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A shop sells cameras among other things. An order is placed each Saturday and the order is delivered to the shop before opening each Monday. If there are any cameras in stock, cameras are not part of the order, and if there are no cameras in stock, three are ordered. If a customers wants a camera and the shop has none in stock, then that sale is lost. Assuming weekly demand follows a Poisson distribution with  
\[\lambda =1\]
, and initially the shop has three cameras in stock find the transition matrix.
The shop has between o and 3 cameras in stock at any time.
Poisson tables are here.
End of Week\End of Next Week 0 1 2 3
0 0.080 0.184 0.368 0.368
1 0.632 0.368 0 0
2 0.264 0.368 0.368 0
3 0.080 0.184 0.368 0.358
This means for example that there is a probability of 0.080 that there will be no cameras at the end of next week given that there are three cameras at the end of this week (no cameras will be ordered, three cameras will be sold next week). The transition matrix is  
\[A= \left( \begin{array}{cccc} 0.080 & 0.184 & 0.368 & 0.368 \\ 0.632 & 0.368 & 0 & 0 \\ 0.264 & 0.368 & 0.368 & 0 \\ 0.080 & 0.184 & 0.368 & 0.368 \end{array} \right)\]
.
This is called the ONE STEP transition matrix - ONE STEP because we are using a week as a unit of time, and go from week to week.
The TWO STEP transition matrix is  
\[A^2= \left( \begin{array}{cccc} 0.249 & 0.286 & 0.300 & 0.165 \\ 0.283 & 0.252 & 0.233 & 0.233 \\ 0.351 & 0.319 & 0.233 & 0.097 \\ 0.249 & 0.286 & 0.300 & 0.165 \end{array} \right)\]
, and predicts the number of cameras in the shop at the end of two weeks.
The N STEP transition matrix is  
\[A^n\]
, and predicts the number of cameras in the shop at the end of  
\[n\]
  weeks.