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with
N
0. The probability that exactly
n
events occur in a time interval
t
can be
determined from
(
0
)
=
p
n
(
t
)
=
Pr
{
N
(
t
)
=
n
}=
F
n
(
t
)
−
F
n
−
1
(
t
),
(3.196)
where the functions on the rhs of (
3.196
) are the probabilities of the sum of (
n
+
1)
events at times exceeding
t
, given by
∞
t
ψ
n
+
1
t
dt
.
F
n
(
t
)
=
Pr
{
τ
1
+
τ
2
+···+
τ
n
+
1
>
t
}=
(3.197)
Inserting the waiting-time distribution function (
3.195
) into the integral in (
3.197
)
allows us to do a straightforward integration and obtain
n
k
(γ
t
)
e
−
γ
t
F
n
(
t
)
=
,
(3.198)
k
!
k
=
1
which is the cumulative distribution for a Poisson process. Consequently, inserting
(
3.198
)into(
3.196
) and taking the indicated difference yields the Poisson distribution
n
)
=
(γ
t
)
e
−
γ
t
p
n
(
t
.
(3.199)
n
!
Thus, an exponential distribution of waiting times implies a Poisson distribution for the
number of events occurring in a given time interval, indicating that the statistics of the
time intervals and the statistics of the number of time intervals are not the same, but are
related.
3.4.2
Statistics with memory
The dynamical process generated using (
3.185
) is renewal because the initial condition
ξ
of (
3.186
) is completely independent of earlier initial conditions. Consequently the
sojourn time
τ
of (
3.189
) is independent of earlier sojourn times as well, the time series
{
τ
j
adopted to produce the memoryless
time series {
t
j
} is not important. So now let us introduce some complexity into this
renewal process by perturbing the rate in the dynamical generator (
3.185
) to obtain
τ
j
} has no memory and the order of the times
(
)
dQ
t
=
γ(
)
(
).
t
Q
t
(3.200)
dt
Thesolutionto(
3.200
) after a time
t
is given by
exp
t
+
t
t
t
)
=
ξ
t
)
dt
Q
(
t
+
γ(
(3.201)
and, using the boundary condition
Q
=
1, (
3.188
) is replaced with
exp
t
+
t
t
)
dt
ξ
=
−
γ(
.
(3.202)
t
