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README
Markov Tutorial
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 Sum Int
.
Probabilities are multiplied each step,
and added when duplicate steps are combined.
We want decimal probabilities, so
we can represent this with Product Rational
.
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 Markov
.
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.