Comment by storus
8 hours ago
It's not really Markov chain as you need full P(x_t|x_{t-1}, x_{t-2}... x_1) instead of just P(x_t|x_{t-1}).
8 hours ago
It's not really Markov chain as you need full P(x_t|x_{t-1}, x_{t-2}... x_1) instead of just P(x_t|x_{t-1}).
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