Lecture
For a simple renewal process we study the distribution of the epoch of the last renewal on the interval [0,t). In this study the epoch t0=0 is taken to be a renewal epoch in the situation where no renewals occurred on the interval [0,t).
Denote this epoch by ζt . Then obviously
P{ζt<0}=0, P{ζt=0}=1-F(t), P{ζtfor x>t . (2.35)
For 0<=t we obtain the equality
,
(2.36)
if we take into account that dH(y) is the probability of a renewal occurring in the neighborhood of the point y, and 1-F(t-y) is the probability that there will be no renewals on (y,t), that is, that the last renewal before t occurred in the neighborhood of the point y. Equalities (2.35) and (2.36) solve the problem posed.
The distribution of the random variable ζt has a jump at zero (the jump size equals 1-F(t)) and is continuous at x=t , since

if we take into account equality (2.16).
From (2.36) we obtain
(2.37)
The backward recurrence time ηt (undershoot time) is defined as the time from the epoch of the last renewal occurring on the interval [0,t) to the epoch t. From the definitions it follows that a functional relation ηt+ζt=t holds between the random variables. Consequently, for
and from (2.35) and (2.36) we obtain
P{ηt <0}=P{ζt>t}=0, P{ηt =t}=P{ζt=0}=1-F(t),
P{ηt ζt>t-x}=1 for x>t, (2.38)
(2.39)
The distribution of the random variable ηt has a jump at x=t, the jump size equals 1-F(t), since from (2.39) we have limx→tP{ηtand is continuous at x=0 , since the distribution (2.36) is continuous at x=t.
The result obtained is easy to explain if we pay attention to the equality of events - for xthe event {ηt>x} means that there are no renewals on the interval (t-x,t), for x=t the event {ηt=x} means that there are no renewals on the interval (0,t).
From (2.39) we obtain for the expectation
(2.40)
The forward recurrence time ξt (overshoot time) is defined as the time from the epoch t to the nearest renewal occurring after t. Note that for any x>0 the event {ξt>x} means that there are no renewals on the interval (t,t+x). We write out the desired distribution using the formula of total probability. For the conditional probabilities we have for x>=0, 0<=y<=t
,
where it is taken into account that there are no renewals on the period [y,t).
Then by the formula of total probability we obtain
or for the distribution function we obtain
(2.41)
From (2.41) we obtain an expression for the expectation
(2.42)
Here it is appropriate to give an expression for the expectation of the interval covering the point t. From equalities (2.40) and (2.42) we obtain the sum
(2.43)
Let us note one important circumstance - the expectation of this interval does not coincide with the expectation of the random variable ξ.
For 0<=x<=t, y>=0 we write out the probabilities P{ξt>=y,ηt>=x}, from which it is easy to obtain the joint distribution P{ξtηt. Indeed,
P{ξt>=y,ηt>=x}+P{ξtηtξt>=y,ηtξtηt>=x}=1
P{ξtξtηtξtηt>=x},
P{ηtξtηtξt>=y,ηt
P{ξt>=y,ηt>=x}+P{ξtηtξtηt
For the event {ξt>=y}⋂{(ηt>=x} to occur, it is necessary and sufficient that there be no renewals on the interval (t-x,t+y). Therefore,
(2.44)
Since the random variable ηt has a positive atom at x=t, we must single out separately the case
P{ξt>=y,ηt=t}=1-F(t+y),
which corresponds to the first term in (2.44).
Finally, for x>t, y>=0 the joint probability equals zero by virtue of equality (2.42).
Equality (2.44) shows the dependence of the random variables ξt and ηt, since the probability cannot be represented as a product of probabilities and .
In conclusion of this section, let us give the distribution of the interval covering an arbitrary epoch t, that is, the distribution of the sum
For x we obtain
(2.45)
For
we obtain
(2.46)
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