<p>I am not from the academic economics community, but from what I have read Peters (2019) seems to be just a rediscovery of Jensen's inequality, with possibly some more nefarious motives to advance his agenda.</p>
<p>I am looking for a proof for the following statement:</p>
<p><em>Let <span class="math-container">$r$</span> be the risk free rate, <span class="math-container">$\mathbb E(X)$</span> denote the ensemble average, and <span class="math-container">$f^*$</span> the optimal fraction of wealth to bet that maximizes time average. Then we have:</em>
<span class="math-container">$$f^*\begin{cases} 
      =0 &amp; \mathbb E(X)\leq r \\
      &gt;0 &amp; \mathbb E(X)&gt; r
   \end{cases}$$</span>
The <span class="math-container">$\mathbb E(X)\leq r$</span> case follows directly from the definitions and AM-GM, though I am struggling to prove the latter.</p>

CC BY-SA 4.0
2024-12-31T04:24:03.200Z
Hi: If the expected value ( on a percent basis ) is less than the risk free rate, then it would not make sense ( for anyone using a reasonable utility function ) to enter the bet because you can expect to make more by putting the money in the bank. In that case, any bet of any positive amount is not on the bettor's efficient frontier, so that's why $f^{*} = 0$ if $E(x) \le r $.
mark leeds
CC BY-SA 4.0
2024-12-31T04:28:11.303Z
Note that AM-GM says that the arithmetic mean of a set of numbers ( returns in this case I assume )  is always greater than the geometric mean but this is not relevant to your question.
mark leeds
CC BY-SA 4.0
2024-12-31T14:31:16.747Z
@markleeds That is a heuristic, not a valid mathematical proof. How would we know if time average is a "reasonable utility function"? Note that time average is calculated as E(\lim_{n\to\infty}(\prod_i^n r(1-f)+fX_i)^{1/n}). Applying AM-GM to this definition yields our first result.
psubodiosa
CC BY-SA 4.0
2024-12-31T14:34:23.700Z
@markleeds I am asking this question because Peters is claiming all economists make the mistake that assuming ergodicity, or that the ensemble average, E(X), is always equal to the time average.
psubodiosa
<expected-utility>
Peters, ergodicity, and optimal betting
psubodiosa
# Peters, ergodicity, and optimal betting

(Asked by: username_0 on 2024-12-27)

I am not from the academic economics community, but from what I have read Peters (2019) seems to be just a rediscovery of Jensen's inequality, with possibly some more nefarious motives to advance his agenda.

I am looking for a proof for the following statement:

*Let $r$ be the risk free rate, $\mathbb E(X)$ denote the ensemble average, and $f^\*$ the optimal fraction of wealth to bet that maximizes time average. Then we have:*
$$f^\*\begin{cases}
=0 & \mathbb E(X)\leq r \\
>0 & \mathbb E(X)> r
\end{cases}$$
The $\mathbb E(X)\leq r$ case follows directly from the definitions and AM-GM, though I am struggling to prove the latter.

**Tags:** expected-utility
> **Comments:**
> **username_1**: Hi: If the expected value ( on a percent basis ) is less than the risk free rate, then it would not make sense ( for anyone using a reasonable utility function ) to enter the bet because you can expect to make more by putting the money in the bank. In that case, any bet of any positive amount is not on the bettor's efficient frontier, so that's why $f^{*} = 0$ if $E(x) \le r $.  
> **username_1**: Note that AM-GM says that the arithmetic mean of a set of numbers ( returns in this case I assume )  is always greater than the geometric mean but this is not relevant to your question.  
> **username_0**: @markleeds That is a heuristic, not a valid mathematical proof. How would we know if time average is a "reasonable utility function"? Note that time average is calculated as E(\lim_{n\to\infty}(\prod_i^n r(1-f)+fX_i)^{1/n}). Applying AM-GM to this definition yields our first result.  
> **username_0**: @markleeds I am asking this question because Peters is claiming all economists make the mistake that assuming ergodicity, or that the ensemble average, E(X), is always equal to the time average.  

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## 0 answers

