## Related calculators

If you find any bug or need any improvements in solution report it here

# Absorbing markov chain calculator

This calculator is for finding the expected number of steps or time for absorbing starting from each of the transient states. This matrix describes the transitions of a Markov chain. This matric is also called as probability matrix, transition matrix, etc.

Enter the Markov chain stochastic matrix Use ',' to separate between values. Use newline for new row: 0.5, 0.5 ,0 ,0 0, 0.5, 0.5, 0 0.5, 0, 0, 0.5 0, 0, 0 ,1

## You can see a sample solution below. Enter your data to get the solution for your question

$$\displaylines{---}$$
$$\displaylines{ \mathbf{\color{Green}{Let's\;call\;All\;possible\;states\;as\;}} \begin{bmatrix} 1 & 2 & 3 & 4 & \end{bmatrix} \\ \\ \mathbf{\color{Green}{First\;we\;have\;to\;create\;Stochastic\;matrix}} \\ \\ Stochastic\;matrix\;=\;P= \begin{bmatrix} * & 1 & 2 & 3 & 4 & \\ \\ 1 & 0.5 & 0.5 & 0.0 & 0.0 & \\ \\ 2 & 0.0 & 0.5 & 0.5 & 0.0 & \\ \\ 3 & 0.5 & 0.0 & 0.0 & 0.5 & \\ \\ 4 & 0.0 & 0.0 & 0.0 & 1.0 & \end{bmatrix} \\ \\ \mathbf{\color{Green}{Absorbing\;states\;are\;}} \mathbf{\color{Red}{4}} \\ \\ \mathbf{\color{Green}{Now\;we\;have\;to\;make\;matrix\;Q}} \\ \\ \mathbf{\color{Green}{Q\;is\;formed\;by\;removing\;all\;absorbing\;states\;from\;original\;matrix}} \\ \\ Q= \begin{bmatrix} * & 1 & 2 & 3 & \\ \\ 1 & 0.5 & 0.5 & 0.0 & \\ \\ 2 & 0.0 & 0.5 & 0.5 & \\ \\ 3 & 0.5 & 0.0 & 0.0 & \end{bmatrix} \\ \\ \mathbf{\color{Green}{Now\;we\;make\;matrix\;N\;using\;below\;formula\;}} \mathbf{\color{Red}{N\;=\;(I-Q)^{-1}}} \\ \\ \mathbf{\color{Green}{Where\;I\;is\;Identity\;matrix}} \\ \\ \Rightarrow \left ( \begin{bmatrix} 1 & 0 & 0 & \\ \\ 0 & 1 & 0 & \\ \\ 0 & 0 & 1 & \end{bmatrix} \mathbf{\color{Red}{-}} \begin{bmatrix} 0.5 & 0.5 & 0.0 & \\ \\ 0.0 & 0.5 & 0.5 & \\ \\ 0.5 & 0.0 & 0.0 & \end{bmatrix} \right ) ^{ \mathbf{\color{Red}{-1}}} \\ \\ \Rightarrow \left ( \begin{bmatrix} 0.5 & -0.5 & 0.0 & \\ \\ 0.0 & 0.5 & -0.5 & \\ \\ -0.5 & 0.0 & 1.0 & \end{bmatrix} \right ) ^{ \mathbf{\color{Red}{-1}}} \\ \\ \Rightarrow \mathbf{\color{Green}{N\;=\;}} \begin{bmatrix} 4.0 & 4.0 & 2.0 & \\ \\ 2.0 & 4.0 & 2.0 & \\ \\ 2.0 & 2.0 & 2.0 & \end{bmatrix} \\ \\ \mathbf{\color{Green}{The\;expected\;number\;of\;steps\;starting\;from\;each\;of\;the\;transient\;states\;is}} \\ \\ \mathbf{\color{Green}{t\;=\;N*}} \begin{bmatrix} 1 \\ \\ 1 \\ \\ 1 \end{bmatrix} \\ \\ \Rightarrow \begin{bmatrix} 4.0 & 4.0 & 2.0 & \\ \\ 2.0 & 4.0 & 2.0 & \\ \\ 2.0 & 2.0 & 2.0 & \end{bmatrix} \mathbf{\color{Red}{*}} \begin{bmatrix} 1 \\ \\ 1 \\ \\ 1 \end{bmatrix} \\ \\ \Rightarrow \begin{bmatrix} 10.0 \\ \\ 8.0 \\ \\ 6.0 \end{bmatrix} \\ \\ \mathbf{\color{Green}{Therefore,\;}} \mathbf{\color{Green}{Expected\;number\;of\;steps\;if\;started\;from\;state\;1\;=\;}} \mathbf{\color{Red}{10.0}} \\ \\ \mathbf{\color{Green}{Expected\;number\;of\;steps\;if\;started\;from\;state\;2\;=\;}} \mathbf{\color{Red}{8.0}} \\ \\ \mathbf{\color{Green}{Expected\;number\;of\;steps\;if\;started\;from\;state\;3\;=\;}} \mathbf{\color{Red}{6.0}} \\ \\ }$$