The equivalences are clear. For $x\in S$,

1. $$\mathbb{P}(x^{-1}X\in A\mid X\in xS)=\mathbb{P}(X\in xA\mid X\in xS)=\frac{\mathbb{P}(X\in xA)}{F(x)},A\in \mathcal{S}$$
2. $$\mathbb{P}(x^{-1}X\in yS\mid X\in xS)=\mathbb{P}(X\in xyS\mid X\in xS)=\frac{\mathbb{P}(X\in xyS)}{\mathbb{P}(X\in xS)}=\frac{F(xy)}{F(x)},y\in S$$

Let $A=yS$ in part (a) of definition [1]. Then $$F(xy)=\mathbb{P}[X\in (xy)S]=\mathbb{P}[X\in x(yS)]=F(x)\mathbb{P}(X\in yS)=F(x)F(y),x,y\in S$$

The proofs rely on basic algebraic properties of the semigroup.

1. Recall that for $x\in S$, the mapping $A\mapsto xA$ takes $\mathcal{S}$ one-to-one and onto the measurable subsets of $xS$. Specifically, if $A\in \mathcal{S}$ then $xA\in \mathcal{S}$ and $xA\subseteq xS$. Conversely, if $B\in \mathcal{S}$ and $B\subseteq xS$ then $B=xA$ where $A=x^{-1}B=\{t\in S:xt\in B\}$. So let $B\in \mathcal{S}$ with $B\subseteq xS$ and let $A=x^{-1}B$. Then $$\mathbb{P}(X\in B\mid X\in xS)=\mathbb{P}(X\in xA\mid X\in xS)=\mathbb{P}(x^{-1}X\in A\mid X\in xS)$$ and $$\mathbb{P}(xX\in B)=\mathbb{P}(X\in x^{-1}B)=\mathbb{P}(X\in A)$$ So the conditional distributon of $X$ given $X\in xS$ is the same as the distribution of $X$ if and only if the conditional distribution of $x^{-1}X$ given $X\in xS$ is the same as the distribution of $X$.
2. Recall that for $x\in S$, the function $t\mapsto xt$ takes $S$ one-to-one onto $xS$, with inverse function $y\mapsto x^{-1}y$. Both functions are measurable. So let $y\in xS$ so that $y=xt$ for unique $t=x^{-1}y\in S$. Let $F$ denote the reliability function of $X$ on $(S,\to )$. Then the conditional reliability function of $X$ given $X\in xS$ at $y$ is $$\mathbb{P}(X\in yS\mid X\in xS)=\mathbb{P}(X\in xtS\mid X\in xS)=\frac{F(xt)}{F(x)}$$ The reliability function of $xX$ at $y$ is $$\mathbb{P}(xX\in yS)=\mathbb{P}(X\in x^{-1}yS)=\mathbb{P}(X\in tS)=F(t)$$ The two functions are the same if and only if $F(xt)=F(x)F(t)$ for all $t\in S$.

Let $\mathit{Y}=(Y_1,Y_2,\dots )$ 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 $(S,\to )$, the conditional distribution of $Y_{n+1}$ given $Y_n=x$ is the same as the distribution of $X$ given $x\to X$ (that is, $X\in xS$). 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 $xX$. Hence the two random walks are the same if and only if the conditional distribution of $X$ given $X\in xS$ is the same as the distribution of $xX$. By [3], this is the case if and only if $X$ has an exponential distribution.

The equivalence of (a) and (b) follows from definition [1]. The equivalence of (a) and (c) follows from proposition [3].

We first show closure. Suppose that $x,y\in S_X$ and $A\in \mathcal{S}$. Then $$\mathbb{P}[X\in (xy)A]=\mathbb{P}[X\in x(yA)]=F(x)\mathbb{P}(X\in yA)=F(x)F(y)\mathbb{P}(X\in A)$$ In particular, letting $A=S$ we have $F(xy)=F(x)F(y)$, so substituting back we have $$\mathbb{P}[X\in (xy)A]=F(xy)\mathbb{P}(X\in A)$$ and so $xy\in S_X$. Next we show completeness. Suppose that $x,y\in S_X$ and that $x\to y$ so that $y=xu$ for some $u\in S$. We need to show that $u=x^{-1}y\in S_X$. Let $A\in \mathcal{S}$. First, since $x\in S_X$ we have $$\mathbb{P}[X\in (xu)A]=\mathbb{P}[X\in x(uA)]=F(x)\mathbb{P}(X\in uA)$$ On the other hand, since $y=xu\in S_X$ we have $$\mathbb{P}[X\in (xu)A]=F(xu)\mathbb{P}(X\in A)$$ Again since $x\in S_X$ we have $$F(xu)=\mathbb{P}[X\in (xu)S]=\mathbb{P}[X\in x(uS)]=F(x)\mathbb{P}(X\in uS)=F(x)F(u)$$ Combining the displayed equations we have $$F(x)\mathbb{P}(X\in uA)=F(x)F(u)\mathbb{P}(X\in A)$$ Since $F(x)>0$ we have $\mathbb{P}(X\in uA)=F(u)\mathbb{P}(X\in A)$, so $u\in S_X$.

Recall that for $x\in S$, the mapping $A\mapsto xA$ from $\mathcal{S}$ into $\mathcal{S}$ preserves all of the set operations. Let $(A_1,A_2,\dots )$ be a sequence of disjoint sets in ${\mathcal{S}}_X$ and let $x\in S$. Then $(x{A}_1,x{A}_2,\dots )$ is a disjoint sequence and $$\begin{array}{rl}\mathbb{P}(X\in x\bigcup _{i=1}^{\mathrm{\infty }}{A}_i)& =\mathbb{P}(X\in \bigcup _{i=1}^{\mathrm{\infty }}x{A}_i)=\sum _{i=1}^{\mathrm{\infty }}\mathbb{P}(X\in x{A}_i)=\sum _{i=1}^{\mathrm{\infty }}F(x)\mathbb{P}(X\in A_i)\\ & =F(x)\sum _{i=1}^{\mathrm{\infty }}\mathbb{P}(X\in A_i)=F(x)\mathbb{P}(X\in \bigcup _{i=1}^{\mathrm{\infty }}{A}_i)\end{array}$$ Hence $\bigcup _{i=1}^{\mathrm{\infty }}{A}_i\in {\mathcal{S}}_X$. Next let $A,B\in {\mathcal{S}}_X$ with $A\subseteq B$ and let $x\in S$. Then $xA\subseteq xB$ and $x(B-A)=xB-xA$. Hence $$\begin{array}{rl}\mathbb{P}[X\in x(B-A)]& =\mathbb{P}[X\in (xB-xA)]=\mathbb{P}(X\in xB)-\mathbb{P}(X\in xA)\\ & =F(x)\mathbb{P}(X\in A)-F(x)\mathbb{P}(X\in B)=F(x)[\mathbb{P}(X\in B)-\mathbb{P}(X\in A)]\\ & =F(x)\mathbb{P}(X\in B-A)\end{array}$$ Hence $B-A\in {\mathcal{S}}_X$. Of course $S\in {\mathcal{S}}_X$ and hence the other results follow.

Let $F$ denote the reliabiltiy function of $X$. Since $X$ is memoryless, $$\mathbb{P}(X\in xyS)=F(xy)=F(x)F(y)=F(x)\mathbb{P}(X\in yS),x,y\in S$$ It follows that $yS\in {\mathcal{S}}_X$ for $y\in S$ and hence $\mathcal{A}\subseteq {\mathcal{S}}_X$. Since $\mathcal{A}$ is closed under intersection, it is a $\pi$-system, again in the standard terminology of measure theory. Since ${\mathcal{S}}_X$ is a $\lambda$-system, a basic result in measure theory states that ${\mathcal{S}}_0=\sigma (\mathcal{A})\subseteq {\mathcal{S}}_X$.

Since $(S,⪯)$ is an upper semi-lattice, the collection $\mathcal{A}$ of right neighbor sets is closed under intersetion: $xS\cap yS=(x\vee y)S$ for $x,y\in S$. Hence the result follows from [8].

Recall that the assumptions mean that $S$ has a locally compact Hausdorff topology with a countable base, and that the mapping $(x,y)\mapsto xy$ is continuous. Suppose that $(x_n:n\in {\mathbb{N}}_{+})$ is a sequence in $S$ converging to $x\in S$. Let $y\in xS$ satisfy the stated assumptions. Then $\{u\in S:u\to y\}$ is a neighborhood of $x$ so $x_n\to y$ for $n$ sufficiently large. Also, $yS$ has nonempty interior so there exists a (nonempty) open neighborhonnd $U\subset yS$. By Urysohn's lemma, there exists $\phi :S\to [0,1]$ with $\phi >0$ on $U$ and $\phi =0$ on $U^c$ (More specifically, pick $u\in U$ and then separaate the closed, disjoint sets $\{u\}$ and $U^c$ with a continuous function.) By the continuity of multiplication, $\underset{n\to \mathrm{\infty }}{lim}{x}_{n}X=xX$ and by the continuity of $\phi$, $\underset{n\to \mathrm{\infty }}{lim}\phi (x_{n}X)=\phi (xX)$. By the bounded convergence theorem, $\underset{n\to \mathrm{\infty }}{lim}\mathbb{E}[\phi (x_{n}X)]=\mathbb{E}[\phi (xX)]$. Since $U\subseteq xS$, and $\phi =0$ on $U^c$, note that $\mathbb{E}[\phi (X),X\in xS]=\mathbb{E}[\phi (X),X\in U]$. Similarly, $\mathbb{E}[\phi (X),X\in x_{n}S]=\mathbb{E}[\phi (X),X\in U]$ for $n$ sufficiently large. Moreover, the expected values are positive since $X$ is supported by $(S,\cdot )$. By the exponential property, $\mathbb{E}[\phi (X),X\in xS]=F(x)\mathbb{E}[\phi (xX)]$ and similarly, $\mathbb{E}[\phi (X),X\in x_{n}S]=F(x_n)\mathbb{E}[\phi (x_{n}S)]$ for $n$ sufficiently large. Therefore we have $$F(x)=\frac{\mathbb{E}[\phi (X),X\in U]}{\mathbb{E}[\phi (xX)]}$$ Similarly, for $n$ sufficently large, $$F(x_n)=\frac{\mathbb{E}[\phi (X),X\in U]}{\mathbb{E}[\phi (x_{n}X)]}$$ Hence $\underset{n\to \mathrm{\infty }}{lim}F(x_n)=F(x)$.

This follows directly from [10] since $\mathbb{P}(X\in xA)=F(x)\mathbb{P}(X\in A)$ for $A\in \mathcal{S}$.

Let $F$ denote the reliability function of $X$ on $(S,\to )$. Suppose first that $X$ has an exponential distribution on $(S,\cdot )$. Then by [2], $X$ has a memoryless distribution. Now let $\mu$ be the $\sigma$-finite measure defined by $$\mu (A)=\mathbb{E}[\frac{1}{F(X)},X\in A],A\in \mathcal{S}$$ Then from Section 1.5, $X$ has density $F$ with respect to $\mu$ and hence has contant rate 1 with respect to $\mu$: $$\mathbb{P}(X\in A)={\int }_{A}F(x)d\mu (x),A\in \mathcal{S}$$ Let $x\in S$ and $A\in \mathcal{S}$. By the integral version of the exponential property in and by the memoryless property, $$\begin{array}{rl}\mu (xA)& =\mathbb{E}[\frac{1}{F(X)},X\in xA]=F(x)\mathbb{E}[\frac{1}{F(xX)},X\in A]\\ & =F(x)\mathbb{E}[\frac{1}{F(x)F(X)},X\in A]=\mathbb{E}[\frac{1}{F(X)},X\in A]=\mu (A)\end{array}$$ so $\mu$ is left invariant on $(S,\cdot )$. Conversely, suppose that the distribution of $X$ is memoryless on $(S,\cdot )$ and has constant rate $\alpha \in (0,\mathrm{\infty })$ on $(S,\to )$ 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 $A\in \mathcal{S}$. Using the memoryless property and the integral version of left invariance in Section 3, $$\begin{array}{rl}\mathbb{P}(X\in xA)& ={\int }_{xA}\alpha F(y)d\lambda (y)={\int }_A\alpha F(xz)d\lambda (z)={\int }_A\alpha F(x)F(z)d\lambda (z)\\ & =F(x){\int }_A\alpha F(z)d\lambda (z)=F(x)\mathbb{P}(X\in A)\end{array}$$ Hence $X$ has an exponential distribution.

Suppose first that $F$ is the reliability function of an exponential distribution for $(S,\cdot )$ that has constant rate $\alpha \in (0,\mathrm{\infty })$ with respect to $\lambda$. By [2], the distribution is memoryless, so (a) holds. Also $\alpha F$ is a probability density function so $${\int }_{S}F(x)d\lambda (x)=\frac{1}{\alpha }<\mathrm{\infty }$$ and hence (b) holds. Conversely, suppose that (a) and (b) hold. Let $f=\alpha F$ where $\alpha =1/{\int }_{S}F(x)d\lambda (x)$. Then by (b), $f$ is a probability density function. Let $X$ be a random variable with density $f$, and let $x\in S$ and $A\in \mathcal{S}$. Using (a) and the integral version of the left invariance property in Section 3, $$\begin{array}{rl}\mathbb{P}(X\in xA)& ={\int }_{xA}f(y)d\lambda (y)={\int }_{xA}\alpha F(y)d\lambda (y)={\int }_A\alpha F(xz)d\lambda (z)\\ & =F(x){\int }_A\alpha F(z)d\lambda (z)=F(x){\int }_{A}f(z)d\lambda (z)=F(x)\mathbb{P}(X\in A)\end{array}$$ 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 $\alpha$.

In particular, this would apply if the associated graph $(S,\to )$ is stochastic, and with the right $\sigma$-algebra ${\mathcal{S}}_0=\mathcal{S}$, the reference $\sigma$-algebra. Compare this result with propositions [8] and [9].

From Section 1.5, $H$ is the reliability function for a distribution with constant rate $\alpha$. On the other hand, $$H(xy)=pF(xy)+(1-p)G(xy)=pF(x)F(y)+(1-p)G(x)G(y),x,y\in S$$ while $$\begin{array}{rl}H(x)H(y)& =[pF(x)+(1-p)G(x)][pF(y)+(1-p)G(y)]\\ & =p^{2}F(x)F(y)+p(1-p)[F(x)G(y)+F(y)G(x)]+(1-p{)}^{2}G(x)G(y),x,y\in S\end{array}$$ So $$H(xy)-H(x)H(y)=p(1-p)[F(x)F(y)-F(x)G(y)-F(y)G(x)+G(x)G(y)],x,y\in S$$ With $x=y$ we have $$H(x^2)-H^2(x)=p(1-p)[F(x)-G(x){]}^2,x\in S$$ So if $F(x)\ne G(x)$ for some $x\in S$ then $H(x^2)\ne H^2(x)$.

Clearly $F^m(xy)=F^m(x)F^m(y)$ for $x,y\in S$. Thus the result follows immediately from [14].

Let $e$ denote the identity element of $(S,\cdot )$. Letting $y=e$ in the equation above gives $G(x)=\alpha f(x)$ where $\alpha =f(e)\in (0,\mathrm{\infty })$. Let $F$ denote the reliability function of $f$ on $(S,\cdot )$. Then using the integral version of the left-invariance property in Section 3, $$\begin{array}{rl}F(x)& ={\int }_{xS}f(y)d\lambda (y)={\int }_{xS}\frac{1}{\alpha }G(y)d\lambda (y)=\frac{1}{\alpha }{\int }_{S}G(xu)d\lambda (u)\\ & =\frac{1}{\alpha }{\int }_{S}f(x)f(u)d\lambda (u)=\frac{1}{\alpha }f(x),x\in S\end{array}$$ Thus the distribution has constant rate $\alpha$. Finally, $$F(xy)=\frac{1}{\alpha }f(xy)=\frac{1}{{\alpha }^2}G(xy)=\frac{1}{{\alpha }^2}f(x)f(y)=F(x)F(y)$$ so the distribution is memoryless. Hence $f$ is the density of an exponential distribution by [12].

These results are simply restatements of results in Section 1.5 and hold since the random walk on the semigroup $(S,\cdot )$ is the same as the random walk on the graph $(S,\to )$ by [4]. Note that the transition density $P$ satisfies $P(x,y)=f(y)/F(x)=f(x^{-1}y)$ for $x\to y$ where $f=\alpha F$ is the density function..

Suppose that $\lambda (S)<\mathrm{\infty }$ and let $X$ have the uniform distribution on $S$. Since the relation associated with $(S,\cdot )$ is complete, (equivalently $xS=S$ for all $x\in S$), the reliability function $F$ of $X$ is the constant 1: $\mathbb{P}(X\in xS)=1$ for $x\in S$. Hence $$\mathbb{P}(X\in xA)=\frac{\lambda (xA)}{\lambda (S)}=\frac{\lambda (A)}{\lambda (S)}=\mathbb{P}(X\in xS)\mathbb{P}(X\in A),x\in S,A\in \mathcal{S}$$

The assumptions mean that $S$ has an LCCB topology and that the mapping $(x,y)\mapsto x^{-1}y$ is continuous on $S^2$. Since $(S,\cdot )$ is a group, the associated relation is the complete reflexive relation $\equiv$.

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 the constant function 1, as noted in [21]. 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 [14]. If $F(xy)=F(x)F(y)$ then $F(y)=F(x)F(y)$ for $x,y\in S$. From our support assumption, $F(y)>0$ 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 $\lambda$) $$\mathbb{P}(X\in A)=\frac{\lambda (A)}{\lambda (S)},A\in \mathcal{S}$$ On the other hand, if $\lambda (S)=\mathrm{\infty }$ 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 $X$: $$\mathbb{E}[\frac{1}{F(X)};X\in A]=\mathbb{P}(X\in A),A\in \mathcal{S}$$ and trivially, $X$ has the uniform distribution with repsect to itself: $$\mathbb{P}(X\in A)=\frac{\mathbb{P}(X\in A)}{\mathbb{P}(X\in S)},A\in \mathcal{S}$$

From Section 3, $\mathrm{\#}$ is the unique left invariant measure for $(S,\cdot )$, up to multiplication by positive constants. From Section 1.5 if $X$ has constant rate $\alpha$ for a discrete partial order graph $(S,⪯)$ then $\alpha \in (0,1]$, and $\alpha =1$ if and only if the graph is $(S,=)$. But a discrete positive semigroup $(S,\cdot )$ has associted partial order graph $(S,=)$ if and only if the semigroup is trivial: $S=\{e\}$.

We assume of course that $I\ne \mathrm{\varnothing }$ so that $(S,\cdot )$ is nontrivial. Let $F$ and $F_1$ denote the reliability functions of $X$ for $(S,⪯)$ and $(S,\uparrow )$ respectively. Then since $X$ has constant rate $\alpha$ for $(S,⪯)$ and is memoryless for $(S,\cdot )$, $$\begin{array}{rl}{F}_1(x)& =\sum _{x\uparrow y}f(y)=\sum _{i\in I}f(xi)=\sum _{i\in I}\alpha F(xi)\\ & =\sum _{i\in I}\alpha F(x)F(i)=\alpha F(x)\sum _{i\in I}F(i)=f(x)\sum _{i\in I}\frac{1}{\alpha }f(i)\end{array}$$

This follows from a general result in Section 1.5