Hey team,
for clarity, I have a question regarding the Tutorial of W2D4 (Telegraph processes), the fishing project:
On slide #33, Xaq states that when prob(stay) = 0.9, the school of fish should have changed sides after latest 10 time steps.
Do I understand it correctly that all time steps are dependent on all others before it? In terms of the fish example: only in the first time step after the fish are on a “new” side is there a prob(stay) = .9 and then the fish get bored over time and are more likely to switch.
Thanks in advance.
Marco
Hi Marco,
In every time step there’s a probability of 0.9 that fish will stay, but each time step t is only dependent on the time step t-1, not the others before it. This is the Markov property: at every step in time fish have a 90% chance of staying and a 10% of switching, and the timesteps before t-1 do not factor in the decision. What he means on slide #33 is that a p(stay) = 0.9 defines a time scale of 10 timesteps in which you expect to lose information about where fish are, and this timescale is much longer if p(stay) is closer to 1.
Here’s another way to think about this: if you simulated this process many, many times, having a p(stay) = 0.9 would mean your ability to guess where fish are on average would deteriorate after about 10 steps, but having a p(stay) = 0.999 means you can still make pretty good guesses at just 10 timesteps if you knew where the fish were in the first timestep.