# 16.2 Problems on conditional independence, given a random vector  (Page 2/2)

The transition matrix P for a homogeneous Markov chain is as follows (in m-file npr16_08.m ):

$P=\left[\begin{array}{ccccccc}0.2& 0.5& 0.3& 0& 0& 0& 0\\ 0.6& 0.1& 0.3& 0& 0& 0& 0\\ 0.2& 0.7& 0.1& 0& 0& 0& 0\\ 0& 0& 0& 0.6& 0.4& 0& 0\\ 0& 0& 0& 0.5& 0.5& 0& 0\\ 0.1& 0.3& 0& 0.2& 0.1& 0.1& 0.2\\ 0.1& 0.2& 0.1& 0.2& 0.2& 0.2& 0\end{array}\right]$
1. Note that the chain has two subchains, with states $\left\{1,2,3\right\}$ and $\left\{4,5\right\}$ . Draw a transition diagram to display the two separate chains. Can any state in one subchain be reached from any state in the other?
2. Check the convergence as in part (a) of [link] . What happens to the state probabilities for states 6 and 7 in the long run? What does thatsignify for these states? Can these states be reached from any state in either of the subchains? How would you classify these states?

Increasing power P n show the probability of being in states 6, 7 go to zero. These states cannot be reached from any of the other states.

The transition matrix P for a homogeneous Markov chain is as follows (in m-file npr16_09.m ):

$P=\left[\begin{array}{ccccccc}0.1& 0.2& 0.1& 0.3& 0.2& 0& 0.1\\ 0& 0.6& 0& 0& 0& 0& 0.4\\ 0& 0& 0.2& 0.5& 0& 0.3& 0\\ 0& 0& 0.6& 0.1& 0& 0.3& 0\\ 0.2& 0.2& 0.1& 0.2& 0& 0.1& 0.2\\ 0& 0& 0.2& 0.7& 0& 0.1& 0\\ 0& 0.5& 0& 0& 0& 0& 0.5\end{array}\right]$
1. Check the transition matrix P for convergence, as in part (a) of [link] . How many steps does it take to reach convergence to four or more decimal places? Does this agree with the theoretical result?
2. Examine the long run transition matrix. Identify transient states.
3. The convergence does not make all rows the same. Note, however, that there are two subgroups of similar rows. Rearrange rows and columns in the longrun Matrix so that identical rows are grouped. This suggests subchains. Rearrange the rows and columns in the transition matrix P and see that this gives a pattern similar to that for the matrix in [link] . Raise the rearranged transition matrix to the power for convergence.

Examination of P 16 suggests sets $\left\{2,7\right\}$ and $\left\{3,4,6\right\}$ of states form subchains. Rearrangement of P may be done as follows:

PA = P([2 7 3 4 6 1 5], [2 7 3 4 6 1 5]) PA =0.6000 0.4000 0 0 0 0 0 0.5000 0.5000 0 0 0 0 00 0 0.2000 0.5000 0.3000 0 0 0 0 0.6000 0.1000 0.3000 0 00 0 0.2000 0.7000 0.1000 0 0 0.2000 0.1000 0.1000 0.3000 0 0.1000 0.20000.2000 0.2000 0.1000 0.2000 0.1000 0.2000 0 PA16 = PA^16PA16 = 0.5556 0.4444 0 0 0 0 00.5556 0.4444 0 0 0 0 0 0 0 0.3571 0.3929 0.2500 0 00 0 0.3571 0.3929 0.2500 0 0 0 0 0.3571 0.3929 0.2500 0 00.2455 0.1964 0.1993 0.2193 0.1395 0.0000 0.0000 0.2713 0.2171 0.1827 0.2010 0.1279 0.0000 0.0000

It is clear that original states 1 and 5 are transient.

Use the m-procedure inventory1 (in m-file inventory1.m ) to obtain the transition matrix for maximum stock $M=8,$ reorder point $m=3$ , and demand $D\sim$ Poisson(4).

1. Suppose initial stock is six. What will the distribution for X n , $n=1,3,5$ (i.e., the stock at the end of periods 1, 3, 5, before restocking)?
2. What will the long run distribution be?
inventory1 Enter value M of maximum stock 8Enter value m of reorder point 3 Enter row vector of demand values 0:20Enter demand probabilities ipoisson(4,0:20) Result is in matrix Pp0 = [0 0 0 0 0 0 1 0 0];p1 = p0*P p1 =Columns 1 through 7 0.2149 0.1563 0.1954 0.1954 0.1465 0.0733 0.0183Columns 8 through 9 0 0p3 = p0*P^3 p3 =Columns 1 through 7 0.2494 0.1115 0.1258 0.1338 0.1331 0.1165 0.0812Columns 8 through 9 0.0391 0.0096p5 = p0*P^5 p5 =Columns 1 through 7 0.2598 0.1124 0.1246 0.1311 0.1300 0.1142 0.0799Columns 8 through 9 0.0386 0.0095a = abs(eig(P))' a =Columns 1 through 7 1.0000 0.4427 0.1979 0.0284 0.0058 0.0005 0.0000Columns 8 through 9 0.0000 0.0000a(2)^16 ans =2.1759e-06 % Convergence to at least five decimals for P^16 pinf = p0*P^16 % Use arbitrary p0, pinf approx p0*P^16pinf = Columns 1 through 7 0.2622 0.1132 0.1251 0.1310 0.1292 0.1130 0.0789Columns 8 through 9 0.0380 0.0093

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