Timeline for Median wealth after repeated iterations of multiplicative game?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 8 at 15:53 | audit | First answers | |||
| Nov 8 at 16:01 | |||||
| Oct 30 at 14:20 | audit | First answers | |||
| Oct 30 at 15:10 | |||||
| Oct 28 at 3:50 | comment | added | btilly | @AndrésMejía You're welcome! | |
| Oct 27 at 20:44 | vote | accept | Andrés Mejía | ||
| Oct 27 at 20:43 | comment | added | Andrés Mejía | Amazing! Thank you very much for this detailed answer. This was extremely helpful and fun! | |
| Oct 27 at 2:18 | comment | added | btilly | @Eric I hadn't thought of that, because my brain was thinking in logs.. I went to irrational probabilities first. Regardless of how you get there, discretization errors can be $O(1)$, | |
| Oct 27 at 1:59 | comment | added | Eric | Ehh, but then it’s still $(0.5+2)/2=1.25\neq 1$. If you really want the geometric mean to also be $\neq 1$, you could do $1/3$ of the time $1/4$ and $2/3$ times $2$ which when $n$ is $1 \mod 3$ give either $\leq 1/4$ or $\geq 2$ | |
| Oct 27 at 1:33 | comment | added | btilly | @Eric That's the one I thought of first, but many people define the median of an even number of things as the average of the middle 2. Which allows the median to land at 1 again. So I had to look for an example where the median couldn't be split that way. | |
| Oct 27 at 1:24 | comment | added | Eric | A simpler counterexample is just flipping a coin and either halving or doubling. Then, when $n$ is odd, $W_n$ is either $\leq 1/2$ or $\geq 2$, so the median will forever be far from $1$ for odd $n$. | |
| Oct 27 at 1:20 | history | edited | Eric | CC BY-SA 4.0 | Style cleanup |
| Oct 27 at 0:39 | history | answered | btilly | CC BY-SA 4.0 |