1
$\begingroup$

This example appears in a different question, but there is something I don't understand. Maybe this question is better suitated for algebra stackexchange.

John’s utility function for food (f) and clothes (c) is given by
$$U(f, c) = (f^\alpha + c^\alpha)^{1/\alpha}$$

Does this function satisfy diminising MRS?
The answer is: no, because if α >1 the graph curve is concave.

I perfectly understand the theory, but how do I actually draw the curve from the given function to test the answer? Algebrically/geometrically/visually what is the given function (f,c) = α(fα+cα)$\frac{1}{α}$ ? Is it actually the equation of the curve? Why does it look so weird? And what is α ? Where would something like α come into play when measuring someone's utility?

$\endgroup$
0
$\begingroup$

Doing this Analytically

Analytically you just need to examine the second derivatives of a function. Following Sydsaeter et al FEMA pp. 466 a multivariate $C^2$ function of two variables $f(x,y)$ is concave when:

$$f^{''}_{11}(x, y) \leq 0, f^{''}_{22}(x, y) \leq 0, \text{ and } f^{''}_{11}(x, y)f^{''}_{22}(x, y) − f^{''}_{12}(x, y)^2 \geq 0$$

So analytically you can just calculate the above. In your case the function is $U(f,c)$ but otherwise you are just applying the above.

If you want to know if the function is concave everywhere then you have to inspect if the concavity is satisfied for all parameter and variable restrictions. In this case the function will be concave everywhere assuming $f>0, c>0$. In economics, we typically assume variables have to be positive or at least non-negative (you cannot consume -1 units of bread). If you do the above you will see that for $f>0; c>0$ the function is concave everywhere, but note if you do not impose any restriction of $f$ and $c$ the function would actually become convex for $f<0;c<0$.

Graphically

You just need to plot the function and look at its shape. The concave shape in 3D will look like cupola of a dome, whereas convex shape looks like the bowl (at least that is how I like to think about it in 3D).

Below you can see the plot for your function without any restrictions made in Wolfram Alpha. You need to do that for some particular parameters here I have chosen $\alpha = 0.5$ but the shape would be similar for any $0<\alpha<1$. As you can see for $f>0$ and $c>0$ the function is clearly concave, and this is what's referenced in the answer (again in economics typically we restrict variables to be positive unless stated otherwise), but you can also see that if you do not make any restrictions on $f$ and $c$ function is convex in III quadrant.

enter image description here

$\endgroup$
2
  • $\begingroup$ Thanks! I am not familiar with the analytical theory you mention, but the graphic explanation is amazing. I have tried to used wolfram before with mixed results. Since I don't know the correct notation and formatting, could you tell me your input to obtain that graph? $\endgroup$ – Janpert Jun 12 at 14:26
  • 1
    $\begingroup$ @Janpert you are welcome. I just used plot (f^0.5+c^0.5)^(1/0.5) . Also, if you think this answer answered your question consider accepting it. $\endgroup$ – 1muflon1 Jun 12 at 15:29
2
$\begingroup$

This is an expansion on @1muflon1's answer.

Motivation and Applications

In the utility function, $\alpha$ allows for an alternative to a linear growth function, especially when the parameters are close to zero. This has practical applications; for instance, the example function you have above is a special case of the CES utility function.

Analytical Determination

To determine convexity of a multivariable function, we need to consider its Hessian. For this example, the utility function is bivariate so the Hessian is given by $$H_{U(f,c)}=\begin{bmatrix}U_{ff}&U_{fc}\\U_{cf}&U_{cc}\end{bmatrix}=\begin{bmatrix}U_{ff}&U_{fc}\\U_{fc}&U_{ff}\end{bmatrix}$$ since $U(f,c)=U(c,f)$. Convexity follows from its positive semi-definiteness; that is, if both $$\det U_{ff}=U_{ff}=(\alpha-1)f^{\alpha-2}c^\alpha(f^\alpha+c^\alpha)^{1/\alpha-2}\ge0$$ and $$\det\begin{bmatrix}U_{ff}&U_{fc}\\U_{fc}&U_{ff}\end{bmatrix}=U_{ff}^2-U_{fc}^2=(\alpha-1)^2f^{2\alpha-4}c^{2\alpha-2}(c^2-f^2)(f^\alpha+c^\alpha)^{2/\alpha-4}\ge0.$$ This is satisfied if both $\alpha\ge1$ and $c\ge f$, but because $U$ is symmetric, the last criterion is redundant! Therefore, $U(f,c)$ is convex for all non-negative $f,c$ (as should be the case in real life) if and only if $\alpha\ge1$.

Here is a shortcut for those mathematically inclined: note that in this special case $U$ is a homogeneous function of order $1$. From Altenberg (2012)1, it is sufficient to show that $U$ is convex in just one variable.

Graphical Visualisation

The Mathematica plot provided above is great, as the contour lines evidently show convexity/concavity at a given set of coordinates. For those who don't have a Mathematica subscription and wish to interact with the contour plot, I have made a quick GeoGebra plot that is shared publicly.

Note that there is a lack of contours on the other three quadrants, as $U(f,c)$ is not defined when $\alpha$ is not an integer.


Reference

[1] Altenberg, L. (2012). Resolvent positive linear operators exhibit the reduction phenomenon. Proceedings of the National Academy of Sciences. 109(10):3705-3710.

$\endgroup$
2
  • 1
    $\begingroup$ +1 but please in the future don’t just give out answer to user who did not showed attempt at solution in question, I withheld full answer on purpose. On this site we give out full solutions only to the questions that show attempt at solution. But I see you are new here so no harm done just in future try to avoid that please $\endgroup$ – 1muflon1 Jun 12 at 21:33
  • $\begingroup$ Wow thanks to both. I am sorry it appears I haven't done my homework, but has been a while since I worked on derivates. Now I don't know who to pick as best answer, you have both been so helpful! About the graph, is this how I am supposed to look at it? Graph $\endgroup$ – Janpert Jun 16 at 0:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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