20092ee132B_1_hw6sol

20092ee132B_1_hw6sol - U CLA Electrical Engineering...

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EE132B HW Set Solution #6 Professor Izhak Rubin Solution to Problem 1 (1) We calculate the stationary distribution by using the following equations: ( 29 , 1 i S P i π = = This set of equations yields 52 21 20 , , 93 93 93 = . (2) Let A 0 ={X 0 =c}, A 1 ={X 1 =b}, A 2 ={X 2 =c}, A 3 ={X 3 =a}, A 4 ={X 4 =c}, A 5 ={X 5 =a}, A 6 ={X 6 =c}, and A 7 ={X 7 =b}. Then we have: ( 29 ( 29 ( 29 ( 29 ( 29 7 6 1 0 7 6 0 6 5 0 2 1 0 1 0 ... | | ... | ... | | P A A A A P A A A P A A A P A A A P A A = ⋅⋅⋅ I I I I I I I I . Since X is a Markov chain, we have ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 7 6 1 0 7 6 6 5 1 0 ... | | | | , , , , , , , 3 2500 P A A A A P A A P A A P A A p c b p b c p c a p a c p c a p a c p c b = ⋅⋅⋅ = = I I I (3) Note that ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 1 2 1 1 2 | , | | | , , 1 | , , 6 k k k k k n S k k k k n S n S k k n S P X j X i P X j X n X i
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20092ee132B_1_hw6sol - U CLA Electrical Engineering...

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