Most characterizations of the exponential distribution (and its generalizations) in the classical setting are based on the equivalence of the time-shifted distribution with the original distribution, in some sense. In the semigroup setting (and particularly in the positive semigroup setting), there are natural generalizations of these concepts. Once again we start with measurable space \((S, \ms S)\) and a measurable semigroup \((S, \cdot)\) as discussed in Section 1. The relation \(\rta\) associated with \((S, \cdot)\) is given by \(x \rta y\) if and only if \(y \in x S\). So if \(X\) is a random variable in \(S\) then the reliability function \(F\) of \(X\) for the graph \((S, \rta)\) is given by \[F(x) = \P(x \rta X) = \P(X \in x S), \quad x \in S\] We assume that \(X\) is supported by \((S, \cdot)\) so that \(F(x) \gt 0\) for \(x \in S\).
Suppose that \(X\) is a random variable in \(S\) with reliability function \(F\) for \((S, \rta)\).
Recall that a positive measure \(\mu\) on a topological group \((S, \cdot)\) is relatively invariant if \(\mu(x A) = F(x) \mu(A)\) for \(x \in S\) and \(A \in \ms{S}\) where \(F: S \to [0, \infty)\) is measurable (see the book by Halmos). So an exponential distribution is simply a relatively invariant probability measure, but on a semigroup rather than a group.
If \(X\) has an exponential distribution on \((S, \cdot)\) then \(X\) has a memoryless distribution on \((S, \cdot)\).
Generalizing \(\rta\), recall that the relation \(\rta_A\) associated with \(A \in \ms S\) is defined by \(x \rta_A y\) if and only if \(y \in A x\), so that \(y = x a \) for some \(a \in A\). In terms of the relations associated with \((S, \cdot)\), the exponential and memoryless properties have the form \begin{align*} \P(x \rta_A X \mid x \rta X) &= \P(X \in A), \quad x \in S, \, A \in \ms S \\ \P(x y \rta X \mid x \rta X) & = \P(y \rta X), \quad x, \, y \in S \end{align*} Sepcializing further, if \((S, \cdot)\) is a positive semigroup, so that the associated relation is a partial order \(\preceq\), the exponential property and memoryless properties have the more familiar form \begin{align*} \P(X \in x A \mid X \succeq x) &= \P(X \in A), \quad x \in S, \, A \in \ms S \\ \P(X \succeq x y \mid X \succeq x) &= \P(X \succeq y), \quad x, \, y \in S \end{align*} As usual, the right zero semigroup provides an extreme example:
Suppose that \((S, \cdot)\) is the right zero semigroup on \(S\), so that \(x y = y\) for \(x, \, y \in S\). It follows that \(x A = A\) for \(x \in S\) and \(A \in \ms S\) so if \(X\) is a random variable in \(S\) then \[\P(X \in x A) = \P(X \in A) = \P(X \in S) \P(X \in A) = \P(X \in x S) \P(X \in A), \quad x \in S, \, A \in \ms S\] So every probability distribution is exponential on \((S, \cdot)\), just as every \(\sigma\)-finite measure is left invariant.
Here are alternative formulations of the exponential and memoryless properties:
Suppose again that \(X\) is a random variable in \(S\).
The proofs rely on basic algebraic properties of the semiroup.
As a simple corollary we can answer the question of when the random walk on semigroup \((S, \cdot)\) associated with random variable \(X\) is the same as the random walk on the graph \((S, \rta)\) associated with \(X\).
The random walk on the semigroup \((S, \cdot)\) associated with \(X\) is the same as the random walk on the graph \((S, \rta)\) associated with \(X\) if and only if \(X\) has an exponential distribution.
Let \(\bs{Y} = (Y_1, Y_2, \ldots)\) be a discrete-time, homogeneous Markov process in \(S\). For both random walks, the distribution of \(Y_1\) is the same as the distribution of \(X\). For the random walk on the graph \((S, \rta)\), the conditional distribution of \(Y_{n + 1}\) given \(Y_n = x\) is the same as the distribution of \(X\) given \(x \rta X\) (that is, \(X \in x S\)). For the random walk on the semigroup \((S, \cdot)\), the conditional distribution of \(Y_{n + 1}\) given \(Y_n = x\) is the same as the distribution of \(x X\). Hence the two random walks are the same if and only if the conditional distribution of \(X\) given \(X \in x S\) is the same as the distribution of \(x X\). By , this is the case if and only if \(X\) has an exponential distribution.
The following two results deal with the set of points and the collection of sets satisfying the exponential property.
Suppose again that \(X\) is a random variable in \(S\) with reliability function \(F\) for \((S, \rta)\). Define \[S_X = \{x \in S: \P(X \in x A) = F(x) \P(X \in A) \text{ for all } A \in \ms S\}\] If \(S_X \ne \emptyset\) then \((S_X, \cdot)\) is a complete sub-semigroup of \((S, \cdot)\).
We first show closure. Suppose that \(x, \, y \in S_X\) and \(A \in \ms S\). Then \[\P[X \in (x y) A] = \P[X \in x (y A)] = F(x) \P(X \in y A) = F(x) F(y) \P(X \in A)\] In particular, letting \(A = S\) we have \(F(x y) = F(x) F(y)\), so substituting back we have \[\P[X \in (x y) A] = F(x y) \P(X \in A)\] and so \(x y \in S_X\). Next we show completeness. Suppose that \(x, \, y \in S_X\) and that \(x \rta y\) so that \(y = x u\) for some \(u \in S\). We need to show that \(u = x^{-1} y \in S_X\). Let \(A \in \ms S\). First, since \(x \in S_X\) we have \[\P[X \in (x u) A] = \P[X \in x (u A)] = F(x) \P(X \in u A)\] On the other hand, since \(y = x u \in S_X\) we have \[\P[X \in (x u) A] = F(x u) \P(X \in A)\] Again since \(x \in S_X\) we have \[F(x u) = \P[X \in (x u) S] = \P[X \in x (u S)] = F(x) \P(X \in u S) = F(x) F(u)\] Combining the displayed equations we have \[F(x) \P(X \in u A) = F(x) F(u) \P(X \in A)\] Since \(F(x) \gt 0\) we have \(\P(X \in u A) = F(u) \P(X \in A)\), so \(u \in S_X\).
In the case that \((S, \cdot)\) is a positive semigroup, note that the identity element \(e \in S_X\) so \((S_X, \cdot)\) is a complete positive sub-semigroup. Of course in general, \(S_X\) may be empty, or in the case of a positive semigroup we could have \(S_X = \{e\}\). These cases aside, every distribution satisfies the exponential property on some complete sub-semigroup.
Suppose again that \(X\) is a random variable in \(S\) with reliability function \(F\) for \((S, \rta)\). Define \[\ms S_X = \{A \in \ms S: \P(X \in x A) = F(x) \P(X \in A) \text{ for all } x \in S\}\] Then \(\ms S_X\) is closed under countable disjoint unions, proper differences, countable increasing unions, and countable decreasing intersections.
Recall that for \(x \in S\), the mapping \(A \mapsto x A\) from \(\ms S\) into \(\ms S\) preserves all of the set operations. Let \((A_1, A_2, \ldots)\) be a sequence of disjoint sets in \(\ms S_X\) and let \(x \in S\). Then \((x A_1, xA_2, \ldots)\) is a disjoint sequence and \begin{align*} \P\left(X \in x \bigcup_{i = 1}^\infty A_i \right) &= \P\left(X \in \bigcup_{i = 1}^\infty x A_i\right) = \sum_{i = 1}^\infty \P(X \in x A_i) = \sum_{i=1}^\infty F(x) \P(X \in A_i)\\ &= F(x) \sum_{i=1}^\infty \P(X \in A_i) = F(x) \P\left(X \in \bigcup_{i=1}^\infty A_i \right) \end{align*} Hence \(\bigcup_{i=1}^\infty A_i \in \ms S_X\). Next let \(A, \, B \in \ms S_X\) with \(A \subseteq B\) and let \(x \in S\). Then \(x A \subseteq x B\) and \(x (B - A) = x B - x A\). Hence \begin{align*} \P[X \in x (B - A)] &= \P[X \in (x B - x A)] = \P(X \in x B) - \P(X \in x A) \\ &= F(x) \P(X \in A) - F(x) \P(X \in B) = F(x)[\P(X \in B) - \P(X \in A)] \\ &= F(x) \P(X \in B - A) \end{align*} Hence \(B - A \in \ms S_X\). Of course \(S \in \ms S_X\) and hence the other results follow.
From it follows that \(\ms S_X\) is a monotone class. From the monotone class theorem, it follows that if \(\ms A\) is a collection of sets that generates \(\ms S\) and \(\ms A \subseteq \ms S_X\) then \(\ms S_X = \ms S\) and hence \(X\) has an exponential distribution. Our next result gives expected value characterizations of the exponential property.
Suppose again that \(X\) is a random variables in \(S\) and reliability function \(F\) for \((S, \cdot)\). The following are equivalent:
If \(X\) has an exponential distribution on \((S, \cdot)\) then (a) and (b) holds more generally for measurable \(g: S \to \R\), assuming that the expected values exist.
Suppose that \((S, \cdot)\) is a topological semigroup with the property tht \(\{t \in S: t \rta x\}\) and \(\{y \in S: x \rta y\}\) have nonempty interiors for each \(x \in S\). If \(X\) has an exponential distribution on \((S, \cdot)\) then the reliability function \(F\) of \(X\) is continuous.
Recall that by definition, \(S\) has a locally compact Hausdorff topology with a countable base, and \((x, y) \mapsto x y\) is continuous. If \((S, \cdot)\) is discrete (so that \(S\) is countable with the discrete topology), then all functions, including \(F\), are continuous. So assume that \(S\) is uncountable. Suppose that \((x_n: n \in \N_+)\) is a sequence in \(S\) converging to \(x \in S\). Let \(y \in x S\) and \(z \in y S\) where \(x, \, y , \, z\) are distinct. Then \(\{t \in S: y \in t S\}\) is a neighborhood of \(x\) so \(y \in x_n S\) for \(n\) sufficiently large. Also, \(y S\) is a neighborhood of \(z\) so there exists a compact neigbhorhood \(K\) of \(z\) and an open neighborhonnd \(U\) of \(z\) with \(K \subset U \subseteq y S\) By Urysohn's lemma, there exists \(\varphi: S \to [0, 1]\) with \(\varphi = 1\) on \(K\) and \(\varphi = 0\) on \(U^c\). By the continuity of multiplication, \(x_n X \to x X\) as \(n \to \infty\) and by the continuity of \(\varphi\), \(\varphi(x_n X) \to \varphi(x X)\) as \(n \to \infty\). By the bounded convergence theorem, \( \E[\varphi(x_n X)] \to \E[\varphi(x X)] \) as \(n \to \infty\). Since \(U \subseteq x S\), and \(\varphi = 0\) on \(U^c\), note that \(\E[\varphi(X), X \in x S] = \E[\varphi(X), X \in U]\). Similarly, \(\E[\varphi(X), X \in x_n S] = \E[\varphi(X), X \in U]\) for \(n\) sufficiently large. By the exponential property, \(\E[\varphi(X), X \in x S] = F(x) \E[\varphi(x X)] \) and similarly, \(\E[\varphi(X) X \in x_n S] = F(x_n) \E[\varphi(x_n S)]\) for \(n\) sufficiently large. Therefore we have \[ F(x) = \frac{\E[\varphi(X), X \in U]}{\E[\varphi(x X)]}\] Similarly, for \(n\) sufficently large, \[ F(x_n) = \frac{\E[\varphi(X), X \in U]}{\E[\varphi(x_n X)]}\] Hence \(F(x_n) \to F(x)\) as \(n \to \infty\).
In the setting of proposition , \(x \mapsto \P(X \in x A)\) is continuous for every \(A \in \ms{S}\).
So far, we have not needed to refer to a reference measure \(\lambda\) on \((S, \ms S)\) or a density function of \(X\) with respect to \(\lambda\), as we did for constant rate distributions. The following theorem bridges the gap and gives one of the main characterization of exponential distributions.
Suppose again that \(X\) is a random variable in \(S\). Then \(X\) has an exponential distribution on \((S, \cdot)\) if and only if \(X\) is memoryless on \((S, \cdot)\) and has constant rate on \((S, \rta)\) with respect to a \(\sigma\)-finite measure that is left invariant for \((S, \cdot)\).
Let \(F\) denote the reliability function of \(X\) on \((S, \rta)\). Suppose first that \(X\) has an exponential distribution on \((S, \cdot)\). Then by , \(X\) has a memoryless distribution. Now let \(\mu\) be the \(\sigma\)-finite measure defined by \[\mu(A) = \E\left[\frac{1}{F(X)}, X \in A\right], \quad A \in \ms S\] Then from Section 1.5, \(X\) has density \(F\) with respect to \(\mu\) and hence has contant rate 1 with respect to \(\mu\): \[\P(X \in A) = \int_A F(x) d \mu(x), \quad A \in \ms S\] Let \(x \in S\) and \(A \in \ms S\). By the integral version of the exponential property in and by the memoryless property, \begin{align*} \mu(x A) &= \E\left[\frac{1}{F(X)}, X \in x A\right] = F(x) \E\left[\frac{1}{F(x X)}, X \in A\right] \\ &= F(x) \E\left[\frac{1}{F(x)F(X)}, X \in A\right] = \E\left[\frac{1}{F(X)}, X \in A\right] = \mu(A) \end{align*} so \(\mu\) is left invariant on \((S, \cdot)\). Conversely, suppose that the distribution of \(X\) is memoryless on \((S, \cdot)\) and has constant rate \(\alpha \in (0, \infty)\) on \((S, \rta)\) with respect to a \(\sigma\)-finite measure \(\lambda\) that is left invairant on \((S, \cdot)\). Thus \(f = \alpha F\) is a density function of \(X\) with respect to \(\lambda\). Let \(x \in S\) and \(A \in \ms S\). Using the memoryless property and the integral version of left invariance in Section 3, \begin{align*} \P(X \in x A) &= \int_{x A} \alpha F(y) d \lambda(y) = \int_A \alpha F(x z) d\lambda(z) = \int_A \alpha F(x) F(z) d \lambda(z) \\ &= F(x) \int_A \alpha F(z) d \lambda(z) = F(x) \P(X \in A) \end{align*} Hence \(X\) has an exponential distribution.
In particular, if \((S, \cdot)\) has an exponential distribution then \((S, \cdot)\) must have a left-invariant measure, not surprising since the existence of an exponential distribution requires somewhat more of the semigroup \((S, \cdot)\) than the existence of a left-invariant measure.
Suppose that \(\lambda\) is the unique left-invariant measure for \((S, \cdot)\) up to multiplication by positive constants. Then \(X\) has an expoential distribution on \((S, \cdot)\) if and only if \(X\) is memoryless and constant rate with respect to \(\lambda\).
In particular, this corollary applies to discrete positive semigroups, with counting measure \(\#\) as the left invariant measure.
Suppose that \((S, \cdot)\) is a discrete positive semigroup with associated partial order \(\preceq\). Then \(X\) has an exponential distribution on \((S, \cdot)\) if and only if \(X\) is memoryless on \((S, \cdot)\) and has constant rate on \((S, \preceq)\), with rate constant \(\alpha \in (0, 1]\).
The following result is another important characterization.
Suppose that \(\lambda\) is a left-invariant measure on \((S, \cdot)\) and that \(F: S \to (0, 1]\) is measurable. Then \(F\) is the reliability function of an exponential distribution on \((S, \cdot)\) that has constant rate with respect to \(\lambda\) if and only if
Suppose first that \(F\) is the reliability function of an exponential distribution for \((S, \cdot)\) that has constant rate \(\alpha \in (0, \infty)\) with respect to \(\lambda\). By , the distribution is memoryless, so (a) holds. Also \(\alpha F\) is a probability density function so \[\int_S F(x) d\lambda(x) = \frac{1}{\alpha} \lt \infty\] and hence (b) holds. Conversely, suppose that (a) and (b) hold. Let \(f = \alpha F\) where \(\alpha = 1 \big / \int_S F(x) d\lambda(x)\). Then by (b), \(f\) is a probability density function. Let \(X\) be a random variable with density \(f\), and let \(x \in S\) and \(A \in \ms S\). Using (a) and the integral version of the left invariance property in Section 3, \begin{align*} \P(X \in x A) &= \int_{x A} f(y) d\lambda(y) = \int_{x A} \alpha F(y) d \lambda(y) = \int_A \alpha F(x z) d \lambda(z) \\ &= F(x) \int_A \alpha F(z) d\lambda(z) = F(x) \int_A f(z) d\lambda(z) = F(x) \P(X \in A) \end{align*} Letting \(A = S\) we see that \(F\) is the reliability function of \(X\), and so it then follows that \(X\) has an expeontnial distribution with rate \(\alpha\).
If \(\lambda\) is the unique left invariant measure for \((S, \cdot)\), up to multiplication by positive constants, then gives a method for finding all exponential distributions. It also follows that the memoryless property almost implies the constant rate property (and hence the full exponential property). More specifically, if \(F\) is a reliability function satisfing (a) and (b) of , then \(f = \alpha F\) is the probability density function of an exponential distribution with reliability function \(F\) (where again, \(\alpha = 1 \big / \int_S F(x) d\lambda(x)\)). But in general, there may be other probability density functions with same reliability function \(F\) that are not multiples of \(F\) (and hence do not have constant rate). It may also be possible for \(F\) to satisfy (a) but with \(\int_S F(x) d\lambda(x) = \infty\). But to emphasize, we do have the following:
Suppose that \((S, \cdot)\) is a semigroup in which the reliability function uniquely determines the underlying distribution. Then a distribution is exponential on \((S, \cdot)\) if and only if it is memoryless.
Section 4.4 gives an example of a discrete, positive semigroup where the reliability function does not determine the distribution and where there are memoryless distributions that are not exponential. The following examples also provides some insight.
Suppose that \((S, \cdot)\) is a semigroup with left-invariant measure \(\lambda\) and whose associated relation is the complete reflexive relation \(\equiv\), so that \(x \equiv y\) for every \(x, \, y \in S\). If \(\lambda(S) \lt \infty\) then the uniform distribution on \(S\) (with respect to \(\lambda\)) is exponential for \((S, \cdot)\).
Suppose that \(\lambda(S) \lt \infty\) and let \(X\) have the uniform distribution on \(S\). Since the relation associated with \((S, \cdot)\) is complete, (equivalently \(x S = S\) for all \(x \in S\)), the reliability function \(F\) of \(X\) is the constant 1: \(\P(X \in x S) = 1\) for \(x \in S\). Hence \[\P(X \in x A) = \frac{\lambda(x A)}{\lambda(S)} = \frac{\lambda(A)}{\lambda(S)} = \P(X \in x S) \P(X \in A), \quad x \in S, \, A \in \ms S \]
Here is a concrete example:
Suppose that \(S\) has an LCCB topology as defined in the Preface and that \((S, \cdot)\) is a topological group. Let \(\lambda\) denote the left invariant measure for \((S, \cdot)\) (unique up to multiplication by positive constants).
Since \((S, \cdot)\) is a group, the associated relation \(\equiv\) is complete.
In particular, the unique exponential distribution for a finite group \((S, \cdot)\) is the uniform distribution on \(S\) (with respect to counting measure \(\#\)). But once again, we see that a full group is not particularly interesting in the reliability context. Rather, a maximal positive sub-semigroup is the appropriate object of study.
Suppose that \((S, \cdot)\) is the right zero semigroup on \(S\), so that \(x y = y\) for \(x, \, y \in S\). Recall that every measure on \(S\) is left invaiant for \((S, \cdot)\) and every distribution on \(S\) is exponential for \((S, \cdot)\). Suppose that \(\lambda\) is a reference measure on \(S\).
The corresponding relation is the complete reflexive relation \(\equiv\) so that \(x \equiv y\) for every \((x, y) \in S^2\). Hence the reliability function on \((S, \equiv)\) of every distribution is simply the constant function 1. So this is a trivial example of a semigroup that has exponential (and hence memoryless) distributions that do not have constant rate with respect to a given left-invariant measure. We can also view this example through the lens of . If \(F(x y) = F(x) F(y)\) then \(F(y) = F(x) F(y)\) for \(x, \, y \in S\). From our support assumption, \(F(y) \gt 0\) so \(F(x) = 1\) for \(x \in S\). So \(\int_S F(x) \, d\lambda(x) = \lambda(S)\) and hence condition (b)) holds if and only if \(\lambda\) is a finite measure, in which case the corresponding constant rate distribution is simply the uniform distribution on \(S\) (with respect to \(\lambda\)) \[\P(X \in A) = \frac{\lambda(A)}{\lambda(S)}, \quad A \in \ms S\] On the other hand, if \(\lambda(S) = \infty\) then there is no distribution that has constant rate with respect to \(\lambda\). So to summarize, a distributionn on \(S\) (which is necessarily exponential for \((S, \cdot)\)) has contant rate with respect to a measure \(\lambda\) (which is necessarily left invariant) if and only if \(\lambda\) is a finite measure, in which case the distribution is uniform with respect to \(\lambda\). This statement is not as restrictive as it might seem. The canonical measure associated with \(X\) is simply the distribution of \(X\): \[\E\left[\frac{1}{F(X)}; X \in A\right] = \P(X \in A), \quad A \in \ms S\] and trivially, \(X\) has the uniform distribution with repsect to itself: \[\P(X \in A) = \frac{\P(X \in A)}{\P(X \in S)}, \quad A \in \ms S\]
Conversely, the constant rate property does not imply the memoryless property. The following general example shows that mixtures of distinct exponential distributions with the same constant rate will still have the constant rate property, but not the memoryless property. The free semigroup studied in Chapter 5 gives a specific example where there are different exponential distributions with the same rate.
Suppose that \((S, \cdot)\) is a semigroup with a fixed left-invariant measure \(\lambda\). Suppose that \(F\) and \(G\) are reliability functions for distinct exponential distributions on \((S, \cdot)\), each having constant rate \(\alpha \in (0, \infty)\) with respect to \(\lambda\). Let \(p \in (0, 1)\) and \(H = p F + (1 - p) G\). Then \(H\) is also the reliability function for a distribution with constant rate \(\alpha\). The distributions corresponding to \(F\) and \(G\) are memoryless, but not the distribution corresponding to \(H\): \[H(x y) = p F(x y) + (1 - p) G(x y) = p F(x) F(y) + (1 - p) G(x) G(y), \quad x, \, y \in S\] while \begin{align*} H(x) H(y) &= [p F(x) + (1 - p) G(x)][p F(y) + (1 - p) G(y)] \\ &= p^2 F(x) F(y) + p (1 - p) [F(x) G(y) + F(y) G(x)] + (1 - p)^2 G(x) G(y), \quad x, \, y \in S \end{align*}
Suppose that \(F\) is the reliability function of an exponential distribution on \((S, \cdot)\) that has constant rate with respect to the left invariant measure \(\lambda\). If \(m \in (0, \infty)\) and \[\frac{1}{\alpha_m} := \int_S F^m(x) d\lambda(x) \lt \infty\] (in particular if \(m \ge 1\)), then \(F^m\) is the reliability function of an exponential distribution on \((S, \cdot)\) that has rate \(\alpha_m\) with respect to \(\lambda\).
Suppose that \((S, \cdot)\) is a positive semigroup with left invariant measure \(\lambda\) and that \(f\) is a probability density function with respect to \(\lambda\) satisfying \[f(x) f(y) = G(x y), \quad x, \, y \in S\] for some measurable function \(G: S \to (0, \infty)\). Then \(f\) is the density of an exponential distribution on \((S, \cdot)\).
Let \(e\) denote the identity element of \((S, \cdot)\). Letting \(y = e\) in the equation above gives \(G(x) = \alpha f(x)\) where \(\alpha = f(e) \in (0, \infty)\). Let \(F\) denote the reliability function of \(f\) on \((S, \cdot)\). Then using the integral version of the left-invariance property in Section 3, \begin{align*} F(x) &= \int_{x S} f(y) d\lambda(y) = \int_{x S} \frac{1}{\alpha} G(y) d\lambda(y) = \frac{1}{\alpha} \int_S G(x u) d\lambda(u) \\ &= \frac{1}{\alpha} \int_S f(x) f(u) d\lambda(u) = \frac{1}{\alpha} f(x), \quad x \in S \end{align*} Thus the distribution has constant rate \(\alpha\). Finally, \[F(x y) = \frac{1}{\alpha} f(x y) = \frac{1}{\alpha^2} G(x y) = \frac{1}{\alpha^2} f(x) f(y) = F(x) F(y)\] so the distribution is memoryless. Hence \(f\) is the density of an exponential distribution by .
The following definition gives the abstract version of the new better than used and new worse than used properties. Once again, we have a measurable semigroup \((S, \cdot)\) with associated relation \(\rta\).
Suppose that \(X\) is a random variable in \(S\).
In terms of the relation \(\rta\), the NBU and NWU properties are, respectively \begin{align*} \P(y \rta x^{-1} X \mid x \rta X) &\le \P(y \rta X), \quad x, \, y \in S\\ \P(y \rta x^{-1} X \mid x \rta X) &\ge \P(y \rta X), \quad x, \, y \in S \end{align*} Once again, in the case of a positive semigroup, with a partial order \(\preceq\) as the relation, these properties take the more recognizable form \begin{align*} \P(x^{-1} X \succeq y \mid X \succeq x) &\le \P(X \succeq y), \quad x, \, y \in S \\ \P(x^{-1} X \succeq y \mid X \succeq x) &\ge \P(X \succeq y), \quad x, \, y \in S \end{align*} We will concentrate on the memoryless and exponential properties in this text.