In [1]:

    

from sympy import *; init_session()
from applpy import *
import numpy as np


    

    




IPython console for SymPy 0.7.7.dev (Python 2.7.10-64-bit) (ground types: python)

These commands were executed:
>>> from __future__ import division
>>> from sympy import *
>>> x, y, z, t = symbols('x y z t')
>>> k, m, n = symbols('k m n', integer=True)
>>> f, g, h = symbols('f g h', cls=Function)
>>> init_printing()

Documentation can be found at http://docs.sympy.org/dev

In [2]:

    

P = np.array([
        [1,0,0,0],
        [Rational(3,10),Rational(4,10),Rational(1,10),Rational(2,10)],
        [0,0,1,0],
        [Rational(2,10),Rational(3,10),Rational(4,10),Rational(1,10)]
    ])


    

In [3]:

    

X = MarkovChain(P, states = ['red','blue','black','green'])


    

In [4]:

    

abs_red = X.absorption_prob('red')
vector_display(abs_red,X.state_space)


    

    
Out[4]:






  

    

      

      
Prob
    
  
  

    

      
red
      
1
    
    

      
blue
      
31/48
    
    

      
black
      
0
    
    

      
green
      
7/16
    
  

In [5]:

    

abs_steps = X.absorption_steps()
vector_display(abs_steps, X.state_space)


    

    
Out[5]:






  

    

      

      
Prob
    
  
  

    

      
red
      
0
    
    

      
blue
      
55/24
    
    

      
black
      
0
    
    

      
green
      
15/8
    
  

In [2]:

    

P2 = np.array([
        [Rational(2,10),0,0,Rational(8,10)],
        [0,1,0,0],
        [Rational(1,10),Rational(4,10),Rational(2,10),Rational(3,10)],
        [Rational(7,10),0,0,Rational(3,10)]         
    ])
X = MarkovChain(P2, states = ['red','blue','black','green'])


    

In [3]:

    

Pi = X.long_run_probs(method = 'rational')
matrix_display(Pi,X.state_space)


    

    




[7/15 8/15]
1/2






    
Out[3]:






  

    

      

      
red
      
blue
      
black
      
green
    
  
  

    

      
red
      
7/15
      
0
      
0
      
8/15
    
    

      
blue
      
0
      
1
      
0
      
0
    
    

      
black
      
7/30
      
1/2
      
0
      
4/15
    
    

      
green
      
7/15
      
0
      
0
      
8/15
    
  

In [9]:

    

P2[0][0]


    

    
Out[9]:





$$\frac{1}{5}$$

In [ ]: