2
$\begingroup$

I wonder why at the beginning sf(k(t)) is steeper than δk(𝑡), but at some point, it starts flatter than δk(t)?

$\endgroup$
2
  • $\begingroup$ The standard chart used to explain the Solow model plots output per worker (vertical) against capital per worker (horizontal). It does not have a time axis. So when you say "at the beginning", a more accurate description would be "if capital per worker is very low". $\endgroup$ Commented Sep 15, 2023 at 10:48
  • $\begingroup$ Hi! Did you know you can accept answers to your questions? $\endgroup$
    – Giskard
    Commented Sep 15, 2023 at 11:10

2 Answers 2

3
$\begingroup$

The answer: it is an assumption
Solow-models usually assume that $f$ fulfills the Inada conditions, points 3. and 4. of which state $$ \begin{equation*} \lim_{k \to 0} f'(k) = \infty \\ \lim_{k \to \infty} f'(k) = 0. \end{equation*} $$

A frequent special case
If the production function $F(K,L)$ is of Cobb-Douglas type with constant returns to scale and has positive parameters, then the derived $f(k) = F(K,L)/L$ function will satisfy the Inada conditions.

Caveat
Based on the graphs they see, students of economics sometimes mistakenly assume that all strictly concave functions have these properties, but this is not the case. $f(k) = \sqrt{k} + k$ does not have the second one, while $f(k) = \sqrt{1+k}$ does not have the first one. Thus properties like $f$ being concave or non-linear are not sufficient, one has to assume the conditions.

$\endgroup$
2
$\begingroup$

Because $\delta k$ is linear function, so the slope of the function is always constant, whereas $sf(k)$ is typically nonlinear function exponential function like $f(k)=k^\alpha$ with $0<\alpha<1$, so mathematically such function starts with steep slope, but this slope always gets smaller (this can be seen from $d/dk= \alpha k^{\alpha -1}$.

However, note depending on value of parameter $\delta$, $\delta k$ could have equal or steeper slope but in such case capital accumulation is impossible because capital depreciates faster than it can be accumulated given $s$.

$\endgroup$
7
  • $\begingroup$ Can you please give me a depreciation parameter $\delta$ at which the slope of $\delta k$ will always be steeper than the slope of, let's say, $0.5\sqrt{k}$? If not, can you please explain what you mean by your second paragraph. $\endgroup$
    – Giskard
    Commented Sep 15, 2023 at 11:08
  • $\begingroup$ @Giskard $\delta$ that satisfies $\delta > \max \frac{1}{\sqrt{k}}$ for k>0. Also savings rate can be just zero and in that case it’s trivial $\endgroup$
    – 1muflon1
    Commented Sep 15, 2023 at 12:52
  • 1
    $\begingroup$ But you specify "always", when you write "always steeper slope", so I don't think a $\delta > \max_{k>0} \frac{1}{\sqrt{k}}$ exists? You are right about the edge case of $s = 0$, but that seems quite specific, not useful for the general case. $\endgroup$
    – Giskard
    Commented Sep 15, 2023 at 13:19
  • $\begingroup$ @Giskard I edited out the word always $\endgroup$
    – 1muflon1
    Commented Sep 15, 2023 at 13:44
  • 1
    $\begingroup$ @1muflon1 Adding to Giskard's point, surely $\lim_{k \to 0} \frac{1}{\sqrt{k}}=\infty$? $\endgroup$ Commented Sep 16, 2023 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.