This section continues the setting of Section 3, but with additional assumption that the underlying space is discrete.
Suppose that \(S\) is countably infinite with counting measure \(\#\) as the reference measure, and that \(\varphi: S \to \N\) induces a partition \(\ms P = \{S_n: n \in \N\}\) of \(S\) into finite, nonempty subsets \(S_n = \varphi^{-1}\{n\}\) for \(n \in \N\). Let \(k_n = \#(S_n) \in \N_+\) for \(n \in \N\). Define the strict partial order \(\prec\) on \(S\) by \(x \prec y\) if and only if \(\varphi(x) \lt \varphi(y)\), and let \(\preceq\) denote the corresponding partial order.
That is, for \((x, y) \in S^2\), \(x \prec y\) if and only if \(x \in S_m\), \(y \in S_n\) for some \(m, \, n \in \N\) with \(m \lt n\). The strict partial order graph \((S, \prec)\) is the graph induced by \((\N, \lt)\) as studied in Section 3, but now we are interested in the partial order graph \((S, \preceq)\). So \(x \preceq y\) if and only if \(x = y\) or \(x \in S_m, \, y \in S_n\) for some \(m, \, n \in \N\) with \(m \lt n\). This graph is not an induced graph, but can be constructed as the lexicographic sum of the graphs \((S_n, =)\) over \((\N, \lt)\) as described in Section 1.8. The elements in \(S_0\) are minimal, and in particular, if \(S_0 = \{e\}\) then \(e\) is the minimum element of \(S\). Note that the covering graph of \((S, \preceq)\) is the graph \((S, \Upa)\) induced by \((\N, \upa)\), the covering graph of \((\N, \le)\). In a special case, as we will see below, \((S, \preceq)\) is the graph associated with a positive semigroup.
The path function for the partial order graph \((S, \preceq)\) is considerably more complicated that for our previous graphs.
The path function \(\gamma_j\) of order \(j \in \N\) for \((S, \preceq)\) is given by \[\gamma_j(x) = \sum_{l=0}^n \binom{j}{l} \sum\left\{\prod_{i \in J} k_i: J \subseteq \{0, 1, \ldots, n - 1\}, \, \#(J) = l\right\}, \quad x \in S_n, \, n \in \N\]
For the proof, it helps to have some additional notation. If \(J \subset \N\) is finite, define \(k_J = \prod_{j \in J} k_j\). For \(n \in \N\) and \(l \in \{0, 1, \ldots, n\}\) define \[\ms J_{n,l} = \{J \subseteq \{0, 1, \ldots, n - 1\}: \#(J) = l\}\] Note that \(J_{n,0} = \{\emptyset\}\). Recall that \(\gamma_0(x) = 1\) for \(x \in S\) and \(\gamma_{j+1}(x) = \sum_{y \preceq x} \gamma_j(y)\) for \(x \in S\) and \(j \in \N\). So we can give a proof by induction on \(j\). The formula is vacuously true when \(j = 0\). Suppose now that the formula holds for a given \(j \in \N\) and \(x \in S\). Then for \(x \in S_n\), \begin{align*} \gamma_{j+1}(x) &= \sum_{y \preceq x} \gamma_j(y) = \gamma_j(x) + \sum_{m=0}^{n-1} \sum_{y \in S_m} \gamma_j(y) = \sum_{l=0}^n \binom{j}{l} \sum_{J \in \ms J_{n,l}} k_J + \sum_{m=0}^{n-1} k_m \sum_{l=0}^m \binom{j}{l} \sum_{J \in \ms J_{m,l}} k_J \\ &= \sum_{l=0}^n \binom{j}{l} \sum_{J \in \ms J_{n,l}} k_J + \sum_{l=0}^{n-1} \binom{j}{l} \sum_{m=l}^{n-1} \sum_{J \in \ms J_{m,l}} k_m k_J = \sum_{l=0}^n \binom{j}{l} \sum_{J \in \ms J_{n,l}} k_J + \sum_{l=0}^{n-1} \binom{j}{l} \sum_{J \in \ms J_{n,l+1}} k_J \\ &= \sum_{l=0}^n \binom{j}{l} \sum_{J \in \ms J_{n,l}} k_J + \sum_{l=1}^n \binom{j}{l-1} \sum_{J \in \ms J_{n,l}} k_J = 1 + \sum_{l=1}^n \left[\binom{j}{l} + \binom{j}{l - 1}\right] \sum_{J \in \ms J_{n,l}} k_J\\ & = \sum_{l=0}^n \binom{j + 1}{l} \sum_{J \in \ms J_{n,l}} k_J \end{align*}
Suppose that \(x \in S_3\). Explicitly compute \(\gamma_j(x)\) for \(j \in \{0, 1, 2, 3, 4\}\).
For \(x \in S_3\), \begin{align*} \gamma_0(x) &= 1 \\ \gamma_1(x) & = 1 + k_0 + k_1 + k_2 \\ \gamma_2(x) & = 1 + 2(k_0 + k_1 + k_2) + (k_0 k_1 + k_0 k_2 + k_1 k_2)\\ \gamma_3(x) & = 1 + 3(k_0 + k_1 + k_2) + 3(k_0 k_1 + k_0 k_2 + k_1 k_2) + k_0 k_1 k_2\\ \gamma_4(x) & = 1 + 4(k_0 + k_1 + k_2) + 6(k_0 k_1 + k_0 k_2 + k_1 k_2) + 4 k_0 k_1 k_2\\ \end{align*}
Next we compute the Möbius function \(M\) for the partial order graph \((S, \preceq)\). Since \(M(x, x) = 1\) for \(x \in S\) and \(M(x, y) = 0\) if \(x \npreceq y\), the only values of interest are \(M(x, y)\) with \(x \in S_m\) and \(y \in S_{m+n}\) with \(m \in \N\) and \(n \in \N_+\). As with the path function, it's not surprising that \(M\) is contant for such \((x, y)\).
The Möbius function \(M\) for \((S, \preceq)\) is given as follows, for \(m \in \N\) and \(n \in \N_+\): \[M(x, y) = \sum_{l=0}^{n-1} (-1)^{l + 1} \sum\left\{\prod_{j \in J} k_{m+j}: J \subseteq \{1, \ldots, n - 1\}, \#(J) = l\right\}, \quad x \in S_m, \, y \in S_{m+n}\]
Suppose that \(m \in \N\) and that \(x \in S_m\). If \(y \in S_{m+1}\) then \[M(x, y) = -\sum_{t \in [x, y)} M(x, t) = -M(x, x) = -1\] Similarly, if \(y \in S_{m+2}\) then \[M(x, y) = -\sum_{t \in [x, y)} M(x, t) = -\left[M(x, x) + \sum_{t \in S_{m+1}} M(x, t)\right] = -[1 - k_{m+1}] = -1 + k_{m+1}\] For \(y \in S_{m+3}\), \begin{align*} M(x, y) &= - \sum_{t \in [x, y)} M(x, t) = -\left[M(x, x) + \sum_{t \in S_{m+1}} M(x, t) + \sum_{t \in S_{m+2}} M(x, t)\right] \\ &= -[1 - k_{m+1} + k_{m+2}(-1 + k_{m+1})] = - 1 + k_{m+1} + k_{m+2} - k_{m+1} k_{m+2} \end{align*} Continuing in this way gives the result. A more formal induction proof over \(n\) is as follows. For \(n \in \N_+\) and \(j \in \{0, 1, \ldots, n - 1\}\), let \(\ms J_{n,l} = \{J \subseteq \{1, 2, \ldots, n - 1\}: \#(J) = l\}\). As shown above, the formula is true when \(m \in \N\) and \(n = 1\). Assume the formula holds for \(m \in \N\) and for positive integers \(r \le n\) for a given \(n \in \N_+\). Let \(x \in S_m\) and \(y \in S_{m + n + 1}\). Then \begin{align*} M(x, y) &= -\sum_{t \in [x, y)} M(x, t) = -\left[M(x, x) + \sum_{r=1}^n \sum_{t \in S_{m+r}} M(x, t)\right] \\ &= -\left[1 + \sum_{r=1}^n k_{m+r} \sum_{l=0}^{r-1} (-1)^{l+1} \sum\left\{\prod_{j \in J} k_{m+j}: J \in \ms J_{r,l}\right\}\right] \\ &= -\left[1 + \sum_{l=0}^{n-1} (-1)^{l+1} \sum_{r=l+1}^n \sum\left\{k_{m+r} \prod_{j \in J} k_{m+j}: J \in \ms J_{r,l}\right\}\right] \\ &= -\left[1 + \sum_{l=0}^{n-1} (-1)^{l+1} \sum\left\{\prod_{j \in J} k_{m+j}: J \in \ms J_{n+1,l+1}\right\}\right] \\ &= -\left[1 + \sum_{l=1}^n (-1)^l \sum\left\{\prod_{j \in J} k_{m+j}: J \in \ms J_{n+1,l}\right\}\right] \\ &= \sum_{l=0}^n (-1)^{l+1} \sum\left\{\prod_{j \in J} k_{m+j}: J \in \ms J_{n+1,l}\right\} \end{align*}
Suppose now that \(X\) is a random variable in \(S\) with probability density function \(f\). Let \(N = \varphi(X)\) denote the corresponding index random variable in \(\N\), so that \(N = n\) if and only if \(X \in S_n\) for \(n \in \N\). Thus \(N\) has density function \((p_n: n \in \N)\) given by \[p_n = \P(N = n) = \P(X \in S_n) = \sum_{x \in S_n} f(x), \quad n \in \N\] Let \(F\) denote the reliability function of \(X\) for the graph \((S, \preceq)\). Then \[F(x) = f(x) + \sum_{m = n + 1}^\infty p_m, \quad n \in \N, \, x \in S_n\]
On the other hand, \(f\) can be recovered from \(F\) via Möbius inversion under certain conditions. For \(m \in \N\) and \(n \in \N_+\), let \(M_{m, m + n}\) denote the constant value of the Möbius function \(M(x, y)\) given in when \(x \in S_m\) and \(y \in S_{m + n}\).
Assuming absolute convergence of the series, \[f(x) = F(x) + \sum_{n = 1}^\infty M_{m, m + n} \sum_{y \in S_{m + n}} F(y), \quad m \in \N, \, x \in S_m\]
Let \(m \in \N\) and \(x \in S_m\). By the Möbius inversion formula, again assuming absolute convergence of the series, \begin{align*} f(x) &= \sum_{x \preceq y} M(x, y) F(y) = F(x) + \sum_{x \prec y} M(x, y) F(y) \\ &= F(x) + \sum_{n = 1}^\infty \sum_{y \in S_{m + n}} M(x, y) F(y) = F(x) + \sum_{n = 1}^\infty M_{m, m + n} \sum_{y \in S_{m + n}} F(y) \end{align*}
We now turn our attention to constant rate distribution for the partial order graph \((S, \preceq)\).
Suppose that \(X\) has constant rate \(\alpha \in (0, 1)\) for \((S, \preceq)\). Then \(X\) has density function \(f\) given by \[f(x) = \frac{\alpha (1 - \alpha)^n}{(1 - \alpha + \alpha k_0) (1 - \alpha + \alpha k_1) \cdots (1 - \alpha + \alpha k_n)}, \quad x \in S_n, \, n \in \N \]
As above, let \(p_n = \P(X \in S_n) = \P(N = n)\) for \(n \in \N\) so that \(n \mapsto p_n\) is the density function of \(N\). Let \(P_n = \sum_{m = n + 1}^\infty p_m\) for \(n \in \N\), so that \(n \mapsto P_n\) is the reliability function of \(N\) for the strict partial order graph \((N, \lt)\). As noted earlier, the reliability function \(F\) of \(X\) for \((S, \preceq)\) is given by \(F(x) = f(x) + P_n\) for \(x \in S_n\) and \(n \in \N\). Hence if \(X\) has constant rate \(\alpha \in (0, 1)\), then \(f = \alpha F\) and so \[F(x) = \frac{1}{1 - \alpha} P_n, \quad x \in S_n, \, n \in \N\] But \(P_n - P_{n+1} = p_{n + 1}\) for \(n \in \N\) and moreover, \[p_m = \P(X \in S_m) = \sum_{x \in S_m} f(x) = \sum_{x \in S_m} \alpha F(x) = \sum_{x \in S_m} \frac{\alpha}{1 - \alpha} P_m = k_m \frac{\alpha}{1 - \alpha} P_m, \quad m \in \N\] Substituting we have \(P_n - P_{n+1} = [\alpha k_{n+1} / (1 - \alpha)] P_{n+1}\) for \(n \in \N\) or equivalently, \[P_{n+1} = \frac{1 - \alpha}{1 - \alpha + \alpha k_{n+1}} P_n, \quad n \in \N\] Solving gives \[P_n = \frac{(1 - \alpha)^n}{(1 - \alpha + \alpha k_1) \cdots (1 - \alpha + \alpha k_n)} P_0, \quad n \in \N\] Finally, \(P_0 = 1 - p_0 = 1 - [\alpha / (1 - \alpha)] P_0\) and so \[P_0 = \frac{1 - \alpha}{1 - \alpha + \alpha k_0}\]
Suppose that \(X\) has constant rate \(\alpha \in (0, 1)\) for \((S, \preceq)\) as in . Let \(N\) denote the index variable of \(X\), so that \(N = n\) if and only if \(X \in S_n\) for \(n \in \N\).
So if \(k_n\) is increasing, or decreasing, or constant in \(n \in \N\), then \(N\) has increasing rate, decreasing rate, or constant rate, respectively, for \((\N, \lt)\). In the last case, of course, \(N\) has a geometric distribution. In the decreasing case, \(k_n\) must eventually be constant in \(n \in \N\).
Part (a) of defines an interesting class of distributions on \(\N\). Here are some special cases:
If \(k_n = k\) for all \(n \in \N\) and \(\alpha \in (0, 1)\) then \(N\) has the geometric distribution on \(\N\) with success parameter \(\alpha k / (1 - \alpha + \alpha k)\),
In this case, \(N\) has constant rate for the graph \((\N, \lt)\), as noted above, but also (with different constants) for the graphs \((\N, \le)\) and \((\N, \upa)\). In particular, if \(k = 1\), \(N\) has the geometric distribution with success parameter \(\alpha\), which of course must be the case since \((S, \preceq)\) is isomorphic to \((\N, \le)\).
Open the simulation of the geometric distribution in . Vary \(k\) and \(\alpha\) with the scrollbars and note the shape of the probability density function. Run the simulation and compare the empirical density function to the probability density function.
Suppose that \(k_n = n + 1\) for \(n \in \N\) and \(\alpha \in (0, 1)\).
Open the simulation of the distribution in . Vary \(\alpha\) with the scrollbars and note the shape of the probability density function. Run the simulation and compare the empirical density function to the probability density function.
Suppose that \(\alpha = \frac 1 2\) and \(k_n \in \N_+\) for \(n \in \N\).
Suppose that \(k_n = n + 1\) for \(n \in \N\) and \(\alpha = \frac 1 2\).
Consider again the distribution in .
Open the simulation of the distribution in by keeping the default setting \(\alpha = 0.5\). Note the shape of the probability density function. Run the simulation and compare the empirical density function to the probability density function.
In the general lexicographic construction considered above, only one case corresponds to a positive semigroup, and that case corresponds to \(k_0 = 1\) and \(k_n = k \in \N_+\) for \(n \in \N_+\). We give a construction that leads to an elegant notation. Let \(I\) be a set with \(k\) elements, which we think of as a set of letters. We start with the free semigroup \((I^*, \cdot)\) as studied in Chapter 5 but then we impose the word equations \(i j = j^2\) for \(i, \, j \in I\). It then follows more generally that if \(x \in I^*\) is a word of length \(n \in \N_+\) ending in the letter \(i \in I\), then \(x = i^n\). Words of this form are distinct: if \(m, \, n \in \N_+\) and \(i, \, j \in I\) then \(i^m = j^n\) if and only if \(i = j\) and \(m = n\).
The semigroup is \((S, \cdot)\) where \(S = \{i^n: n \in \N, \, i \in I\}\) and with the operation \(\cdot\) defined on \(S_+ = \{i^n: n \in \N_+, \, i \in I\}\) by \[i^m j^n = j^{m + n}, \quad i^m, \, j^n \in S_+\]
Of course \(i^0 = e\) for \(i \in I\) where \(e\) is the empty word, which functions as the identity just as it does in \((I^*, \cdot)\). The partial order \(\preceq\) associated with \((S, \cdot)\) is given by \(i^m \prec j^n\) if and only if \(m \lt n\). Hence \((S, \preceq)\) is a special case of the partial order graph considered above, with \(S_0 = \{e\}\) and \(S_n = \{i^n: i \in I\}\) for \(n \in \N_+\).
Suppose that \(X\) is a random variable in \(S\) and that \(X\) is memoryless for \((S, \cdot)\). Then \(X\) has reliability function \(F\) for \((S, \cdot)\) given by \(F(i^n) = q^n\) for \(i^n \in S\), where \(q \in (0, 1)\) is a parameter. The probability density function \(f\) of \(X\) is given as follows:
The memoryless property of an reliability function \(F\) for \((S, \cdot)\) is \[F(i^m) F(j^n) = F(j^{m+n}), \quad i^m, \, j^n \in S_+\] So \(F(i^m) = F(j^{m+n}) / F(j^n)\) and it follows that \(F(i^m)\) is constant in \(i \in I\) for each \(m \in \N_+\). Another application of the memoryless property then implies that \(F(i^n) = q^n\) for \(i^n \in S_+\) where \(q = F(i)\). Since \(F(e) = 1\), it follows that \(F(i^n) = q^n\) for all \(i^n \in S\).
Now let \(f\) denote the probability density function of \(X\). For \(n \in \N_+\) and \(i, \, j \in I\), \begin{align*} F(i^n) &= f(i^n) + \sum_{m = n + 1}^\infty \sum_{u \in I} f(u^m)\\ F(j^n) &= f(j^n) + \sum_{m = n + 1}^\infty \sum_{u \in I} f(u^m) \end{align*} Since \(F(i^n)\) is constant in \(i \in I\) for each \(n \in \N_+\), it follows that \(f\) has this property as well. So, let \(\varphi(n)\) denote the common value of \(f(i^n)\) for \(i^n \in S_+\) and let \(\varphi(0) = f(e)\). It follows that \[q^n = \varphi(n) + k \sum_{m = n + 1}^\infty \varphi(m), \quad n \in \N\] Subtracting the equation with \(n + 1\) from the equation with \(n\) gives \[(1 - q) q^n = \varphi(n) + (k - 1)\varphi(n + 1)\] Using this result recursively gives the equation in part (b). Finally, the only requirement on \(\varphi(0) = f(e)\) is that \(\varphi(n) \ge 0\) for \(n \in \N\). Solving the various inequalities gives \(f(e) = (1 - q) / (1 - q + k q)\) if \(q \lt 1 / (k - 1)\), and \(0 \le f(e) \le 1 - q\) if \(q \ge 1 / (k - 1)\).
Suppose again that \(X\) is a random variable in \(S\), Then \(X\) has an exponential distribution for \((S, \cdot)\) if and only if the probability density function \(f\) of \(X\) has the form \[f(i^n) = \frac{(1 - q) q^n}{1 - q + k q}, \quad i^n \in S\] where \(q \in (0, 1)\) is a parameter. The reliability function \(F\) of \(X\) for \((S, \cdot)\) is given by \(F(i^n) = q^n\) for \(i^n \in S\), and \(X\) has constant rate \((1 - q) / (1 - q + k q)\) for \((S, \cdot)\).
Random variable \(X\) has an exponential distribution for \((S, \cdot)\) if an only if \(X\) is memoryless and has constant rate with respect to counting measure \(\#\) (the only left-invariant measure, up to multiplcation by positive constants). The result then follows from propositions and .
In the case \(q \lt 1 / (k - 1)\), the only memoryless distribution is the exponential distribution. In the case \(q \ge 1 / (k - 1)\), the exponential probability density function corresponds to choosing \(f(e) = (1 - q) /(1 - q + k q)\) in part (b) of . But of course, this choice of \(f(e)\) is not the only possible one. Indeed, any choice of \(f(e)\) with \(0 \le f(e) \le 1 - q\) will lead to a probability density function whose corresponding reliability function \(F\) is \(F(i^n) = q^x\) for \(i^n \in S\), corresponding to a memoryless distribution. In particular, the reliability function does not uniquely specify the distribution, and there are probability distributions that are memoryless but not exponential. The following exercise explores a concrete example.
Suppose that \(k = 4\) and that \(q = \frac 1 2\), so that all of the distributions defined by in part (b) of are memoryless for \((S, \cdot)\), with reliability function \(F\) given by \(F(i^n) = \left(\frac 1 2 \right)^n\) for \(i^n \in S\). Find the probability density function \(f\) for each of the following choices of \(f(e)\):
On the other hand, for this positive semigroup, the constant rate property implies the full exponential property.
Suppose again that \(X\) is a random variable in \(S\). If \(X\) has constant rate for \((S, \preceq)\) then \(X\) is exponential for \((S, \cdot)\).
Suppose that \(F\) is the reliability function of a distribution which has constant rate \(\alpha \in (0,\,1)\) for \((S, \preceq)\). Then from \[F(i^n) = \frac{(1 - \alpha)^n}{(1)(1 - \alpha + \alpha k)^n} = \left(\frac{1 - \alpha}{1 - \alpha + \alpha k}\right)^n, \quad i^n \in S\] Hence the distribution is memoryless as well, and hence exponential for \((S, \cdot)\).
So to review, every distribution with constant rate for \((S, \preceq)\) is memoryless (and hence exponential) for \((S, \cdot)\), but conversely, there are memoryless distributions that do not have constant rate. Of course, if \(k = 1\) then \((S, \cdot)\) is isomorphic to \((\N, +)\) and so the constant rate distribution is the geometric distribution with rate \(\alpha\).
We continue with the semigroup \((S, \cdot)\) in the last subsection to illustrate some of the results in Section 2.8 on quotient spaces.
Fix \(j \in I\) and let \(S_j = \{j^n: n \in \N\}\) so that \((S_j, \cdot)\) is the complete sub-semigroup of \((S, \cdot)\) generated by the element \(j\). The corresponding quotient space is \[S / S_j = \{e\} \cup (I - \{j\})\] The basic assumptions are satisifed, so that \(i^n \in S\) has a unique factoring over \(S_j\) and \(S / S_j\). Specifically, the non-trivial factoring is \[i^n = j^{n-1} i, \quad n \in \N_+, \, i \in I - \{j\}\]
If \(X\) is a random variable in \(S\) then \(X = j^N Y\) where \(N \in \N\) and \(Y \in S / S_j\). Let \(f\) denote the probability density function of \(X\). Then the probability density function of \((N, Y)\) is given by \begin{align*} \P(N = n, Y = e) &= f(j^n), \quad n \in \N_+ \\ \P(N = n, Y = i) &= f(i^{n+1}), \quad n \in \N, \, i \in I - \{j\} \end{align*}
Suppose that \(X\) has the exponential distribution given in . Then
The fact that \(N\) has an exponential distribution on \((\N, +)\) and that \(N\) and \(Y\) are independent follows from the general theory in Section 2.8. The particular details in this theorem follow from the density function of \((N, Y)\) and standard techniques.
\begin{align*} \P(N = n, Y = e) &= \P(X = j^n) = \frac{(1 - q) q^n}{1 - q + k q}, \quad n \in \N \\ \P(N = n, Y = i) &= \P(X = i^{n+1}) = \frac{(1 - q) q^{n+1}}{1 - q + k q}, \quad n \in \N, \, i \in I - \{j\} \end{align*}Our next discussion gives a counterexample about conditional exponential distributions. We allow the set \(I\) to be countably infinite.
Let \(p_i \in (0, 1)\) for each \(i \in I\) and assume that \(\sum_{i \in I} (1 - p_i) / p_i \lt \infty\). Suppose that random variable \(X \in S\) has probability density function \(f\) given by \(f(i^n) = d (1 - p_i)^n\) for \(i^n \in S\) where \[d = \frac{1}{1 + \sum_{i \in I} (1 - p_i) / p_i} \] Then the conditional distribution of \(X\) given \(X \in S_x\) is exponential for \((S_x, \cdot)\) for each \(x \in S\), but \(X\) does not have a memoryless distribution for \((S, \cdot)\) unless \(I\) is finite and \(p_i\) is constant in \(i \in I\).
First we verify that \(f\) is a valid density function on \(S\): \[\sum_{x \in S} f(x) = f(e) + \sum_{i \in I} \sum_{n = 1}^\infty f(i^n) = d + d \sum_{i \in I} \sum_{n = 1}^\infty (1 - p_i)^n = d\left[1 + \sum_{i \in I} (1 - p_i) / p_i \right] = 1\] by definition of \(d\). Next note that need only prove the theorem when \(x = i \in I\), since every other semigroup \(S_x\) is a sub-semigroup of one of these. Towards that end, \[\P(X \in S_i) = \sum_{n = 0}^\infty f(i^n) = d + d \sum_{n = 1}^\infty (1 - p_i)^n = d \left[1 + (1 - p_i) / p_i\right] = d / p_i\] Hence \[\P(X = i^n \mid X \in S_i) = \frac{f(i^n)}{\P(X \in S_i))} = p_i (1 - p_i)^n, \quad n \in \N\] which is the geometric distribution with rate \(p_i\) under the isomorphism \(i^n \mapsto n\) for \(n \in \N\). But clearly \(X\) does not have a memoryless distribution for \((S, \cdot)\) unless \(S\) is finite and \(p_i = p \in (0, 1)\) for \(i \in I\). In this case, \(d = p / [p + \#(I) (1 - p)]\) and so \(f(i^n) = d (1 - p)^n\) for \(i^n \in S\), which agrees with the exponential distribution in .
Suppose that \(X\) has the distribution in . Fix \(j \in I\) and consider the decomposition \(X = j^N Y\) where \(N \in \N\) and \(Y \in S / S_j\). Then \((N, Y)\) has probability density function defined as follows: \begin{align*} \P(N = n, Y = e) &= \frac{(1 - p_j)^n}{{1 + \sum_{y \in I} (1 - p_y) / p_y}}, \quad n \in \N \\ \P(N = n, Y = i) &= \frac{(1 - p_i)^{n + 1}}{{1 + \sum_{y \in I} (1 - p_y) / p_y}}, \quad n \in \N, \, i \in I - \{j\} \end{align*}
First \[\P(N = n, Y = e) = \P(X = j^n) = d(1 - p_j)^n, \quad n \in \N\] Next, \[\P(N = n, Y = i) = \P(X = i^{n + 1}) = d(1 - p_i)^{n + 1}, \quad n \in \N, \, i \in I - \{j\}\]
Suppose again that \(X\) has the distribution in . Fix \(j \in I\) and consider the decomposition \(X = j^N Y\) where \(N \in \N\) and \(Y \in S / S_j\). Then
In particular, note from the last two results that \(N\) and \(Y\) are dependent.