A general result regarding non-occurrence of the inspection paradox can be derived on the basis of Representation (
5) with a random inspection time. This is realised in Theorem 1 and the result is applied to the special cases of equality in (
3) and (
4); see Sub
Section 4.2 and Remark 3, respectively. Either result can be used to decide whether a strict inequality holds for specific choices of distribution functions
F and
G. In particular, the condition for equality is easy to check and can therefore be utilized without calculating, e.g., the expected value of
explicitly.
4.1. General Results
This subsection is concerned with determining distributions for and T with distribution functions F and G, respectively, such that holds, given that is not a constant function. Lemma 2 serves as a key component in the discussion and states that given the support endpoints of the interarrival times, we can explicitly calculate the smallest index n for which the succeeding occurrence times have overlapping supports.
Throughout, is called an atom of (the distribution of) if .
Lemma 2. Let the interarrival times have a left support endpoint and right support endpoint . Then, there exists a natural number at which the supports of two consecutive occurrence times and , , overlap in the sense that . In particular, κ is given by Proof. Assuming that there is no point of overlap for any
, i.e., that
leads to
which is a contradiction to the assumption that
. Therefore, there exists a natural number for which the supports of successive occurrence times with large enough indices overlap. If
for some
, then
If the right support endpoint is of the form for some and or is an atom of , then the first point of overlap is at , i.e., the supports of and touch (). If neither nor are atoms of , then the first overlap happens for the supports of and (i.e., ). □
The trivial case is excluded in Lemma 2 as the support of , covered by the interval (or ), is a subset of the support of , covered by (or ), for all . In this case, the supports of all occurrence times overlap.
We will now determine cases with equality of the expected values in (
5), i.e., non-occurrence of the inspection paradox, under the general assumption that
G is a left-continuous distribution function.
Theorem 1. Let the interarrival times have left support endpoint , finite right support endpoint and let κ be defined as in Lemma 2.
Let be a measurable, monotone non-decreasing function such that all expected values and integrals are well-defined and exist finitely. Furthermore, assume that φ is not constant on -almost surely.
Let T be a non-negative random variable with left-continuous distribution function G that is independent of the interarrival times .
Then, is equivalent towhere are constants and , , are functions such that G is a left-continuous distribution function. If , then is equivalent to , , if α is not an atom of , and to , , , if α is an atom of .
Proof. With Equation (
5),
is equivalent to
since
and
G are both monotone non-decreasing, and thus all covariances are non-negative. For
, applying Lemma 1 yields
for some constant
due to the assumption that
is not constant on
-almost surely. First, assume that
, i.e.,
is not an atom of
. Since
G as a distribution function is monotone non-decreasing and assumed to be left-continuous, we have
for all
. With the same arguments, we obtain for a general
for a constant
by using Lemma 1. This implies
for all
, due to the monotonicity and continuity of the distribution function
G. Furthermore, the sequence of constants
needs to satisfy
for all
(since
G is monotone non-decreasing) and
(since
).
With Lemma 2,
is the index where the supports overlap, and thus
for all
. If
, then
, and thus the distribution function
G has to be of the form
,
. Otherwise, the distribution function
G has to be of the form
where
and
,
, are functions such that
for all
,
, and such that
are monotone non-decreasing and left-continuous.
If
, then
necessarily has to hold for all
,
. Thus, for
G as in (
7) to be left-continuous at the points
, the additional assumption
,
, is needed.
For the opposite implication, noticing that
G as in (
7) is constant on the support of all occurrence times and applying Lemma 1, we see that the covariances
are all equal to zero. This leads to equality of the expected values. □
We note that the constant as introduced in Lemma 2 indicates the kind of support the interarrival times have. For example, can correspond to the case , where may be infinite. In this case, the distribution of the random inspection time T can be chosen as , or among others. On the other hand, must be finite for .
Remark 1. The case for (which would correspond to in Theorem 1) is excluded, because it is already well-known from the literature (cf. Section 2) that and that the expected values are always equal, i.e., The equality remains when introducing a random variable T. This can also be derived with the following argument: If , then for all implies for all and for any choice of the distribution function G, leading to equality for any φ.
Theorem 1 can be applied to any interarrival distribution. In particular, only the left support endpoint and the right support endpoint of the interarrival times are of interest. Given a specific interarrival distribution and a distribution function G, Theorem 1 establishes whether or not the inspection paradox occurs.
Example 3. The case is a particular one (see Lemma 2 with ). According to Theorem 1 where α is supposed to not be an atom of , i.e., , the inspection paradox does not occur if G is given bywith and chosen such that G is left-continuous. Here, for , T may have a two-point distribution on and may be uniformly distributed on the interval .
In the case of α being an atom of , i.e., , G is given byand has to be fulfilled. Furthermore, the following example derived from the results of Theorem 1 shows that, in the degenerate time case with
, equality in (
1) or (
2) can also take place for non-degenerate interarrival times. Nevertheless, this situation is irrelevant, since the inspection time coincides with the lower bound of the support of
.
Example 4. For absolutely continuous interarrival times with a left endpoint (i.e., α is not an atom of ) and right endpoint , which corresponds to the case , we have equality for any φ satisfying the assumptions of Theorem 1, since , , is of the form (7) with . Therefore, choosing and for in Theorem 1, we obtain an example for whichholds even though the interarrival times have a distribution other than the degenerate distribution. Consequently, requiring equality for a single z and t is not sufficient in general to derive a characterization for the distribution of the interarrival times. This case will be considered in detail in Section 5. The case that was not included in Theorem 1 is studied separately in the following theorem.
Theorem 2. Let the non-degenerate interarrival times have a left support endpoint , where . Let be a measurable, monotone non-decreasing function such that all expected values and integrals are well-defined and exist finitely. Furthermore, assume that φ is not constant on -almost surely.
Let T be a non-negative random variable with a left-continuous distribution function G that is independent of the interarrival times .
Then, holds if and only if T has a degenerate distribution in 0 for every .
Proof. Analogously to the proof of Theorem 1, must hold for all and for all , wherefore we obtain for all . Thus, G is a distribution function of the degenerate distribution in 0. The opposite direction follows directly from an application of Lemma 1. □
Remark 2. If in Theorem 2 and is the left-continuous version of the distribution function of the degenerate distribution in 0, then the first covariance is positive and an inspection paradox occurs. This is the case as G is not constant -almost surely on the support of the interarrival times.
In the above situation with and , we findand thus . 4.2. Equality of the Survival Functions
Choosing
,
,
, in Theorem 1 yields distributions with equality in the inspection paradox, in the sense that
. In general, a random inspection time
T having a distribution function of the particular form (
7) is sufficient for equality in the inspection paradox. More precisely, the inspection paradox does not appear in this case and both random variables
and
are identically distributed, as stated in the following corollary.
Corollary 1. Let the interarrival times have a left support endpoint and right support endpoint . If T has a distribution function of the form (7), theni.e., . Proof. Since the
G given in (
7) is constant on the supports of
and of
for all
, applying Lemma 1 yields
for all
and for all
as in the proof of Theorem 1. Therefore, with
and are identically distributed. □
In the characterization of Theorem 1, the function is assumed to be non-constant on -almost surely. Therefore, for the function to take both values 0 and 1 on the support of with positive probability, we require .
Corollary 2. Let the interarrival times have a left support endpoint , right support endpoint and let . Let T be a non-negative random variable with left-continuous distribution function G that is independent of the interarrival times . If there is a such that takes both values 0 and 1 on with positive probability andthen G is of the form (7). Proof. This follows from Theorem 1 for the choice , . □
Corollary 2 shows that equality for an appropriate value of
z is enough to determine the general form of the distribution function
G (on finite intervals). Thus, if we have equality of the survival functions of
and
for this
, then
T is of the form (
7) and Corollary 1 yields
.
Remark 3. The inspection paradox is also discussed in terms of expected values, as we have seen in Section 2 and Section 3. With the choice , , it follows under the assumptions of Theorem 1 that equality of the expected values holds if and only if G is of the form (7). Similarly, equality of the moments, i.e., , also determines the distribution of T. This can be obtained from Theorem 1 via the choice , , .