Looking for a particular function

In summary, the professor is discussing a function that has the following properties: it is continuous, has a continuous derivative, and takes the values 2, 2^2, 2^3, 2^4,..., 2^n. He asks for someone to help him construct the function.
  • #1
Frank Einstein
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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.
 
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  • #2
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.
 
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  • #3
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.
 

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