Looking for a particular function

[Frank Einstein]

TL;DR Summary: I want to find a function with f'>0, f''<0 and takes the values 2, 2^2, 2^3, 2^4,..., 2^n

Hello everyone.

A professor explained the St. Petersburgh paradox in class and the concept of utility function U used to explain why someone won't play a betting game with an infinite expected value.

Then he talked about Kar Meyer's finding of a bounded utility function and still infinite payoff and told us to find a bounded function with the following properties:

f'>0, f''<0 and takes the values 2, 2^2, 2^3, 2^4, ..., 2^n

The values 2, 2^2, 2^3, 2^4, ..., 2^n have to be taken for x=1,2,3,4,..., n.

I have been thinking and reading about this but I have found no answer. I have even read Meyer's article and there he says that if the amount gained per bet is exp(2^n) then the expected value of the logarithm of the gain is infinite.

Thus, I think that either my professor has made a mistake or is trolling my class. However, before writing an email telling him that what he asks is impossible I would like to see if someone here agrees or disagrees with me.

Any answer is appreciated.
Best regards.

 

[Gaussian97]

Well, there are a few assumptions one needs to make about $f$. What I understand you mean is:
$f: [1,\infty) \to \mathbb{R}$ is a function with the property $f(n)=2^n$. Furthermore $f'(x)>0$ and $f''(x)<0$ in all the domain.
Then, $f$ is continuous with a continuous derivative and we can use the Mean Value Theorem:
Lets choose $a_0=1, a_1=2, a_3=3$, applying the MVT twice we get
$$f'(b_0) = \frac{f(a_1)-f(a_0)}{a_1-a_0} = \frac{f(2)-f(1)}{2-1} = 2^2 - 2^1 = 2$$
$$f'(b_1) = \frac{f(a_2)-f(a_1)}{a_2-a_1} = \frac{f(3)-f(2)}{3-2} = 2^3 - 2^2 = 4$$
with $b_0\in (1,2)$ and $b_1 \in (2,3)$. Now, whatever these values are we can write
$$f''(c) = \frac{f'(b_1)- f'(b_0)}{b_1-b_0} = \frac{2}{b_1-b_0} > 0$$
which is a contradiction to the satatement $f''(x)<0$.

Now, notice that if the conditions $f'>0$ and $f''<0$ are not applied to the whole domain, and (for example) are only valid for the natural numbers, then the answer is trivial since you can construct functions $f_n(x)$ with the conditions $f(n)=2^n, f'(n)=1, f''(n)=-1$. Then f(x) can be defined as a piecewise function and will fulfil all the requirements.

 

[Frank Einstein]

Thanks for the answer. I will try to understand it.

However, could you please explain to me how to construct the function $f$?

Best regards.