Show HN: Markov chains explained visually
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Re: Show HN: Markov chains explained visually
#92Re: Show HN: Markov chains explained visually
#93The one thing to add to this is that usually each state doesn't emit a single token ("I am in state 1" then "I am in state 2") but instead you assume that each state has a range of possible actions and the likelihood of a choice of action varies with state. So if might not be that your model is sunny versus rainy but instead cold front v warm front. Since rain is more likely during a cold front your observation of ra…
Re: Show HN: Markov chains explained visually
#94But this tutorial, both visually attractive and expained with real life examples, make me want to re-learn this topic. Just quote one passage:
[if you made a Markov chain model of a baby's behavior, you might include "playing," "eating", "sleeping," and "crying" as states, which together with other behaviors could form a 'state space': a list of all possible states.]
Thanks for sharing!
Re: Show HN: Markov chains explained visually
#95Earlier quoted context omitted.
Why isn't your example Markov?
It is in fact Markov; Markov just means that the probability distribution of the future depends only on the present, and so the past adds no additional information in conjunction with the present. That's certainly the case here. This is an example of a Markov chain that is not aperiodic; what that means is that, given a starting node, at any point in time in the future, it will always be the case that it is impossibl…
Re: Show HN: Markov chains explained visually
#96Earlier quoted context omitted.
It is in fact Markov; Markov just means that the probability distribution of the future depends only on the present, and so the past adds no additional information in conjunction with the present. That's certainly the case here. This is an example of a Markov chain that is not aperiodic; what that means is that, given a starting node, at any point in time in the future, it will always be the case that it is impossibl…
fix some n in n. if my starting state is [1; 0] then the probability of the occupancy being [1; 0] after n cycles is either 1 or 0. If the starting state is [0; 1] then the probabilty of the occupancy being [1; 0] is exactly the opposite, so for a fixed point in the future, the probability is tightly past-dependent.