## Transcribed Text

2. Show the observations a) and b) in page 176.
From page 176:
or rewriting things P(Tn+1 - t > s|Nt = n) = e-às. Let I' = TN(t)+1 - t,
and T1/2 = TN(t)+k - TN(t)+k-1 for k > 2. The last computation shows that
I' is independent of Nt. If we observe that
P(Tn u, Intk - In+k-1 =
K
= P(Tn t, In+1 > u) II P(&n+k > Vk)
k=2
then it follows that
(a) T1,T2,
are i.i.d. and independent of Nt.
The last observation shows that the arrivals after time t are independent
of Nt and have the same distribution as the original sequence. From this
it follows easily that:
(b) If 0 = =to cts < tn then N (ti) - N(ti-1), i = 1 n are indepen-
dent.
To see this, observe that the vector (N(t2) - N ((t1)
N(tn) - N(tn-1))
is o (TK, k > 1) measurable and hence is independent of N(t1). Then use
induction to conclude
n
P(N(ti)-N(ti-1)=ki,i=1, - = =
= A(ti-ti-1)
Ki!
Remark. The key to the proof of (b) is the lack of memory property of
the exponential distribution:
P(T > t + S T > t) = P P(T > s)
(3.7.1)
which implies that the location of the first arrival after t is independent
of what occurred before time t and has an exponential distribution.

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