Analysis of quantities determining the critical inverse temperature in the annealed Potts model with Pareto vertex weights

TL;DR

This paper analyzes the critical inverse temperature in the annealed Potts model with Pareto weights on sparse graphs, deriving bounds and asymptotic behaviors for key quantities that determine phase transitions.

Contribution

It provides a detailed formulation and bounds for the quantities $t_c$, $t_c'$, and $t_c''$ that determine the critical temperature in the model, extending previous results to general parameters.

Findings

Derived bounds for $t_c$, $t_c'$, and $t_c''$ in terms of $q$ and $ au$

Established inequalities $0<t_c''<t_c'<t_c$ and their sharpness

Analyzed asymptotic behavior as $ au o \infty$

Abstract

We consider in this work the crucial quantity $t_c$ that determines the critical inverse temperature $\beta_c$ in the $q$-state Potts model on sparse rank-1 random graphs where the vertices are equipped with a Pareto weight density $(\tau-1)\,w^{-\tau}\,{\cal X}_{[1,\infty)}(w)$. It is shown in \cite{ref1} that this $t_c$ is the unique positive zero of a function ${\cal K}$ that is obtained by an appropriate combination of the stationarity condition and the criticality condition for the case the external field $B$ equals 0 and that $q\geq3$ and $\tau\geq4$, see \cite{ref1}, Theorem~1.14 and Theorem ~1.21 and their proofs in \cite{ref1}, Section~7.1 and Section~7.3. From the proof of \cite{ref1}, Theorem~1.14, it is seen that ${\cal K}'$ and ${\cal K}''$ also have a unique positive zero, $t_c'$ and $t_c''$, respectively, and $t_c'=t_b$ and $t_c''=t_{\ast}$, where $t_b$ and $t_{\ast}$ are the unique positive zeros of ${\cal F}_0(t)-t\,{\cal F}_0'(t)$ and ${\cal F}_0''(t)$, respectively. Here, ${\cal F}_0(t)=E\,[W(e^{tW}-1)/(E\,[W]\,(e^{tW}+q-1))]$, and $t_c$, $t_b$ and $t_{\ast}$ play a key role in the graphical analysis of \cite{ref1}, Section~5.1 and Figure~1. Furthermore, $\gamma_c=\exp(\beta_c)-1$ and $t_c$ are related according to $\gamma_c=t_c/{\cal F}_0(t_c)$. We analyse $t_c$, $t_c'$ and $t_c''$ for general real $\tau\geq4$ and general real $q>2$ by an appropriate formulation of their defining equations ${\cal K}(t_c)={\cal K}'(t_c')={\cal K}''(t_c'')=0$. Thus we find, along with the inequality $0<t_c''<t_c'<t_c<\infty$, the simple upper bounds $t_c<2\,{\rm ln}(q-1)$, $t_c'<\frac32\,{\rm ln}(q-1)$, $t_c''<{\rm ln}(q-1)$, as well as certain sharpenings of these simple bounds and counterparts about the large-$q$ behaviour of $t_c$, $t_c$ and $t_c''$. We show that these bounds are sharp in the sense that they hold with equality for the limiting homogeneous case $\tau\to\infty$.

Equations

$${\mathcal{F}}_0(t)=E\left[\frac{W}{E[W]}\frac{e^{tW}-1}{e^{tW}+q-1}\right],t\ge 0.$$

$$p(t,\gamma )=E[\ln (e^{tW}+q-1)]-E[W]\frac{\gamma }{2q}((q-1)(\frac{t}{\gamma }{)}^2+2\frac{t}{\gamma }-1).$$

$${\mathcal{F}}_0(t)=\frac{t}{\gamma }$$

$$p(t,\gamma )=p(0,\gamma )$$

$$\frac{1}{E[W]}E\left[\ln \right(\frac{e^{tW}+q-1}{q}\left)\right]-\frac{q-1}{2q}t\cdot \frac{t}{\gamma }-\frac{t}{q}=0.$$

$$\mathcal{K}(t)=\frac{1}{E[W]}E\left[\ln \right(\frac{e^{tW}+q-1}{q}\left)\right]-\frac{q-1}{2q}t{\mathcal{F}}_0(t)-\frac{t}{q},t\ge 0,$$

$${\beta }_c=\ln (1+\frac{t_c}{{\mathcal{F}}_0(t_c)}).$$

$${\mathcal{K}}^{\prime }(t)=\frac{q-1}{2q}[{\mathcal{F}}_0(t)-t{\mathcal{F}}_0^{\prime }(t)],t\ge 0,$$

$${\mathcal{K}}^{\prime \prime }(t)=-\frac{q-1}{2q}t{\mathcal{F}}_0^{\prime \prime }(t),t\ge 0.$$

$${\mathcal{F}}_0^{\prime \prime }(t_{\ast })=0,{\mathcal{F}}_0(t_b)-t_b{\mathcal{F}}_0^{\prime }(t_b)=0.$$

$$t_c=t_c,t_c^{\prime }=t_b,t_c^{\prime \prime }=t_{\ast };0<t_c^{\prime \prime }<t_c^{\prime }<t_c<\infty ,$$

$$\mathcal{K}(t)=\frac{q-1}{q}\left(\underset{0}{\overset{t}{\int }}{\mathcal{F}}_0(s)ds-\frac{1}{2}t{\mathcal{F}}_0(t)\right),t\ge 0.$$

$$\underset{0}{\overset{t_c}{\int }}{\mathcal{F}}_0(s)ds=\frac{1}{2}{t}_c{\mathcal{F}}_0(t_c)=\underset{0}{\overset{t_c}{\int }}s{\mathcal{F}}_0(t_c)/t_{c}ds,$$

$$t<t_c\iff \mathcal{K}(t)<0;t>t_c\iff \mathcal{K}(t)>0,$$

$$t<t_c^{\prime }\iff {\mathcal{K}}^{\prime }(t)<0;t>t_c^{\prime }\iff {\mathcal{K}}^{\prime }(t)>0,$$

$$t<t_c^{\prime \prime }\iff {\mathcal{K}}^{\prime \prime }(t)<0;t>t_c^{\prime \prime }\iff {\mathcal{K}}^{\prime \prime }(t)>0.$$

$$f_W(w)=(\tau -1)w^{-\tau }{\mathcal{X}}_{[1,\infty )}(w),w\ge 0,$$

$$\mathcal{K}(t)=\frac{\tau -2}{\tau -1}\ln \left(\frac{e^t+q-1}{q}\right)+(\frac{1}{\tau -1}-\frac{q+1}{2q})t+\frac{(\tau -2)(\tau -3)}{2(\tau -1)}t(q-1)D,$$

$${\mathcal{K}}^{\prime }(t)=\frac{1}{2}(q-1)(\frac{1}{q}-\frac{\tau -2}{e^t+q-1}+(\tau -2)(\tau -3)D),$$

$${\mathcal{K}}^{\prime \prime }(t)=\frac{(q-1)(\tau -2)}{2t}(\frac{t{e}^t}{(e^t+q-1{)}^2}-\frac{\tau -3}{e^t+q-1}+(\tau -2)(\tau -3)D),$$

$$D=D(t)=\underset{1}{\overset{\infty }{\int }}\frac{w^{-\tau +1}dw}{e^{tw}+q-1}.$$

$$t_c<2\ln (q-1),t_c^{\prime }<\frac{3}{2}\ln (q-1),t_c^{\prime \prime }<\ln (q-1),$$

$$t_c^{\prime \prime }=t_{\ast }<t_c^{\prime }=t_b<t_c,$$

$$t_c<2\frac{\tau -2}{\tau -1}\ln (q-1).$$

$$\frac{\mathcal{K}(2\ln (q-1))}{{\mathcal{K}}^{\prime }(2\ln (q-1))}=\frac{2\ln (q-1)}{\tau -1},$$

$$t_c<2\ln (q-1)-\frac{\mathcal{K}(2\ln (q-1))}{{\mathcal{K}}^{\prime }(2\ln (q-1))},$$

$$\frac{t{e}^t}{(e^t+q-1{)}^2}+\frac{1}{e^t+q-1}-\frac{1}{q}=0.$$

$$t_c^{\prime }<T<\frac{3}{2}\ln (q-1).$$

$$\frac{{\mathcal{K}}^{\prime }(T)}{{\mathcal{K}}^{\prime \prime }(T)}=\frac{T}{\tau -2}.$$

$$T-\frac{{\mathcal{K}}^{\prime }(T)}{{\mathcal{K}}^{\prime \prime }(T)}=\frac{\tau -3}{\tau -2}T,$$

Full text

Analysis of quantities determining the critical inverse temperature in the annealed Potts model with Pareto vertex weights

A.J.E.M. Janssen

Eindhoven University of Technology

Department of Mathematics and Computer Science

Abstract.

We consider in this work the crucial quantity $t_{c}$ that determines the critical inverse temperature $\beta_{c}$ in the $q$-state Potts model on sparse rank-1 random graphs where the vertices are equipped with a Pareto weight density $(\tau-1)\,w^{-\tau}\,{\cal X}_{[1,\infty)}(w)$. It is shown in [1] that this $t_{c}$ is the unique positive zero of a function ${\cal K}$ that is obtained by an appropriate combination of the stationarity condition and the criticality condition for the case the external field $B$ equals 0 and that $q\geq 3$ and $\tau\geq 4$, see [1], Theorem 1.14 and Theorem  1.21 and their proofs in [1], Section 7.1 and Section 7.3. From the proof of [1], Theorem 1.14, it is seen that ${\cal K}^{\prime}$ and ${\cal K}^{\prime\prime}$ also have a unique positive zero, $t_{c}^{\prime}$ and $t_{c}^{\prime\prime}$, respectively, and $t_{c}^{\prime}=t_{b}$ and $t_{c}^{\prime\prime}=t_{\ast}$, where $t_{b}$ and $t_{\ast}$ are the unique positive zeros of ${\cal F}_{0}(t)-t\,{\cal F}_{0}^{\prime}(t)$ and ${\cal F}_{0}^{\prime\prime}(t)$, respectively. Here, ${\cal F}_{0}(t)={{E}}\,[W(e^{tW}-1)/({{E}}\,[W]\,(e^{tW}+q-1))]$, and $t_{c}$, $t_{b}$ and $t_{\ast}$ play a key role in the graphical analysis of [1], Section 5.1 and Figure 1. Furthermore, $\gamma_{c}=\exp(\beta_{c})-1$ and $t_{c}$ are related according to $\gamma_{c}=t_{c}/{\cal F}_{0}(t_{c})$.

We analyse $t_{c}$, $t_{c}^{\prime}$ and $t_{c}^{\prime\prime}$ for general real $\tau\geq 4$ and general real $q>2$ by an appropriate formulation of their defining equations ${\cal K}(t_{c})={\cal K}^{\prime}(t_{c}^{\prime})={\cal K}^{\prime\prime}(t_{c}^{\prime\prime})=0$. Thus we find, along with the inequality $0<t_{c}^{\prime\prime}<t_{c}^{\prime}<t_{c}<\infty$, the simple upper bounds $t_{c}<2\,{\rm ln}(q-1)$, $t_{c}^{\prime}<\frac{3}{2}\,{\rm ln}(q-1)$, $t_{c}^{\prime\prime}<{\rm ln}(q-1)$, as well as certain sharpenings of these simple bounds and counterparts about the large-$q$ behaviour of $t_{c}$, $t_{c}^{\prime}$ and $t_{c}^{\prime\prime}$. We show that these (sharpened) bounds are sharp in the sense that they hold with equality for the limiting case $\tau\rightarrow\infty$ (homogeneous case). These results have consequences for $\gamma_{c}$ and $\beta_{c}$ via the equation $\gamma_{c}=t_{c}/{\cal F}_{0}(t_{c})$. While the large-$q$ behaviour of $t_{c}$, $t_{c}^{\prime}$ and $t_{c}^{\prime\prime}$ is qualitatively the same for all $\tau\geq 4$, such a thing does not hold for the behaviour when $q\downarrow 2$. It appears that one has to distinguish between the cases $\tau=4$, $4<\tau<5$, $\tau=5$ and $\tau>5$ with decay behaviour as $q\downarrow 2$ ranging from exponentially small in $q-2$ to linear in $q-2$.

1 Introduction

In [1] a comprehensive study has been made of the annealed ferromagnetic $q$-state Potts model on sparse rank-1 random graphs, where vertices are equipped with a vertex weight and where the probability of an edge is proportional to the product of the weights. Under a rather general condition on the weight distribution, phase transition is first order, with a unique discontinuity of the order parameter, see [1], Abstract and Section 1. Let $\beta>0$ denote the inverse temperature, and let $\gamma=e^{\beta}-1$ (in [1] one writes $\beta^{\prime}$ instead of $\gamma$, but in the present work the prime ′ has been reserved to indicate differentiation). Assume that the external field $B$ equals 0 and that ${{E}}\,[W^{2}]<\infty$ ($W$ is the random weight variable). Consider the function

[TABLE]

The function ${\cal F}_{0}(t)$ is non-negative, strictly increasing and ${\cal F}_{0}(t)\rightarrow 1$, $t\rightarrow\infty$. Assume that ${\cal F}_{0}^{\prime\prime}(t)$ is first positive and then negative, so that ${\cal F}_{0}(t)$ has a unique inflection point $t_{\ast}>0$ with ${\cal F}_{0}^{\prime\prime}(t_{\ast})=0$ and ${\cal F}_{0}^{\prime\prime}(t)>0$, $0\leq t<t_{\ast}$, and ${\cal F}_{0}^{\prime\prime}(t)<0$, $t>t_{\ast}$. Also consider the annealed pressure

[TABLE]

Under the above condition on ${\cal F}_{0}$, there is a unique pair $(t_{c},\gamma_{c})$ with $t_{c}>0$, $\gamma_{c}>0$ such that simultaneously the stationarity condition

[TABLE]

and the criticality condition

[TABLE]

is satisfied by $(t,\gamma)=(t_{c},\gamma_{c})$. The criticality condition can be written as

[TABLE]

In Theorem 1.14 of [1] about the critical inverse temperature $\beta_{c}={\rm ln}(1+\gamma_{c})$, there is considered the function

[TABLE]

so that $\gamma$ is eliminated from the left-hand side of (5) by using (3). By [1], Theorem 1.14, $t_{c}$ is the unique $t>0$ such that ${\cal K}(t)=0$, and

[TABLE]

The proof of [1], Theorem 1.14 furthermore shows that

[TABLE]

[TABLE]

According to [1], Section 5.1, for the case $B=0$, there are unique numbers $t_{\ast}$, $t_{b}>0$ such that

[TABLE]

Denoting the unique positive zeros of ${\cal K}$, ${\cal K}^{\prime}$ and ${\cal K}^{\prime\prime}$ by $t_{c}$, $t_{c}^{\prime}$ and $t_{c}^{\prime\prime}$ respectively, we thus see that

[TABLE]

where we also refer to [1], Figure 1. Hence, $t_{c}^{\prime\prime}=t_{\ast}$ and $t_{c}^{\prime}=t_{b}$ have a geometric meaning in terms of the graph of ${\cal F}_{0}$ (inflection point and point determining the non-trivial tangent line that passes through the origin, respectively). We can find for $t_{c}$ such a geometric meaning as well from (8). Integrating (8) from 0 to $t$, using ${\cal K}(0)=0$ and partial integration, we get

[TABLE]

Hence, the condition ${\cal K}(t_{c})=0$ means that

[TABLE]

so that the sets $\{(s,v)\,|\,0\leq s\leq t_{c},~{\cal F}_{0}(s)\leq v\leq s\,{\cal F}_{0}(t_{c})/t_{c}\}$ and $\{(s,v)\,|\,0\leq s\leq t_{c},~s\,{\cal F}_{0}(t_{c})/t_{c}\leq v\leq{\cal F}_{0}(s)\}$ have equal area. Recall that $1/\gamma_{c}={\cal F}_{0}(t_{c})/t_{c}$. We illustrate all this in Figure 1 below, where we display for the homogeneous case $W\equiv 1$ with $q=100$ the graph of ${\cal F}_{0}(t)$, $0\leq t\leq 10$ (this example was suggested by C. Giardinà and C. Giberti). In this case we have ${\cal F}_{0}(t)=1-100/(e^{t}+99)$, and $t_{c}^{\prime\prime}={\rm ln}(q-1)=4.5951...$, $t_{c}=2\,{\rm ln}(q-1)=9.1902...$, while $t_{c}^{\prime}$ can be determined numerically as $t_{c}^{\prime}=1.3648\,...\,{\rm ln}(q-1)=6.2716...\,$. In this Figure 1, we also have indicated on the $t$-axis the points $t_{c}$, $t_{c}^{\prime}$ and $t_{c}^{\prime\prime}$, as well as the tangent line passing through $(0,0)$ and $(t_{c}^{\prime},{\cal F}_{0}^{\prime}(t_{c}^{\prime}))$ and the line $\{s\,{\cal F}_{0}(t_{c})/t_{c}\,|\,0\leq s\leq t_{c}\}$ that nicely realizes the equal area condition (compare [1], Proposition 5.6).

We have ${\cal K}(0)={\cal K}^{\prime}(0)={\cal K}^{\prime\prime}(0)=0$ while ${\cal K}(t)$ is negative for small positive $t$ and positive for large positive $t$. Therefore, ${\cal K}$, ${\cal K}^{\prime}$ and ${\cal K}^{\prime\prime}$ are first negative and then positive, and for all $t>0$ we have

[TABLE]

[TABLE]

[TABLE]

2 Main results

We are interested in bounds and approximations of $t_{c}$, $t_{c}^{\prime}$ and $t_{c}^{\prime\prime}$ in the particular case that $W$ has Pareto density

[TABLE]

with $\tau\geq 4$. Here ${\cal X}_{[1,\infty)}(w)$ equals 1 for $1\leq w<\infty$ and 0 for $0\leq w<1$. The restriction $\tau\geq 4$ ensures that ${\cal F}_{0}$ in (1) and ${\cal K}$ in (6) have for all $q>2$ a unique, positive inflection point $t_{\ast}=t_{c}^{\prime\prime}$, see [1], Theorem 1.21. In [2], a similar effort has been done for $t_{c}$ in the case that $W$ has an exponential density $f_{W}(w)=\exp({-}w)\,{\cal X}_{[0,\infty)}(w)$, $w\geq 0$. It appears that the results we obtain in the present work are, compared with those obtained in [2], more transparent, complete and appealing and hardly require numerical efforts.

In Section 3, we shall show that for $t\geq 0$, $q>2$, $\tau\geq 4$

[TABLE]

[TABLE]

[TABLE]

where

[TABLE]

It happens to be a very convenient fact that in (18), (19) and (20) the same integral $D$ appears.

In Section 4, we shall show that for all $\tau\geq 4$ and $q>2$

[TABLE]

where we recall that

[TABLE]

see [1], Figure 1. The proof of (22) uses the criteria in (14), (15) and (16), the representations (18), (19) and (20) of ${\cal K}$, ${\cal K}^{\prime}$ and ${\cal K}^{\prime\prime}$ and a convexity property of the function $(e^{tw}+q-1)^{-1}$, see (21), as a function of $w\geq 1$.

In Section 5, we shall improve the bounds on $t_{c}$ and $t_{c}^{\prime}$ as follows. We have for $\tau\geq 4$ and $q>2$

[TABLE]

The proof of this uses the observation that

[TABLE]

so that by $t_{c}<2\,{\rm ln}(q-1)$ and convexity of ${\cal K}(t)$ in $t\geq t_{c}$

[TABLE]

the right-hand side of (26) being the first Newton iterand $t^{(1)}$ to compute $t_{c}$ starting with $t^{(0)}=2\,{\rm ln}(q-1)$. From (25), (26) we get (24).

To improve the bound on $t_{c}^{\prime}$, we let $T$ be the unique positive solution $t$ of the equation

[TABLE]

Then it appears that

[TABLE]

Furthermore, the choice of $T$ is such that

[TABLE]

Hence, we have

[TABLE]

where the left-hand side of (30) is the first Newton iterand $t^{(1)}$ to compute $t_{c}^{\prime}$ starting with $t^{(0)}=T$. While ${\cal K}(t)$ is convex in $t\geq t_{c}$, convexity of ${\cal K}^{\prime}(t)$ in $t\in[t_{c}^{\prime},T]$ does not always hold, and there are indeed cases that $\frac{\tau-3}{\tau-2}\,T<t_{c}^{\prime}$, i.e., the right-hand side of (30) is not necessarily an upper bound of $t_{c}^{\prime}$.

In Section 6, we show the following counterparts to the inequalities (22) and (24). We have for any $\tau\geq 4$

[TABLE]

Furthermore, it is shown that $t_{c}^{\prime}<{\rm ln}(q-1)$ when $\tau=4$, $q>2$, and that $t_{c}^{\prime}>{\rm ln}(q-1)$ when $\tau>4$ and $q$ is sufficiently large. We also discuss in Section 6 the conjecture that for $\tau\geq 4$ and $q>2$

[TABLE]

By the first limit relation in (31), we have that (32) holds for $\tau\geq 4$ and $q$ sufficiently large. We also give in Section 6 a consequence of the first limit relation in (31) for $\gamma_{c}=\exp(\beta_{c})-1$, with $\beta_{c}$ the critical inverse temperature.

In Section 7, we consider the limiting behaviour of $t_{c}^{\prime\prime}$ as $q\downarrow 2$. This is based on the observation that $t_{c}^{\prime\prime}=t_{\ast}$ is the unique positive zero of ${\cal F}_{0}^{\prime\prime}$, so that by [1], Section 7.2, $t_{c}^{\prime\prime}$ is the unique positive zero $t$ of

[TABLE]

The function $a(x)$ is positive for $x\in[0,{\rm ln}(q-1))$ and negative for $x\in({\rm ln}(q-1),\infty)$, where ${\rm ln}(q-1)\downarrow 0$ as $q\downarrow 2$. Moreover, $\Phi(t)\rightarrow\infty$ as $t\downarrow 0$ in a $\tau$-dependent way. It thus appears that one has to distinguish between the cases

[TABLE]

with respective decay behaviour of $t_{c}^{\prime\prime}$ as $b={\rm ln}(q-1)\downarrow 0$

[TABLE]

where $K_{1}$, $K_{2}$, $K_{3}$, $K_{4}$ depend on $\tau$, and, interestingly, $K_{4}=\frac{\tau-5}{\tau-4}$.

Finally, in Section 8, we consider the homogeneous case that arises when $\tau\rightarrow\infty$, so that $f_{W}(w)\rightarrow\delta(w-1)$. Denote by $t_{H,c}$, $t_{H,c}^{\prime}$ and $t_{H,c}^{\prime\prime}$ the unique positive zeros of ${\cal K}_{H}$, ${\cal K}_{H}^{\prime}$ and ${\cal K}_{H}^{\prime\prime}$, respectively, with ${\cal K}_{H}$ the ${\cal K}$-function corresponding to this homogeneous case. As is well known, we have $t_{H,c}=2\,{\rm ln}(q-1)$. Now it turns out that $t_{H,c}^{\prime}=T$, with $T$ the unique positive solution $t$ of (27). Furthermore, we have $t_{H,c}^{\prime\prime}={\rm ln}(q-1)$. Hence, the inequalities $t_{c}\leq 2\,{\rm ln}(q-1)$, $t_{c}^{\prime}\leq T$ and $t_{c}^{\prime\prime}<{\rm ln}(q-1)$, holding for the finite-$\tau$ Pareto case, are sharp in the sense that they hold with equality in the limit case $\tau\rightarrow\infty$. While $t_{H,c}$ and $t_{H,c}^{\prime\prime}$ have simple closed forms, as just noted, such a thing does not hold for $t_{H,c}^{\prime}$, and so a further analysis is required for the latter case. It is shown in Section 9 that $t_{H,c}^{\prime}=T(q)$ increases in $q>2$ from 0 to $\infty$, and that $T(q)/{\rm ln}(q-1)$ decreases in $q>2$ from $3/2$ to 1 with $T(q)/{\rm ln}(q-1)\sim 1+{\rm ln}({\rm ln}(q-1))/{\rm ln}(q-1)$ as $q\rightarrow\infty$.

3 Proof of the representations (18), (19) and (20)

The proof of the representation (18) of ${\cal K}$ has been given in [1], Appendix C by using the definition (6) of ${\cal K}$. Thus, by partial integration,

[TABLE]

Furthermore,

[TABLE]

and the remaining integrals in (3) and (37) equal the $D$ in (21).

For proving the representation (19) for ${\cal K}^{\prime}$, one differentiates the expression (18) for ${\cal K}$ with respect to $t$, and then one is left with the $D$-integral itself and its derivative with respect to $t$. As to the latter, we have

[TABLE]

Finally, for proving the representation (20) of ${\cal K}^{\prime\prime}$, one differentiates the expression (19) for ${\cal K}^{\prime}$ and uses (3) once more.

The fact that in the representation of ${\cal K}$, ${\cal K}^{\prime}$ and ${\cal K}^{\prime\prime}$ only one and the same integral $D$ is required, is quite convenient, both from an analytic, see the following sections, and a numerical point of view. As to the latter point, both $t_{c}$ and $t_{c}^{\prime}$ can be computed by a Newton iteration

[TABLE]

where an appropriate starting value $t^{(0)}$ should be chosen. For ${\cal L}={\cal K}$, any upper bound of $t_{c}$ can be taken since ${\cal K}(t)$ is convex in $t\in(t_{c}^{\prime\prime},\infty)\supset[t_{c},\infty)$. The computation of $t_{c}$ by Newton iteration has also been considered in [1], Appendix C. For the Pareto case, there is given in [1], (C.10), a considerably more complicated expression for ${\cal K}^{\prime}$ than the one we presently have in (19). Moreover, the starting value $t^{(0)}$ that is used in [1], Appendix C, is $t^{(0)}=\frac{\tau-2}{\tau-1}~\frac{2q\,{\rm ln}\,q}{q-1}$ which exceeds $2\,\frac{\tau-2}{\tau-1}\,{\rm ln}(q-1)$, see (24).

We can compute, in principle, also $t_{c}^{\prime}$ by Newton iteration, but there is here the complication that ${\cal K}^{\prime}$ is not convex on the whole range $[t_{c}^{\prime},\infty)$. For it is seen from the expression for ${\cal K}^{\prime\prime}(t)$ in (20) that ${\cal K}^{\prime\prime}(t)\rightarrow 0$ as $t\rightarrow\infty$, and so ${\cal K}^{\prime\prime\prime}(t)<0$ for certain large $t$. With $T$ the upper bound of $t_{c}^{\prime}$ obtained in (27), (28), it appears (at least numerically) that ${\cal K}^{\prime}(t)$ is convex in $t\in[t_{c}^{\prime},T]$ when $q$ is not very large compared to $\tau$.

4 Proof of the simple bounds in (22)

In this section we show that

[TABLE]

for all $\tau\geq 4$ and $q>2$. According to the criteria in (14), (15), (16), we have that (40) is equivalent with

[TABLE]

To establish (41), we consider for $\alpha\geq 1$ and $q>2$, $t=\alpha\,{\rm ln}(q-1)$ the expression

[TABLE]

where $D=D(t)$ with $t=\alpha\,{\rm ln}(q-1)$ in (21). The function

[TABLE]

is positive, increasing and strictly convex in $w>1$. Indeed, we have $\alpha\geq 1$, ${\rm ln}(q-1)>0$ and so $z:=(\alpha w-1)\,{\rm ln}(q-1)>0$ for $w>1$, while

[TABLE]

It follows that

[TABLE]

and so, by straightforward integration,

[TABLE]

Now we show the first inequality in (41) for which we make the choice $\alpha=2$. We have from (18)

[TABLE]

With $\alpha=2$ in (43), we have $f(w)=((q-1)^{2w-1}+1)^{-1}$, and so

[TABLE]

Hence, from (46),

[TABLE]

Then from (4)

[TABLE]

since $x^{2}-1-2x\,{\rm ln}\,x>0$ for $x=q-1>1$.

We next show the second inequality in (41), for which we make the choice $\alpha=3/2$. We have from (19) with $t=\frac{3}{2}\,{\rm ln}(q-1)$, $\exp(t)=(q-1)^{3/2}$

[TABLE]

According to (46)

[TABLE]

where now $f(w)=((q-1)^{\frac{3}{2}w-1}+1)^{-1}$. Then from

[TABLE]

we see that

[TABLE]

Hence, we have ${\cal K}^{\prime}(\frac{3}{2}\,{\rm ln}(q-1))>0$ when we can show that

[TABLE]

Setting $y=x^{1/2}>1$, we consider

[TABLE]

Then (55) is equivalent with $\varphi(y)>0$, $y>1$. We have

[TABLE]

and so $\varphi(y)>0$, $y>1$ indeed holds.

We finally show the third inequality in (41), for which we make the choice $\alpha=1$. We have from (20) with $t={\rm ln}(q-1)$, $\exp(t)=q-1$

[TABLE]

According to (46)

[TABLE]

where now $f(w)=((q-1)^{w-1}+1)^{-1}$. Then from

[TABLE]

we see that

[TABLE]

as required.

A quick proof of $t_{c}^{\prime\prime}<{\rm ln}(q-1)$ follows from the characterization of $t_{c}^{\prime\prime}$ as the unique positive zero of $\Phi(t)$ in (33) which is obviously negative for $t\geq{\rm ln}(q-1)$ since $a(x)<0$ for $x>{\rm ln}(q-1)$.

5 Sharpening of the simple bounds for $t_{c}$ and $t_{c}^{\prime}$

We start by showing that

[TABLE]

for all $\tau\geq 4$ and $q>2$. The second inequality is obvious. Since ${\cal K}(t)$ is convex in $t\geq t_{c}>t_{c}^{\prime\prime}$, we have for any upper bound $T$ of $t_{c}$ that

[TABLE]

where the middle member is the first Newton iterand $t^{(1)}$ to compute $t_{c}$ starting with $t^{(0)}=T$. We take $T=2\,{\rm ln}(q-1)$, and we have from (4)

[TABLE]

At the same time, we compute from (19)

[TABLE]

in which the two $D$’s in (64) and (65) are equal. Hence,

[TABLE]

and so, by (63),

[TABLE]

We next consider sharpening the simple upper bound $\frac{3}{2}\,{\rm ln}(q-1)$ of $t_{c}^{\prime}$. Our initial aim is to find an upper bound $T$ of $t_{c}^{\prime}$ such that ${\cal K}^{\prime}(T)/{\cal K}^{\prime\prime}(T)$ takes a simple form, just as we succeeded in doing so to improve the simple upper bound $2\,{\rm ln}(q-1)$ of $t_{c}$. Such a simple form for ${\cal K}^{\prime}(T)/{\cal K}^{\prime\prime}(T)$ arises from (19) and (20) when we manage to take $T=t>0$ such that

[TABLE]

the two $D$’s in (19) and (20) being the same. Thus, with this $T=t$, we get

[TABLE]

The first Newton iterand $t^{(1)}$ to compute $t_{c}^{\prime}$ starting with $t^{(0)}=T$ is given by

[TABLE]

We now detail this approach further. The equation in (68) is the same as

[TABLE]

from which $\tau$ has disappeared altogether. Multiplying through by $(e^{t}+q-1)^{2}$, we get from (71) the equation

[TABLE]

in which

[TABLE]

We shall show that $\psi(y)$ has a unique zero $Y>1$. We have, since $q>2$,

[TABLE]

and

[TABLE]

Therefore, $\psi^{\prime\prime}(y)>0$ for $y\in[1,\frac{1}{2}\,q)$ and $\psi^{\prime\prime}(y)<0$ for $y\in(\frac{1}{2}\,q,\infty)$. Thus, $\psi(y)$ is positive, increasing and convex in $y\in(1,\frac{1}{2}\,q)$, and $\psi(y)$ is concave in $y\in(\frac{1}{2}\,q,\infty)$, while we see from (73) that $\psi(y)\rightarrow{-}\infty$ as $y\rightarrow\infty$. We conclude that $\psi(y)$ has a unique positive zero $Y$ with $Y>\frac{1}{2}\,q>1$.

We shall show in the appendix that

[TABLE]

Hence

[TABLE]

Therefore, $T={\rm ln}\,Y$ is the unique positive solution $t$ of (72), and

[TABLE]

From the bound $t_{c}^{\prime\prime}<{\rm ln}(q-1)$, we have $t_{c}^{\prime\prime}<T$, and so, from (16),

[TABLE]

The choice of $T$ is such that the two quantities in the brackets (  ) at the right-hand sides of the representations (19) and (20) of ${\cal K}^{\prime}$ and ${\cal K}^{\prime\prime}$ are the same, see (68), for $t=T$. From ${\cal K}^{\prime\prime}(T)>0$, we then conclude that ${\cal K}^{\prime}(T)>0$ as well, and so, by (15), $t_{c}^{\prime}<T$. Therefore,

[TABLE]

and so $T$ indeed sharpens $\frac{3}{2}\,{\rm ln}(q-1)$ as an upper bound of $t_{c}^{\prime}$.

It is now tempting to ask whether $\frac{\tau-3}{\tau-2}\,T$, is an upper bound of $t_{c}^{\prime}$ as well. This would be the case when ${\cal K}^{\prime}(t)$ were convex in $t\in[t_{c}^{\prime},T]$, i.e., when ${\cal K}^{\prime\prime}(t)$ were increasing in $t\in[t_{c}^{\prime},T]$. However, this does not hold always, as we illustrate now.

[TABLE]

6 Limiting behaviour of $t_{c}$, $t_{c}^{\prime}$ and $t_{c}^{\prime\prime}$ as $q\rightarrow\infty$

We shall now establish the limit relations in (31). To show the first one of these, viz. $t_{c}/{\rm ln}(q-1)\rightarrow 2(\tau-2)/(\tau-1)$ as $q\rightarrow\infty$, it is by (14) and (24) sufficient to show that for any $\alpha\in(1,2(\tau-2)/(\tau-1))$ we have

[TABLE]

when $q$ is sufficiently large. So let $\alpha\in(1,2(\tau-2)/(\tau-1))$. We use (18) with $t=\alpha\,{\rm ln}(q-1)$, $\exp(t)=(q-1)^{\alpha}$, and we write

[TABLE]

Furthermore,

[TABLE]

Therefore,

[TABLE]

Since $1<\alpha<2(\tau-2)/(\tau-1)$, we see that ${\cal K}(\alpha\,{\rm ln}(q-1))<0$ indeed, when $q$ is sufficiently large.

We conjecture the lower bound for $\tau\geq 5$ and all $q>2$

[TABLE]

and this seems to be a very sharp lower bound when $\tau$ is not too small and $q$ is not too large. From $t_{c}/{\rm ln}(q-1)\rightarrow 2(\tau-2)/(\tau-1)$ as $q\rightarrow\infty$ and $(\tau-2)/(\tau-1)>(\tau-5)/(\tau-4)$, we see that (85) holds when $\tau\geq 4$ and $q$ is sufficiently large. For $\tau=6$, we would have $t_{c}>{\rm ln}(q-1)$.

In fact, a limited numerical exploration with some other pdfs (like the exponential $\exp({-}w)\,{\cal X}_{[0,\infty)}(w)$ and ${\cal X}_{[0,1)}(w)$) shows that there may hold more generally

[TABLE]

where $\mu_{n}$ is the $n^{{\rm th}}$ moment,

[TABLE]

of the pdf $f_{W}$. Note that by a continuous version of Lehmer’s inequality, we have $\mu_{p-1}/\mu_{p}>\mu_{q-1}/\mu_{q}$ when $p<q$ (see [3]), whence $\mu_{3}/\mu_{4}<\mu_{0}/\mu_{1}$.

We now give a consequence of the limit relation satisfied by $t_{c}$ for $\gamma_{c}=\exp(\beta_{c})-1$ with $\beta_{c}$ the critical inverse temperature. We have

[TABLE]

where ${\cal F}_{0}$ is given by (1). In the Pareto case, we can elaborate ${\cal F}_{0}(t)$ as

[TABLE]

Therefore,

[TABLE]

and so

[TABLE]

With $\alpha\in(1,(\tau-2)/(\tau-1))$ fixed, we have

[TABLE]

when $q$ is sufficiently large, and so

[TABLE]

We next consider the limiting behaviour of $t_{c}^{\prime}$ when $\tau\geq 4$ and $q\rightarrow\infty$. We first show that for any $\alpha>1$

[TABLE]

when $q$ is sufficiently large. Hence, by (15), we have for any $\alpha>1$ that $t_{c}^{\prime}<\alpha\,{\rm ln}(q-1)$ when $q$ is sufficiently large. So let $\alpha>1$. We have from (19), (21)

[TABLE]

since $\alpha>1$ and $D$ in (21) is positive. Hence, (94) holds when $\alpha>1$ and $q$ is sufficiently large.

Now let $\alpha\in(0,1)$. We shall show that

[TABLE]

when $q$ is sufficiently large. For this, we use the result of the third limit relation in (31), to be shown below, so that $t_{c}^{\prime\prime}>\alpha\,{\rm ln}(q-1)$ when $q$ is sufficiently large. Since $t_{c}^{\prime}>t_{c}^{\prime\prime}$, we thus get (96) when $\alpha\in(0,1)$ and $q$ is sufficiently large. From this and the above, it is concluded that $t_{c}^{\prime}/{\rm ln}(q-1)\rightarrow 1$ as $q\rightarrow\infty$ for any $\tau\geq 4$, and this is the second limit relation in (31).

We shall now consider ${\cal K}^{\prime}({\rm ln}(q-1))$, and we shall show that

[TABLE]

when $q$ is sufficiently large, and that

[TABLE]

for all $q>2$. We compute from (19) and (21) for $\tau\geq 4$ and $q>2$

[TABLE]

We have by dominated convergence

[TABLE]

Hence, when $\tau>4$, we have

[TABLE]

and so (97) holds when $q$ is sufficiently large.

When $\tau=4$, the limit in (101) equals 0, and we have to proceed more carefully. For $\tau=4$, we have from (6)

[TABLE]

With $t={\rm ln}(q-1)$, $q=\exp(t)+1$, we have

[TABLE]

and we shall show that this exceeds 1 for all $q>2$, so that ${\cal K}^{\prime}({\rm ln}(q-1))>0$ and thus $t_{c}^{\prime}<{\rm ln}(q-1)$ for all $q>2$ and $\tau=4$. The function $g(y):=(e^{t}+1)/(e^{ty}+1)$, $y\geq 0$, is for any $t>0$ positive, decreasing and convex in $y\geq 0$, and we have

[TABLE]

Hence,

[TABLE]

We then have

[TABLE]

We finally show the third limit relation in (31). Since $t_{c}^{\prime\prime}<{\rm ln}(q-1)$, it is sufficient to let $\tau\geq 4$ and $\alpha\in(0,1)$, and to show that ${\cal K}^{\prime\prime}(\alpha\,{\rm ln}(q-1))<0$ when $q$ is sufficiently large, see (16). With $t=\alpha\,{\rm ln}(q-1)$, $\exp(t)=(q-1)^{\alpha}$, the expression in the brackets (  ) at the right-hand side of (20) becomes

[TABLE]

As to the integral, we have

[TABLE]

Hence, by dominated convergence,

[TABLE]

Since $\alpha<1$, we have that $(q-1)^{\alpha-1}\rightarrow 0$ as $q\rightarrow\infty$, and so

[TABLE]

Furthermore, since $\alpha\in(0,1)$,

[TABLE]

Therefore, from (6) and (111) we have that the expression in (6) is negative when $q$ is sufficiently large, i.e., ${\cal K}^{\prime\prime}(\alpha\,{\rm ln}(q-1))<0$ when $q$ is sufficiently large.

7 Limiting behaviour of $t_{c}^{\prime\prime}$ as $q\downarrow 2$

The characterization of $t_{c}^{\prime\prime}$ as the unique positive zero $t=t_{\ast}$ of ${\cal F}_{0}^{\prime\prime}(t)$ gives a convenient handle to study the behaviour of $t_{c}^{\prime\prime}$ as $q\downarrow 2$; such a characterization is not available for $t_{c}$ or $t_{c}^{\prime}$. By [1], Sections 7.2 and 7.3, we have for $\tau\geq 4$ that $t_{\ast}$ is the unique positive zero $t$ of the function

[TABLE]

where

[TABLE]

Note that $a(x)$ has exponential decay as $x\rightarrow\infty$, and that for any $x\geq 0$

[TABLE]

where

[TABLE]

Hence, $\Phi(t)<0$ for $t\geq b$, and so $t_{c}^{\prime\prime}=t_{\ast}<b={\rm ln}(q-1)$ since $\Phi(t_{\ast})=0$.

We write for $t\in(0,b)$

[TABLE]

where $T_{1}$ and $T_{2}$ are positive. We consider this for $b\downarrow 0$, and we should find $t$ such that $T_{1}=T_{2}$. We shall solve the equation $T_{1}=T_{2}$ approximately by finding the leading-order behaviour of $T_{1}$ and $T_{2}$ as $0<t\leq b$, $b\downarrow 0$. As to $T_{1}$, we use for this

[TABLE]

and we therefore approximate $T_{1}$ by

[TABLE]

As to $T_{2}$, we have

[TABLE]

where we have used the substitutions $s=e^{x}\geq q-1$ and $v=s-(q-1)\geq 0$, together with (115).

We first consider $4\leq\tau<5$. Then we have

[TABLE]

where $C(\tau)$ is finite and positive. For $\tilde{T}_{1}$ we set $t=\alpha b$ with $\alpha\in(0,1]$, and we have by the substitution $y=x/b\in[\alpha,1]$

[TABLE]

When $\alpha$ would be bounded away from 0 as $b\downarrow 0$, we would have $\tilde{T}_{1}=O(b^{-\tau+5})\rightarrow 0$ as $b\downarrow 0$ while $\tilde{T}_{2}\rightarrow C(\tau)>0$ as $b\downarrow 0$. Therefore, we have $\alpha\rightarrow 0$ when we equate $\tilde{T}_{1}$ and $\tilde{T}_{2}$ as $b\downarrow 0$, and then we have

[TABLE]

In the case that $\tau=4$, we have

[TABLE]

and equating this to $\tilde{T}_{2}$, we get from (120) the limiting behaviour

[TABLE]

showing exponential decay of $\alpha$ and $t$ as $b\downarrow 0$.

In the case that $4<\tau<5$, we have

[TABLE]

and equating this to $\tilde{T}_{2}$, we get from (120) the limiting behaviour

[TABLE]

as $b\downarrow 0$.

Next, we consider the case $\tau=5$. Then we have from (122)

[TABLE]

From (119) and ${\rm ln}(1+v+b)=(v+b)(1+O(v+b))$, we have

[TABLE]

Equating $\tilde{T}_{1}$ and $\tilde{T}_{2}$, we get the limiting behaviour

[TABLE]

as $b\downarrow 0$.

We finally consider the case $\tau>5$. From (119) we have

[TABLE]

By partial integration

[TABLE]

In a similar fashion

[TABLE]

in which the integral on the right-hand side is $O(1)$ when $5<\tau<6$, $O({\rm ln}(\frac{1}{b}))$ when $\tau=6$ and $O(b^{-\tau+6})$ when $\tau>6$. Thus, in all cases where $\tau>5$, we have

[TABLE]

As to $\tilde{T}_{1}$, we now return to (121), and equating $\tilde{T}_{2}$ from (133) and $\tilde{T}_{1}$ from (121), we get for $\alpha\in(0,1)$ the equation

[TABLE]

Now we have

[TABLE]

Therefore, (134) can be solved in closed form for $\alpha$, and we get the limiting behaviour, as $b\downarrow 0$,

[TABLE]

As said, we do not have a characerization of $t_{c}$ or $t_{c}^{\prime}$ of the type that we have of $t_{c}^{\prime\prime}$ via (112), (113). Nevertheless, from a numerical investigation, solving ${\cal K}(t)=0$, ${\cal K}^{\prime}(t)=0$ using the representations (18), (19), a picture, similar to the one for $t_{c}^{\prime\prime}$ as $b\downarrow 0$, arises for $t_{c}$ and $t_{c}^{\prime}$ as $b\downarrow 0$.

8 The homogeneous case as limiting Pareto case with $\tau\rightarrow\infty$

We now consider the case that the Pareto pdf $f_{W}(w)$ in (17) is replaced by $\delta(w-1)$, also see [1], Appendix C. This is the homogeneous case that arises from the Pareto case when we take $\tau\rightarrow\infty$. The ${\cal F}_{0}$ of (1) and the ${\cal K}$ of (6) for the homogeneous case are given by

[TABLE]

and

[TABLE]

respectively. By straightforward differentiation of (138), we get

[TABLE]

and

[TABLE]

It is easy to show that ${\cal K}_{H}(t)$ vanishes at $t=2\,{\rm ln}(q-1)$. Furthermore, it is seen at once that ${\cal K}_{H}^{\prime}(t)$ vanishes at $t=T$, with $t=T$ the unique positive solution of the equation (27). Finally, we have that ${\cal K}_{H}^{\prime\prime}(t)$ vanishes at $t={\rm ln}(q-1)$. Therefore, the inequalities $t_{c}\leq 2\,{\rm ln}(q-1)$, $t_{c}^{\prime}\leq T$ and $t_{c}^{\prime\prime}\leq{\rm ln}(q-1)$, that hold for the Pareto case with finite $\tau$, are sharp in the sense that there is equality in the limit case $\tau\rightarrow\infty$.

Observe that from (18), (19) and (20) we have

[TABLE]

9 Analysis of $T=t_{H,c}^{\prime}$

Denoting by $t_{H,c}$, $t_{H,c}^{\prime}$ and $t_{H,c}^{\prime\prime}$ the unique positive zeros of ${\cal K}_{H}(t)$, ${\cal K}_{H}^{\prime}(t)$ and ${\cal K}_{H}^{\prime\prime}(t)$, respectively, with ${\cal K}_{H}$ the ${\cal K}$-function for the homogeneous case, we have seen in Section 8 that $t_{H,c}$ and $t_{H,c}^{\prime\prime}$ have an elementary closed form. Such a thing does not hold for $t_{H,c}^{\prime}$ since the unique positive solution $T=t_{H,c}^{\prime}$ of the equation (27) does not have a closed form. We shall, therefore, present a further analysis of $T$.

We shall show that $T(q)$ is an increasing function of $q\geq 2$, with $T(2)=0$ and $T(\infty)=\infty$, and we shall show that $T(q)/{\rm ln}(q-1)$ decreases from $3/2$ at $q=2$ to 1 at $q=\infty$. Furthermore, we shall show that

[TABLE]

where we observe that

[TABLE]

This agrees with the upper bound $T(q)<\frac{3}{2}\,{\rm ln}(q-1)$ in (78). In addition, we shall show that

[TABLE]

and this agrees asymptotically with the upper bound $T(q)\geq{\rm ln}(q-1)$ in (78).

The equation (27) for $t=T=T(q)$ can be written as

[TABLE]

and this can be worked out further to

[TABLE]

The quantity at the left-hand side of (146) decreases in $q\geq 2$ for all $t>0$, it vanishes uniquely at $t=T(q)$, and is negative for $0<t<T(q)$ and positive for $t>T(q)$. It follows that $T(q)$ is an increasing function of $q\geq 2$.

The proof that $T(q)/{\rm ln}(q-1)$ decreases in $q\geq 2$ is considerably more complicated. With $\alpha\in(1,3/2)$, we consider

[TABLE]

where we have set

[TABLE]

so that $x^{\alpha}=e^{t}$ and $t=\alpha\,{\rm ln}\,x$. Our goal is to show that $V(x,\alpha)$ has a unique zero $x(\alpha)>1$ and that $x(\alpha)$ decreases strictly in $\alpha\in(1,3/2)$, Once this goal is reached, we have from $V(x(\alpha),\alpha)=0$ and (148) that

[TABLE]

Denoting by $\alpha(q)$, $q>2$, the inverse function of $q(\alpha)$, $\alpha\in(1,3/2)$, we then have that

[TABLE]

is a decreasing function of $q>2$.

To show existence and uniqueness of a zero $x>1$ of $V(x,\alpha)$ when $\alpha\in(1,3/2)$, we write

[TABLE]

where we have set

[TABLE]

With an argument quite similar to the proof of (76), we have

[TABLE]

so that the right-hand side of (151) is zero-free in $y>1$ when $\beta\geq 1$ or $\beta\leq 2/3$. Hence, we concentrate on the case $\beta\in(2/3,1)$.

Lemma 9.1**.**

We have for $y-1$ small positive

[TABLE]

and for $y-1$ large positive

[TABLE]

Proof.  The left-hand side of (155) tends to $\infty$ as $y\rightarrow\infty$ since $\beta<1$, and is therefore positive when $y$ is large.

As to (154), we write $y=1+t$ with $t$ small positive. Then

[TABLE]

We have

[TABLE]

At the same time, we have

[TABLE]

Hence,

[TABLE]

for small positive $t$ since $\beta>2/3$. This completes the proof.

We conclude from Lemma 9.1 that the right-hand side of (151) has at least one zero $y>1$. As to uniqueness of this zero, we write

[TABLE]

where

[TABLE]

Lemma 9.2**.**

We have $p(1)=0$, $p(y)<0$ for small positive $y-1$, and $p(y)>0$ for large positive $y-1$. Furthermore, $p(y)$, $y>1$, has a unique inflection point, that is, there is a unique $\hat{y}>1$ such that $p(y)$ is concave in $y\in(1,\hat{y})$ and convex in $y\in(\hat{y},\infty)$.

Proof.  The first statement follows from Lemma 9.1. As to the second statement, we compute

[TABLE]

where

[TABLE]

We have

[TABLE]

Furthermore,

[TABLE]

since $\beta>2/3$. Finally,

[TABLE]

since $\beta<1$. Therefore, $r(y)$ is strictly convex in $y\geq 1$. From (163) we see that $r(y)>0$ for large $y>1$ since $\beta<1$, and from (164) and (165) we see that $r(y)<0$ for small positive $y-1$. Hence, $r(y)$ has at least one zero $>\,1$, and this zero is unique by strict convexity of $r(y)$. Denote this zero by $\hat{y}$.

We conclude from (162) that $p(y)$ is strictly concave in $y\in(1,\hat{y})$ and strictly convex in $y\in(\hat{y},\infty)$. This completes the proof.

We conclude from Lemma 9.2 that $p(y)$ has precisely one zero in $(\hat{y},\infty)$. According to (151) and (9), we see that $V(x,\alpha)$ has precisely one zero $x(\alpha)$ in $(1,\infty)$ for any $\alpha\in(1,3/2)$.

We next show that $x(\alpha)$ decreases strictly in $\alpha\in(1,3/2)$. It follows from Lemma 9.1 that $V(x,\alpha)<0$ for small positive $x-1$ and that $V(x,\alpha)>0$ for large positive $x-1$. Hence we have $V(x,\alpha)<0$ for $x\in(1,x(\alpha))$ and $V(x,\alpha)>0$ for $x\in(x(\alpha),\infty)$, while $V(x(\alpha),\alpha)=0$.

Lemma 9.3**.**

We have for $\alpha\in(1,3/2)$ that

[TABLE]

Proof.  We have from (147)

[TABLE]

Now assume that $x\geq x(\alpha)$ so that $V(x,\alpha)\geq 0$, i.e.,

[TABLE]

Therefore,

[TABLE]

Hence $\tfrac{\partial V}{\partial\alpha}\,(x,\alpha)>0$ when

[TABLE]

i.e., when

[TABLE]

Hence, with $y=x^{\alpha}>1$, we should show that $(y+1)\,{\rm ln}\,y/(y-1)>2$. This indeed holds, see below (153), and this completes the proof.

Let $\alpha\in(1,3/2)$. We have from (167) that $V(x(\alpha),\alpha_{1})>0$ when $\alpha_{1}-\alpha$ is small positive. Therefore $x(\alpha)>x(\alpha_{1})$ when $\alpha_{1}-\alpha$ is small positive. That is, $x(\alpha)$ decreases in $\alpha\in(1,3/2)$.

We next show the result (142) on the behaviour of $T(q)$ when $q-2\downarrow 0$. To this end, we use the functions $a_{n}(q)$, introduced in [2], Section 5, to expand

[TABLE]

We have, by definition [2], (75),

[TABLE]

Since

[TABLE]

we find (using $a_{0}(q)=1/q$)

[TABLE]

We proceed as in [2], Section 7, where the unique zero $t_{c}(q)$ of the ${\cal K}$-function corresponding to the exponential pdf $\exp({-}w)\,{\cal X}_{[0,\infty)}(w)$ has been approximated as $q-2\downarrow 0$ using the functions $a_{n}(q)$. In the present case, the details are much simpler since no awkward truncation analysis is needed, the expansion in (176) being valid for $|t|<\pi$. We get for $t=T(q)$ from ${\cal K}_{H}^{\prime}(t)=0$ the equaton

[TABLE]

so that

[TABLE]

We have by [2], (94), (95) and (166)

[TABLE]

Furthermore, from [2], (79), we have $a_{2}(2)=a_{4}(2)=...=0$. Using [2], (90),

[TABLE]

we then get

[TABLE]

Finally, from [2], (95) and (97),

[TABLE]

Then combining (178), (179), (181) and (182) and using that $t=T(q)=O({\rm ln}(q-1))=O(v)$, we get

[TABLE]

Using that $t=O(v)$, we first get $t=\frac{3}{2}\,v+O(v^{2})$ from (183), and inserting this into the second line of (183), we get

[TABLE]

and this is (142).

We finally show the result (144) on the asymptotics of $T(q)$ as $q\rightarrow\infty$. We start by rewriting the equation (145) for $t=T(q)$ as

[TABLE]

This is achieved by working out $(e^{t}+q-1)^{2}$ as $e^{2t}+2(q-1)\,e^{t}+(q-1)^{2}$, collecting terms and division by $q$. Thus we get the equation

[TABLE]

Divide by $s$, set $B=q/e$, so that

[TABLE]

and we get the equation

[TABLE]

The left-hand side of (188) is increasing in $s>1/2$ for any $q>2$.

We aim in (144) at a result for $t=T(q)={\rm ln}(q\,s)$ in which a term ${\rm ln}({\rm ln}\,B)$ appears. We want here ${\rm ln}({\rm ln}\,B)$ to be well-defined and positive, and thus we require

[TABLE]

The left-hand side of (188) increases in $s>1/2$ and is less than 1 at $s=1$. Since ${\rm ln}\,B>1$, we therefore restrict to $s>1$. Then ${\rm ln}\,s>0$ and so we have $s>{\rm ln}\,B$ with ${\rm ln}\,B\rightarrow\infty$ as $q\rightarrow\infty$.

We continue by considering, instead of (188), the equation

[TABLE]

Hence, we have deleted from (188) the term

[TABLE]

The function $s\geq 1\mapsto s-{\rm ln}\,s$ is strictly increasing, and both equations (188) and (190) have unique solutions when ${\rm ln}\,B>1$. The solution of (188) exceeds the solution of (190) in which ${\rm ln}\,B$ is replaced by ${\rm ln}\,B+\tfrac{2}{q}$ and is less than the solution of (190) in which ${\rm ln}\,B$ is replaced by ${\rm ln}\,B+\tfrac{3}{q}$, see (191). These two solutions of (190) have absolute difference of the order $1/q$, since $(s-{\rm ln}\,s)^{\prime}=1-1/s$, which is therefore exponentially small in ${\rm ln}\,B$.

With ${\rm ln}\,B>1$, the solution $s>1$ of (190) can be shown to satisfy

[TABLE]

The first inequality in (192) is obvious and for the second inequality we use convexity of the function $s\geq 1\mapsto s-{\rm ln}\,s$, so that

[TABLE]

We shall now approximate the solution $s$ of (190) by iterating (190) a few times. Thus, we have

[TABLE]

Now from (192),

[TABLE]

Hence,

[TABLE]

where the implicit constant in $O$ is of the order unity.

Recalling that $T={\rm ln}(qs)$, with exponentially small error in ${\rm ln}\,B$, we have

[TABLE]

and this is (144).

Appendix A. Proof of (76)

We show that for $q>2$

[TABLE]

where

[TABLE]

As to the first inequality in (A1), we let $x=q-1>1$, and we compute

[TABLE]

It is easy to show that $(x+1)\,{\rm ln}\,x-2x+2>0$ when $x>1$.

As to the second inequality in (A1), we let $y=(q-1)^{3/2}$, $q=1+y^{2/3}$, with $q>2$, $y>1$. We compute

[TABLE]

Setting $y=v^{3}$, $v=y^{1/3}>1$, we shall show that

[TABLE]

Dividing through by $v^{2}$, we have that (A5) holds when

[TABLE]

We have $\xi(v)={-}\varphi(v)$, see (56), and so (A6) follows.