# 8140 of the time example 416 mm1 queue in this case q

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Unformatted text preview: e? The detailed balance condition implies that the ﬂows of sand between each pair of sites are balanced, which then implies that the net amount of sand ﬂowing into each vertex is 0, i.e., ⇡ Q = 0. Proof. Summing 4.16 over all k 6= j and recalling the deﬁnition of X X ⇡ (k )q (k, j ) = ⇡ (j ) q (j, k ) = ⇡ (j ) j k6=j k6=j Rearranging we have (⇡ Q)j = X k6=j ⇡ (k )q (k, j ) ⇡ (j ) j =0 j gives 131 4.3. LIMITING BEHAVIOR As in discrete time, (4.16) is much easier to check but does not always hold. In Example 4.10 ⇡ (2)q (2, 1) = 0 &lt; ⇡ (1)q (1, 2) As in discrete time, detailed balance holds for Example 4.12. Birth and death chains. Suppose that S = {0, 1, . . . , N } with N 1 and q (n, n + 1) = q (n, n for n &lt; N n 1) = µn for n &gt; 0 Here n represents the birth rate when there are n individuals in the system, and µn denotes the death rate in that case. If we suppose that all the n and µn listed above are positive then the birth and death chain is irreducible, and we can divide to write the detailed balance condition as n1 ⇡ (n) = ⇡ (n 1) (4.17) µn Using this again we have ⇡ (n ⇡ (n) = 1) = ( n1 µn · n 2...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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