markov-realization alternatives and similar packages
Based on the "Statistics" category.
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Let Xₙ denote the nth state of a Markov chain with state space ℕ. For x ≠ 0 define transition probabilities
p(x,0) = q,
p(x,x) = r, and
p(x,x+1) = s.
When x = 0, let
p(x,0) = q+r,
p(x,x+1) = s.
Let p(x,y) = 0 in all other cases.
Suppose we wanted to find
P[Xₙ = j and d = k],
where d denotes the number of transitions from a positive integer to zero.
There are three values we need to track —
extinctions, probability, and state.
Extinctions add a value to a counter each time they happen
and the counter takes integral values,
so they can be represented by
Probabilities are multiplied each step,
and added when duplicate steps are combined.
We want decimal probabilities, so
we can represent this with
We will make a new type for the state.
newtype Extinction = Extinction Int deriving newtype (Eq, Num, Ord, Show)
Combining identical states should not change the state,
so we make an instance of
Combine as follows.
instance Combine Extinction where combine = const
All that remains is to make an instance of
instance Markov ((,) (Sum Int, Product Rational)) Extinction where transition = \case 0 -> [ 0 >*< (q+r) >*< id , 0 >*< s >*< (+1) ] _ -> [ 1 >*< q >*< const 0 , 0 >*< r >*< id , 0 >*< s >*< (+1) ] where q = 0.1; r = 0.3; s = 0.6
We can now see a list of states, extinctions, and the probabilities.
> chain [pure 0 :: Sum Int :* Product Rational :* Extinction] !! 3
[ ((0,8 % 125),0) , ((0,111 % 500),1) , ((1,51 % 500),0) , ((0,9 % 25),2) , ((1,9 % 250),1) , ((0,27 % 125),3) ]
This means that starting from a state of zero, after three time steps there is a 51/500 chance that the state is zero and there has been one extinction.