Lexicographic Sums

In this section we generalize the lexicographic product of graphs studied in Section 8.

General Theory

Graph Basics

Suppose that \((S, \rta)\) is a discrete, irreflexive graph (so a graph without loops in the combinatorial sense) and that \((T_x, \upa_x)\) is a measurable graph for each \(x \in S\). Underlying \((S, \rta)\) is the discrete measure space \((S, \ms S, \#)\) of course, where \(\ms S = \ms P(S)\) is the power set of \(S\) and \(\#\) is counting measure. Underlying \((T_x, \upa_x)\) is a \(\sigma\)-finite measure space \((T_x, \ms T_x, \mu_x)\), as usual. Let \(T = \bigcup_{x \in S} T_x\). Here is the new space we will need:

Define the \(\sigma\)-finite measure space \((U, \ms U, \lambda)\) as follows:

\[U = \bigcup_{x \in S} \left(\{x\} \times T_x\right)\]
\[\ms U = \left\{\bigcup_{x \in S} \left(\{x\} \times A_x\right): A_x \in \ms T_x \text{ for all } x \in S\right\}\]
\[\lambda\left[\bigcup_{x \in S} \left(\{x\} \times A_x\right)\right] = \sum_{x \in S} \mu_x(A_x), \quad A_x \in \ms T_x \text{ for } x \in S\]

Note that \(U \subseteq S \times T\). In the special case that \((T_x, \ms T_x, \mu)\) is the common measure space \((T, \ms T, \mu)\) for all \(x \in S\), the measure space \((U, \ms U, \lambda)\) is the product space \((S \times T, \ms S \times \ms T, \# \times \mu)\). Next is the graph we will study.

The lexicographic sum of the graphs \((T_x, \upa_x)\) over \((S, \rta)\) is the graph \((U, \nea)\) where for \((u, v), \, (x, y) \in U\), \[(u, v) \nea (x, y) \text{ if and only if } u \rta x \text{ or } (u = x \text{ and } v \upa_x y)\]

Details:

Since \((S, \rta)\) is irreflexive, the two conditions defining \(\nea\) are mutually exclusive. The measurability of \((U, \nea)\) is clear since \(\nea\) can be written as a countable union of measurable sets in \(U^2\).

In the special case that \((T_x, \upa_x)\) is the common graph \((T, \upa)\) (on the common measure space) for all \(x \in S\), the graph \((S \times T, \nea)\) is the lexicographic product of \((S, \rta)\) and \((T, \upa)\) as studied in Section 8. The lexicographic sum (and particularly the lexicographic product) is a common construction in ordinary combinatorial graph theory. It is also common in the study of partial orders, and it is in this context that the graph gets its name.

Suppose that \((S, \prec_0)\) is a discrete, strict partial order graph.

If \((T_x, \prec_x)\) is a strict partial order graph for each \(x \in S\), then the lexicographic sum \((U, \prec)\) is also a strict partial order graph: \[(u, v) \prec (x, y) \text{ if and only if } u \prec_0 x \text{ or } (u = x \text{ and } v \prec_x y)\]
If \((T_x, \preceq_x)\) is a partial order graph for each \(x \in S\), then the lexicographic sum \((U, \preceq)\) is also a partial order graph: \[(u, v) \preceq (x, y) \text{ if and only if } u \prec_0 x \text{ or } (u = x \text{ and } v \preceq_x y)\]

Details:

The proofs are simple from the various cases in the defintion.

First \((x, y) \not \prec (x, y)\) for \(x \in S, \, y \in T_x\), so \(\prec\) is irreflexive. If \((u, v) \prec (x, y)\) and \((x, y) \prec (z, w)\) then \((u, v) \prec (z, w)\) for \((u, v), \, (x, y), \, (z, w) \in U\), so \(\prec\) is transitive.
It's clear that \(\preceq\) is the partial order associated with the strict partial order \(\prec\) in (a), when \(\preceq_x\) is the partial order associated with the strict partial order \(\prec_x\) for each \(x \in S\).

Returning to the general setting, there is a simple relationship between right neigbor sets.

Let \(A_x\) denote the set of right neighbors of \(x \in S\) for the graph \((S, \rta)\) and for \(x \in S\) and \(y \in T_x\) let \(B_{(x, y)}\) denote the set of right neighbors of \(y\) for the graph \((T_x, \upa)\). Then the set of right neighbors of \((x, y) \in U\) for the graph \((U, \nea)\) is \[C_{(x, y)} = \left(A_x \times T_x\right) \cup \left(\{x\} \times B_{(x, y)}\right)\]

Details:

This follows from the definition of the graph \((U, \nea)\) given in ,

Recall that the \(\sigma\)-algebra associated with each graph is the \(\sigma\)-algebra generated by the collection of right neighbor sets. There is also a simple relationship between the walk functions of order 1. The relationship is more complicated for higher orders, but the order 1 relationship is important for the constant rate distributions that we will discuss below. For \(x \in S\) let \(a_x\) denote the walk functions of order 1 for \((T_x, \upa_x)\), and let \(b\) and \(c\) denote the walk functions of order 1 for \((S, \rta)\) and \((U, \nea)\), respectively.

The left walk functions of order 1 are related as follows: \[c(x, y) = \sum_{u \rta x} \mu_u(T_u) + a_x(y), \quad x \in S, \, y \in T_x\]

Details:

This follows from the definition of the walk function. \begin{align*} c(x, y) &= \lambda\{(u, v) \in U: (u, v) \nea (x, y)\}\\ &= \lambda\{(u, v): u \in S, v \in T_u, u \rta x\} + \lambda\{(x, v): v \in T_x, v \upa y\}\\ &= \sum_{u \rta x} \mu_u(T_u) + a_x(y), \quad x \in S, \, y \in T_x \end{align*}

Note that \(c(x, y) \lt \infty\) implies \(a_x(y) \lt \infty\) and \(\sum_{u \rta x} \mu_u(T_u) \lt \infty\). Conversely, if \(a_x(y) \lt \infty\), \(b(x) \lt \infty\) and \(\mu_u(T_u) \lt \infty\) for every \(u \rta x\) then \(c(x, y) \lt \infty\).

Probability

Suppose now that \((X, Y)\) is a random variable in \(U\), so that \(X\) has values in \(S\), and given \(X = x\), random variable \(Y\) has values in \(T_x\). Unconditionally, \(Y\) has values in \(T = \bigcup_{x \in S} T_x\). To setup the notation, let \(F\) denote the reliability function of \(X\) for \((S, \rta)\), and for \(x \in S\), let \(G(\cdot \mid x)\) denote the conditional reliability function of \(Y\) for \((T_x, \upa_x)\), given that \(X = x\). Let \(H\) denote the reliability function of \((X, Y)\) for lexicographic sum graph \((U, \nea)\). Finally, let \(f\) denote the density function of \(X\), so that \(f(x) = \P(X = x)\) for \(x \in S\). We assume that \(f(x) \gt 0\) for \(x \in S\).

The reliability functions are related as follows: \[H(x, y) = F(x) + f(x) G(y \mid x), \quad x \in S, \, y \in T_x\]

Details:

This follows from the definition of the lexicographic relation \(\nea\): \begin{align*} H(x, y) &= \P[(x, y) \nea (X, Y)] = \P(x \rta X) + \P(X = x, y \upa Y) \\ &=\P(x \rta X) + \P(X = x) \P(y \upa Y \mid X = x) = F(x) + f(x) G(y \mid x), \quad x \in S, \, y \in T_x \end{align*} Note that since the discrete graph \((S, \rta)\) is irreflexive, the events \(\{x \rta X\}\) and \(\{X = x\}\) are disjoint.

The reason for requiring that the graph \((S, \rta)\) be discrete is clear: If \(X\) has a continuous distribution on an uncountable set \(S\) then \(\P(X = x) = 0\) for \(x \in S\) and so the lexicographic relation is irrelevant from a probabilistic viewpoint. The following results on the density function and rate function of \((X, Y)\) is a simple corollary. As usual, we have special interest in constant rate distributions.

Suppose that given \(X = x \in S\), random variable \(Y\) has density function \(g( \cdot \mid x)\) on \(T_x\).

\((X, Y)\) has density \(h\) given by \(h(x, y) = f(x) g( y \mid x)\) for \(x \in S, \, y \in T_x\).
\((X, Y)\) has rate function for \((U, \nea)\) given by \[ (x, y) \mapsto \frac{f(x) g(y \mid x)}{F(x) + f(x) G(y \mid x)}, \quad x \in S, \, y \in T_x \]
The condition for \((X, Y)\) to have constant rate \(\alpha \in (0, \infty)\) is \[f(x) g(y \mid x) = \alpha [F(x) + f(x) G(y \mid x)], \quad x \in S, \, y \in T_x\]

The condition in part (c) if can be written as \[g(y \mid x) - \alpha G(y \mid x) = \alpha \frac{F(x)}{f(x)}, \quad x \in S, \, y \in T_x\] So in particular, the left side of the displayed equation does not depend on \(y \in T_x\), while the right side is \(\alpha / r(x)\) where \(r\) is the rate function of \(X\) for \((S, \rta)\).

Suppose that \(X\) has constant rate \(\alpha \in (0, \infty)\) for \((S, \rta)\), and that given \(X = x \in S\), the conditional density function of \(Y\) and the conditional reliability function of \(Y\) for \((T_x, \upa_x)\) are related by \[g(y \mid x) = \beta[1 + \alpha G(y \mid x)], \quad y \in T_x\] for some \(\beta \in (0, \infty)\). Then \((X, Y)\) has constant rate \(\alpha \beta\) for \((U, \nea)\).

Details:

The assumptions are \(f(x) = \alpha F(x)\) for \(x \in S\) and \(g(y \mid x) = \beta[1 + \alpha G(y \mid x)]\) for \(x \in S\) and \(y \in T_x\). The result then follows from

For the following result, recall the notation for the walk functions given in . Recall also that the existence of a constant rate distribution for a graph implies that the walk function is finite.

Suppose that \((X, Y)\) has constant rate \(\alpha \in (0, \infty)\) for \((U, \nea)\). Then

The density function \(f\) of \(X\) is given by \[f(x) = \frac{\alpha F(x) \mu_x(T_x)}{1 - \alpha \E[a_x(Y) \mid X = x]}, \quad x \in S\]
For \(x \in S\), a conditional density of \(Y\) given \(X = x\) is defined by \[g(y \mid x) = \alpha \left[\frac{F(x)}{f(x)} + G(y \mid x)\right], \quad y \in T_x\]
\(Y\) has density function \(g\) defined by \[g(y) = \alpha \left(\E[b(X)] + \E[G(y \mid X)]\right), \quad y \in T\]

Details:

By assumption, \(h = \alpha H\) is a probability density function of \((X, Y)\).

This follows the basic moment result in Section 3: \begin{align*} f(x) &= \int_{T_x} h(x, y) \, d\mu_x(y) = \alpha F(x) \mu_x(T_x) + \alpha f(x) \int_{T_x} G(y \mid x) d\mu_x(y)\\ &= \alpha F(x) \mu_x(T_x) + \alpha f(x) \E[a_x(Y) \mid X = x], \quad x \in S \end{align*} solving for \(f(x)\) gives the result.
For \(x \in S\), \[g(y \mid x) = \frac{h(x, y)}{f(x)} = \alpha \frac{F(x) + f(x) G(y \mid x)}{f(x)}, \quad y \in T_x\] Simplifying gives the result.
Again we use the basic moment result in Section 3. \[g(y) = \sum_{x \in S} h(x, y) = \alpha \sum_{x \in S} F(x) + \alpha \sum_{x \in S} f(x) G( y \mid x) = \alpha \E[b(X)] + \alpha \E[G(y \mid X)], \quad y \in T\]

Examples

Standard Graphs

In this subsection, we generalize the standard example of the lexicographic product in Section 8. We use similar notation to make the results easier to compare.

Let \((U, \preceq)\) denote the lexicographic sum of \(([n, n + 1), \le)\) over \((\N, \lt)\). So \((m, x) \preceq (n, y)\) if and only if \(m \lt n\) or (\(m = n\) and \(x \le y\)) for \(m, \, n \in \N\), \(x \in [m, m +1)\) and \(y \in [n, n + 1)\). The relation \(\preceq\) is a total order on \(U\).

Details:

Of course, \((\N, \lt)\) has the usual discrete measure structure while \(([n, n + 1), \le)\) has the usual Lebesgue measure structure. From , \(\preceq\) is a partial order. Suppose that \((m, x), \, (n, y) \in U\) so that \(m, \, n \in \N\), \(x \in [m, m + 1)\), and \(y \in [n, n + 1)\). Either \(m \lt n\) or \(n \lt m\) or \(m = n\). In the last case, either \(x \le y\) or \(y \le x\). So in all four cases, either \((m, x) \preceq (n, y)\) or \((n, y) \preceq (m, x)\).

For \(\alpha \in (0, \infty)\), there exists a unique distribution with constant rate \(\alpha\) on \((S, \preceq)\). If \((X, N)\) has this distribution then

The conditional distribution of \(X\) given \(N = n \in \N\) has density function \(g(\cdot \mid n)\) and reliability function \(G(\cdot \mid n)\) for \(([n, n + 1), \le)\) defined by \begin{align*} g(x \mid n) & = \frac{\alpha}{1 - e^{-\alpha}} e^{-\alpha (x - n)}, \quad x \in [n, n + 1)\\ G(x \mid n) & = \frac{1}{1 - e^{-\alpha}} \left[e^{-\alpha (x - n)} - e^{-\alpha}\right], \quad x \in [n, n + 1) \end{align*}
\(N\) has the geoemtric distribution with success parameter \(1 - e^{-\alpha}\), with density function \(f\) defined by \(f(n) = (1 - e^{-\alpha}) e^{-\alpha n}\) for \(n \in \N\).
The (unconditional) distribution of \(X\) is exponential on \([0, \infty)\) with parameter \(\alpha\), and so with density function \(g\) defined by \(g(x) = \alpha e^{-\alpha x}\) for \(x \in [0, \infty)\).

Details:

Note that \(G(x \mid n) = \int_x^{n + 1} g(y \mid n) \, dy\) for \(n \in \N\) and \(x \in [n, n + 1)\). So from the comment after we have \(g^\prime(x \mid n) + \alpha g(x \mid n) = 0\) for \(n \in \N\) and \(x \in [n, n + 1)\). Solving the exponential differential equation and applying the normalizing condition \(\int_n^{n + 1} g(x \mid n) \, dx = 1\) leads to the first displayed equation in (a). Integrating then leads to the second displayed equation.
From part (a), \[g(x \mid n) - \alpha G(x \mid n) = \alpha \frac{e^{-\alpha}}{1 - e^{-\alpha}}, \quad n \in \N, x \in [n, n + 1)\] and so returning to the reliability function \(F\) of \(N\) for \((\N, \lt)\) satisfies \(F(n) = \beta f(n)\) for \(n \in \N\) wherre \(\beta = e^{-\alpha} / (1 - e^{-\alpha})\). But \(F(n) - F(n + 1) = f(n + 1)\) so \(f\) satisfies the difference equation \[f(n + 1) = \frac{\beta}{\beta + 1} f(n) = e^{-\alpha} f(n), \quad n \in \N\] Solving and applying the normalizing condition \(\sum_{n = 0}^\infty f(n) = 1\) gives the geometric distribution in part (b). More generally, note that \(N\) has constant rate for \((\N, \lt)\). As shown in Section 4.2 this implies that \(N\) has the given geometric distribution.
Note that \(X = x \in [0, \infty)\) implies that \(N = n\) where \(n = \lfloor x \rfloor\) so that \(x \in [n, n + 1)\). Hence \(g(x) = f(n) g(x \mid n)\) for \(x \in [0, \infty)\) where again \(n = \lfloor x \rfloor\). Substituting from parts (a) and (b) and simplifying gives \(g(x) = \alpha e^{-\alpha x}\) for \(x \in [0, \infty)\).

Of course, \(N\) has constant rate \(1 - e^{-\alpha}\) for the standard discrete graph \((\N, \le)\) and (the unconditional distribution of) \(X\) has constant rate \(\alpha\) for the standard continuous graph \(([0, \infty), \le)\). On the other hand, the conditional distribution of \(X\) given \(N = n \in \N\) has rate function \(r(\cdot \mid n)\) for \(([n, n + 1), \le)\) defined by \[r(x \mid n) = \frac{\alpha e^{-\alpha (x - n)}}{e^{-\alpha (x - n)} - e^{-\alpha}}, \quad x \in [n, n + 1)\] This rate function increases from \(\alpha / (1 - e^{-\alpha})\) to \(\infty\) as \(x\) increases from \(n\) to \(n + 1\). If the results in look familiar, it's because they are variations on some standard results.

Suppose that \(X\) has the exponential distribution on \([0, \infty)\) with parameter \(\alpha \in (0, \infty)\).Then

\(N = \lfloor X \rfloor\) has the geometric distribution on \(\N\) with success parameter \(1 - e^{-\alpha}\).
The conditional distribution of \(X\) given \(N = n \in \N\) has density function \(g(\cdot \mid n)\) defined by \[g(x \mid n) = \frac{\alpha}{1 - e^{-\alpha}} e^{-\alpha (x - n)}, \quad x \in [n, n + 1)\]

Details:

These are standard results. \(X\) has density function \(g\) and has reliability function \(G\) for \(([0, \infty), \le)\) defined by \(g(x) = \alpha e^{-\alpha x}\) and \(G(x) = e^{-\alpha x}\) for \(x \in [0, \infty)\).

Note that \(N = n\) if and only if \(n \le X \lt n + 1\) for \(n \in \N\) so \(\P(N = n) = G(n) - G(n + 1) = (1 - e^{-\alpha}) e^{-\alpha n}\).
The conditional density of \(X\) given \(N = n\) is \[g(x \mid n) = \frac{g(x)}{\P(N = n)} = \frac{\alpha}{1 - e^{-\alpha}} e^{-\alpha (x - n)}, \quad x \in [n, n + 1)\]

In the context of , if we let \(Y = X - N\) so that \(Y\) is a random variable in \([0, 1)\), then we are in the setting of the example in Section 8: \((N, Y)\) has constant rate \(\alpha\) for the lexicographic product of \((\N, \lt)\) and \(([0, 1), \le)\), and this product graph in turn is isomorphic to \(([0, \infty), \le)\).

Sums Over Equality Graphs

For the simple examples in this subsection and the next, we assume that \(T_x\) is finite and let \(k_x = \#(T_x) \in \N_+\) for \(x \in S\). For this subsection, let \((U, \nea)\) denote the lexicographic sum of the equality graphs \((T_x, =)\) over the discrete, irreflexive graph \((S, \rta)\). So \((u, v) \nea (x, y)\) if and only if either \(u \rta x\) or \((u, v) = (x, y)\).

For the graph \((U, \nea)\),

The set of right neighbors of \((x, y) \in U\) is \(C_{(x, y)} = (A_x \times T_x) \cup \{(x, y)\}\) where \(A_x\) is the set of right neighbors of \(x \in S\) for the graph \((S, \rta)\).
The walk function \(c\) is given by \[c(x, y) = \sum_{u \rta x} k_u + 1, \quad (x, y) \in U\]

Details:

This follows from .
This follows from .

Random variable \((X, Y)\) in \(U\) has constant rate \(\alpha \in (0, 1)\) for \((U, \nea)\) if and only if

\(X\) has rate function \(r\) for \((S, \rta)\) defined by \[r(x) = \frac{\alpha}{1 - \alpha} k_x, \quad x \in S\]
The conditional distribution of \(Y\) given \(X = x\) is uniform on \(T_x\) for \(x \in S\).

Details:

Note that the conditional reliability function of \(Y\) for \((T_x, =)\) is \(G(y \mid x) = g(y \mid x)\) for \(x \in S\) and \(y \in T_x\). So the results follow from

In particular, if \(k_x = k \in \N_+\) for all \(x \in S\) then \(X\) has constant rate \(\alpha k / (1 - \alpha)\). This can also be derived from . We now specialize a bit further.

Suppose that the base graph \((S, \rta)\) is \((\N, \rta)\) where \(n \rta n + 1\) for \(n \in \N\) (so that \(\rta\) is the covering relation for the standard order \(\le\)). Random variable \((X, Y)\) in \(U\) has constant rate \(\alpha \in (0, 1)\) for \((U, \nea)\) if and only if

\(X\) has density function \(f\) defined as follows, where \(f(0)\) is the normalizing constant: \[f(n) = f(0) \left(\frac{1 - \alpha}{\alpha}\right)^n \frac{1}{k_0 k_1 \cdots k_{n - 1}}, \quad n \in N\]
The conditional distribution of \(Y\) given \(X = n\) is uniform on \(T_n\) for \(x \in S\).

Details:

In this case, the reliability function \(F\) of \(X\) for \((\N, \rta)\) is \(F(n) = f(n + 1)\). So from part (a) of we have \[f(n + 1) = \frac{1 - \alpha}{\alpha} \frac{1}{k_n} f(n), \quad n \in \N\] So \(f\) has the representation given in part (a). The normalizing constant is \(f(0) = 1 / C\) where \[C = \sum_{n = 0}^\infty \left(\frac{1 - \alpha}{\alpha}\right)^n \frac{1}{k_0 k_1 \cdots k_{n - 1}}\] Note that this sum converges regardless of the value of \(k_n \in \N_+\) for \(n \in \N\).

In the context of ,

If \(k_n = k \in \N_+\) for \(n \in \N\) then \(X\) has the geometric distribution on \(\N\) with success parameter \(1 - (1 - \alpha)/(\alpha k)\).
If \(k_n = n + 1\) for \(n \in \N\) then \(X\) has the Poisson distribution on \(\N\) with parameter \((1 - \alpha) / \alpha\).

Sums Over Complete Graphs

Next let \((U, \nea)\) denote the lexicographic sum of the complete reflexive graphs \((T_x, \equiv)\) over the discrete, irreflexive graph \((S, \rta)\). So \((u, v) \nea (x, y)\) if and only if either \(u \rta x\) or \(u = x\). Note that the second coordinates play no role in the relation \(\nea\), and in terms of the first coordinate, the relation is simply the reflexive closure of \(\rta\).

For the graph \((U, \nea)\),

The set of right neighbors of \((x, y) \in U\) is \(C_{(x, y)} = (A_x \times T_x) \cup (\{x\} \times T_x)\) where \(A_x\) is the set of right neighbors of \(x \in S\) for the graph \((S, \rta)\).
The walk function \(c\) is given by \[c(x, y) = \sum_{u \rta x} k_u + k_x, \quad (x, y) \in U\]

Details:

This follows from .
This follows from .

Random variable \((X, Y)\) in \(U\) has constant rate \(\alpha \in (0, 1)\) for \((U, \nea)\) if and only if

\(\alpha \lt 1 / k_x\) for all \(x \in S\) and \(X\) has rate function \(r\) for \((S, \rta)\) defined by \[r(x) = \frac{\alpha k_x}{1 - \alpha k_x}, \quad x \in S\]
The conditional distribution of \(Y\) given \(X = x\) is uniform on \(T_x\) for \(x \in S\).

Details:

Note that the conditional reliability function of \(Y\) for \((T_x, =)\) is \(G(y \mid x) = 1\) for \(x \in S\) and \(y \in T_x\). So the results follow from

In particular, \(k_x \in \N_+\) must be bounded in \(x \in S\) for the existence of a constant rate distribution. If \(k_x = k \in \N_+\) for all \(x \in S\) then \(X\) has constant rate \(\alpha k / (1 - \alpha k)\).

\(X\) has density function \(f\) defined as follows, where \(f(0)\) is the normalizing constant: \[f(n) = f(0) \left(\frac{1 - \alpha k_0}{k_0}\right) \left(\frac{1 - \alpha k_1}{k_1}\right) \cdots \left(\frac{1 - \alpha k_{n - 1}}{k_{n - 1}}\right), \quad n \in N\]
The conditional distribution of \(Y\) given \(X = n\) is uniform on \(T_n\) for \(x \in S\).

Details:

In this case, the reliability function \(F\) of \(X\) for \((\N, \rta)\) is \(F(n) = f(n + 1)\). So from part (a) of we have \[f(n + 1) = \frac{1 - \alpha k_n}{k_n} f(n), \quad n \in \N\] So \(f\) has the representation given in part (a). The normalizing constant is \(f(0) = 1 / C\) where \[C = \sum_{n = 0}^\infty \left(\frac{1 - \alpha k_0}{k_0}\right) \left(\frac{1 - \alpha k_1}{k_1}\right) \cdots \left(\frac{1 - \alpha k_{n - 1}}{k_{n - 1}}\right)\] Note that this sum converges regardless of the value of \(k_n \in \N_+\) for \(n \in \N\): If \(k_n = 1\) then \((1 - \alpha k_n) / k_n = 1 - \alpha\). If \(k_n \ge 2\) then \((1 - \alpha k_n) / k_n \lt 1 / 2\). So the sum is dominated by a convergent geometric series.

In the special case that \(k_n = k \in \N_+\) for all \(n \in \N\), \(X\) has a geometric distribution with success parameter \(1 - (1 - \alpha k) / k\). Another interesting example of lexicographic sum in the context of the standard discrete space is given in Section 3.5.