2. Show the observations a) and b) in page 176. From page 176: or...

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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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