2 n 1 2 n n n 1 so we can pick 0 to make

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Unformatted text preview: /µn 1 )⇡ (n n2 µn 1 · ⇡ (n 2) and it follows that 2) Repeating the last reasoning leads to ⇡ (n) = n 1 · n 2 ··· 0 ⇡ (0) µn · µn 1 · · · µ1 (4.18) To check this formula and help remember it, note that (i) there are n terms in the numerator and in the denominator, and (ii) if the state space was {0, 1, . . . , n}, then µ0 = 0 and n = 0, so these terms cannot appear in the formula. To illustrate the use of (4.18) we will consider several concrete examples. Example 4.13. Two state chains. Suppose that the state space is {1, 2}, q (1, 2) = , and q (2, 1) = µ, where both rates are positive. The equations ⇡ Q = 0 can be written as ✓ ◆ ⇡1 ⇡ 2 =0 0 µ µ The first equation says ⇡1 + µ⇡2 = 0. Taking into account that we must have ⇡1 + ⇡2 = 1, it follows that ⇡1 = µ +µ and ⇡2 = +µ Example 4.14. Barbershop. A barber can cut hair at rate 3, where the units are people per hour, i.e., each haircut requires an exponentially distributed amount of time with mean 20 minutes. Suppo...
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