User:IssaRice/Ergodicity: Difference between revisions

Revision as of 22:56, 12 February 2020

Here's my suggested path to learning about this, assuming you just know basic probability and statistics and calculus but you know nothing about ergodicity:

read http://squidarth.com/math/2018/11/27/ergodicity.html and http://squidarth.com/math/2019/04/13/ergodicity-animated.html
- Really understand the 1.5x/0.6x game
- Try to formalize the game using random variables
- Come up with a formal expression for the time-average winnings vs ensemble winnings. I think you come up with something like $\sum_{k = 0}^{n} (\binom{n}{k}) 1 . 5^{k} 0 . 6^{n - k} \frac{1}{2^{n}}$
- figure out what percentage of flips need to be heads in order for you to have positive winnings. you want $1 . 5^{k} 0 . 6^{n - k} > 1$ so the break-even value of k is $k = - n \log 0.6 / \log (1.5 / 0.6)$ so the fraction of heads is $k / n \approx 0.557$ . Understand that by law of large numbers, your fraction of heads will be super close to 0.5 in the long run, so you will almost never get anywhere close to having 56% of your flips be heads, so you almost certainly lose.
read https://ergodicityeconomics.files.wordpress.com/2018/06/ergodicity_economics.pdf sections 1.1, 1.2, 2.1, 2.2, 2.3, 2.4, 2.5. something like that.
read https://royalsocietypublishing.org/doi/pdf/10.1098/rsta.2011.0065
read https://www.greaterwrong.com/posts/gptXmhJxFiEwuPN98/meetup-notes-ole-peters-on-ergodicity
- don't miss comments by abram demski and me

@@ Line 1: / Line 1: @@
+Here's my suggested path to learning about this, assuming you just know basic probability and statistics and calculus but you know nothing about ergodicity:
+* read http://squidarth.com/math/2018/11/27/ergodicity.html and http://squidarth.com/math/2019/04/13/ergodicity-animated.html
+** Really understand the 1.5x/0.6x game
+** Try to formalize the game using random variables
+** Come up with a formal expression for the time-average winnings vs ensemble winnings. I think you come up with something like <math>\sum_{k=0}^n \binom nk 1.5^k 0.6^{n-k} \frac1{2^n}</math>
+** figure out what percentage of flips need to be heads in order for you to have positive winnings. you want <math>1.5^k 0.6^{n-k} > 1</math> so the break-even value of k is <math>k = -n\log0.6/\log(1.5/0.6)</math> so the fraction of heads is <math>k/n \approx 0.557</math>. Understand that by law of large numbers, your fraction of heads will be super close to 0.5 in the long run, so you will almost never get anywhere close to having 56% of your flips be heads, so you almost certainly lose.
+* read https://ergodicityeconomics.files.wordpress.com/2018/06/ergodicity_economics.pdf sections 1.1, 1.2, 2.1, 2.2, 2.3, 2.4, 2.5. something like that.
+* read https://royalsocietypublishing.org/doi/pdf/10.1098/rsta.2011.0065
+* read https://www.greaterwrong.com/posts/gptXmhJxFiEwuPN98/meetup-notes-ole-peters-on-ergodicity
+** don't miss comments by abram demski and me
 ==See also==
 * https://machinelearning.subwiki.org/wiki/User:IssaRice/Multiplicative_process