Unformatted text preview: C =
y 2A ⇡ (y ). Then
⌫ (x) = ⇡ (x)/C is a stationary distribution for Yt . 139 4.5. MARKOVIAN QUEUES Proof. If x, y 2 A then detailed balance for Xt implies ⇡ (x)q (x, y ) = ⇡ (y )q (y, x).
From this it follows that ⌫ (x)¯(x, y ) = ⌫ (y )¯(y, x) so ⌫ satisﬁes the detailed
balance condition for Yt .
It follows from Lemma 4.7 that
⇡ (n) =
µ for 1 n N To compute the normalizing constant, we recall that if ✓ 6= 1, then
X ✓n = n=0 Suppose now that 1 ✓N +1
1✓ (4.24) 6= µ. Using (4.24), we see that
C= 1 ( /µ)N +1
/µ so the stationary distribution is given by
⇡ (n) =
1 ( /µ)N +1 µ for 0 n N (4.25) The new formula is similar to the old one in (4.23) and when < µ reduces
to it as N ! 1. Of course, when the waiting room is ﬁnite, the state space
is ﬁnite and we always have a stationary distribution, even when > µ. The
analysis above has been restricted to 6= µ. However, it is easy to see that
when = µ the stationary distribution is ⇡ (n) = 1/(N + 1) for 0 n N .
To check formula (4.25), we note that the barbershop chain, Example 4.14,
has this form with N = 3,...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
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