\delta}_{t+1}}{\partial \delta} = \frac {-2 \alpha^2 A (\beta - \delta)^{(\alpha-1)}(1 - \sigma^2)} {\beta - 2\delta + \sqrt{(\beta - 2\delta)^2 + 4(1 - \sigma^2)\delta(\beta-\delta)}} + \frac {2 \alpha A (\beta - \delta)^\alpha(1 - \sigma^2) \left( -2 + \frac{-4(\beta - 2\delta) + 4(1-\sigma^2)\beta - 8(1-\sigma^2)\beta \delta}{2 \sqrt{(\beta - 2\delta)^2 + 4(1 - \sigma^2)\delta(\beta-\delta)}} \right)} {\left( \beta - 2\delta + \sqrt{(\beta - 2\delta)^2 + 4(1 - \sigma^2)\delta(\beta-\delta)} \right)^2} $

(5.4) $ \frac{\partial K^{*\delta}}{\partial \delta} = -1 $

$ R^{f*\delta}_{t+1} = \frac {2 \alpha A (\beta - \delta)^\alpha(1 - \sigma^2)} {\beta - 2\delta + \sqrt{(\beta - 2\delta)^2 + 4(1 - \sigma^2)\delta(\beta-\delta)}} $

$ \partial $