Semigroup Products

In this section we consider semigroups relative to underlying measurable spaces

(S, S)

and

(T, T)

. Recall that the product measurable space is

(S \times T, S \times T)

. We will use

\cdot

(and concatenation) generically as the semigroup operator, regardless of the underlying base set, but the semigroup under consideration should be clear from context.

Suppose that $(S, \cdot)$ and $(T, \cdot)$ are measurable semigroups. The direct product is the semigroup $(S \times T, \cdot)$ with the binary operation $\cdot$ defined by $(x, y) (u, v) = (x u, y v)$

Details:

We need to show that the product space is a semigroup and satisfies the assumptions we have imposed. Let $(u, v), (x, y), (z, w) \in S \times T$ . Then $[(x, y) (u, v)] (z, w) = (x u, y v) (z, w) = (x u z, y v w) = (x, y) (u v, z w) = (x, y) [(u, v) (z, w)]$ so the associative property holds. Next suppose that $(u, v) (x, y) = (u, v) (z, w)$ . Then $u x = u z$ and $v y = v w$ . Hence $x = z$ and $y = w$ so $(x, y) = (z, w)$ . Therefore the left cancellation law holds. Finally, $[(x, y), (u, v)] \mapsto (x, y) (u, v) = (x u, y v)$ is measurable since $(x, u) \mapsto x u$ and $(y, v) \mapsto y v$ are measurable. The graph $(S \times T, =)$ is measurable since $(S, =)$ and $(T, =)$ are measurable.

Note that if

(x, y) \in S \times T

A \in S

, and

B \in T

then

\begin{aligned} (x, y) (A \times B) & = (x A) \times (y B) \\ (x, y)^{- 1} (A \times B) & = (x^{- 1} A) \times (y^{- 1} B) \end{aligned}

Of special importance is the case where

(S, S, \cdot) = (T, T, \cdot)

so that the direct product

(S^{2}, \cdot)

is the direct power of order 2.

Suppose again that $(S, \cdot)$ and $(T, \cdot)$ are semigroups with associated graphs $(S, \to)$ and $(T, ↑)$ , respectively. Then the graph $(S \times T, ↗)$ associated with the product semigroup $(S \times T, \cdot)$ is the direct product of the graphs $(S, \to)$ and $(T, ↑)$ .

Details:

Let $(u, v), (x, y) \in S \times T$ . By definition, $(u, v) ↗ (x, y)$ if and only if $(x, y) \in (u, v) (S \times T) = (u S) \times (v T)$ if and only if $x \in u S$ and $y \in v T$ if and only if $u \to x$ and $v ↑ y$ .

So all of the results in Section 1.8 on direct products of graphs apply to direct products of semigroups.

Suppose again that $(S, \cdot)$ and $(T, \cdot)$ are semigroups with direct product $(S \times T, \cdot)$ .

If $(S, \cdot)$ and $(T, \cdot)$ are positive semigroups, then so is $(S \times T, \cdot)$ .
If $(S, \cdot)$ or $(T, \cdot)$ is a strict positive semigroup, then so is $(S \times T, \cdot)$ .

Details:

Let $e \in S$ and $ϵ \in T$ denote the identity elements of $(S, \cdot)$ and $(T, \cdot)$ respectively. Then for $(x, y) \in S \times T$ , $(x, y) (e, ϵ) = (x e, y ϵ) = (x, y)$ and $(e, ϵ) (x, y) = (e x, ϵ y) = (x, y)$ . So $(e, ϵ)$ is the identity for $(S \times T, \cdot)$ . Suppose now that $(u, v), (x, y) \in (S \times T)_{+} = (S \times T) ∖ {(e, ϵ)}$ . Then either $x \neq e$ or $y \neq ϵ$ . In the first case, $u x \neq u$ and in the second case $v y \neq v$ . In both cases, $(u, v) (x, y) = (u x, v y) \neq (u, v)$ .
Suppose again that $(x, y), (u, v) \in S \times T$ . Then $(u, v) (x, y) = (u x, v y)$ . Since one of the semigroups is strictly positive, either $u x \neq u$ or $v y \neq v$ . In both cases, $(u, v) (x, y) \neq (u, v)$ .

In part (b), the strict positive semigroup

(S \times T, \cdot)

can be made into a positive semigroup with the addition of an identity element

(e, ϵ)

as described in Section 1.

If $(S, \cdot)$ and $(T, \cdot)$ are right zero semigroups then so is the direct product $(S \times T, \cdot)$ .

Details:

By definition, $(u, v) (x, y) = (u x, v y) = (x, y), (u, v), (x, y) \in S \times T$

For the next result, suppose that

μ

and

ν

are

σ

-finite measures on

(S, S)

and

(T, T)

respectively. Recall that

μ \times ν

denotes the product measure on

(S \times T, S \times T)

, also

σ

-finite.

If $μ$ and $ν$ are left-invariant for $(S, \cdot)$ and $(T, \cdot)$ , respectively, then $μ \times ν$ is left invariant for $(S \times T, \cdot)$ .

Details:

For $x \in S$ , $y \in T$ , $A \in S$ , and $B \in T$ , $\begin{aligned} (μ \times ν) [(x, y) (A \times B)] & = (μ \times ν) (x A \times y B) \\ = μ (x A) ν (y B) = μ (A) ν (B) = (μ \times ν) (A \times B) \end{aligned}$ Therefore, for fixed $(x, y) \in S \times T$ , the measures $C \mapsto (μ \times ν) [(x, y) C]$ and $C \mapsto (μ \times ν) (C)$ on $(S \times T, S \times T)$ agree on the measurable rectangles $A \times B$ where $A \in S$ and $B \in T$ . Hence, these measures must agree on all of $S \times T$ , and hence $μ \times ν$ is left-invariant for $(S \times T, \cdot)$ .

Suppose now that $(S, \cdot)$ and $(T, \cdot)$ are positive semigroups with identity elements $e$ and $ϵ$ , respectively, and that the left-invariant measures $μ$ and $ν$ are unique, up to multiplication by positive constants. We show that $μ \times ν$ has the same property. Let $C (T) = {B \in T : ν (T) \in (0, \infty)}$ and suppose that $λ$ is a $σ$ -finite, left-invariant measure for $(S \times T, \cdot)$ . For $C \in C (T)$ , define $μ_{C} (A) = λ (A \times C), A \in S$ Then $μ_{C}$ is a regular measure on $S$ (although it may not have support $S$ ). Moreover, for $x \in S$ and $A \in S$ , $μ_{C} (x A) = λ (x A \times C) = λ [(x, ϵ) (A \times C)] = λ (A \times C) = μ_{C} (A)$ so $μ_{C}$ is left invariant for $(S, \cdot)$ . It follows that for each $C \in C (T)$ , there exists $ρ (C) \in [0, \infty)$ such that $μ_{C} = ρ (C) μ$ ; that is, $λ (A \times C) = μ (A) ρ (C), A \in B (S), C \in C (T)$ Fix $A \in S$ with $μ (A) \in (0, \infty)$ . If $C, D \in C (T)$ and $C \subseteq D$ then $μ (A) ρ (C) = λ (A \times C) \leq λ (A \times D) = μ (A) ρ (D)$ so $ρ (C) \leq ρ (D)$ . If $C, D \in C (T)$ are disjoint then $\begin{aligned} μ (A) ρ (C \cup D) & = λ [A \times (C \cup D)] = λ [(A \times C) \cup (A \times D)] \\ = λ (A \times C) + λ (A \times D) = μ (A) ρ (C) + μ (A) ρ (D) \end{aligned}$ so $ρ (C \cup D) = ρ (C) + ρ (D)$ . If $C, D \in C (T)$ then $\begin{aligned} μ (A) ρ (C \cup D) & = λ [A \times (C \cup D)] = λ [(A \times C) \cup (A \times D)] \\ \leq λ (A \times C) + λ (A \times D) = μ (A) ρ (C) + μ (A) ρ (D) \end{aligned}$ so $ρ (C \cup D) \leq ρ (C) + ρ (D)$ . Thus, $ρ$ is a content in the sense of Halmos, and hence can be extended to a regular measure on $T$ (which we will continue to call $ρ$ ). Thus from the equation above we have $λ (A \times C) = (μ \times ρ) (A \times C), A \in S, B \in C (T)$ By regularity, it follows that $λ = μ \times ρ$ . Again fix $A \in S$ with $0 < μ (A) < \infty$ . If $y \in T$ and $B \in T$ then $μ (A) ρ (y B) = λ (A \times y B) = λ [(e, y) (A \times B)] = λ (A \times B) = μ (A) ρ (B)$ so it follows that $ρ (y B) = ρ (B)$ and hence $ρ$ is left-invariant for $(T, \cdot)$ . Thus, $ρ = c ν$ for some positive constant $c$ and so $λ = c (μ \times ν)$ . Therefore $μ \times ν$ is the unique left-invariant measure for $(S \times T, \cdot)$ , up to multiplication by positive constants.

The following example is the semigroup version of the graph first studied in Section 1.2 and then again in Section 1.8.

Suppose that $I$ is a set with $k \in N_{+}$ elements. Let $(N_{+} \times I, \cdot)$ be the direct product of $(N_{+}, +)$ and the right zero semigroup $(I, \cdot)$ . That is, $(m, i) (n, j) = (m + n, j)$ for $(m, i), (n, j) \in N_{+} \times I$ . Then

$(N_{+} \times I, \cdot)$ is a strict positive semigroup.
The associated strict partial order graph $(N_{+} \times I, ≺)$ is the direct product of $(N_{+}, <)$ and $(I, \equiv)$ , the complete reflexive graph on $I$ . That is $(m, i) ≺ (n, j)$ if and only if $m < n$ for $(m, i), (n, j) \in N_{+} \times I$ .

Details:

The results follow from [2] and [3]. Note that $(N_{+}, <)$ is a strict positive semigroup.

Probability

Naturally our interest is the relationship between memoryless and exponential distributions for the individual semigroups

(S, \cdot)

and

(T, \cdot)

, and for the product semigroup

(S \times T, \cdot)

Suppose that random variable $X$ has an exponential distribution for $(S, \cdot)$ , random variable $Y$ has an exponential distribution for $(T, \cdot)$ , and that $X$ and $Y$ are independent. Then $(X, Y)$ has an exponential distribution for the product semigroup $(S \times T, \cdot)$

Details:

If $A \in S$ , $B \in T$ , and $(x, y) \in S \times T$ then $\begin{aligned} P [(X, Y) \in (x, y) (A \times B)] & = P (X \in x A, Y \in y B) = P (X \in x A) P (Y \in y B) \\ = P (X \in x S) P (X \in A) P (Y \in y T) P (Y \in B) \\ = P (X \in x S, Y \in y T) P (X \in A, Y \in B) \\ = P [(X, Y) \in (x, y) (S \times T)] P [(X, Y) \in A \times B], (x, y) \in S \times T \end{aligned}$ Hence for fixed $(x, y) \in S \times T$ , the finite measures on $S \times T$ given by $\begin{aligned} C & \mapsto P [(X, Y) \in (x, y) C] \\ C & \mapsto P [(X, Y) \in (x, y) (S \times T)] P [(X, Y) \in C] \end{aligned}$ agree on the measurable rectangles $A \times B$ where $A \in S$ and $B \in B$ . Hence these measures agree on $S \times T$ and so $(X, Y)$ is exponential for $(S \times T, \cdot)$ .

Suppose that $(S, \cdot)$ has identity element $e$ and that $(T, \cdot)$ has identity element $ϵ$ . Then $(X, Y)$ is memoryless for $(S \times T, \cdot)$ if and only if $X$ is memoryless for $(S, \cdot)$ , $Y$ is memoryless for $(T, \cdot)$ , and $X, Y$ are right independent.

Details:

Let $F$ , $G$ , and $H$ denote the reliability functions for $X$ , $Y$ , and $(X, Y)$ with respect to the semigroups $(S, \cdot)$ , $(T, \cdot)$ and $(S \times T, \cdot)$ respectively. Suppose first that $(X, Y)$ is memoryless for $(S \times T, \cdot)$ . Then from Section 1.8, $F (u x) = H (u x, ϵ) = H [(u, ϵ) (x, ϵ)] = H (u, ϵ) H (x . ϵ) = F (u) F (x), u, x \in S$ So $X$ is memoryless for $(S, \cdot)$ By a symmetric argument, $Y$ is memoryless for $(T, \cdot)$ . Next note that $H (x, y) = H [(x, ϵ) (e, y)] = H (x, ϵ) H (e, y) = F (x) G (y), x \in S, y \in T$ so $X$ and $Y$ are right independent. Conversely, suppose that $X$ and $Y$ are memoryless and are right independent. Then $\begin{aligned} H [(u, v) (x, y)] & = H (u x, v y) = F (u x) G (v y) = F (u) F (x) G (v) G (y) \\ = [F (u) G (v)] [F (x) G (y)] = H (u, v) H (x, y), (u, v), (x, y) \in S \times T \end{aligned}$ Hence $(X, Y)$ is memoryless for $(S \times T, \cdot)$ .

Suppose again that $(S, \cdot)$ has identity element $e$ and that $(T, \cdot)$ has identity element $ϵ$ . If $(X, Y)$ is exponential for $(S \times T, \cdot)$ then $X$ is exponential for $(S, \cdot)$ , $Y$ is exponential for $(T, \cdot)$ , and $X$ and $Y$ are right independent.

Details:

Since $(X, Y)$ is exponential for $(S \times T, \cdot)$ , $\begin{aligned} P (X \in x A) & = P [(X, Y) \in x A \times T] = P [(X, Y) \in (x, ϵ) (A \times T)] \\ = P [(X, Y) \in (x, ϵ) (S \times T)] P [(X, Y) \in A \times T] = P (X \in x S) P (X \in A), x \in S, A \in S \end{aligned}$ Hence $X$ is exponential for $(S, \cdot)$ . By a symmetric argument, $Y$ is exponential for $(T, \cdot)$ . Finally, since $(X, Y)$ is exponential for $(S \times T, \cdot)$ , it is also memeoryless and hence $X$ and $Y$ are right independent by [8].

From Section 5, the random variable

(X, Y)

in [9] has constant rate

δ \in (0, \infty)

for

(S \times T, \cdot)

with respect to a left-invariant measure

λ

(S \times T, S \times T)

. Hence

(X, Y)

has density function

h

given by

h (x, y) = δ H (x, y) = δ F (x) G (y)

. But we cannot conclude that that

X

and

Y

are fully independent since we don't know that

λ

is a product measure on

(S \times T, S \times T)

. Note that the canonical such measure

λ

is given by

λ (A \times B) = E [\frac{1}{H (X, Y)}; (X, Y) \in A \times B] = E [\frac{1}{F (X) G (Y)}; X \in A, Y \in B]

But we cannot factor the expression further without full independence of

X

and

Y

. However, we have the following corollary:

In the setting of [9], suppose that $λ = μ \times ν$ is the unique left-invariant measure for $(S \times T, S \times T)$ , up to multiplication by positive constants, where $μ$ is left invariant for $(S, \cdot)$ and $ν$ is left invariant for $(T, \cdot)$ . Then $(X, Y)$ is exponential for $(S \times T, \cdot)$ if and only if $X$ is exponential for $(S, \cdot)$ , $Y$ is exponential for $(T, \cdot)$ , and $X$ and $Y$ are fully independent.

Details:

Suppose that $(X, Y)$ has an exponential distribution for $(S \times T, \cdot)$ . Then by [9], $X$ is exponential for $(S, \cdot)$ , $Y$ is exponential for $(T, \cdot)$ , and $X, Y$ are right independent. But as in the remarks above, $(X, Y)$ has joint density $h$ with respect to $λ = μ \times ν$ of the from $h (x, y) = δ F (x) G (y)$ where $δ \in (0, \infty)$ and where $F$ and $G$ are the reliability functions of $X$ and $Y$ for $(S, \cdot)$ and $(T, \cdot)$ , respectively. From the factorization theorem, $X$ and $Y$ are independent.

The direct product

(S \times T, \cdot)

has several natural sub-semigroups. First,

({(x, ϵ) : x \in S}, \cdot)

is a complete sub-semigroup isomorphic to

(S, \cdot)

. Similalry,

({(e, y) : y \in T}, \cdot)

is a complete sub-semigroup isomorphic to

(T, \cdot)

. If

(S, \cdot) = (T, \cdot)

, then the diagonal

({(x, x) : x \in S}, \cdot)

is a complete sub-semigroup isomorphic to

(S, \cdot)

. The results of this subsection apply to positive semigroups of course since such semigroups have identities. Next we continue with example [6] and give the exponential version of the constant rate distributions studied in Section 1.8.

In the setting of example [6], suppose that random variable $X$ has the geometric distribution on $N_{+}$ with success parameter $p \in (0, 1)$ , random variable $Y$ is uniformly distributed on $I$ , and that $X$ and $Y$ are independent. Then $(X, Y)$ has an exponential distribution for $(N_{+} \times I, \cdot)$ . The rate constant for $(N_{+} \times I, ≺)$ is $p / [k (1 - p)]$ .

Details:

Random variable $X$ has an exponential distributin for $(N_{+}, +)$ and has rate $p / (1 - p)$ for the associated graph $(N_{+}, <)$ . Similarly, random variable $Y$ has an exponential distribution for $(I, \cdot)$ and has rate $1 / k$ for the associated graph $(I, \equiv)$ . So the result follows from [7].

Higher Order Products

Naturally, the results above can be extended to the direct product of

n

semigroups

(S_{1}, \cdot), (S_{2}, \cdot), \dots (S_{n}, \cdot)

for

n \in N_{+}

, and in particular to the

n

-fold direct power

(S^{n}, \cdot)

of a semigroup

(S, \cdot)

. In the latter case, if

λ

is left invariant for

(S, \cdot)

then

λ^{n}

is left invariant for

(S^{n}, \cdot)

for each

n \in N_{+}

. The following definition gives an infinite construction that will be useful.

Suppose that $(S_{i}, \cdot)$ is a discrete semigroup with identity element $e_{i}$ for $i \in N_{+}$ . Let $T = {(x_{1}, x_{2}, \dots) : x_{i} \in S_{i} for each i and x_{i} = e_{i} for all but finitely many i \in N_{+}}$ As before, we define the component-wise operation: $x \cdot y = (x_{1} y_{1}, x_{2} y_{2}, \dots), x = (x_{1}, x_{2}, \dots), y = (y_{1}, y_{2}, \dots) \in T$ Then $(T, \cdot)$ is a discrete semigroup with identity $e = (e_{1}, e_{2}, \dots)$ .

Details:

Note that $T$ is closed under the operation, and is countable by the requirement that $x_{i} = e_{i}$ for all but finitely many $i \in N_{+}$ for $x = (x_{1}, x_{2}, \dots) \in T$ . The other semigroup properties follow just as in [1].

In particular, if

(S_{i}, \cdot)

is a positive semigroup for each

i \in N_{+}

then

(T, \cdot)

is also a positive semigroup.

Marshall-Olkin Distributions

In this subsection we generalize the multivariate exponential distribution defined and studied by Marshall and Olkin. To set the stage, suppose that

(S, \cdot)

is a positive semigroup with identity

e

whose associated partial order graph

(S, ⪯)

is a lattice. For

n \in N_{+}

(S^{n}, \cdot)

denotes the power semigroup of

(S, \cdot)

of order

n

, whose partial order graph

(S^{n}, ⪯_{n})

is the power of

(S, ⪯)

of order

n

, also a lattice. Once again,

(S, \cdot)

is measurable with respect to underlying reference space

(S, S)

, so that

(S^{n}, \cdot)

and the graph

(S^{n}, ⪯_{n})

are measurable with respect to

(S^{n}, S^{n})

The Bivariate Case

Suppose that $U$ , $V$ , and $W$ are right independent and have memoryless distributions on $(S, \cdot)$ . Let $X = U \land W$ and $Y = V \land W$ . Then $(X, Y)$ has a Marshall-Olkin distribution on $(S^{2}, \cdot)$ .

Our first result follows immediatley from the definition and a basic result from Section 6.

Suppose that $(X, Y)$ has a Marshall-Olkin distribution on $(S^{2}, \cdot)$ as in definition [13]. Let $F_{1}$ , $F_{2}$ , and $F_{3}$ denote the reliability functions of $U$ , $V$ , and $W$ on $(S, \cdot)$ respectively. Then

$X = U \land W$ is memoryless with reliability function $F_{1} F_{3}$ .
$Y = V \land W$ is memoryless with reliability function $F_{2} F_{3}$ .
$X \land Y = U \land V \land W$ is memoryless with reliability function $F_{1} F_{2} F_{3}$ .

But of course,

X

and

Y

are dependent. Moreover, a Marshall-Olkin distribution places positive probability on the diagonal.

Suppose that $(X, Y)$ has a Marshall-Olkin distribution on $(S^{2}, \cdot)$ . Then $P (X = Y) > 0$ .

Details:

Suppose that $X = U \land W$ and $Y = V \land W$ as in definition [13]. Then $W ⪯ U, W ⪯ V ⟹ X = W, Y = W ⟹ X = Y$ Hence $P (X = Y) \geq P (U ⪰ W, V ⪰ V) = E [P (U ⪰ W, V ⪰ W ∣ W)] = E [P (U ⪰ W ∣ W) P (V ⪰ W ∣ W)] = E [F_{1} (W) F_{2} (W)]$ From our usual support assumption, the right hand side is positive.

Suppose that

λ

is the left-invariant reference measure for

(S, \cdot)

, so that

λ^{2}

is the left-invariant measure for

(S^{2}, \cdot)

. In the continuous case with

S

uncountable, we typically have

λ^{2} {(x, x) : x \in S} = 0

, so a Marshall-Olkin distribution has an absolutely continuous part and a singular part.

Suppose that $(X, Y)$ has a Marshall-Olkin distribution on $(S^{2}, \cdot)$ as in definition [13], with reliability function $H$ . Then $H (x, y) = F_{1} (x) F_{2} (y) F_{3} (x \lor y), (x, y) \in S^{2}$

Details

By definition and right independence $\begin{aligned} H (x, y) & = P (U \land W ⪰ x, V \land W ⪰ y) = P (U ⪰ x, W ⪰ x, V ⪰ y, W ⪰ y) \\ = P (U ⪰ x, V ⪰ y, W ⪰ x \lor y) = P (U ⪰ x) P (V ⪰ y) P (W ⪰ x \lor y) = F_{1} (x) F_{2} (y) F_{3} (x \lor y), (x, y) \in S^{2} \end{aligned}$

Our next result is the abstract version of one of the original characterizations of the Marshall-Olkin distribution.

Suppose again that $(X, Y)$ has a Marshall-Olkin distribution on $(S^{2}, \cdot)$ as in definition [13], with reliability function $H$ . Then $(X, Y)$ satisfies the partial memoryless property $H [(t, t) (x, y)] = H (t, t) H (x, y), x, y, t \in S$

Details

Let $t, x, y \in S$ . Using [15] we have $H [(t, t) (x, y)] = H (t x, t y) = F_{1} (t x) F_{2} (t y) F_{3} [(t x) \lor (t y)] = F_{1} (t x) F_{2} (t y) F_{3} [t (x \lor y)]$ But since $U$ , $V$ , and $W$ are memoryless, $H [(t, t) (x, y)] = F_{1} (t) F_{1} (x) F_{2} (t) F_{2} (y) F_{3} (t) F_{3} (x \lor y) = H (t, t) H (x, y)$

Stated in terms of conditional probability, the partial memoryless property has the form

P (X ⪰ t x, Y ⪰ t y ∣ X ⪰ t, Y ⪰ t) = P (X ⪰ x, Y ⪰ y), x, y, t \in S

The General Multivariate Case

The extension of the Marshall-Olkin distribution to higher dimensions is a bit complicated and requires some additional notation to state the definition and results cleanly. For

n \in N_{+}

let

B_{n}

denote the set of bit strings of length

n

, excluding the 0 string

00 \dots 0

Suppose that $n \in N_{+}$ and that ${Z_{b} : b \in B_{n}}$ is a collection of right independent variables, each memoryless on $(S, \cdot)$ . Define $X_{i} = inf {Z_{b} : b \in B_{n}, b_{i} = 1}, i \in {1, 2, \dots, n}$ Then $(X_{1}, X_{2}, \dots, X_{n})$ has the Marshall-Olkin distribution on $(S^{n}, \cdot)$ .

So a collection of

2^{n} - 1

right independent, memoryless variables on

(S, \cdot)

is required for the construction of the Marshall-Olkin variable on

(S^{n}, \cdot)

. The marginal distributions are of the same type. For the following results, let

F_{b}

denote the reliability function of

Z_{b}

for

b \in B_{n}

Suppose again that $n \in N_{+}$ and that $X = (X_{1}, X_{2}, \dots, X_{n})$ has a Marshall-Olkin distribution on $(S^{n}, \cdot)$ as in definition [18]. For $k \in N_{+}$ with $k \leq n$ , let $(j_{1}, j_{2}, \dots, j_{k})$ be a subsequence of $(1, 2, \dots, n)$ . Then

$(X_{j_{1}}, X_{j_{2}}, \dots, X_{j_{k}})$ has a Marshall-Olkin distribution on $(S^{k}, \cdot)$ .
$inf {X_{j_{1}}, X_{j_{2}}, \dots, X_{j_{k}}}$ is memoryless on $(S, \cdot)$ with reliability function $\prod {F_{b} : b \in B_{n}, b_{j_{1}} = b_{j_{2}} = \dots = b_{j_{k}} = 1}$

Suppose again that $n \in N_{+}$ and that $X = (X_{1}, X_{2}, \dots, X_{n})$ has a Marshall-Olkin distribution on $(S^{n}, \cdot)$ as in definition [18]. Let $H$ denote the reliability function of $X$ on $(S^{n}, \cdot)$ . Then $H (x_{1}, x_{2}, \dots, x_{n}) = \prod_{b \in B_{n}} F_{b} (sup {x_{i} : b_{i} = 1}), (x_{1}, x_{2}, \dots, x_{n}) \in S^{n}$

Suppose again that $n \in N_{+}$ and that $X = (X_{1}, X_{2}, \dots, X_{n})$ has a Marshall-Olkin distribution on $(S^{n}, \cdot)$ as in definition [18]. Then $X$ has the partial memoryless property $H [(t, t, \dots, t) (x_{1}, x_{2}, \dots, x_{n})] = H (t, t, \dots t) H (x_{1}, x_{2}, \dots, x_{n}), t \in S, (x_{1}, x_{2}, \dots, x_{n}) \in S^{n}$

We will revisit Marshall-Olkin distributions for the standard continous semigroup

([0, \infty), +)

in Section 3.4.

7. Semigroup Products

Basic Theory

Probability

Higher Order Products

Marshall-Olkin Distributions

The Bivariate Case

The General Multivariate Case