The various models in this section and the one that follows are a good source of counterexamples and are good illustrations of some of the general theory. For the basic setup, we suppose that \((S, \ms S, \lambda)\) is a measure space (with a measurable diagonal) and that \(S\) has a measurable partition \(\ms P = \{S_n: n \in \N\}\) with \(\beta_n = \lambda(S_n) \in (0, \infty)\) for each \(n \in \N\) (so in particular, the measure space is \(\sigma\)-finite). Define the index function \(\varphi: S \to \N\) by \(\varphi(x) = n\) if \(x \in S_n\). In the discrete case, \(S\) is countable and \(\beta_n = \#(S_n) \in \N_+\) is for each \(n \in \N\). Functions on \(\N\) are usually given in subscript form. As usual, we also have a probability space \((\Omega, \ms F, \P)\) in the background.
In the following subsections, we will consider graphs that are induced, in the sense of Section 1.7, by the standard graphs on \(\N\) studied in Section 1 and Section 2. Here is a partial review, with the notation that we will use.
Suppose that \(X\) is a random variable in \(S\) with density function \(f\) with respect to \(\lambda\). Let \(N = \varphi(X)\) denote the corresponding index variable in \(\N\), so that by \(N = n\) if and only if \(X \in S_n\) for \(n \in \N\). Let \(p\) denote the discrete density of \(N\). Then \[p_n = \P(N = n) = \P(X \in S_n) = \int_{S_n} f(x) \, d\lambda(x), \quad n \in \N\]
A relation on \(\N\) induces a relation on \(S\) in a natural way.
If \(\rta\) is a relation on \(\N\) then the induced relation \(\Rta\) on \(S\) is defined by \(x \Rta y\) if and only if \(\varphi(x) \rta \varphi(y)\) for \(x, \, y \in S\). That is, \(x \Rta y\) if and only if \(x \in S_m\) and \(y \in S_n\) for some \(m, \, n \in \N\) with \(m \rta n\).
The left walk function \(u_m\) of order \(m \in \N_+\) is given by \[u_m(x) = \sum_{n_1 \rta n_2 \rta \cdots \rta n_m \rta n} \beta_{n_1} \beta_{n_2} \cdots \beta_{n_m}, \quad n \in \N, \, x \in S_n\]
For the induced graphs on \(S\) studied below, the walk and generating funnctions do not have simple closed forms. However, if \(\beta_n = \beta \in (0, \infty)\) for \(n \in \N\) then \(u_m(x) = \beta^m v_m(n)\) for \(n \in \N\) and \(x \in S_n\) where \(v_m\) is the walk function of order \(m \in \N\) for \((\N, \rta)\). Similarly, \(U(x, t) = V(n, \beta t)\) for \(n \in \N\), \(x \in S_n\) and \(t \in \R\), where \(U\) and \(V\) are the generating functions for \((S, \Rta)\) and \((\N, \rta)\) respectively.
Suppose again that \(X\) is a random variable in \(S\). Let \(F\) denote the reliability function of \(X\) with respect to the graph \((S, \Rta)\) and \(P\) the reliability function of \(N\) with respect to \((\N, \rta)\). Then \[F(x) = \P(x \Rta X) = \P(n \rta N) = P_n, \quad n \in \N, \, x \in S_n\]
As usual, constant rate distributions are of special interest. Here is the main result from Section 1.7:
Random variable \(X\) in \(S\) has constant rate \(\alpha \in (0, \infty)\) if and only if \(p_n = \alpha \beta_n P_n\) for \(n \in \N\). In this case,
So in particular, \(N\) has increasing, decreasing, or constant rate for \((\N, \rta)\) if \(\beta_n\) is increasing, decreasing, or constant in \(n \in \N\), respectively. In the last case, if the weight factors are constant, so that \(\beta_n = \beta \in (0, \infty)\) for \(n \in \N\), then \(X\) has constant rate \(\alpha \in (0, \infty)\) for \((S, \Rta)\) if and only if \(N\) has constant rate \(\alpha \beta\) for \((N, \rta)\). In the discrete case, note that if \(\beta_n \in \N_+\) is decreasing in \(n \in \N\) then \(\beta_n\) is eventually constant in \(n \in \N\). Because of parts (a) and (b), our emphasis in this section will be on the distribution of the index variable \(N\). As was pointed out in Section 1.7, and is particularly true in this setting, it's best to think of \((S, \Rta)\) as a weighted version of \((\N, \rta)\) where \(n\) is given weight \(\beta_n\) for \(n \in \N\). Here is a general result on measure theory:
For each of the induced graphs on \(S\) considered below,
Both parts follows from the general theory in Section 1.7. In each case, the \(\sigma\)-algebra associated with the graph on \(\N\) is the reference \(\sigma\)-algebra \(\ms P(\N)\), and each graph on \(\N\) is stochastic.
As before, let \((\N, \upa)\) denote the covering graph for the standard discrete total order graph \((\N, \le)\). The corresponding induced graph is \((S, \Upa)\) so that \(x \Upa y\) if and only if \(x \in S_n\) and \(y \in S_{n + 1}\) for some \(n \in \N\). Our first goal is to compute the left walk functions for \((S, \Upa)\). As with all partitioned graphs, the walk functions are constant on the partition sets \(S_n\) for \(n \in \N\).
Relative to the graph \((S, \Upa)\),
Part (a) follows from the walk function for \((\N, \upa)\) in Section 1.2 and . Part (b) follows from part (a) since \(U(x, t) = \sum_{m = 0}^\infty u_m(x) t^m\) for \(x \in S\) and \(t \in \R\).
Here is our main result on the existence of constant rate distributions.
Random variable \(X\) in \(S\) has constant rate \(\alpha \in (0, \infty)\) for \((S, \Upa)\) if and only if \[ \frac{1}{p_0} = \sum_{n = 0}^\infty \frac{1}{\alpha^n \beta_0 \beta_1 \cdots \beta_{n - 1}} \lt \infty \] In this case, the probability density function of the index variable \(N\) is given by \[\P(N = n) = \frac{1}{\alpha^n \beta_0 \beta_1 \ldots \beta_{n - 1}} p_0, \quad n \in \N\]
The condition for a constant rate distribution in is \[p_{n + 1} = \frac{1}{\alpha \beta_n} p_n, \quad n \in \N\] Solving gives \[p_n = \frac{1}{\alpha^n \beta_0 \beta_1 \cdots \beta_{n - 1}} p_0, \quad n \in \N\] The condition \(\sum_{n = 0}^\infty p_n = 1\) then requires that \[\frac{1}{p_0} = \sum_{n = 0}^\infty \frac{1}{\alpha^n \beta_0 \beta_1 \cdots \beta_{n - 1}} \lt \infty\]
In particular, a constant rate distribution exists if \(\beta_n\) is bounded away from 0 for \(n \in \N\).
If \(\beta_n \ge \beta\) for all \(n \in \N\) where \(\beta \in (0, \infty)\) then a distribution with constant rate \(\alpha\) for \((S, \Upa)\) exists for all \(\alpha \in (1 / \beta, \infty)\).
Since \(\beta_n \ge \beta \gt 0\) for all \(n \in \N\) we have \[\sum_{n = 0}^\infty \frac{1}{\alpha^n \beta_0 \cdots \beta_n} \le \sum_{n = 0}^\infty \left(\frac{1}{\alpha \beta}\right)^n \] The sum converges if \(\alpha \beta \gt 1\).
In the discrete case, \(\beta_n \in \N_+\) for each \(n \in \N\) and hence a constant rate distribution exists for every \(\alpha \in (1, \infty)\). The distributions defined in are interesting, and include geometric and Poisson distributions as special cases, as we will see next.
Suppose that \(\beta_n = \beta \in (0, \infty)\) for all \(n \in \N\) and that \(\alpha \gt 1 / \beta\). If \(X\) has constant rate \(\alpha\) for \((S, \Upa)\) then \(N\) has the geometric distribution on \(\N\) with success parameter \(1 - 1 / \alpha \beta\): \[\P(N = n) = \left(1 - \frac{1}{\alpha \beta}\right) \left(\frac{1}{\alpha \beta}\right)^n, \quad n \in \N\] \(N\) has constant rate \(\alpha \beta\) for \((\N, \upa)\).
This follows from the remarks after and results in Section 2: If \(X\) has constant rate \(\alpha\) then \(N\) has constant rate \(\alpha \beta\) and hence \(N\) has the geometric distribution with success parameter \(1 - 1 / \alpha \beta\). Of course, the result also follows easily by direct substitution in .
Suppose that \(\beta_n = n + 1\) for \(n \in \N\) and that \(\alpha \gt 0\). If \(X\) has constant rate \(\alpha\) for \((S, \Upa)\) then \(N\) has the Poisson distribution with parameter \(1 / \alpha\):
\[\P(N = n) = e^{-1 / \alpha} \frac{1}{\alpha^n n!}, \quad n \in \N\]
\(N\) has increasing rate for \((\N, \upa)\).
Note that \(p_0 = e^{-1 / \alpha}\).Details:
If \(N\) has the distribution in then \(\E(N) = \var(N) = 1 / \alpha\) and the ordinary probability generating function of \(N\) is \(\E(t^N) = \exp[(t - 1) / \alpha] \) for \(t \in \R\). But of course, were are also interested in the graph moments and generating function.
Suppose that \(N\) has the distributionn in . Then relative to the graph \((\N, \upa)\),
Both results follow from more general results in Section 2.
As before, let \((\N, \rta)\) denote the reflexive closure of the graph \((\N, \upa)\) so that \(n \rta n\) and \(n \rta n + 1\) for each \(n \in \N\). The graph \((S, \Rta)\) induced by \((\N, \rta)\) is given by \(x \Rta y\) if and only if for some \(n \in \N\), either \(x, \, y \in S_n\) or \(x \in S_n\) and \(y \in S_{n + 1}\). Note that \((S, \Rta)\) is not the reflexive closure of the graph \((S, \Upa)\) studied in the last section.
Random variable \(X\) in \(S\) has constant rate \(\alpha\) for \((S, \Rta)\) if and only if \(\alpha \beta_n \lt 1\) for all \(n \in \N\) and
\[\frac{1}{p_0} := \sum_{n = 0}^\infty \prod_{k = 0}^{n - 1} \left(\frac{1}{\alpha \beta_k} - 1\right) \lt \infty\]
In this case, the probability density function of the index variable \(N\) is given by
\[\P(N = n) = p_0 \prod_{k = 0}^{n - 1} \left(\frac{1}{\alpha \beta_k} - 1\right), \quad n \in \N\]
Relative to the grpah \((\N, \rta)\), the index variable \(N\) has reliability function \(P_n = p_n + p_{n + 1}\) for \(n \in \N\). So the condition for a constant rate distribution on \((S, \Rta)\) in is \(p_n = \alpha \beta_n (p_n + p_{n + 1})\) for \(n \in \N\) or equivalently,
\[p_{n + 1} = \left(\frac{1}{\alpha \beta_n} - 1\right) p_n, \quad n \in \N\]
Solving gives
\[p_n = p_0 \prod_{k = 0}^{n - 1} \left(\frac{1}{\alpha \beta_k} - 1\right), \quad n \in \N\]
Hence a constant rate distribution exists if and only if \(\alpha \beta_n \lt 1\) for all \(n \in \N\) and
\[\frac{1}{p_0} = \sum_{n = 0}^\infty \prod_{k = 0}^{n - 1} \left(\frac{1}{\alpha \beta_k} - 1\right) \lt \infty\]
in which case the discrete density function on \(\N\) is given by
\[p_n = p_0 \prod_{k = 0}^{n - 1} \left(\frac{1}{\alpha \beta_k} - 1\right), \quad n \in \N\]
Details:
In particular, if \(\beta_n\) is bounded away from 0 and \(\infty\) in \(n \in \N\), in a certain way, then a constant rate distribution exists.
Suppose that \(a \le \beta_n \le b\) for all \(n \in \N\) where \(0 \lt a \lt b \lt 2 a \lt \infty\). Then a distribution with constant rate \(\alpha\) for \((S, \Rta)\) exists for all \(\alpha \in \left(\frac{1}{2 a}, \frac{1}{b}\right)\).
First, \(\alpha \beta_n \lt \frac{1}{b} b = 1\) for all \(n \in \N\). Next, \(\alpha \gt \frac{1}{2 a}\) so \(\alpha \ge \frac{1}{(1 + r) a}\) for some \(r \in (0, 1)\). Hence \(\alpha \beta_n \ge \frac{1}{(1 + r) a}{a} = \frac{1}{1 + r}\) and so \(\frac{1}{\alpha \beta_n} - 1 \le r\). Therefore \[\sum_{n = 0}^\infty \prod_{k = 0}^{n - 1} \left(\frac{1}{\alpha \beta_k} - 1\right) \le \sum_{k = 0}^\infty r^n \lt \infty\]
Once again, when \(\beta_n\) is constant in \(n \in \N\), the index variable \(N\) has a geometric distribution.
Suppose that \(\beta_n = \beta \in (0, \infty)\) for \(n \in \N\) and that random variable \(X\) in \(S\) has constant rate \(\alpha \in \left(\frac{1}{2 \beta}, \frac{1}{\beta}\right) \) for \((S, \Rta)\). Then \(N\) has the geometric distribution on \(\N\) with success parameter \(2 - \frac{1}{\alpha \beta}\): \[ \P(N = n) = \left(2 - \frac{1}{\alpha \beta}\right) \left(\frac{1}{\alpha \beta} - 1\right)^n, \quad n \in \N \] \(N\) has constant rate \(\alpha \beta\) for \((\N, \rta)\).
This follows from the remarks after and results in Section 2: If \(X\) has constant rate \(\alpha\) then \(N\) has constant rate \(\alpha \beta\) and hence \(N\) has the geometric distribution with success parameter \(2 - 1 / \alpha \beta\). Of course, the result also follows easily by direct substitution in .
The index variable \(N\) can have Poisson distributions as well.
Suppose that \(\beta_n = (n + 1) / (n + 2)\) for \(n \in \N\). If \(X\) has constant rate 1 for \((S, \Rta)\) then \(N\) has the Poisson distribution on \(\N\) with parameter 1: \[\P(N = n) = e^{-1} \frac{1}{n!}, \quad n \in \N\] \(N\) has decreasing rate for \((\N, \rta)\).
In this case, \[\prod_{k = 0}^{n - 1} \left(\frac{1}{\alpha \beta_k} - 1\right) = \prod_{k = 0}^{n - 1} \frac{1}{k + 1} = \frac{1}{n!}\] and \(p_0 = e^{-1}\).
Of course, for the index variable \(N\) in , \(\E(N) = \var(N) = 1\) and \(\E(t^N) = e^{t - 1}\) for \(t \in \R\). But once again, we are also interested in the graph moments and generating function.
Suppose that \(N\) has the Poisson distribution in . Then relative to the graph \((\N, \rta)\),
The results follow from the general results in Section 2.
Next we consider the graph \((S, \prec)\) induced by \((\N, \lt)\). So \(x \prec y\) if and only if \(x \in S_m\) and \(y \in S_n\) for some \(m, \, n \in \N\) with \(m \lt n\). As the notation suggests, \(\prec\) is a strict partial order on \(S\), since clearly \(\prec\) is anti-reflexive and transitive. With \(S\) is discrete, the corresponding partial order graph will be studied in Section 4. As usual, we start with the left walk functions.
The left walk function \(u_k\) of order \(k \in \N_+\) for \((S, \prec)\) is given by \[u_k(x) = \sum_{n_1 \lt n_2 \lt \cdots \lt n_k \lt n} \beta_{n_1} \beta_{n_2} \cdots \beta_{n_k}, \quad x \in S_n, \, n \in \N\]
This follows directly from the general theory in Section 1.7.
So the walk function does not have a simple closed form, except in special cases.
Suppose that \(\beta_n = \beta \in (0, \infty)\) for each \(n \in \N\). Then for \(k \in \N_+\), \[u_k(x) = \binom{n}{k} \beta^k, \quad x \in S_n, \quad n \in \N\]
Random variable \(X\) in \(S\) has constant rate \(\alpha \in (0, \infty)\) for \((S, \prec)\) if and only if \[\lim_{n \to \infty} \frac{1}{(1 + \alpha \beta_0) \cdots (1 + \alpha \beta_n)} = 0\] In this case, the probability density function of the index variable \(N\) is given by \begin{align*} \P(N = n) & = \frac{\alpha \beta_n}{(1 + \alpha \beta_0) \cdots (1 + \alpha \beta_n)} \\ & = \frac{1}{(1 + \alpha \beta_0) \cdots (1 + \alpha \beta_{n - 1})} - \frac{1}{(1 + \alpha \beta_0) \cdots (1 + \alpha \beta_n)}, \quad n \in \N \end{align*}
Note that \(P\) is related to \(p\) by \(P_n = \sum_{k = n + 1}^\infty p_k\). Equivalently \(p_0 = 1 - P_0\) and \(p_n = P_{n - 1} - P_n\) for \(n \in \N_+\). Again, the basic condition for the existence of a constant rate distribution in is \(p_n = \alpha \beta_n P_n\) for \(n \in \N\). Solving we have \begin{align*} P_0 & = \frac{1}{1 + \alpha \beta_0} \\ P_n & = \frac{1}{1 + \alpha \beta_n} P_{n - 1}, \quad n \in \N_+ \end{align*} Hence \[P_n = \frac{1}{(1 + \alpha \beta_0) \cdots (1 + \alpha \beta_n)}, \quad n \in \N\] Note that \(P_n\) is decreasing in \(n \in \N\) so if \(P_n \to 0\) as \(n \to \infty\) then \(P\) is a valid reliability function for \((\N, \lt)\). Then the corresponding discrete probability density function \(p\) is given by \[p_n = \frac{\alpha \beta_n}{(1 + \alpha \beta_0) \cdots (1 + \alpha \beta_n)}, \quad n \in \N\] The second representation follows by simple algebra. Of course, the first term is 1 if \(n = 0\) (Recall that in general, a product over an empty index set is interpreted as 1.)
So the distributions in are defined by telescoping density functions. In particular, a constant rate distribution for \((S, \prec)\) exists if \(\beta_n\) is bounded away from 0 in \(n \in \N\).
If \(\beta_n \ge \beta\) for all \(n \in \N\) where \(\beta \in (0, \infty)\) then a constant rate distribution for for the graph \((S, \prec)\) induced by \((\N, \lt)\) exists for all \(\alpha \in (0, \infty)\).
If \(\beta_n \ge \beta \gt 0\) for all \(n \in \N\) then \[P_n \le \frac{1}{(1 + \alpha \beta)^n} \to 0 \text{ as } n \to \infty \]
Once again, when \(\beta_n\) is constant in \(n \in \N\), the index variable \(N\) has a geometric distribution.
Suppose that \(\beta_n = \beta \in (0, \infty)\) for \(n \in \N\) and that random variable \(X\) in \(S\) has constant rate \(\alpha \in (0, \infty) \) for \((S, \prec)\). Then \(N\) has the geometric distribution on \(\N\) with success parameter \(\alpha \beta / (1 + \alpha \beta)\): \[ \P(N = n) = \frac{\alpha \beta}{1 + \alpha \beta} \left(\frac{1}{1 + \alpha \beta}\right)^n, \quad n \in \N \] \(N\) has constant rate \(\alpha \beta\) for \((\N, \lt)\).
This follows from the remarks after and results in Section 2: If \(X\) has constant rate \(\alpha\) for \((S, \prec)\) then \(N\) has constant rate \(\alpha \beta\) for \((\N, \lt)\) and hence \(N\) has the geometric distribution with success parameter \(\alpha \beta / (1 + \alpha \beta)\). Of course, the result also follows by direct substitution in .
Here is another simple special case:
Suppose that \(\beta_n = n + 1\) for \(n \in \N\) and that \(X\) has constant rate \(\alpha = 1\) for \((S, \prec)\). Then \(N\) has density function given by \[\P(N = n) = \frac{1}{(n + 1)!} - \frac{1}{(n + 2)!}, \quad n \in \N\] \(N\) has increasing rate for \((\N, \lt)\).
Next are the ordinary mean, variance and probability generating function of \(N\).
Suppose that \(N\) has the distribution in . Then
Parts (a) and (b) follow from a few simple compuatations: \begin{align*} \sum_{n = 1}^\infty \frac{n}{(n + 1)!} & = 1 \\ \sum_{n = 1}^\infty \frac{n}{(n + 2)!} & = 3 - e \\ \sum_{n = 1}^\infty \frac{n^2}{(n + 1)!} & = e - 2 \\ \sum_{n = 1}^\infty \frac{n^2}{(n + 2)!} & = 2 e - 5 \end{align*} Part (c) also follows from simple calculus. Of course, the generating function is 1 at \(t = 0\).
But of course, we are also interested in the graph moments and generating function.
Suppose again that \(N\) has the distribution in . Then relative to the graph \((\N, \lt)\),
Suppose next that \((S, \Rrightarrow)\) is the graph induced by \((\N, \le)\), so \(x \Rrightarrow y\) if and only if \(x \in S_m\) and \(y \in S_n\) with \(m \le n\). Note that \((S, \Rrightarrow)\) is not a partial order graph since \(\Rrightarrow\) is not antisymmetric.
Random variable \(X\) in \(S\) has constant rate \(\alpha \in (0, \infty)\) for \((S, \Rrightarrow)\) if and only if \(0 \lt \alpha \beta_k \lt 1\) for each \(k \in \N\) and \(\prod_{k = 0}^\infty (1 - \alpha \beta_k) = 0\). In this case the density function of the index variable \(N\) is given by \[\P(N = n) = \alpha \beta_n \prod_{k = 0}^{n - 1} (1 - \alpha \beta_k); \quad n \in \N\]
The condition for the existence of a distribution with constant rate \(\alpha \in (0, \infty)\) for \((S, \Rrightarrow)\) is \(p_n = \alpha \beta_n P_n\) for \(n \in \N\). But in this case, \(p_n = P_n - P_{n + 1}\) and so we have \(P_{n + 1} = (1 - \alpha \beta_n) P_n\) for \(n \in \N\). Solving gives \[P_n = \prod_{k = 0}^{n - 1} (1 - \alpha \beta_k), \; p_n = \alpha \beta_n \prod_{k = 0}^{n - 1} (1 - \alpha \beta_k); \quad n \in \N\] \(P\) is a valid reliability function for \((\N, \le)\) if \(\alpha\) satisfies \(0 \lt \alpha \beta_k \lt 1\) for each \(k \in \N\) and \(\prod_{k = 0}^\infty (1 - \alpha \beta_k) = 0\).
In particular, if \(\beta_n\) is bounded away from 0 and \(\infty\) in \(n \in \N\), then there exist constant rate distributions.
Suppose that \(a \le \beta_n \le b\) for \(n \in \N\) where \(0 \lt a \lt b \lt \infty\). Then there a distribution with constant rate \(\alpha\) for the graph \((S, \Rrightarrow)\) exists for each \(\alpha \in (0, 1 / b)\).
Under the conditions, \(0 \lt \alpha \beta_n \lt b / b = 1\) and \(0 \lt 1 - \alpha \beta_n \lt 1 - \alpha a \lt 1\) for \(n \in \N\). Hence \(\prod_{k = 0}^\infty (1 - \alpha \beta_k) \lt \prod_{k = 0}^\infty (1 - \alpha a) = 0\).
Once again, when \(\beta_n\) is constant in \(n \in \N\), the index variable \(N\) has a geometric distribution.
Suppose that \(\beta_n = \beta \in (0, \infty)\) for \(n \in \N\) and that random variable \(X\) in \(S\) has constant rate \(\alpha \in (0, 1 / \beta) \) for \((S, \Rrightarrow)\). Then \(N\) has the geometric distribution on \(\N\) with success parameter \(\alpha \beta\): \[ \P(N = n) = \alpha \beta (1 - \alpha \beta)^n, \quad n \in \N \] \(N\) has constant rate \(\alpha \beta\) for \((\N, \le)\).
This follows from the remarks after and results in Section 2: If \(X\) has constant rate \(\alpha\) for \((S, \Rrightarrow)\) then \(N\) has constant rate \(\alpha \beta\) for \((\N, \le)\) and hence \(N\) has the geometric distribution with success parameter \(\alpha \beta\). Of course, the result also follows by direct substitution in
Another special case, with the same partition sizes as in , leads to the same distribution as in .
Suppose that \(\beta_n = (n + 1) / (n + 2)\) and that \(X\) has constant rate \(\alpha = 1\) for \((S, \Rrightarrow)\). Then \(N\) has density function given by \[\P(N = n) = \frac{1}{(n + 1)!} - \frac{1}{(n + 2)!}, \quad n \in \N\] \(N\) has decreasing rate for \((\N, \le)\).
The ordinary mean, variance, and probability generating functtion of \(N\) are given in . But once again, we are interested in the graph moments and generating function.
Suppose that \(N\) has the distribution in . Then relative to the graph \((\N, \le)\),
Both results follow from general results in Section 1. Recall that