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Unformatted text preview: s the ﬁrst holding interval (U ). Thus,
E[V X0 = i, X1 = i + 1] = E[U1 X0 = i, X1 = i + 1] = 1/(λ + µ) Conditional on {X0 = i, Xi+1 = i − 1}, we know that the ﬁrst transition is a departure.So
the time until the ﬁrst arrival is sum of the time for ﬁrst transition (i.e., a departure) and
the time until the next arrival. The second term is exponentially distributed with rate λ,
so we have: 5 E[V X0 = i, X1 = i − 1] = E[U1 X0 = i, X1 = i − 1] + E[V X1 = i − 1] = 1
1
+
λ+µ λ d) Using the total expectation lemma, we have:
E[V X0 = i] = E[V X0 = i, X1 = i + 1]Pr {X1 = i + 1X0 = i} +
E[V X0 = i, X1 = i − 1]Pr {X1 = i − 1X0 = i}
�
�
1
λ
1
1
µ
1
=
+
+
=
λ+µλ+µ
λ+µ λ λ+µ
λ
Since this is true for any choice of i > 0, and it was assumed that X0 = i, for i > 0,
E[V ] = 1/λ.
Exercise 6.2:
The transition diagram for the embedded chain is:
�� ��
��0��� 1
3/5 �� � ��
��1��� 2/5
3/5 �� � ��
��2��� 2/5
3/5 �� � ��
��3��� 2/5
3/5 �� � ��
��4��� 2/5 � ... 3/5 3
a) The steady state probabilities satisfy π0 = 3 π1 , 2 πi−1 = 5 πi for i ≥ 2. Iterating on
5
5
these equations, � �i−1
��
2
5 2 i−1
π1 =
π0 , for i ≥ 1
3
33
⎤
⎡
� 5 � 2 �i−1
�
⎦ = 6π0
1=
πi = π0 ⎣1 +
33 2
πi...
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 Spring '11
 RobertGallager

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