1 Continuous Time Processes - IPM

Therefore
P [Am | Nt = m] = t1(t2 − t1) · · · (tm − tm−1) m!
. (1.2.6)
tm To obtain the conditional joint density of T1, · · · , Tm given Nt = m we apply
the differential operator ∂m to (1.2.6) to obtain
∂t1···∂tm
fT |N(t1, · · · , tm) = m!
t m , 0 ≤ t1 < t2 < · · · < tm ≤ t. (1.2.7)
We deduce the following remarkable fact:
Proposition 1.2.1 With the above notation and hypotheses, the conditional
joint density of T1, · · · , Tm given Nt = m is identical with that of the order
statistics of m uniform random variables from [0, t].
Proof - Follows from lemma 1.2.1 and (1.2.7). ♣.
A simple consequence of Proposition 1.2.1 is a proof of the independence
of increments for a Poisson process. To do so let t1 < t2 ≤ t3 < t4, U =
Nt2 − Nt1 and V = Nt4 − Nt3. Then
E[ξ U η V ] = E[E[ξ U η V | Nt4]]
From Proposition 1.2.1 we know that conditioned on Nt4 = M, the times
t1, t2 and t3 have the same distribution as the order statistics of three uniform
random variables on [0, t4]. In particular, the number of arrivals in an interval
(a, b) ⊂ (0, t4) is proportional to the length b − a. Therefore the conditional
expectation E[ξ U η V | Nt4] is easily computed (see example ??)
E[ξ U η V | Nt4] =
t2 − t1
t4
ξ +
t4 − t3
t4
η +
t1 + t3 − t2
t4
M
(1.2.8)
Since for fixed t, Nt is a Poisson random variable with parameter µt, we can
calculate the outer expectation to obtain
E[ξ U η V ] = e −µ[ξ(t2−t1)+η(t4−t3)+(t1+t3−t2−t4)]
= e −µ(t2−t1)(ξ−1) e −µ(t4−t3)(η−1)
= E[ξ U ]E[η V ],
which proves the independence of increments of a Poisson process. Summarizing
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