As in Section 3, we start with an arithmetic semigroup \((S, \cdot)\) with the countable set \(I\) as the set of prime elements. But now, we assume that we have a norm \(|\cdot|\) for \((S, \cdot)\) as defiined in Section 2, and therefore also the Dirichlet series corresponding to an arithmetic function and the Mangoldt function.
A probability distributions can easily be defined in terms of a Dirichlet series.
Suppose that \(g\) is a positive arithmetic function for \((S, \cdot)\) and that the corresponding Dirichlet series \(G\) converges on the interval \((t_0, \infty)\) for some \(t_0 \in (0, \infty)\). The Dirichlet distribution on \((S, \cdot)\) corresponding to the function \(g\) (or the series \(G\)), with parameter \(t \in (t_0, \infty)\), has probability density function \(f\) given by \[f(x) = \frac{g(x)}{|x|^t G(t)}, \quad x \in S\]
Recall that \[ G(t) = \sum_{x \in S} \frac{g(x)}{|x|^t}, \quad t \in (t_0, \infty) \]
So \(f\) is proportional to the function \(x \mapsto g(x) / |x|^t\), and of course the normalizing constant must then be \(G(t)\). A given positive arithmetic function defines a one parameter family of Dirichlet distributions. The most famous Dirichlet distribution corresponds to the constant function 1 on \(S\).
The zeta distribution for \((S, \cdot)\) with parameter \(t\) in the interval of convergence \((t_0, \infty)\) has probability density function \(f\) given by \[f(x) = \frac{1}{|x|^t \zeta(t)}, \quad x \in S\]
Recall that \(\zeta\) is the Drichlet series corresponding to the constant function 1 on \(S\) so \[\zeta(t) = \sum_{x \in S} \frac{1}{|x|^t}\]
For the standard arithmetic semigroup \((\N_+, \cdot)\), the Dirichlet distributions, and in particular the zeta distribution, are the standard ones.
We can now characterize expoonential distributions in terms of Dirichlet distributions.
Random variable \(X\) has an exponential distribution on \((S, \cdot)\) if and only if \(X\) has a Dirichlet distributions corresponding to a completely multiplicative arithmetic function \(g\). The rate constant is \(1 / G(t)\) where \(G\) is the series function and \(t\) is the parameter of the distribution.
Suppose that \(g\) is completely multiplicative arithmetic function for \((S, \cdot)\) and that \(X\) has the Dirichlet distribution corresponding to \(g\) with parameter \(t\) in the interval of convergence of the Dirichlet series \(G\). That is, the density function \(f\) of \(X\) is given by \(f(x) = g(x) / [|x|^t G(t)]\) for \(x \in S\). So the reliability function \(F\) of \(X\) is given by \begin{align*} F(x) &= \P(X \succeq x) = \sum_{y \succeq x} \frac{g(y)}{|y|^t G(t)} = \sum_{z \in S} \frac{g(x z)}{|x z|^t G(t)} = \sum_{z \in S} \frac{g(x) g(z)}{|x|^t |z|^t G(t)} \\ &= \frac{g(x)}{|x|^t G(t)} \sum_{z \in S} \frac{g(z)}{|z|^t} = \frac{g(x)}{|x|^t G(t)} G(t) = \frac{g(x)}{|x|^t}, \quad x \in S \end{align*} Hence \(X\) has constant rate \(1 / G(t)\). Also, \(X\) is memoryless, since \(g\) is completely multiplicative: \[F(x y) = \P(X \succeq x y) = \frac{g(x y)}{|x y|^t} = \frac{g(x) g(y)}{|x|^t |y|^t} = \frac{g(x)}{|x|^t} \frac{g(y)}{|y|^t} = \P(X \succeq x)\P(X \succeq y)\] Therefore \(X\) has an exponential distribution.
The converse is trivially true. Suppose that \(X\) has an exponential distribution with reliability function \(F\). For fixed \(t \in (0, \infty)\), let \(g(x) = |x|^t F(x)\) for \(x \in S\), and let \[G(s) = \sum_{x \in S} \frac{g(x)}{|x|^s}\] Then \(g\) is a completely multiplicative function and \(G\) is the corresponding Dirichlet series. Moreover, \(t\) is in the interval of convergence since \[G(t) = \sum_{x \in S} \frac{g(x)}{|x|^t} = \sum_{x \in S} F(x) \lt \infty\] The probability density function \(f\) of \(X\) is given by \[f(x) = \P(X = x) = \frac{F(x)}{\sum_{y \in S} F(y)} = \frac{g(x)}{|x|^t G(t)}, \quad x \in S\] and so \(X\) has the Dirichlet distribution corresponding to \(g\) with parameter \(t\). Note that since \(g\) is completely multiplicative, all members of this Dirichlet family are exponential, from the first part of the theorem.
So in particular, the zeta distribution in is an exponential distribution on \((S, \cdot)\). From Section 3 recall that random variable \(X\) has an exponential distribution for \((S, \cdot)\) if and only if the prime exponents are independent and have geometric distributions on \(\N\). Proposition below shows how the geometric parameters of the random prime powers are are related to the arithmetic function. Recall that \(\ms P\) is the collection of functions \(\bs{p} = (p_i: i \in I)\) where \(p_i \in (0, 1)\) for \(i \in I\) and \(\prod_{i \in I} p_i \gt 0\).
Suppose that \(X\) has an exponential distribution on \((S, \cdot)\), so that \(X\) has two representations:
Then \(p_i = 1 - g(i) / |i|^t\) for \(i \in I\).
In the first representation, \(\P(X \succeq i) = 1 - p_i\) for \(i \in I\) and in the second representation, representation, \(\P(X \succeq i) = g(i) / |i|^t\) for \(i \in I\)
For the standard case \((\N_+, \cdot)\), this result was considered surprising in the paper by Lin, but is quite natural in the context of positive semigroups. As a corollary to , we get a probabilistic proof of the product formula for the Dirichlet series of a completely multiplicative function.
Suppose that \(g\) is a positive, completely multiplicative function for \((S, \cdot)\) and that the corresponding Dirichlet series converges on \((t_0, \infty)\) for some \(t_0 \in (0, \infty)\). Then \[G(t) = \prod_{i \in I} \frac{|i|^t}{|i|^t - g(i)}, \quad t \in (t_0, \infty)\] In particular \[\zeta(t) = \prod_{i \in I} \frac{|i|^t}{|i|^t - 1}, \quad t \in (t_0, \infty)\]
Consider the exponential distribution on \((S, \cdot)\) corresponding to \(g\) and the parameter \(t\). As noted in , the geometric parameter for the prime element \(i \in I\) is \[p_i = \frac{|i|^t - g(i)}{|i|^t}\] From Section 3, the reliability function \(F\) satisfies \[\sum_{x \in S} F(x) = \prod_{i \in I} \frac{1}{p_i} = \prod_{i \in I} \frac{|i|^t} {|i|^t - g(i)}\] But as shown in the details of , \(\sum_{x \in S} F(x) = G(t)\).
The app below is simulation of the standard zeta distribution on \(\N_+\). The parameter \(t\) cam be varied with the scrollbar and the value of the rate constant \(\alpha = 1 / \zeta(t)\) is also shown.
The standard moment results for an exponential distribution on \((S, \cdot)\) can be rephrased in terms of Dirichlet distributions. Recall that \(\tau_k\) denotes the left walk function (divisor function) of order \(k \in \N\).
If \(X\) has the Dirichlet distribution for \((S, \cdot)\) with completely multiplicative function \(g\) and parameter \(t\) in the interval of convergence, then \(\E[\tau_k(X)] = G^k(t)\) for \(k \in \N\). In particular, if \(X\) has the zeta distribution for \((S, \cdot)\) with parameter \(t\) in the interval of convergence then \(\E[\tau_k(X)] = \zeta^k(t)\) for \(k \in \N\).
This follows from the general theory of constant rate distribution in Section 1.5 since the rate constant is \(1 / G(t)\).
Next we obtain a result from the paper by Lin. Our proof is better because it takes advantage of the general theory developed in this text. Let \(L\) denote the adjacency kernel of \((S, \cdot)\). Suppose that \(g: \N_+ \to [0, \infty)\) is a nonnegative arithmetic function, not identically zero, and let \(G\) be the corresponding Dirichlet series.
Suppose that \(X\) has the zeta distribution for \((S, \cdot)\) with parameter \(t\) and that \(g\) is a nonnegative arithmetic function with Dirichlet series \(G\). If \(t\) is in the interval of convergence of \(\zeta\) and \(G\) then \[\E[ (g L)(X)] = G(t)\]
It follows immediately from the basic moment result in Section 1.5 that \[\E[(g L)(X)] = \zeta(t) \E[g(X)]\] since \(1 / \zeta(t)\) is the rate constant of the exponential distribution of \(X\). But \[\zeta(t) \E[g(X)] = \zeta(t) \sum_{x \in S} \frac{g(x)}{|x|^t \zeta(t)} = \sum_{x \in S} \frac{g(x)}{|x|^t} = G(t)\]
Our next results concern moments the norm of an exponential variable.
Suppose that \(X\) has the exponential distribution on \((S, \cdot)\) with parameter \(\bs{p} = (p_i: i \in I) \in \ms P\). Suppose also that \(k \in (0, \infty)\) and that \(p_i \gt 1 - 1 / |i|^k\) for \(i \in I\). Then \[\E\left(|X|^k\right) = \prod_{i \in I} \frac{p_i}{1 - |i|^k (1 - p_i)}\]
This follows from the moment result in Section 3 since \(x \mapsto |x|^k\) is completely multiplicative, and the probability generating function of the geometric distribution with success parameter \(p \in (0, 1)\) is \(t \mapsto p / [1 - t(1 - p)]\) for \(|t| \lt 1 / (1 - p)\).
Here is a restatement of in terms of Dirichlet distributions.
Suppose that \(X\) has the Dirichlet distribution on \((S, \cdot)\) corresponding to a positive, completely multiplicative function \(g\) and Dirichlet series \(G\), with parameter \(t\) in the interval of convergence \((t_0, \infty)\). If \(k \lt t - t_0\) then \[\E(|X|^k) = \frac{G(t - k)}{G(t)}\]
In terms of the Dirichlet formulation, the density function \(f\) of \(X\) is given by \(f(x) = g(x) / [|x|^t G(t)]\) for \(x \in S\). Hence \[\E(|X|^k) = \sum_{x \in S} |x|^k \frac{g(x)}{|x|^t G(t)} = \frac{1}{G(t)} \sum_{x \in S} \frac{g(x)}{|x|^{t - k}} = \frac{G(t - k)}{G(t)}\] assuming that \(t - k \gt t_0\). The result also follows from and the product expansion of the Dirichlet series \(G\) since \(p_i = 1 - g(i) / |i|^t\) for \(i \in I\).
Next is the basic entropy result in terms of Dirchlet distributions.
Suppose that \(X\) has the Dirichlet distribution on \((S, \cdot)\) corresponding to a positive, completely multiplicative function \(g\) and parameter \(t\) in the interval of convergence of the Dirichlet seriest \(G\). Then \(X\) maximizes entropy over all random variables \(Y \in S\) with \(\E(\ln |Y|) = \E(\ln |X|)\) and \(\E[\ln g(Y)] = \E[\ln g(X)]\). The maximum entropy is \[H(X) = \ln G(t) - \E[\ln g(X)] + t \E[\ln |X|]\]
Let \(F\) denote the reliability function of \(X\). From the main entropy result in Section 1.5, \(X\) maximizes entropy over all random variables \(Y\) in \(S\) with \(\E[\ln F(Y)] = \E[\ln F(X)]\), and then the maximum entropy is \(H(X) = -\ln \alpha - \E[\ln F(X)]\) where \(\alpha \in (0, \infty)\) is the rate constant. But \(F(x) = g(x) / |x|^t\) for \(x \in S\). Hence \(\E[\ln F(X)] = \E[\ln g(X)] - t \E(\ln |X|)\) and similarly \(\E[\ln F(Y)] = \E[\ln g(Y)] - t \E(\ln |Y|)\). So if \(\E(\ln |Y|) = \E(\ln |X|)\) and \(\E[\ln g(Y)] = \E[\ln g(X)]\) then \(\E[\ln F(Y)] = \E[\ln F(X)]\). Finally, the rate constant is \(1 / G(t)\).
From the paper by Gut, \[\E(\ln |X|) = \frac{1}{G(t)} \sum_{x \in S} \ln(|x|) \frac{g(x)}{|x|^t} = \frac{G^\prime (t)}{G(t)}\]
The next two results are restatements of results in Section 3, but now in terms of Dirichlet distributions. Both use the Mangoldt function \(\Lambda\) defined in Section 2.
Suppose that \(X\) has the Dirichlet distribution for \((S, \cdot)\) with completely multiplicative coefficient function \(g\) and parameter \(t\). Then \[X = \prod_{x \in S_+} x^{V_x}\] where \((V_x: x \in S_+)\) is a sequence of independent variables, and for \(x \in S_+\), \(V_x\) has the Poisson distribution with parameter \[\lambda_x = \frac{g(x) \Lambda(x)}{|x|^t \ln |x|}\]
From the corresponding result in Section 3, \(X\) has the representation given in the theorem, with \(\lambda_x = (1 - p_i) / n\) if \(x = i^n\) for some \(i \in I\) and \(n \in \N_+\) and \(\lambda_x = 0\) otherwise, where \(\bs{p} = (p_i: i \in I)\) is the sequence of prime exponents. But from , \(p_i = 1 - g(i) / |i|^t\) for \(i \in I\). Suppose that \(x = i^n\) for some \(i \in I\) and \(n \in \N_+\). Then \[\lambda_x = \frac{g^n(i)}{n |i|^{n t}} = \frac{g(i^n)}{n |i^n|^t} = \frac{g(x)}{n |x|^t}\] Next, note that \(\ln |x| = n \ln |i|\) and by definition \(\Lambda(x) = \ln |i|\). Hence \(\lambda_x\) has the representation given in the theorem. On the other hand, if \(x\) is not of the form \(i^n\) for some \(i \in I\) and \(n \in \N_+\), then again by definition, \(\Lambda(x) = 0\).
Next is the compound Poisson result.
Suppose that \(X\) has the Dirichlet distribution on \((S, \cdot)\) corresponding to a positive, completely multiplicative function \(g\) and Dirichlet series \(G\), with parameter \(t\) in the interval of convergence. Then \(X\) has a compound Poisson distribution. Specifically \[X = V_1 V_2 \cdots V_M\] where \(\bs V = (V_1, V_2, \ldots)\) is a sequence of independent copies of \(V\), with probability density function given by \[\P(V = x) = \frac{g(x) \Lambda(x)}{\ln |x| |x|^t \ln G(t)}, \quad x \in S\] The random index \(M\) is independent of \(\bs V\) and has the Poisson distribution with parameter \(\ln G(t)\).
From the corresponding result in Section 3, \(X\) can be represented as the random product given in the theorem, where \(V\) takes values in the set of prime powers with \[\P(V = i^n) = -\frac{(1 - p_i)^n}{n \sum_{j \in I} \ln p_j}, \quad i \in I, \, n \in \N_+\] and where \(M\) is independent of \(\bs V\) and has the Poisson distribution with parameter \(-\sum_{i \in I} \ln p_i\). But as shown in the details of , \[\frac{(1 - p_i)^n}{n} = \frac{g(x) \Lambda(x)}{|x|^t \ln |x|}, \quad i \in I, \, n \in \N_+\] From the product formula in , \(-\sum_{i \in I} \ln p_i = \ln G(t)\).
As noted in the details, \(V\) takes values in the set of prime powers \(\{i^n: i \in I, \, n \in \N_+\}\).
Suppose now that \(\bs X = (X_1, X_2, \ldots)\) is a sequence of independent copies of \(X\) where \(X\) has the Dirichlet distribution for \((S, \cdot)\) corresponding to the completely multiplicative function \(g\), Dirichlet series \(G\), and parameter \(t\) in the interval of convergence. Let \(Y_n = X_1 X_2 \cdots X_n\) for \(n \in \N_+\) so that \(\bs Y = (Y_1, Y_2, \ldots)\) is the random walk on \((S, \cdot)\) associated with \(X\). Recall again that \(\tau_n\) denotes the left walk function (divisor function) of order \(n \in \N\).
For \(n \in \N_+\), \(Y_n\) has density function \(f_n\) given by \[f_n(x) = \frac{\tau_{n - 1}(x) g(x)}{G^n(t) |x|^t}, \quad x \in S\]
This follows from the general formula in Section 1.5 since \(\tau_{n - 1}\) is the walk function of order \(n - 1\), \(1 / G(t)\) is the rate constant, and \(x \mapsto g(x) / |x|^t\) is the reliability function.
Thus, \(Y_n\) also has a Dirichlet distribution for \(n \in \N\), but corresponding to a multiplicative function instead of a completely multiplicative function. As a corollary, we have the following result in analytic number theory that seemingly has nothing to do with probability.
Suppose that \(g\) is a completely multiplicative arithmetic function for \((S, \cdot)\) with Dirichlet series \(G\) that has interval of convergence \((t_0, \infty)\). Then \[\sum_{x \in S} \frac{\tau_{n - 1}(x) g(x)}{|x|^t} = G^n(t), \quad t \in (t_0, \infty), \, n \in \N_+\]
Consider the special case of the zeta distribution for \((S, \cdot)\) with parameter \(t\) in the interval of convergence. For \(n \in \N_+\), the density function \(f_n\) of \(Y_n\) is given by \[f_n(x) = \frac{\tau_{n - 1}(x)}{\zeta^n(t) |x|^t}, \quad x \in S\] So in this special case it follows that \[\sum_{x \in S} \frac{\tau_{n - 1}(x)}{|x|^t} = \zeta^n(t), \quad n \in \N_+\]
As noted in Section 3, and exponential distribution on \((S, \cdot)\) not only has constant rate for the corresponding partial order graph \((S, \preceq)\), but also for the other natural graphs associated with the semigroup.
Suppose that \(X\) has the Dirichlet distribution on \((S, \cdot)\) corresponding to the completely multiplicative function \(g\), Dirichlet series \(G\) and parameter \(t\) in the interval of convergence.
The results follow from the corresponding results in Section 3, and propositions and .