I’m having a hard time trying to prove that a univariate function is (strictly) quasi-concave (sorry for the long post).

I think the main reason for my struggle is:

(1) that the function is quite complex (not complicated but very large) and

(2) That I can’t find a good example where someone is actually going through a proof of quasi-concavity (except for something like $f(x,y)=x^2+y^2$).

Therefore even an actual example on how to prove quasi-concavity would be quite useful for me.

In the next few lines I will present the function and its properties.

I have a function $f(x)$ that consists of three linear functions $g(x), h(x), k(x)$ and that looks like this: $$f(x)= \sum_{i=1}^n g(x)(1-h(x))^i \left ( {(1+k(x)^n -(1+k(x))^{i-1} \over (1+k(x))^n-1} k(x)\right ) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -m\sum_{i=1}^ng(x)(1-h(x))^{i-1} \left ( {(1+k(x)^n \ -(1+k(x))^{i-1} \over (1+k(x))^n-1}\right )~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-\sum_{i=1}^ng(x)h(x)(1-h(x))^{i-1} \left ( {(1+k(x)^n -(1+k(x))^{i-1} \over (1+k(x))^n-1} \right )~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$ with $n \ge1, n \in \mathbb{N}$ and $x \in \mathbb{R}^{+}$, as well as $f(x)\ge 0$.

All the sub functions of $f(x)$ are linear functions with the following properties (one can also assume that they are continuous and differentiable): $$g(x)>0; ~ g'(x)>0; g''(x)=0$$ $$0<h(x)<1; ~h('x)>0; h''(x)=0$$ $$ 0<k(x)<1; ~k'(x)<0; k''(x)=0$$

Using some properties of the sub functions one can see that:

$0<(1-h(x))^i<1 $ and because of $0<h(x)<1$ the term $0<(1-h(x))^i<1 $ is always decreasing.

It is also clear that $( {(1+k(x)^n -(1+k(x))^{i-1} \over (1+k(x))^n-1} )$ is decreasing with an increase in $x$ which is also true for $( {(1+k(x)^n -(1+k(x))^{i-1} \over (1+k(x))^n-1} )k(x)$ using the product rule.

Apart from that all of the three expressions after the sigma signs are positive (with this I mean that I'm only interested in positive values). Using this I want to show that $f(x)$ is either decreasing over $x$ or is inverted U-shaped and therefore is strictly quasi concave.

In the next part I will give my impression on $f(x)$ with two ideas on how to proof the statement.

Looking at strictly quasi concavity there are two statements that I think could be useful in proving that $f(x)$ actually is strictly quasi concave:

(1) $ f(\lambda x_1+(1-\lambda)x_2) > min {f(x_1);f(x_2)} $ with $0<\lambda<1$


(2) $f'(x_2)(x_1-x_2)>0$ for $f(x_1)>f(x_2)$

One idea I had is to proof the statement by induction:

  1. Use $n=1$ and show that in this case the function is quasi-concave
  2. Assume that for n=l the function is quasi-concave
  3. Proof that the function is quasi-concave for l+1 using the hypothesis from 2.

Step 1: For n=1 $f(x)$ is simply:


Using the statement (1) for strict quasi-concavity one can see that this is true because of the linearity of $g(x)$, $h(x)$, and $k(x)$ (But this is again a point where an example of someone actually doing something similarly would be nice).

While step 2 is just $f(x)$ with $l$ I have some trouble going through step 3.

Another solution might be to use the linear properties of $g(x), h(x)$ and $k(x)$. The proof goes something like this: Because $g(x)$ is linear $g(\lambda x_1 + (1-\lambda) x_2) > min (g(x_1); g(x_2))$ is always true. Apart from that it is clear that because of $(1-h(x))$ $f(x)$ can not be increasing all the time is has to decrease at a certain point. The same statement can be made about $h(x)$ as well as $k(x)$.

Therefore the first expression $$\sum_{i=1}^n g(x)(1-h(x))^i \left ( {(1+k(x)^n -(1+k(x))^{i-1} \over (1+k(x))^n-1} k(x)\right )$$ has the same property of being bigger than the $min f(x_1), f(x_2)$ and is therefore strictly quasi concave.
But I don't think this is a definite proof. Apart from that I have some trouble trying to wrap my head about the connection between the three expressions (what is if they are all three strictly quasi concave? Does this also mean the the sum of them is quasi concave?)

Sorry for the long and messy post. Any help is appreciated.



Your Answer

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

Browse other questions tagged or ask your own question.