## The Transition Matrix for a sales Probability Distribution Example

A company makes expensive products for which the monthly demand\[D_i\]

is low and intermittent. The total monthly demand is a random variable with the following probability distribution. \[D_i\] | 0 | 1 | 2 | 3 |

\[P(D_i)\] | 1/9 | 6/9 | 1/9 | 1/9 |

We need to complete the table.

Month \[n+1\] \Month \[n\] | 0 | 1 | 2 | 3 |

0 | ||||

1 | ||||

2 | ||||

3 |

\[n +1\]

is zero, and if demand is zero, (with probability 1/9) the state in month \[n+1\]

is 1. The state in month \[n+1\]

cannot be 2 or 3, so the entries must be zero. We can complete the first column.Month \[n+1\] \Month \[n\] | 0 | 1 | 2 | 3 |

0 | 8/9 | |||

1 | 1/9 | |||

2 | 0 | |||

3 | 0 |

\[n +1\]

is zero, and if demand is 1, (with probability 6/9) the state in month \[n+1\]

is 1. If demand is zero (with probability 1/9) the state in month \[n+1\]

is 2. The state in month \[n+1\]

cannot be 3, so the entry must be zero. We can complete the second column.Month \[n+1\] \Month \[n\] | 0 | 1 | 2 | 3 |

0 | 8/9 | 2/9 | ||

1 | 1/9 | 6/9 | ||

2 | 0 | 1/9 | ||

3 | 0 | 0 |

\[n +1\]

is zero, and if demand is 2, (with probability 1/9) the state in month \[n+1\]

is 1. If demand is 1 (with probability 6/9) the state in month \[n+1\]

is 2. If demand is 0 (with probability 1/9) the state in month \[n+1\]

is 3. We can complete the third column.Month \[n+1\] \Month \[n\] | 0 | 1 | 2 | 3 |

0 | 8/9 | 2/9 | 1/9 | |

1 | 1/9 | 6/9 | 1/9 | |

2 | 0 | 1/9 | 6/9 | |

3 | 0 | 0 | 1/9 |

\[n +1\]

is zero, and if demand is 2, (with probability 1/9) the state in month \[n+1\]

is 1. If demand is 1 (with probability 6/9) the state in month \[n+1\]

is 2. If demand is 0 (with probability 1/9) the state in month \[n+1\]

is 3. We can complete the last column.Month \[n+1\] \Month \[n\] | 0 | 1 | 2 | 3 |

0 | 8/9 | 2/9 | 1/9 | 1/9 |

1 | 1/9 | 6/9 | 1/9 | 1/9 |

2 | 0 | 1/9 | 6/9 | 6/9 |

3 | 0 | 0 | 1/9 | 1/9 |