3 ms·
Although these notions are standard language, I still agree with you that the author made it a bit more complicated than it needs to be. Ergodicity for a stocha
by nonameiguess 6y ago
Although these notions are standard language, I still agree with you that the author made it a bit more complicated than it needs to be. Ergodicity for a stochastic process just means the joint distribution of random variables that make up the sample space is time invariant.
For a coin toss example like this, the distribution of heads and tails in each trial is ergodic. The distribution of earnings is not. This isn't because of any difference between time average versus ensemble average. It's because the probability of winning each toss is time invariant but the amount you stand to win or lose isn't because it's a function of both the probability of winning and your current bankroll, and current bankroll is not time invariant.
Although, ironically, because of the numbers he picked, all bankrolls tend to zero eventually, so over a large enough number of trials, wealth eventually becomes an ergodic process as well. Graphing out his scenario over more trials gives a sort of heat death of the universe plot, where some players stay alive longer than others, but in the long run, the enemy always wins.
- coldtea 6y ago>I still agree with you that the author made it a bit more complicated than it needs to be. Ergodicity for a stochastic process just means the joint distribution of random variables that make up the sample space is time invariant. It's funny how you say "the author made it a bit more complicated than it needs to be" and then proceed to explain it with even more jargon ("Ergodicity for a stochastic process just means the joint distribution of random variables that make up the sample space is time invariant").