Shape of indifference curves and quasi-concavity of utility

So my professor told that quasi-concave utilities lead to convex preferences/indifference curves. I have some conceptual problems understanding this statement.

1. Indifference curves are plotted from utility functions and they are convex, so how can the function be quasi-concave and the plotted curve be convex?
2. Quasi-concavity implies concavity so I can just prove utility is concave to prove ID curves are convex? Concavity can be proven through a Hessian matrix of second derivatives right?
3. I understand the difference between concavity vs strictly concave. What is the meaning of quasi-concave in a simpler language?

Thank you so much. I understand that these doubts are lame but I want to understand the concept correctly.

• You can watch this playlist: youtube.com/playlist?list=PLUJGfL_499TLYEd-9IO1DmQKwm6Mtdt4u
– Amit
Oct 13, 2022 at 11:34
• Thank you @Amit, your videos are a hidden gem. I hope more people discover them and benefit from your amazing content. I started watching 8 hours ago and have constantly been thinking about them. Oct 13, 2022 at 20:02
• Thank you @EconNoob
– Amit
Oct 14, 2022 at 10:14

May be, all you need is in the videos on Youtube Amit quoted above, but I would add some concise observations, mainly through examples, that I hope could be useful to you and other contributors.

Let's begin from Point 3 of your question.

Point 3.

To have a simple and intuitive grasp of the concept of quasi-concavity, it could be convenient to see the symmetric concept of quasi-convexity for functions from $$\mathbb{R}$$ to $$\mathbb{R}$$.

(Using quasi concavity would be the same, as quasi-concavity is quasi-convexity with the minus sign before. I use quasi-convexity because I think that pictures and formulas are clearer).

The formal definition is: a function $$f$$ from $$\mathbb{R}$$ to $$\mathbb{R}$$ is quasi-convex if, $$\forall \alpha \in \mathbb{R}$$, the set

$$\{x\in D: f(x)<\alpha\},$$

where $$D$$ is the domani of $$f$$, is convex.

In the picture below there is an example of a non quasi-convex function and of a quasi-convex function which is not convex.

In the first picture we have a non quasi-convex function, $$f(x)= x^2(x^2-2)$$: the set of points for which $$f(x)$$ is under the green line is the union of the two red intervals, and is not convex.

The second picture shows the function $$f(x)=\sqrt|x|$$, which is quasi-convex, but not convex.

Point 1.

how can the function be quasi-concave and the plotted curve be convex?

Remember that indifference curves are level curves of a utility function. A quasi-concave function can ,of course, have convex level curves.

Let's see an example, of a function from $$\mathbb{R^2}$$ to $$\mathbb{R}$$.

Consider a Cobb-Douglas utility function, $$U(x,y)= kx^\alpha y ^{1-\alpha}$$, $$x,y\geq0$$, $$k \in R_+$$ and $$\alpha \in (0,1)$$.

The Cobb Douglas is strictly concave, so it is concave, so it is quasi-concave. The indifference curves are convex.

You can have an intuitive idea from the following picture, where the graphs of the Cobb Douglas in three dimensions, and of its level curves (indifference curves) are represented.

If you calculate analytically the level curves of the Cobb-Douglas, you can see that they are convex.

For example, for $$k=1$$ and $$\alpha=1/2$$, we have the hyperbolas

$$y=c^2/x,$$ $$c\in R$$, for $$x>0$$.

Point 2.

Quasi-concavity implies concavity so I can just prove utility is concave to prove ID curves are convex? Concavity can be proven through a Hessian matrix of second derivatives right?

I think you made a typo, the reverse of what you wrote is true: concavity implies quasi-concavity (quasi-concavity is a weaker property), so it is sufficient to prove concavity.

Concavity can be proved through second derivatives for functions from $$\mathbb{R}$$ to $$\mathbb{R}$$, or through the Hessian matrix, for functions from $$\mathbb{R^n}$$ to $$\mathbb{R}$$, but provided that the utility function is differentiable as needed.

For example, one can’t prove the convexity of $$f(x))= |x|$$ using the derivatives, because this function is not differentiable in all $$\mathbb{R}$$ ( it is not differentiable in $$0$$).

• Honestly 50% of the time I read a great answer on this site, I scroll to the end and it's you @BakerStreet! I have a question related to this one here if you're interested: economics.stackexchange.com/questions/54438/… Feb 14 at 17:49
• Also in your first diagram surely its the humped part of the curve above the green line, between the two read lines that fails to be convex and is actually a concave interval? Feb 14 at 17:50
• Thank you very much @CormJack! You are my advertising agent :-). You are very kind. If you consider that part alone, it is a part of the function that is concave. This part of the function is concave, not the interval on the abscissa, of course. An interval is a convex set. Feb 14 at 18:56
• Hahhaha yeah we'll start a campaign to vote for you as a moderator! So just to clarify then, the hump above the green line between 1/2 and -1/2 is concave...And then the function from -1/2 to - infinity, and 1/2 to infinity is convex? I only ask because I think you wrote the opposite in your post, or am I misreading your quote: "In the first picture we have a non quasi-convex function....the set of points for which 𝑓(𝑥)...is under the green line is the union of the two red intervals, and is not convex." Feb 14 at 19:11
• You have to considre the function as a whole, to say if it is convex and so on. If you take a single part it is different. In the phrase you quote I say that the union of the two red intervals is not convex, meaning that it is not a convex set. Don't confuse convex set with convex function, they are different, even if related, concepts. Feb 14 at 19:54