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Unformatted text preview: 0 µ = µ e µt t e The probability of 0 customers in equilibrium is 1 ( /µ) by (4.23). This
implies the probability of 1 customer is /µ, so combining the two cases:
fD (t) = µ
µ · µ e µ µt t e + µ · µe µt At this point cancellations occur to produce the answer we claimed:
e µt e t +e µt =e t We leave it to the adventurous reader to try to repeat the last calculation for the
M/M/s queue with s > 1 where there is not a neat formula for the stationary
distribution.
Proof of Theorem 4.8. By repeating the proof of (1.13) one can show
Lemma 4.9. Fix T and let Ys = XT
chain with transition probability
pt (i, j ) =
ˆ s for 0 s T . Then Ys is a Markov ⇡ (j )pt (j, i)
⇡ (i) Proof. If s + t T then
P (Ys+t = j Ys = i) =
= P (XT P (XT (s+t) = j, XT s = i)
P (Ys+t = j, Ys = i)
=
P (Ys = i)
P (XT s = i)
= j )P (XT s = iXT (s+t) = j )
⇡ (j )pt (j, i)
(s+t)
=
⇡ (i)
⇡ (i) which is the desired result.
If ⇡ satisﬁes the detailed balance condition ⇡ (i)q (i, j ) = ⇡ (j )q (j, i),...
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 Spring '10
 DURRETT
 The Land

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