<p>Several thoughts in this area:</p>

<p>1) Capital stock lasts over time. I may trade 1 euro for 1 unit of capital. This capital returns 0.01 euros every period forever.  It pays itself off eventually. </p>

<p>2) If one has an incredible high rate of return and a sufficiently low discount rate, one might have divergent consumption paths, ex: I spend all of my money on capital, each unit of capital returns \$2 for the next period. I repeat forever, but never consume anything, because any consumption represents a loss of \$2 tomorrow.  After 1000 periods, I have a huge amount of capital but no consumption.</p>

<p>3) Cobb-Douglass is normalized so the exponents sum to one. This led to some off-the-cuff work. It has derivatives: </p>

<p>$MPK =\alpha_k * (\frac{L}{K})^{1-\alpha_k}=\alpha_k k^{-(1-\alpha_k)}$</p>

<p>$MPL= (1-\alpha_k) (\frac{K}{L})^{\alpha_k} = (1-\alpha_k) k^{\alpha_k}$ </p>

<p>In optimal long-run state, these two will be equal. </p>

<p>$\alpha_k k^{-(1-\alpha_k)} = (1-\alpha_k) k^{\alpha_k}$</p>

<p>$\alpha_k = (1-\alpha_k) k^{\alpha_k --(1-\alpha_k)}$</p>

<p>$\alpha_k = (1-\alpha_k) k^{1}$</p>

<p>$k^{1} =  \frac{\alpha_k}{1-\alpha_k}$</p>

<p>Plug this back in and you get:</p>

<p>$MPL^{*} = \alpha_k \frac{\alpha_k}{1-\alpha_k}^{-(1-\alpha_k)}$</p>

<p>$MPL^{*} = \alpha_k \frac{\alpha_k}{1-\alpha_k}^{\alpha_k-1}$</p>

<p>$MPL^{*} = \alpha_k \frac{1-\alpha_k}{\alpha_k} * \frac{\alpha_k}{1-\alpha_k}^{\alpha_k}$</p>

<p>$MPL^{*} = (1-\alpha_k) * \frac{\alpha_k}{1-\alpha_k}^{\alpha_k}$</p>

<p>$MPL^{*} = (1-\alpha_k) * \frac{\alpha_k^{\alpha_k}}{(1-\alpha_k)^{\alpha_k}}$</p>

<p>$MPL^{*} = \frac{\alpha_k^{\alpha_k}}{(1-\alpha_k)^{\alpha_k - 1}}$</p>

<p>$MPL^{*} = \alpha_k^{\alpha_k}*(1-\alpha_k)^{1-\alpha_k}$</p>

<p>This looks like it's some sort of fraction. A quick screenshot on <a href="https://www.wolframalpha.com/input/?i=y%20%3D%20(x%5Ex)*((1-x)%5E(1-x));%20x%3E0%20%26x%3C%3D1" rel="nofollow noreferrer">wolfram alpha</a> suggests indeed that is the case.</p>

<p><a href="https://i.sstatic.net/JQg48.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/JQg48.png" alt="graph_of_function"></a></p>

CC BY-SA 4.0
2018-06-12T08:13:13.033Z
I think (1) is the most fundamental point.  Output is measured per unit of time, whereas capital is a stock.
Adam Bailey
CC BY-SA 4.0
2018-06-11T20:53:57.610Z
2018-06-11T20:53:57.610Z
RegressForward
<p>Let's assume you have a production function, f,  and you want to know how the output changes with respect to capital, everything else constant (ceteris paribus), so you want to know the marginal productivity or return on capital.</p>

<p>This is done by taking the partial derivative of capital to output.</p>

<p>$$  \frac{\partial y_t}{\partial k_t}  $$</p>

<p>Usually, in most models, the result will be a value between 0 and 1. However, this means that, all other variables being equal, for one extra euro (which comes out of nowhere, I guess, given that all other variables should be the same: ceteris paribus condition), output increases with less than this one euro..., right? I don't get the intuition that this value is less than one, I would it expect to be higher than one, at least. I mean, did 80% of it just vanish?</p>

<p>On wikipedia it is defined as: The marginal product of capital (MPK) is the additional output resulting, ceteris paribus ("all things being equal"), from the use of an additional unit of physical capital. Mathematically, it is the partial derivative of the production function with respect to capital. </p>

<p>I'm having some difficulty with the meaning of "additional" in this definition. It's not additional to this one extra euro of capital out of nowhere, right? This one euro is really transformed in only like e.g. half a euro in output: e.g. capital +1 ==> output + 0.5.</p>

<p>In contrast to this marginal productivity of capital, you can also calculate the direct partial effect of capital on output, which e.g. for a Cobb-Douglas function is given by the elasticity (relative change of output given 1% relative change of capital) parameter for capital (it's power). 
Yet another way to express the influence of a change in capital is the full effect: in some models, also other variables change when capital changes, e.g. labour, technology and so on, this is called the full effect.</p>

<p>I just mention these effects to contrast them with the calculation and interpreatation of the return on capital.</p>

<macroeconomics><microeconomics><production-function><capital-returns><marginal>
intuitive interpretation of the marginal return/productivity of capital being less than one
babipsylon
# intuitive interpretation of the marginal return/productivity of capital being less than one

(Asked by: username_0 on 2018-06-10)

Let's assume you have a production function, f, and you want to know how the output changes with respect to capital, everything else constant (ceteris paribus), so you want to know the marginal productivity or return on capital.

This is done by taking the partial derivative of capital to output.

$$ \frac{\partial y\_t}{\partial k\_t} $$

Usually, in most models, the result will be a value between 0 and 1. However, this means that, all other variables being equal, for one extra euro (which comes out of nowhere, I guess, given that all other variables should be the same: ceteris paribus condition), output increases with less than this one euro..., right? I don't get the intuition that this value is less than one, I would it expect to be higher than one, at least. I mean, did 80% of it just vanish?

On wikipedia it is defined as: The marginal product of capital (MPK) is the additional output resulting, ceteris paribus ("all things being equal"), from the use of an additional unit of physical capital. Mathematically, it is the partial derivative of the production function with respect to capital.

I'm having some difficulty with the meaning of "additional" in this definition. It's not additional to this one extra euro of capital out of nowhere, right? This one euro is really transformed in only like e.g. half a euro in output: e.g. capital +1 ==> output + 0.5.

In contrast to this marginal productivity of capital, you can also calculate the direct partial effect of capital on output, which e.g. for a Cobb-Douglas function is given by the elasticity (relative change of output given 1% relative change of capital) parameter for capital (it's power).
Yet another way to express the influence of a change in capital is the full effect: in some models, also other variables change when capital changes, e.g. labour, technology and so on, this is called the full effect.

I just mention these effects to contrast them with the calculation and interpreatation of the return on capital.

**Tags:** macroeconomics, microeconomics, production-function, capital-returns, marginal

---

## 1 answers

### Answer by username_1 on 2018-06-11. Score: 1

Several thoughts in this area:

1) Capital stock lasts over time. I may trade 1 euro for 1 unit of capital. This capital returns 0.01 euros every period forever. It pays itself off eventually.

2) If one has an incredible high rate of return and a sufficiently low discount rate, one might have divergent consumption paths, ex: I spend all of my money on capital, each unit of capital returns \$2 for the next period. I repeat forever, but never consume anything, because any consumption represents a loss of \$2 tomorrow. After 1000 periods, I have a huge amount of capital but no consumption.

3) Cobb-Douglass is normalized so the exponents sum to one. This led to some off-the-cuff work. It has derivatives:

$MPK =\alpha\_k \* (\frac{L}{K})^{1-\alpha\_k}=\alpha\_k k^{-(1-\alpha\_k)}$

$MPL= (1-\alpha\_k) (\frac{K}{L})^{\alpha\_k} = (1-\alpha\_k) k^{\alpha\_k}$

In optimal long-run state, these two will be equal.

$\alpha\_k k^{-(1-\alpha\_k)} = (1-\alpha\_k) k^{\alpha\_k}$

$\alpha\_k = (1-\alpha\_k) k^{\alpha\_k --(1-\alpha\_k)}$

$\alpha\_k = (1-\alpha\_k) k^{1}$

$k^{1} = \frac{\alpha\_k}{1-\alpha\_k}$

Plug this back in and you get:

$MPL^{\*} = \alpha\_k \frac{\alpha\_k}{1-\alpha\_k}^{-(1-\alpha\_k)}$

$MPL^{\*} = \alpha\_k \frac{\alpha\_k}{1-\alpha\_k}^{\alpha\_k-1}$

$MPL^{\*} = \alpha\_k \frac{1-\alpha\_k}{\alpha\_k} \* \frac{\alpha\_k}{1-\alpha\_k}^{\alpha\_k}$

$MPL^{\*} = (1-\alpha\_k) \* \frac{\alpha\_k}{1-\alpha\_k}^{\alpha\_k}$

$MPL^{\*} = (1-\alpha\_k) \* \frac{\alpha\_k^{\alpha\_k}}{(1-\alpha\_k)^{\alpha\_k}}$

$MPL^{\*} = \frac{\alpha\_k^{\alpha\_k}}{(1-\alpha\_k)^{\alpha\_k - 1}}$

$MPL^{\*} = \alpha\_k^{\alpha\_k}\*(1-\alpha\_k)^{1-\alpha\_k}$

This looks like it's some sort of fraction. A quick screenshot on [wolfram alpha](https://www.wolframalpha.com/input/?i=y%20%3D%20(x%5Ex)*((1-x)%5E(1-x));%20x%3E0%20%26x%3C%3D1) suggests indeed that is the case.

[![graph_of_function](https://i.sstatic.net/JQg48.png)](https://i.sstatic.net/JQg48.png)

> **Comments:**
> **username_2**: I think (1) is the most fundamental point.  Output is measured per unit of time, whereas capital is a stock.  

---

