Question  1.  (2 points)    Solow model in continuous time. Consider the Solow model in continuous time with production function y = f (k) satisfying the usual properties, constant savings rate s, depreciation rate 6 , productivity growth g and employment growth n.

(a) Use the implicit function theorem to show how an increase in s affects the steady state values k* , y* , c* . Does this change in s increase or decrease long run output and consumption per worker? Explain.

Answer:

Steady state capital k*  solves

sf(k* ) = (6 + g + n)k*

This implicitly determines k*  as a function of the savings rate s, say k*  = k(s). Write this

sf(k(s)) = (6 + g + n)k(s)

Di↵erentiating both sides with respect to s gives

f(k(s)) + sf\ (k(s))k\ (s) = (6 + g + n)k\ (s)

Solving for k\ (s) then gives

k\ (s) = -                   > 0

and

c\ (s) = (1 - s)y\ (s) - y(s)

Since a higher savings rate increases steady state capital, it also increases steady state output. The e↵ect on consumption is ambiguous and depends whether the savings rate s is greater or lower than the‘golden rule’level (as discussed in class). 

Now consider the special case of a Cobb-Douglas production function f(k) = ka .

(b) Derive expressions for the elasticities of capital and output with respect to the

savings rate

d log k*   d log s ,

How do these depend on the curvature of the production function α? Explain.

Answer:

With y = f(k) = k↵ we have the solutions

1

k*  = ✓ ◆ 1 一↵

and                                              ↵ 

y*  = ✓ ◆ 1 一↵

Hence

dlog k*           1  

dlog s   1 一 ↵

and

dlog s   1 一 ↵

(c) Derive an exact solution for the time path k(t) of capital per effective worker.

Answer:

With y = f(k) = k↵ the capital stock k(t) solves the nonlinear di↵erential equation k˙ (t) = sk(t)↵ · (6 + g + n)k(t)

But this implies a liuear di↵erential equation in the capital/output ratio x(t) ⌘ k(t)/y(t) This is a stable linear di↵erential equation in x(t) with steady state

k*            s   

y*        6 + g + n

x(t) = e入tx(0) + (1 - e入t)x* ,    λ 三 -(1 - ↵)(6 + g + n) < 0

So the solution for k(t) is

1

k(t) = ⇣e入tk(0)1一↵ +(1 - e入t)k* 1一↵ ⌘ 1 一↵

Now consider the specific numerical values α = 0.3, s = 0.2, δ = 0.05,g = 0.02, n = 0.03.

(d)  Calculate and plot the time paths of k(t), y(t), c(t) starting from the initial condi- tion k(0) = k* /2. How long is the half-life of convergence?

Answer:

Figure 1 below shows the transitional dynamics of k(t),y(t),c(t) for the given parameter values. Notice that k*  = 2.6918, y*  = 1.3459 (so the capital/output ratio is x*  = s/(6 + g + n) = 2) and c*  = 1.0767 so that c* /y*  = 1 - s = 0.80.

The half-life of convergence is the time t*  it takes to close half of the initial deviation from steady state. Since the di↵erential equation for the capital/output ratio is linear, we have

a simple formula for the half-life in the capital/output ratio. Using the solution x(t) = e入tx(0) + (1 - e入t)x*  = x* + e入t(x(0) - x* )

we look for the value of t such that x(t) - x*  = (x(0) - x* )/2. This is given by

log 2          2       

λ    (1 - ↵)(6 + g + n)

With the given parameters, this works out to be

0.69

t*  =                             = 9.90

(measured in years, if the growth rates are annual growth rates).

NoTE∶ does this give the half-life for k(t) too? Or y(t)? Not in general, because of the

1

transformation k(t) = x(t)1 一↵ , the gap k(t) - k*  is not half of k(0) - k*  when x(t) - x* is half of x(0) - x* . But near the steady state k*  (i.e., for small deviations), the speed of adjustment in k(t) is given by

log 2                 2       

sf\ (k* ) - (6 + g + n)   (1 - ↵)(6 + g + n)

just as for the capital/output ratio.

(e) Now suppose that we are in steady state k(0) = k*  when the savings rate suddenly increases to s = 0.22.   Calculate and plot the time paths of k(t), y(t), c(t) in response to this change.  Explain both the short-run and long-run dynamics of k(t), y(t), c(t).  What if instead the savings rate had increased to s = 0.30?  Do you think these are large or small effects on output? Explain.

Figure 1: Time paths of k(t), y(t), c(t)

Question  2.   (3 points)     Natural resource depletion in the  Solow model.

Consider a Solow model in continuous time where output is given by the CRS production function

Y (t) = K(t)aR(t)φ (A(t)L(t))1一a一φ , 0 < α, ϕ < 1, α + ϕ < 1,

where R(t) denotes a stock of natural resources that depletes at rate θ > 0 R˙ (t) = −θR(t),

from some initial R(0). The rest of the model is as standard with constant savings rate s, depreciation rate δ, productivity growth g and employment growth n.

(a) Let gY (t) and gK (t) denote the growth rates of output and the capital stock. Derive a formula for gY (t) in terms of gK (t).

Answer:

Taking logs of the production function

log Y (t) = ↵ log K(t)+ o log R(t)+(1 - ↵ - o)(log A(t)+log L(t))

Then di↵erentiating with respect to t

 = ↵  + o ++(1 - ↵ - o)   + !

Plugging in the given growth rates we then have

gY (t) = ↵gK (t) - o9 +(1 - ↵ - o)(g + n)

(b) Let gY(*) and gK(*) denote the growth rates of output and the capital stock along a balanced growth path.  Show that along any balanced growth path gY(*) = gK(*) . Solve for this growth rate in terms of model parameters.

Answer:

Since the savings rate is constant we can write

K˙ (t) = sY (t) 二 6K(t)

or

gK (t) =     = s        二 6

Hence along any balanced growth path where capital grows at a constant rate gK(*) we must

have

Y(t)

K(t)

That is, along a balanced growth path the capital/output ratio x(t) ⌘ K(t)/Y(t) ratio must be constant, in other words output must be growing at the same rate as the capital

stock, gY(*) = gK(*) . Let this common growth rate be g* . From part (a) this g* satisfies g* = ↵g* ━ 。9 +(1 ━ ↵ ━ 。)(g + n)

which solves for

g* =  ( ━。9 +(1 ━ ↵ ━ 。)(g + n))

Observe that for any t the capital/output ratio x(t) is strictly decreasing in the growth

rate of the capital stock

K(t)      s   

Y(t)   gK (t)+ 6

And along a balanced growth path

s  

g* + 6

Importantly, we will have x(t) > x* if and only if gK (t) < g* .

(c) Does the economy necessarily converge to a balanced growth path? Explain. This subsection is not graded

(d) Now suppose instead that resources R(t) grew in line with population,  R˙ (t) = nR(t). Compare the long-run growth rate of the economy with resource depletion from part (b) to the long growth rate of this alternative economy without resource depletion. What would make this gap between the growth rates large? Explain.

Answer:

Now let 9 = 二n. This makes the growth rate

1                            1

gˆ⇤ = 1 二 ↵ (二φ9 +(1 二 ↵ 二 φ)(g + n))1✓ =一n

= 1 1二 ↵ (φn +(1 二 ↵ 二 φ)(g + n))