**Monthly Downloads**: 15

**Programming language**: Haskell

**License**: BSD 3-clause "New" or "Revised" License

**Tags**: Statistics

**Latest version**: v0.4

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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.