# Validation of the Replica Trick for Simple Models

Takashi Shinzato<EMAIL_ADDRESS>Mori Arinori Center for
Higher Education and Global Mobility, Hitotsubashi University, Tokyo, 1868601,
Japan.

###### Abstract

We discuss replica analytic continuation using several simple models in order
to prove mathematically the validity of replica analysis, which is used in a
wide range of fields related to large scale complex systems. While replica
analysis consists of two analytical techniques, the replica trick (or replica
analytic continuation) and the thermodynamical limit (and/or order parameter
expansion), we focus our study on replica analytic continuation, which is the
mathematical basis of the replica trick. We apply replica analysis to solve a
variety of analytical models, and examine the properties of replica analytic
continuation. Based on the positive results for these models we propose that
replica analytic continuation is a robust procedure in replica analysis.

PACS number(s)

89.90.+n, 75.50.Lk

###### pacs:

89.90.+n

###### pacs:

75.50.Lk

††preprint: APS/123-QED

##  I Introduction

Replica analysis is a widely used analytic approach, developed for the
analysis of large scale complex systems in cross-disciplinary fields.
Applications of replica analysis include the analysis of the physical
properties of disordered magnetic materials, evaluation of the learning
performance of neural networks, estimation of the transmission rate of channel
coding and optimization of the investment risk of diversification investments
Mezard1987 ; Nishimori ; Mezard ; EA ; SK ; KS ; Amit1 ; Amit2 ; Gardner ;
Kabashima2004 ; Tanaka2002 ; Ciliberti ; Shinzato-SA2015 . Replica analysis
has a long history and has been discovered (and rediscovered) in several
research fields Kac ; Lin ; Edwards1970 ; Edwards1971 ; Emery ; Jasnow ;
Grinstein . In one of the pioneering works on replica analysis, Edwards et al.
investigated the singular properties of disordered magnetic materials EA ,
generally called spin glass. They first derived and used replica analysis to
examine the physical properties of spin glass. Based on the observation that
2-body interactions between localized spins can be modeled by the
Ruderma–Kittel–Kasuya–Yoshida interaction, they assumed that 2-body
interactions between localized spins can be randomly assigned and can be used
to analyze the singular physical properties of spin glass. Subsequently,
Sherrington et al. developed this spin glass model into a fully connected spin
glass model by applying the mean field approximation, since it is well known
that solutions based on the mean field approximation are the most rigorous SK
; KS .

Recently, following its success in the analysis of the physical properties of
spin glass, replica analysis has been applied to the mathematical structures
and mathematical analogies of other large scale complex systems. For instance,
for the analysis of the learning performance of neural networks, Amit et al.
applied replica analysis to the associative storage model and examined the
mechanism for recalling stored memories, and evaluated the learning capacity
of the associative storage model Amit1 ; Amit2 . Further, Gardner investigated
the learning performance of perceptron, one of the most simplified neural
network models, using replica analysis, and found that it is consistent with
the findings based on combinatorics in the previous work of Cover et al.
Gardner ; Cover ; Venkatesh . As part of an evaluation of the transmission
rate of communication technology, Kabashima et al. used a sparse connected
spin glass model to resolve the decoding problem of the binary linear Low
Density Parity Check code (LDPC) and guarantee that the communication
performance of its linear code is close to the Shannon limit using replica
analysis Kabashima2004 . In addition, Tanaka used the posterior probability as
a type of perceptron learning to resolve the maximizer of the posterior
marginal estimation problem for the third generation of radio communication
technology Code Division Multiple Access (CDMA) and analyzed the bit error
rate and channel capacity of CDMA using replica analysis Tanaka2002 . To
assess the risk of diversification investments, Ciliberti et al. dissected the
minimal investment risks of the absolute deviation model and expected
shortfall model in the zero temperature limit using replica analysis Ciliberti
. In a later work, Shinzato proved that the minimal investment risk and the
investment concentration of the mean variance model are satisfied by the
property of self-averaging using replica analysis and the Chernoff inequality
Shinzato-SA2015 .

Replica analysis consists of two main steps, the replica trick operation and
the application of the thermodynamical limit. While several previous works
have verified the effectiveness of replica analysis, in our own use of the
replica trick, for convenience of analysis, we use a calculation technique
where a number which is assumed to be an integer is replaced by a real or
complex number in order to implement the limit operation, and for this reason
replica analysis is not always guaranteed mathematically. With this in mind,
we compared the results estimated by replica analysis and other approaches
such as the Markov chain Monte Carlo method and steepest descent method, and
found that the effectiveness of replica analysis was partly validated
Shinzato-Kabashima . However, in several previous works, problems have arisen
regarding negative entropy and the convexity of the order parameter expansion
of free energy. To address these problems, the results of replica analysis
were refined by the 1st step (or full step) replica symmetry breaking solution
or Almeida–Thouless analysis, an approach that can provide a partial fix, but
does not necessarily verify the universal reliability of replica analysis
Mezard1987 ; Nishimori ; AT ; Parisi1 ; Parisi2 . With respect to these
efforts to refine replica analysis, Ogure et. al. analytically evaluated the
expectation of the complex power of the partition function of the grand
canonical discrete random energy model using complex analysis, and then
compared three solutions obtained from replica analysis, the paramagnetic
solution, ferromagnetic solution, and first replica symmetry breaking solution
(spin glass solution) Ogure1 ; Ogure2 . Further, they derived the cases where
the three solutions, composed of the $n$th moments of the partition function,
are not analytic at $n=n_{c}\in(0,1)$ ($n$ is called the replica number), and
verified their results using numerical simulations. In addition, Tanaka
rearranged the problem implied in replica analysis using the framework of the
moment problem, examining several counterexamples of the moment problem and
summarizing the sufficient condition that the distribution of the partition
function is uniquely determined from the moment sequences of the partition
function Tanaka2007 . Previous works have shown that we need to analyze the
Helmholtz free energy in a large scale complex system in order to implement a
quenched analysis of the system. Below we discuss the mathematical basis of
the expectation of the logarithm of the partition function for the analysis of
Helmholtz free energy. One of the disputed issues in replica analysis is the
assumption that for an integer replica number there is an analytical
continuation to the real or complex number case, as this is not always valid
for replica numbers near $0$. In order to validate the reliability of replica
analysis mathematically, we need to examine the mathematical structure of
replica analysis through the models that can be applied to the two distinct
approaches for analyzing the $n$th moment of the partition function in a
similar way to the argument discussed by Ogure et al.

In the present work, as the first step for resolving the validity of replica
analysis and rigorously interpreting the findings obtained from replica
analysis, we separate the two distinct analytical approaches, (1) the replica
trick and (2) the thermodynamical limit (and/or the order parameter
expansion), and focus our study on the first of the techniques, the replica
trick, or equivalently replica analytic continuation. Namely, our goal in this
paper is to discuss the properties of replica analysis by examining replica
analytic continuation.

This paper is organized as follows. In the following section, we explain the
analytical techniques of replica analysis, the replica trick and replica
analytic continuation, and consider the importance of the order of integration
of two distinct variables. In Section III we test replica analytic
continuation using several simple models and propose a conjecture regarding
replica analytic continuation. The final section is devoted to a summary and
plans for future work.

##  II From Replica Trick to Replica Analytic Continuation

###  II.1 Why Do We Need Replica Analysis?

Based on several studies of the disordered magnetic material model, which is
represented by the Edwards–Anderson model and the Sherrington–Kirkpatrik model
EA ; SK ; the neural network model, which is represented by the Hopfield model
and perceptron model Amit1 ; Amit2 ; Gardner ; the channel coding model, which
is represented by LDPC and CDMA Kabashima2004 ; Tanaka2002 ; and the portfolio
optimization problem, which is represented by the mean-variance model and
absolute deviation model Ciliberti ; Shinzato-SA2015 , it is well known that
we can examine some indicators which characterize a system, for instance, the
average magnetization, learning rate, and encoding rate, by evaluating the
Helmholtz free energy of the system $F=-\frac{1}{\beta}\log Z$ or its
expectation $E[F]=-\frac{1}{\beta}E[\log Z]$, where $\beta$ is the inverse
temperature and $Z$ is the partition function. In several analyses of quenched
disordered systems, replica analysis has been used constructively to assess
the expectation of the Helmholtz free energy. However, the validity of several
methods used in replica analysis have not been guaranteed theoretically. For
this, by comparing the results obtained by replica analysis and other
approaches, such as the Markov chain Monte Carlo method and the steepest
descent method, we can partly verify the validity of replica analysis. In this
work, as a first step for confirming the effectiveness of replica analysis, we
examine the replica trick, one of the two analytical approaches which
constitute replica analysis.

###  II.2 The Replica Trick

In general, the replica trick consists of the following identity equation for
any positive random variable $Z$,

$\displaystyle\log Z$ $\displaystyle=$ $\displaystyle\lim_{n\to
0}\frac{Z^{n}-1}{n}.$ (1)

If the random variable $Z$ is positive and its probability density function
$P(Z)$ is known, using the replica trick and the expectation of the $n$th
power of $Z$, $E[Z^{n}]=\int_{0}^{\infty}dZP(Z)Z^{n}$, the expectation of the
logarithm of $Z$, $E[\log Z]=\int_{0}^{\infty}dZP(Z)\log Z$, is estimated as
follows:

$\displaystyle E[\log Z]$ $\displaystyle=$ $\displaystyle\lim_{n\to
0}\left\\{\begin{array}[]{l}\frac{E[Z^{n}]-1}{n}\\\ \frac{1}{n}\log
E[Z^{n}]\\\ \frac{\partial}{\partial n}\log E[Z^{n}]\end{array}\right.,$ (5)

where the three terms on the right hand side are consistent with each other,
which can be easily confirmed by using L’H$\hat{\rm o}$pital’s rule. $n$ of
$E[Z^{n}]$ is the replica number. However, a necessary condition for Eq. (5)
is that $n$ must be a real or complex number. In other words the result
obtained for $n\in{\bf Z}$ is regarded as $n\in{\bf R}$ and/or $n\in{\bf C}$.
However, the validity of this assumption with respect to analytic continuation
of the replica number is not always guaranteed.

It should be noted that in general if we can accurately assess $E[Z^{n}]$
without approximation when the replica number is real or complex, then we can
assess $E[\log Z]$ using Eq. (5), and therefore we are not limited to
analyzing quenched disordered systems. In this case, there is no question of
the validity of the replica trick of Eq. (5). We consider that the assumption
that the result for $E[Z^{n}]$ in $n\in{\bf Z}$ can be solved using the
thermodynamical limit and expanding the Helmholtz free energy on order
parameters and its approximation can be regarded as the result for $E[Z^{n}]$
in $n\in{\bf R}$ or $n\in{\bf C}$ is the basis of the ambiguity in replica
analysis.

###  II.3 Two Types of Integral Assessments

In the analysis of quenched disordered systems, which are represented by the
Sherrington–Kirkpatrick model in spin glass theory and the Hopfield model in
associative memory, it is well-known that by analyzing the expectation of the
Helmholtz free energy $F=-\frac{1}{\beta}\log Z$, $E[F]=-\frac{1}{\beta}E[\log
Z]$, we can assess the typical behavior of the system. To average the free
energy, we use the replica trick of Eq. (5) and analytically evaluate the
$n$th power function of the partition function, $E[Z^{n}]$, in the right hand
side of Eq. (5). This requires the evaluation of two distinct integrals of the
internal variable of the partition function, $S$, and the external variable of
the partition function, $J$. Here, we consider the relationship between the
order of the two integrations of $E[Z^{n}]$ and $n$.

We first consider the procedure first-S-next-J (hereafter abbreviated FSNJ)
for which when evaluating $E[Z^{n}]$, the integral or sum of the
thermodynamical variable or internal variable of the partition function $S$ is
first implemented, followed by the integral or sum (or the expectation) over
the configuration variable or external variable of the partition function $J$.
Namely, for FSNJ the expectation of $Z^{n}$, $E[Z^{n}]$, is derived as
follows:

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\sum_{J}P(J)\left(\sum_{S}e^{-\beta{\cal H}(S,J)}\right)^{n},$
(6)

where ${\cal H}(S,J)$ is the Hamiltonian of the system,
$Z=\sum_{S}e^{-\beta{\cal H}(S,J)}$ is the partition function of the canonical
ensemble of the inverse temperature $\beta$, $\sum_{J}$ and $\sum_{S}$ are the
integrals or sums over the whole configuration of $J$ and $S$, respectively,
and $P(J)$ describes the probability function of the external variable $J$.
Since the integral or sum with respect to the internal variable $S$ is
implemented first, we allow that the replica number $n$ in Eq. (6) is regarded
as a real or complex number.

In the alternate procedure first-J-next-S (FJNS), the integral or sum of the
external variable of the partition function is first implemented, followed by
the integral or sum of the internal variable. That is,

$\displaystyle
E[Z^{n}]=\sum_{S_{1}}\sum_{S_{2}}\cdots\sum_{S_{n}}\left(\sum_{J}P(J)e^{-\beta\sum_{a=1}^{n}{\cal
H}(S_{a},J)}\right).\qquad$ (7)

We first evaluate Eq. (7) with respect to the external variable $J$ and next
evaluate it with respect to the internal variables $S_{1},\cdots,S_{n}$. The
intuitive advantage of FJNS is that for any $A,B>0$, the factor expansion
$(A+B)^{2.1}$ can be described by the infinite series expansion and
$(A+B)^{2}<(A+B)^{2.1}<(A+B)^{3}$ holds, that is, since $(A+B)^{n}$ is
described as a finite series expansion when the power number is an integer, we
can evaluate each term of $(A+B)^{2}=A^{2}+2AB+B^{2}$ and
$(A+B)^{3}=A^{3}+3A^{2}B+3AB^{2}+B^{3}$ and then approximately assess
$(A+B)^{2.1}$ using a sort of false position method from the results of
$(A+B)^{2}$ and $(A+B)^{3}$. A peculiarity of this method is that if the
accuracy of the approximation is reduced, negative entropy and/or convexity of
the order parameter expansion of the Helmholtz free energy may occur, casting
doubt on the veracity of the replica analysis.

$n=0$$n=2$$n=4$$n=1$$n=3$$n=5$$n\in{\bf
C}$$F\left(3.5+5\sqrt{-1}\right)=E\left[Z^{3.5+5\sqrt{-1}}\right]$$G(4)=E[Z^{4}]$$n\in{\bf
Z}$ Figure 1:  Relationship between $F(n)=E[Z^{n}],(n\in{\bf C})$ for FSNJ and
$G(n)=E[Z^{n}],(n\in{\bf Z})$ for FJNS on the complex plane.

###  II.4 Formulation of Replica Analytic Continuation

We set the expectation of the $n$th power of the partition function derived by
FSNJ as $F(n)=E[Z^{n}],(n\in{\bf C})$ and for FJNS as $G(n)=E[Z^{n}],(n\in{\bf
Z})$, shown in Fig. 1. From Eq. (7), the assessment of $G(n)$ when $n\in{\bf
Z}$ requires changing the order of the two integrations.

If we regard $G(n)$ as an analytical function with respect to $n$ and assume
that $G(n)$ can also be described as an analytical function of $n$ when
$n\not\in{\bf Z}$, then $F(n)=G(n)$ always holds in the case $n\in{\bf Z}$. We
can then propose that this system demonstrates replica analytic continuation.
By contrast, for $F(n)=G(n),(n\in{\bf Z})$ and $F(n)\neq G(n),(n\not\in{\bf
Z})$, the system does not satisfy replica analytic continuation. For instance,
when $n\in{\bf C}$, $F(n)=E[Z^{n}]=An^{2}+Bn+C+D\sin(2\pi n)$ (where $D\neq
0$) or $n\in{\bf Z}$, $G(n)=E[Z^{n}]=An^{2}+Bn+C$, since $F(n)\neq
G(n),(n\not\in{\bf Z})$, replica analytic continuation does not apply.

If replica analytic continuation holds for the system under consideration,
from the replica trick of Eq. (5), the expectation of the logarithm of the
partition function, $E[\log Z]$, can be accurately analyzed. We examine
replica analytical continuation by applying it to physical or statistical
models which can be evaluated for both the FSNJ and FJNS approaches, so that
we can compare the results of the two approaches. We have selected models for
which the thermodynamical limit need not be applied, so that we can test the
accuracy of replica analytical continuation alone, as part of our goal to
focus on the replica trick. Note that several previous works have used
$G(n)=E[Z^{n}],(n\in{\bf Z})$, for which it is important to estimate $E[\log
Z]$ accurately, and hence we need to consider the accuracy of $n$ in the
neighborhood of $0$. However, we do not try to evaluate the expectation of the
logarithm of the partition function consistently, since we are interested
mainly in the issue of replica analytic continuation, which is not limited to
the neighborhood $n=0$.

##  III Simple Examples

###  III.1 Classical Harmonic Oscillator Model with Rayleigh Distribution

First we discuss whether replica analytic continuation holds by applying the
well known analytical model of the classical harmonic oscillator model. The
Hamiltonian of the classical harmonic oscillator model, ${\cal H}(p,q)$, is
defined as

$\displaystyle{\cal H}(p,q)$ $\displaystyle=$
$\displaystyle\frac{p^{2}}{2m}+\frac{m\omega^{2}}{2}q^{2},$ (8)

where $p$ and $q$ are the momentum and the position of the harmonic
oscillator, respectively, $m$ is the mass of the particle and $\omega$ is the
angular frequency of the harmonic oscillator. The partition function $Z$ is
solved as follows:

$\displaystyle Z$ $\displaystyle=$
$\displaystyle\frac{1}{h}\int_{-\infty}^{\infty}dpdqe^{-\beta{\cal H}(p,q)}$
(9) $\displaystyle=$ $\displaystyle\frac{1}{\hbar\beta\omega},$

using the Dirac constant $\hbar=\frac{h}{2\pi}$. Since the partition function
of this model is analytically and easily evaluated, it is suitable for
discussing the mathematical structure of replica analytic continuation.

We assume that the angular frequency of the harmonic oscillator which models
the vibration phenomena $\omega$ has a Rayleigh distribution:

$\displaystyle P(\omega)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\omega
e^{-\frac{\omega^{2}}{2}}&\omega>0\\\ 0&\text{otherwise}\end{array}\right..$
(12)

From this, the FSNJ case with $n\in{\bf C}$ can be analyzed directly. The
$n$th moment of $Z$, $F(n)=E[Z^{n}]$, is derived as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}d\omega
P(\omega)\left(\frac{1}{\hbar\beta\omega}\right)^{n}$ (13) $\displaystyle=$
$\displaystyle\frac{2^{-\frac{n}{2}}}{(\hbar\beta)^{n}}\Gamma\left(1-\frac{n}{2}\right).$

Note that hereafter the Gamma function
$\Gamma(s)=\int_{0}^{\infty}dtt^{s-1}e^{-t},(s\in{\bf C})$ is often used in
our discussion Artin ; Andrews ; Levedev . It is a meromorphic function and
has a pole at the origin and at negative integer numbers, so if
$F(n)=E[Z^{n}]$ is described using the Gamma function, then we will omit
$n\in{\bf C}$ which is related to a pole of Gamma in the discussion on replica
analytic continuation.

For the FJNS case, since we can permutate the integral order, the $n$th moment
of $Z$, $G(n)=E[Z^{n}]$, is calculated as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\frac{1}{h^{n}}\int_{0}^{\infty}d\omega
P(\omega)\int_{-\infty}^{\infty}\prod_{a=1}^{n}dp_{a}dq_{a}$ (14)
$\displaystyle\exp\left(-\frac{\beta}{2m}\sum_{a=1}^{n}p_{a}^{2}-\frac{\beta
m\omega^{2}}{2}\sum_{a=1}^{n}q_{a}^{2}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{h^{n}}\left(\frac{2\pi
m}{\beta}\right)^{\frac{n}{2}}\int_{-\infty}^{\infty}\prod_{a=1}^{n}dq_{a}\frac{1}{1+\beta
m\sum_{a=1}^{n}q_{a}^{2}}$ $\displaystyle=$
$\displaystyle\left(\frac{2\pi}{h^{2}\beta^{2}}\right)^{\frac{n}{2}}\int_{-\infty}^{\infty}\prod_{a=1}^{n}dr_{a}\frac{1}{1+\sum_{a=1}^{n}r_{a}^{2}}$
$\displaystyle=$
$\displaystyle\left(\frac{2\pi}{h^{2}\beta^{2}}\right)^{\frac{n}{2}}\frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}\int_{0}^{\infty}dR\frac{R^{n-1}}{1+R^{2}}.$

In addition, we substitute the identity

$\displaystyle 2\int_{0}^{\infty}dR\frac{R^{n-1}}{1+R^{2}}$ $\displaystyle=$
$\displaystyle 2\int_{0}^{\frac{\pi}{2}}d\theta\tan^{n-1}\theta$ (15)
$\displaystyle=$ $\displaystyle B\left(\frac{n}{2},1-\frac{n}{2}\right)$
$\displaystyle=$
$\displaystyle\frac{\Gamma\left(\frac{n}{2}\right)\Gamma\left(1-\frac{n}{2}\right)}{\Gamma\left(\frac{n}{2}+1-\frac{n}{2}\right)},$

into Eq. (14), giving

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\frac{2^{-\frac{n}{2}}}{\hbar^{n}\beta^{n}}\Gamma\left(1-\frac{n}{2}\right),$
(16)

where $r_{a}=q_{a}\sqrt{\beta m}$, $R^{2}=\sum_{a=1}^{n}r_{a}^{2}$ and
$B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$ are used. Furthermore, with
respect to the integral domain ${\cal
D}=\left\\{(r_{1},\cdots,r_{n})|\sum_{a=1}^{n}r_{a}^{2}<A^{2}\right\\}$, from
the identical integral

$\displaystyle\int_{\cal
D}\prod_{a=1}^{n}dr_{a}f\left(\sum_{a=1}^{n}r_{a}^{2}\right)=\frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}\int_{0}^{A}dRR^{n-1}f(R^{2}),\qquad$
(17)

the $n$-dimensional integral of $r_{1},\cdots,r_{n}$ can be rewritten as an
$n$-dimensional hyperspherical volume integral. From this, we find that
$F(n)=E[Z^{n}],(n\in{\bf C})$ in Eq. (13) derived by FSNJ is consistent with
$G(n)=E[Z^{n}],(n\in{\bf Z})$ in Eq. (16) by FJNS. Namely, it is verified that
replica analytic continuation holds for the classical harmonic oscillator.

###  III.2  Classical Harmonic Oscillator Model with Positive Gaussian
Distribution

Similar to our discussion for the classical harmonic oscillator, we now
discuss the case where the angular frequency $\omega$ is drawn from a positive
normal distribution, $P_{\sigma}(\omega)$, defined as

$\displaystyle P_{\sigma}(\omega)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\frac{2}{\sqrt{2\pi\sigma^{2}}}e^{-\frac{\omega^{2}}{2\sigma^{2}}}&\omega>0\\\
0&\text{otherwise}\end{array}\right..$ (20)

For FSNJ, when $n\in{\bf C}$, the expectation of the $n$th power of $Z$,
$F(n)=E[Z^{n}]$, can be assessed directly:

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}d\omega
P_{\sigma}(\omega)\left(\frac{1}{\hbar\beta\omega}\right)^{n}$ (21)
$\displaystyle=$
$\displaystyle\frac{2^{-\frac{n}{2}}}{(\hbar\beta\sigma)^{n}\sqrt{\pi}}\Gamma\left(\frac{1-n}{2}\right).$

For FJNS, when $n\in{\bf Z}$, since we can permutate the integral order,
$G(n)=E[Z^{n}]$ is assessed as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\frac{1}{h^{n}}\left(\frac{2\pi
m}{\beta}\right)^{\frac{n}{2}}\int_{-\infty}^{\infty}\prod_{a=1}^{n}dq_{a}$
(22)
$\displaystyle\int_{0}^{\infty}d\omega\frac{2}{\sqrt{2\pi\sigma^{2}}}e^{-\frac{\omega^{2}}{2}\left(\frac{1}{\sigma^{2}}+\beta
m\sum_{a=1}^{n}q_{a}^{2}\right)}$ $\displaystyle=$
$\displaystyle\frac{(2\pi)^{\frac{n}{2}}}{(h\beta\sigma)^{n}}\int_{-\infty}^{\infty}\prod_{a=1}^{n}dr_{a}\frac{1}{\sqrt{1+\sum_{a=1}^{n}r_{a}^{2}}}$
$\displaystyle=$
$\displaystyle\frac{(2\pi)^{\frac{n}{2}}}{(h\beta\sigma)^{n}}\frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}\int_{0}^{\infty}dR\frac{R^{n-1}}{\sqrt{1+R^{2}}}$
$\displaystyle=$
$\displaystyle\frac{(2\pi)^{\frac{n}{2}}}{(h\beta\sigma)^{n}}\frac{\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}B\left(\frac{n}{2},\frac{1-n}{2}\right)$
$\displaystyle=$
$\displaystyle\frac{2^{-\frac{n}{2}}}{(\hbar\beta\sigma)^{n}\sqrt{\pi}}\Gamma\left(\frac{1-n}{2}\right).$

Since $F(n)=E[Z^{n}],(n\in{\bf C})$ in Eq. (21) is derived by FSNJ and
$G(n)=E[Z^{n}],(n\in{\bf Z})$ in Eq. (22) by FJNS are also consistent, the
replica analytic continuation is validated in this model where in Eq. (22)
$r_{a}=q_{a}\sqrt{\beta m}$, $R^{2}=\sum_{a=1}^{n}r_{a}^{2}$,
$B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$ and Eq. (17) are used.

###  III.3 Classical Harmonic Oscillator Model with Chi Distribution

We consider also the case where $\omega$ is distributed with a $\chi$
distribution, $P_{\chi}(\omega)$, with degrees of freedom $s$, defined as

$\displaystyle P_{\chi}(\omega)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\frac{\omega^{s-1}}{2^{\frac{s-2}{2}}\Gamma\left(\frac{s}{2}\right)}e^{-\frac{\omega^{2}}{2}}&\omega>0\\\
0&\text{otherwise}\end{array}\right..$ (25)

For FSNJ, when $n\in{\bf C}$, $F(n)=E[Z^{n}]$, is evaluated as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}d\omega
P_{\chi}(\omega)\left(\frac{1}{\hbar\beta\omega}\right)^{n}$ (26)
$\displaystyle=$
$\displaystyle\frac{2^{-\frac{n}{2}}}{(\hbar\beta)^{n}}\frac{\Gamma\left(\frac{s-n}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}.$

For FJNS, when $n\in{\bf Z}$, $G(n)=E[Z^{n}]$ is calculated as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\frac{1}{h^{n}}\left(\frac{2\pi
m}{\beta}\right)^{\frac{n}{2}}\int_{-\infty}^{\infty}\prod_{a=1}^{n}dq_{a}$
(27)
$\displaystyle\int_{0}^{\infty}d\omega\frac{\omega^{s-1}}{2^{\frac{s-2}{2}}\Gamma\left(\frac{s}{2}\right)}e^{-\frac{\omega^{2}}{2}\left(1+\beta
m\sum_{a=1}^{n}q_{a}^{2}\right)}$ $\displaystyle=$
$\displaystyle\frac{(2\pi)^{\frac{n}{2}}}{(h\beta)^{n}}\int_{-\infty}^{\infty}\prod_{a=1}^{n}dr_{a}\frac{1}{\left(1+\sum_{a=1}^{n}r_{a}^{2}\right)^{\frac{s}{2}}}$
$\displaystyle=$
$\displaystyle\frac{(2\pi)^{\frac{n}{2}}}{(h\beta)^{n}}\frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}\int_{0}^{\infty}dR\frac{R^{n-1}}{(1+R^{2})^{\frac{s}{2}}}$
$\displaystyle=$
$\displaystyle\frac{2^{-\frac{n}{2}}}{(\hbar\beta)^{n}{\Gamma\left(\frac{n}{2}\right)}}B\left(\frac{n}{2},\frac{s-n}{2}\right)$
$\displaystyle=$
$\displaystyle\frac{2^{-\frac{n}{2}}}{(\hbar\beta)^{n}}\frac{\Gamma\left(\frac{s-n}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}.$

$F(n)=E[Z^{n}],(n\in{\bf C})$ in Eq. (26) analyzed by FSNJ is consistent with
$G(n)=E[Z^{n}],(n\in{\bf Z})$ in Eq. (27) analyzed by FJNS since these
descriptions are consistent with $n\not\in{\bf Z}$, and thus replica analytic
continuation may hold in this model. Note that this model includes the models
in Subsections III.1 and III.2 as special cases.

###  III.4 Quantum Harmonic Oscillator Model with Exponential Distribution

The models for which replica analytic continuation is satisfied are not
limited to the classical harmonic oscillator model. Here, we discuss another
commonly used analytical model, the quantum harmonic oscillator model. The
Hamiltonian of the quantum harmonic oscillator model, ${\cal H}(k)$, is
defined as

$\displaystyle{\cal H}(k)$ $\displaystyle=$
$\displaystyle\hbar\omega\left(k+\frac{1}{2}\right),$ (28)

where $k$ is the quantum number and $\omega$ is the positive angular
frequency. In the case of a Boson, $k=0,1,2,3,\cdots$ and the partition
function $Z$ in this model is calculated as

$\displaystyle Z$ $\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}e^{-\beta{\cal H}(k)}$ (29) $\displaystyle=$
$\displaystyle\frac{e^{-\frac{\hbar\beta\omega}{2}}}{1-e^{-\hbar\beta\omega}}.$

Moreover, it is assumed that the angular frequency $\omega$ is distributed
with the exponential distribution $P_{\mu}(\omega)$:

$\displaystyle P_{\mu}(\omega)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\frac{1}{\mu}e^{-\frac{\omega}{\mu}}&\omega>0\\\
0&\text{otherwise}\end{array}\right..$ (32)

From this, for FSNJ, when $n\in{\bf C}$, $F(n)=E[Z^{n}]$ is evaluated as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}d\omega
P_{\mu}(\omega)\frac{e^{-\frac{n\hbar\beta\omega}{2}}}{(1-e^{-\hbar\beta\omega})^{n}}$
(33) $\displaystyle=$
$\displaystyle\frac{1}{\hbar\beta\mu}\int_{0}^{1}dtt^{\frac{n}{2}+\frac{1}{\hbar\beta\mu}-1}(1-t)^{1-n-1}$
$\displaystyle=$
$\displaystyle\frac{1}{\hbar\beta\mu}B\left(\frac{n}{2}+\frac{1}{\hbar\beta\mu},1-n\right),$

where $t=e^{-\hbar\beta\omega}$ is used.

For FJNS, when $n\in{\bf Z}$, $G(n)=E[Z^{n}]$ is calculated as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\sum_{\vec{k}=0}^{\infty}\int_{0}^{\infty}d\omega
P_{\mu}(\omega)e^{-\hbar\beta\omega\sum_{a=1}^{n}(k_{a}+\frac{1}{2})}$
$\displaystyle=$
$\displaystyle\sum_{\vec{k}=0}^{\infty}\frac{1}{\hbar\beta\mu\left(\frac{1}{\hbar\beta\mu}+\frac{n}{2}+\sum_{a=1}^{n}k_{a}\right)}$
$\displaystyle=$
$\displaystyle\sum_{m=0}^{\infty}\frac{1}{\hbar\beta\mu(\frac{1}{\hbar\beta\mu}+\frac{n}{2}+m)}\left(\begin{array}[]{c}n-1+m\\\
m\end{array}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{\hbar\beta\mu\Gamma(n)}\int_{0}^{\infty}ds\int_{0}^{\infty}duu^{n-1}e^{-s(\frac{n}{2}+\frac{1}{\hbar\beta\mu})-u+ue^{-s}}$
$\displaystyle=$
$\displaystyle\frac{1}{\hbar\beta\mu}\int_{0}^{1}dtt^{\frac{n}{2}+\frac{1}{\hbar\beta\mu}-1}(1-t)^{1-n-1}$
$\displaystyle=$
$\displaystyle\frac{1}{\hbar\beta\mu}B\left(\frac{n}{2}+\frac{1}{\hbar\beta\mu},1-n\right),$
(37)

where $\vec{k}=(k_{1},k_{2},\cdots,k_{n})^{\rm T}\in{\bf Z}^{n}$ and the
Kronecker delta $\delta(c,d)$ are used. Further

$\displaystyle\sum_{\vec{k}=0}^{\infty}\delta\left(m,\sum_{a=1}^{n}k_{a}\right)$
$\displaystyle=$ $\displaystyle\left(\begin{array}[]{c}n-1+m\\\
m\end{array}\right),$ (40)

and $t=e^{-s}$ are employed. From this, $F(n)=E[Z^{n}],(n\in{\bf C})$ in Eq.
(33) analyzed by FSNJ is consistent with $G(n)=E[Z^{n}],(n\in{\bf Z})$ in Eq.
(III.4) calculated by FJNS, indicating that replica analytic continuation also
holds in this model.

###  III.5 Random Field Ising Model

In this subsection, Fermi’s quantum harmonic oscillator, the Fermion, with
$k=0,1$, is discussed using the Ising model as the equivalent representation.
For convenience, we here accept the Ising model, the Hamiltonian of which is
defined as

$\displaystyle{\cal H}(S)$ $\displaystyle=$ $\displaystyle-hS,$ (41)

where $S(=\pm 1)$ indicates the Ising spin and $h$ is the external magnetic
field. Then, the partition function $Z$ is estimated easily as

$\displaystyle Z$ $\displaystyle=$ $\displaystyle\sum_{S=\pm 1}e^{-\beta{\cal
H}(S)}$ (42) $\displaystyle=$ $\displaystyle 2\cosh\beta h.$

Furthermore, it is presumed that the external magnetic field $h$ is
distributed with a Gaussian distribution with mean $\mu$ and variance
$\sigma^{2}$, $P_{\mu,\sigma^{2}}(h)$:

$\displaystyle P_{\mu,\sigma^{2}}(h)$ $\displaystyle=$
$\displaystyle\frac{e^{-\frac{(h-\mu)^{2}}{2\sigma^{2}}}}{\sqrt{2\pi\sigma^{2}}}.$
(43)

Then, for FSNJ, in the case of $n\in{\bf C}$, $F(n)=E[Z^{n}]$ is solved as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}dhP_{\mu,\sigma^{2}}(h)\left(2\cosh\beta
h\right)^{n}$ (44) $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}Dx\left(2\cosh(\beta\mu+\beta\sigma
x)\right)^{n},$

where $Dx=\frac{dx}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}$ is employed.

For FJNS, when $n\in{\bf Z}$, $G(n)=E[Z^{n}]$ is derived as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\sum_{\vec{S}\in\left\\{\pm
1\right\\}^{n}}\int_{-\infty}^{\infty}dhP_{\mu,\sigma^{2}}(h)e^{\beta
h\sum_{a=1}^{n}S_{a}}$ $\displaystyle=$
$\displaystyle\sum_{\vec{S}\in\left\\{\pm
1\right\\}^{n}}\exp\left(\beta\mu\sum_{a=1}^{n}S_{a}+\frac{\beta^{2}\sigma^{2}}{2}\left(\sum_{a=1}^{n}S_{a}\right)^{2}\right)$
$\displaystyle=$
$\displaystyle\sum_{m=0}^{n}e^{\beta\mu(2m-n)+\frac{\beta^{2}\sigma^{2}(2m-n)^{2}}{2}}\left(\begin{array}[]{c}n\\\
m\end{array}\right)$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}Dxe^{-n\beta\mu-n\beta\sigma
x}(1+e^{2\beta\mu+2\beta\sigma x})^{n}$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}Dx\left(2\cosh(\beta\mu+\beta\sigma
x)\right)^{n},$ (48)

where $\vec{S}=(S_{1},S_{2},\cdots,S_{n})^{\rm T}\in\left\\{\pm
1\right\\}^{n}$ is used and the notation $\sum_{\vec{S}\in\left\\{\pm
1\right\\}^{n}}$ means a sum over the whole configuration of $\vec{S}$.
Furthermore, $\int_{-\infty}^{\infty}Dxe^{\theta x}=e^{\frac{\theta^{2}}{2}}$
and

$\displaystyle\sum_{\vec{S}\in\left\\{\pm
1\right\\}^{n}}\delta\left(2m-n,\sum_{a=1}^{n}S_{a}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{c}n\\\ m\end{array}\right),$ (51)

are used. From Eq. (44) and Eq. (48), $F(n)=E[Z^{n}],(n\in{\bf C})$ for FSNJ
and $G(n)=E[Z^{n}],(n\in{\bf Z})$ for FJNS are equivalent, showing that
replica analytic continuation holds for this model.

###  III.6 Potential Energy for Near Earth Gravity

In this subsection, we discuss whether replica analytic continuation holds
using a classical physics model that is not often used in statistical physics,
the potential energy for near Earth gravity. For $g$, the gravitational
acceleration, the Hamiltonian ${\cal H}(x)$ is defined by the potential energy
of the particle (its mass $m$) at position $x$, $mgx$, where the potential
energy at origin is regarded as $0$:

$\displaystyle{\cal H}(x)$ $\displaystyle=$ $\displaystyle mgx.$ (52)

We here consider the canonical ensemble of this model whose range is the set
of non-negative real numbers. The partition function of this model $Z$ is
calculated as

$\displaystyle Z$ $\displaystyle=$
$\displaystyle\frac{1}{h}\int_{0}^{\infty}dxe^{-\beta{\cal H}(x)}$ (53)
$\displaystyle=$ $\displaystyle\frac{1}{h\beta mg}.$

Here it is assumed that the mass $m$ is randomly distributed with the Gamma
distribution with shape parameter $\tau\in{\bf R}$, $P_{\tau}(m)$:

$\displaystyle P_{\tau}(m)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\frac{m^{\tau-1}e^{-m}}{\Gamma(\tau)}&m>0\\\
0&\text{otherwise}\end{array}\right..$ (56)

Using this distribution, for FSNJ, in the case of $n\in{\bf C}$,
$F(n)=E[Z^{n}]$ is solved as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}dmP_{\tau}(m)\left(\frac{1}{h\beta
gm}\right)^{n}$ (57) $\displaystyle=$ $\displaystyle\frac{1}{(h\beta
g)^{n}}\frac{\Gamma(\tau-n)}{\Gamma(\tau)}.$

For FJNS, when $n\in{\bf Z}$, $G(n)=E[Z^{n}]$ is described as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\frac{1}{h^{n}}\int_{0}^{\infty}\prod_{a=1}^{n}dx_{a}\int_{0}^{\infty}dmP_{\tau}(m)e^{-\beta
mg\sum_{a=1}^{n}x_{a}}$ (58) $\displaystyle=$ $\displaystyle\frac{1}{(h\beta
g)^{n}}\int_{0}^{\infty}\prod_{a=1}^{n}dr_{a}\frac{1}{\left(1+\sum_{a=1}^{n}r_{a}\right)^{\tau}},$

where $r_{a}=\beta gx_{a}$ is used. Moreover, when $k>1$,

$\displaystyle\int_{0}^{\infty}\frac{dr}{(\alpha+r)^{k}}$ $\displaystyle=$
$\displaystyle\frac{1}{(k-1)\alpha^{k-1}},$ (59)

is obtained. Using this formula, we can recursively solve $G(n)$ as

$\displaystyle E[Z^{n}]$ $\displaystyle=$ $\displaystyle\frac{1}{(h\beta
g)^{n}}\frac{1}{(\tau-1)}\frac{1}{(\tau-2)}\cdots\frac{1}{(\tau-l)}$ (60)
$\displaystyle\int_{0}^{\infty}\prod_{a=1}^{n-l}dr_{a}\frac{1}{\left(1+\sum_{a=1}^{n-l}r_{a}\right)^{\tau-l}}$
$\displaystyle=$ $\displaystyle\frac{1}{(h\beta
g)^{n}}\frac{1}{(\tau-1)}\frac{1}{(\tau-2)}\cdots\frac{1}{(\tau-n)}$
$\displaystyle=$ $\displaystyle\frac{1}{(h\beta
g)^{n}}\frac{\Gamma(\tau-n)}{\Gamma(\tau)}.$

$F(n)=E[Z^{n}],(n\in{\bf C})$ in Eq. (57) analyzed by FSNJ and
$G(n)=E[Z^{n}],(n\in{\bf Z})$ in Eq. (60) analyzed by FJNS are consistent, and
thus replica analytic continuation is guaranteed in this model.

###  III.7 Moment of Chi-Squared Distribution

In this subsection we discuss whether replica analytic continuation holds for
a model which is not used in physics, namely, where the random variable $Z$ is
distributed with a $\chi^{2}$ distribution with degrees of freedom $m$,
$P_{\chi^{2}(m)}(Z)$:

$\displaystyle P_{\chi^{2}(m)}(Z)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\frac{1}{2^{\frac{m}{2}}\Gamma\left(\frac{m}{2}\right)}Z^{\frac{m}{2}-1}e^{-\frac{Z}{2}}&Z>0\\\
0&\text{otherwise}\end{array}\right.,$ (63)

where $Z$ of this $\chi^{2}$ distribution is the sum of the squares of $m$
random variables, $x_{1},\cdots,x_{m}$, which are independently and
identically drawn from the standard normalized distribution, that is,
$Z=\sum_{i=1}^{m}x_{i}^{2}$. We regard the sum of squares
$Z=\sum_{i=1}^{m}x_{i}^{2}$ to be the partition function.

For FSNJ, since the distribution of $Z$ is already known, that is, we can
consider the thermodynamical average to be determined, when $n\in{\bf C}$,
$F(n)=E[Z^{n}]$, is evaluated as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}dZP_{\chi^{2}(m)}(Z)Z^{n}$ (64)
$\displaystyle=$
$\displaystyle\frac{2^{n}\Gamma\left(\frac{m}{2}+n\right)}{\Gamma\left(\frac{m}{2}\right)}.$

For FJNS, since $Z$ is constructed from the sum of $m$ independently and
identically distributed variables, that is, $Z=\sum_{i=1}^{m}x_{i}^{2}$, when
$n\in{\bf Z}$, $G(n)=E[Z^{n}]$ is expanded as

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}\prod_{i=1}^{m}Dx_{i}\left(\sum_{i=1}^{m}x_{i}^{2}\right)^{n}$
(65) $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}\prod_{i=1}^{m}Dx_{i}\sum_{\vec{k}=0}^{n}\frac{n!}{\prod_{i=1}^{m}k_{i}!}\prod_{i=1}^{m}x_{i}^{2k_{i}}\delta\left(n,\sum_{i=1}^{m}k_{i}\right)$
$\displaystyle=$ $\displaystyle
2^{n}n!\sum_{\vec{k}=0}^{n}\delta\left(n,\sum_{i=1}^{m}k_{i}\right)\prod_{i=1}^{m}\frac{\Gamma\left(k_{i}+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma(k_{i}+1)},$

where $\vec{k}=(k_{1},\cdots,k_{m})^{\rm
T}\in\left\\{0,1,2,\cdots,n\right\\}^{m}$ and

$\displaystyle\int_{-\infty}^{\infty}Dxx^{2k}$ $\displaystyle=$
$\displaystyle\frac{2^{k}}{\Gamma\left(\frac{1}{2}\right)}\Gamma\left(k+\frac{1}{2}\right),$
(66)

are used. We can verify that Eq. (64) and Eq. (65) are consistent by using
mathematical induction. For convenience, the right hand sides of Eq. (64) and
Eq. (65) are rewritten as

$\displaystyle F(m,n)$ $\displaystyle=$
$\displaystyle\frac{2^{n}\Gamma\left(\frac{m}{2}+n\right)}{\Gamma\left(\frac{m}{2}\right)},$
(67) $\displaystyle G(m,n)$ $\displaystyle=$ $\displaystyle
2^{n}n!\sum_{\vec{k}=0}^{n}\delta\left(n,\sum_{i=1}^{m}k_{i}\right)\prod_{i=1}^{m}\frac{\Gamma\left(k_{i}+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma(k_{i}+1)}.$

When $m=1$, for any $n\in{\bf Z}$,

$\displaystyle F(1,n)$ $\displaystyle=$
$\displaystyle\frac{2^{n}\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)},$
(69) $\displaystyle G(1,n)$ $\displaystyle=$ $\displaystyle
2^{n}n!\sum_{k_{1}=0}^{n}\delta\left(n,k_{1}\right)\frac{\Gamma\left(k_{1}+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma(k_{1}+1)}$
(70) $\displaystyle=$ $\displaystyle
2^{n}n!\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma(n+1)}$
$\displaystyle=$ $\displaystyle
2^{n}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)}.$

This shows that $F(1,n)=G(1,n)$ holds.

Next, we set

$\displaystyle F(s,t)$ $\displaystyle=$
$\displaystyle\frac{2^{t}\Gamma\left(\frac{s}{2}+t\right)}{\Gamma\left(\frac{s}{2}\right)},$
(71) $\displaystyle G(s,t)$ $\displaystyle=$ $\displaystyle
2^{t}t!\sum_{\vec{k}=0}^{t}\delta\left(t,\sum_{i=1}^{s}k_{i}\right)\prod_{i=1}^{s}\frac{\Gamma\left(k_{i}+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma(k_{i}+1)},$

and assume that $F(s,t)=G(s,t)$ holds when $m=s\in{\bf Z}$ for any $t\in{\bf
Z}$. Then, when $m=s+1$,

$\displaystyle G(s+1,t)$ (73) $\displaystyle=$ $\displaystyle
2^{t}t!\sum_{k_{s+1}=0}^{t}\sum_{k_{s}=0}^{t}\sum_{k_{s-1}=0}^{t}\cdots\sum_{k_{1}=0}^{t}$
$\displaystyle\delta\left(t,\sum_{i=1}^{s}k_{i}+k_{s+1}\right)\frac{\Gamma\left(k_{s+1}+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(k_{s+1}+1\right)}$
$\displaystyle\prod_{i=1}^{s}\frac{\Gamma\left(k_{i}+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(k_{i}+1\right)}$
$\displaystyle=$ $\displaystyle
2^{t}t!\sum_{k_{s+1}=0}^{t}\frac{\Gamma\left(k_{s+1}+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(k_{s+1}+1\right)}\frac{2^{k_{s+1}-t}}{\Gamma(t-k_{s+1}+1)}$
$\displaystyle\left\\{2^{t-k_{s+1}}(t-k_{s+1})!\sum_{k_{s}=0}^{t-k_{s+1}}\sum_{k_{s-1}=0}^{t-k_{s+1}}\cdots\sum_{k_{1}=0}^{t-k_{s+1}}\right.$
$\displaystyle\left.\delta\left(t-k_{s+1},\sum_{i=1}^{s}k_{i}\right)\prod_{i=1}^{s}\frac{\Gamma\left(k_{i}+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(k_{i}+1\right)}\right\\}$
$\displaystyle=$ $\displaystyle
2^{t}t!\sum_{k_{s+1}=0}^{t}\frac{\Gamma\left(k_{s+1}+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(k_{s+1}+1\right)}\frac{2^{k_{s+1}-t}}{\Gamma(t-k_{s+1}+1)}$
$\displaystyle G(s,t-k_{s+1}).$

Further, from the assumption of mathematical induction, since
$G(s,t-k_{s+1})=F(s,t-k_{s+1})$ holds, we have

$\displaystyle G(s+1,t)$ (74) $\displaystyle=$ $\displaystyle
2^{t}t!\sum_{k_{s+1}=0}^{t}\frac{\Gamma\left(k_{s+1}+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(k_{s+1}+1\right)}\frac{2^{k_{s+1}-t}}{\Gamma(t-k_{s+1}+1)}$
$\displaystyle
2^{t-k_{s+1}}\frac{\Gamma\left(\frac{s}{2}+t-k_{s+1}\right)}{\Gamma\left(\frac{s}{2}\right)}$
$\displaystyle=$
$\displaystyle\frac{2^{t}}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{s}{2}\right)}\sum_{k_{s+1}=0}^{t}\frac{t!}{k_{s+1}!(t-k_{s+1})!}$
$\displaystyle\Gamma\left(k_{s+1}+\frac{1}{2}\right)\Gamma\left(t-k_{s+1}+\frac{s}{2}\right)$
$\displaystyle=$
$\displaystyle\frac{2^{t}}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{s}{2}\right)}\sum_{k=0}^{t}\frac{t!}{k!(t-k)!}$
$\displaystyle\int_{0}^{\infty}duu^{k+\frac{1}{2}-1}e^{-u}\int_{0}^{\infty}dvv^{t-k+\frac{s}{2}-1}e^{-v}$
$\displaystyle=$
$\displaystyle\frac{2^{t}}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{s}{2}\right)}\int_{0}^{\infty}duu^{\frac{1}{2}-1}e^{-u}\int_{0}^{\infty}dvv^{\frac{s}{2}-1}e^{-v}$
$\displaystyle\sum_{k=0}^{t}\frac{t!}{k!(t-k)!}u^{k}v^{t-k}$ $\displaystyle=$
$\displaystyle
2^{t}\int_{0}^{\infty}du\frac{u^{\frac{1}{2}-1}e^{-u}}{\Gamma\left(\frac{1}{2}\right)}\int_{0}^{\infty}dv\frac{v^{\frac{s}{2}-1}e^{-v}}{\Gamma\left(\frac{s}{2}\right)}$
$\displaystyle(u+v)^{t}.$

Setting $u=\frac{x}{2}$ and $v=\frac{y}{2}$ we obtain

$\displaystyle G(s+1,t)$ (75) $\displaystyle=$ $\displaystyle
2^{t}\int_{0}^{\infty}dx\frac{x^{\frac{1}{2}-1}e^{-\frac{x}{2}}}{\Gamma\left(\frac{1}{2}\right)2^{\frac{1}{2}}}\int_{0}^{\infty}dy\frac{y^{\frac{s}{2}-1}e^{-\frac{y}{2}}}{\Gamma\left(\frac{s}{2}\right)2^{\frac{s}{2}}}\left(\frac{x+y}{2}\right)^{t}$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}dxP_{\chi^{2}(1)}(x)\int_{0}^{\infty}dyP_{\chi^{2}(s)}(y)(x+y)^{t}$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}dzP_{\chi^{2}(s+1)}(z)z^{t}$
$\displaystyle=$
$\displaystyle\frac{2^{t}\Gamma\left(\frac{s+1}{2}+t\right)}{\Gamma\left(\frac{s+1}{2}\right)}$
$\displaystyle=$ $\displaystyle F(s+1,t).$

Thus, $F(s+1,t)=G(s+1,t)$ holds. Note that we use the property that the sum of
the random variable $x$, which is drawn from the $\chi^{2}$ distribution with
$1$ degree of freedom, and the random variable $y$, which is drawn from the
$\chi^{2}$ distribution with degrees of freedom $s$, so that $z=x+y$ is
distributed with a $\chi^{2}$ distribution with degrees of freedom $s+1$. From
this induction, it is verified that $F(m,n)=G(m,n)$ holds. Finally, since both
descriptions, $F(n)=E[Z^{n}],(n\in{\bf C})$ for FSNJ and
$G(n)=E[Z^{n}],(n\in{\bf Z})$ for FJNS, are also consistent with each other,
we have shown that replica analytic continuation holds in this model.

###  III.8 Other Toy Models

Lastly, we introduce other toy models which satisfy replica analytic
continuation:

  1. 1.

From the partition function
$Z=\int_{-\infty}^{\infty}Dxe^{x\omega}=e^{\frac{\omega^{2}}{2}}$ and
$P_{\sigma}(\omega)$ in Eq. (20),
$F(n)=E[Z^{n}]=(1-n\sigma^{2})^{-\frac{1}{2}}$ is obtained. Since $x_{a}\sim
N(0,1),(a=1,\cdots,n)$, it is well known that $s=\sigma\sum_{a=1}^{n}x_{a}$ is
distributed with a Gaussian distribution with mean $0$ and variance
$n\sigma^{2}$. Using $s$, we can also easily analyze $G(n)$ for FJNS.

  2. 2.

We consider a random variable $Z$ from the $F$ distribution with degrees of
freedom $s$ and $t$:

$\displaystyle P(Z)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\frac{s^{\frac{s}{2}}t^{\frac{t}{2}}}{B\left(\frac{s}{2},\frac{t}{2}\right)}\frac{Z^{\frac{s-2}{2}}}{(sZ+t)^{\frac{s+t}{2}}}&Z>0\\\
0&Z\leq 0\end{array}\right..$ (78)

Then, $F(n)=E[Z^{n}],(n\in{\bf C})$ is calculated as follows.

$\displaystyle E[Z^{n}]$ $\displaystyle=$
$\displaystyle\left(\frac{t}{s}\right)^{n}\frac{\Gamma\left(\frac{s}{2}+n\right)\Gamma\left(\frac{t}{2}-n\right)}{\Gamma\left(\frac{s}{2}\right)\Gamma\left(\frac{t}{2}\right)}$
(79)

We can regard the random variable $Z$ as being the ratio between the squared
average of $s$ random variables $x_{k},(k=1,\cdots,s)$ which are independently
and identically distributed with a standard normal distribution, and the
squared average of $t$ random variables $y_{l},(l=1,\cdots,t)$ which are also
independently and identically distributed with a standard normal distribution
as follows:

$\displaystyle Z$ $\displaystyle=$
$\displaystyle\frac{\frac{1}{s}\sum_{k=1}^{s}x_{k}^{2}}{\frac{1}{t}\sum_{l=1}^{t}y_{l}^{2}}.$
(80)

Thus, $G(n)$ is also easily solved using the mathematical induction in
Subsection III.7.

###  III.9 Discussion and Open Problems

As stated above, in this study we have examined replica analytic continuation,
the mathematical basis behind the replica trick, one of the two steps that
comprise replica analysis. Some of the concerns about the reliability of
replica analysis are based on doubts about the accuracy of the replica trick.
When applying replica analysis, both the replica trick and the thermodynamical
limit are used, and hence it can be hard to unravel any inaccuracies in
replica analysis. Therefore, we have applied replica analysis to some physical
or statistical models for which the thermodynamical limit need not be applied,
so that we can test the accuracy of replica analytical continuation alone,
thereby definitively addressing part of the basis of replica analysis.

In this paper, although we considered only a few examples due to the
limitations of space, it is clear from the results of the models for which we
can apply both analytical procedures, FSNJ and FJNS, that we can assess
$F(n)=E[Z^{n}],(n\in{\bf C})$ and $G(n)=E[Z^{n}],(n\in{\bf Z})$ without
approximation and can verify that replica analytic continuation is guaranteed
in these models. We note that a more complete analysis would examine a
counterexample to the case of replica analytical continuation in mathematical
models which can be analyzed without approximation. Thus, while the analysis
in this paper does not provide a universal mathematical proof of replica
analysis, the positive examples detailed in this paper lead us to the
following conjecture:

Conjecture on Replica Analytic Continuation

With respect to models where we can assess the FSNJ and FJNS approaches
without approximation, for $F(n)=E[Z^{n}],(n\in{\bf C})$ analyzed by the FSNJ
procedure and $G(n)=E[Z^{n}],(n\in{\bf Z})$ analyzed by the FJNS procedure,
when $n\in{\bf Z}$, $F(n)=G(n)$ is always satisfied. In addition, $G(n)$ is
regarded as an analytical function of the replica number $n$, and we assume
that $G(n)$ is also described as an analytical function of $n$ when
$n\not\in{\bf Z}$, so then if $F(n)=G(n),(n\not\in{\bf Z})$ holds, the replica
analytic continuation for this model is guaranteed.

Namely, given the positive results of our study, it can now be expected that
any doubts about the accuracy of replica analysis will not be directed at
replica analytic continuation. Hence any questions regarding replica analysis
reported in previous works are likely to be based on more complicated aspects
of replica analysis, particularly when the thermodynamical limit and the
expansion on order parameters are included. This paper has taken the approach
of decomposing replica analysis into its composite components, allowing the
mathematical properties of each component to be investigated in order to
clarify the mathematical structure of replica analysis. However, it is also
important to examine problems where the components are used in combination.

##  IV Conclusion and Future Works

In this work, to explore the roots of questions on the validity of replica
analysis, we applied replica analytic continuation to analyze harmonic
oscillator models (classical and quantum) and the Ising model, which do not
need consideration of the thermodynamical limit, and validated the
effectiveness of replica analytic continuation. Although the present paper
handles only a few examples, we discussed whether replica analytic
continuation holds with respect to toy models for which it is possible to
solve $F(n)=E[Z^{n}],(n\in{\bf C})$ and $G(n)=E[Z^{n}],(n\in{\bf Z})$ using
the FSNJ and FJNS approaches without approximations. Although we cannot
provide a universal proof on the validity of replica analysis in this paper,
our goal was to formulate a conjecture for replica analytic continuation. We
do not discuss the thermodynamical limit and order parameter expansion, though
these have already been guaranteed theoretically and practically, and are not
particularly under question. However, considering more complicated cases will
be difficult Anderson  as even if each analytical technique can be proved
individually, they are complicated when combined.

In future work we need to discuss whether replica analytic continuation holds
for physical or statistical models of many-body systems using FSNJ and FJNS
without approximations. Although we considered a few positive examples in this
paper, it is preferable to prove the conjecture of replica analytic
continuation in a general form. Furthermore, although we discussed replica
analytic continuation using the versatile properties of the Gamma function,
other special functions could further support the proof of replica analytic
continuation.

## Acknowledgements

The author appreciates the helpful comments offered by Y. Kabashima, T. Tanaka
and K. Kobayashi. The work is partly supported by Grant-in-Aid Nos. 24710169
and 15K20999; the President Project for Young Scientists at Akita Prefectural
University; grant No. 50 of the National Institute of Informatics, Japan;
grant No. 5 of the Japan Institute of Life Insurance; a grant from the
Institute of Economic Research Foundation; Kyoto University; grant No. 1414 of
the Zengin Foundation for Studies in Economics and Finance; grant No. 2068 of
the Institute of Statistical Mathematics; research project of Mitsubishi UFJ
Trust Scholarship Foundation; and grant No. 2 of the Kampo Foundation.

##  Appendix A Calculation of $E[\log Z]$

Here we discuss the expectation of the logarithm of the partition function
$E[\log Z]$. It is well-known that the Gamma function of $z$,
$\Gamma(z)=\int_{0}^{\infty}dtt^{z-1}e^{-t}$ is described as

$\displaystyle\Gamma(z)$ $\displaystyle=$ $\displaystyle\frac{1}{z}e^{-\gamma
z}\prod_{k=1}^{\infty}\left(\frac{k}{k+z}\right)e^{\frac{z}{k}}.$ (81)

Using this description, the derivative of the logarithm function of the Gamma
function with respect to $z$,

$\displaystyle\psi(z)$ $\displaystyle=$ $\displaystyle\frac{\partial}{\partial
z}\log\Gamma(z)$ (82) $\displaystyle=$
$\displaystyle-\gamma-\frac{1}{z}+\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+z}\right),$

is easily calculated where the Euler constant $\gamma$ is defined as

$\displaystyle\gamma$ $\displaystyle=$
$\displaystyle\lim_{L\to\infty}\left(\sum_{k=1}^{L}\frac{1}{k}-\log L\right)$
(83) $\displaystyle\simeq$ $\displaystyle 0.57721.$

We call $\psi(z)$ the digamma function. Especially for $z\in{\bf Z}$, the
digamma function is formulated as follows;

$\displaystyle\psi(z)$ $\displaystyle=$
$\displaystyle-\gamma+\sum_{k=1}^{z-1}\frac{1}{k},$ (84)
$\displaystyle\psi\left(z+\frac{1}{2}\right)$ $\displaystyle=$
$\displaystyle-\gamma-2\log 2+\sum_{k=1}^{z}\frac{1}{k-\frac{1}{2}}.$ (85)

From which we have,

$\displaystyle\psi(1)$ $\displaystyle=$ $\displaystyle-\gamma,$ (86)
$\displaystyle\psi\left(\frac{1}{2}\right)$ $\displaystyle=$
$\displaystyle-\gamma-2\log 2.$ (87)

From these settings, the expectation of the logarithm of the partition
function $E[\log Z]$ in each case is solved as follows:

###  A.1 Expectation of $\log Z$ in Section III.1

$\displaystyle E[\log Z]$ $\displaystyle=$
$\displaystyle\frac{\gamma}{2}-\frac{1}{2}\log 2-\log(\hbar\beta),$ (88)

###  A.2 Expectation of $\log Z$ in Section III.2

$\displaystyle E[\log Z]$ $\displaystyle=$
$\displaystyle\frac{\gamma}{2}+\frac{1}{2}\log 2-\log(\hbar\beta\sigma),$ (89)

###  A.3 Expectation of $\log Z$ in Section III.3

$\displaystyle E[\log Z]$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\psi\left(\frac{s}{2}\right)-\frac{1}{2}\log
2-\log(\hbar\beta),$ (90)

###  A.4 Expectation of $\log Z$ in Section III.4

$\displaystyle E[\log Z]$ $\displaystyle=$
$\displaystyle\frac{1}{2}\psi\left(\frac{1}{\hbar\beta\mu}\right)+\frac{1}{2}\psi\left(1+\frac{1}{\hbar\beta\mu}\right)-\psi(1)$
(91) $\displaystyle=$
$\displaystyle-\frac{\hbar\beta\mu}{2}+\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+\frac{1}{\hbar\beta\mu}}\right),$

###  A.5 Expectation of $\log Z$ in Section III.5

$\displaystyle E[\log Z]$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}Dx\log 2\cosh\left(\beta\mu+\beta\sigma
x\right),\qquad$ (92)

###  A.6 Expectation of $\log Z$ in Section III.6

$\displaystyle E[\log Z]$ $\displaystyle=$ $\displaystyle-\log(h\beta
g)-\psi(\tau),$ (93)

###  A.7 Expectation of $\log Z$ in Section III.7

$\displaystyle E[\log Z]$ $\displaystyle=$ $\displaystyle\log
2+\psi\left(\frac{m}{2}\right),$ (94)

###  A.8 Expectation of $\log Z$ of 1st Example in Section III.8

$\displaystyle E[\log Z]$ $\displaystyle=$
$\displaystyle\frac{\sigma^{2}}{2},$ (95)

###  A.9 Expectation of $\log Z$ of 2nd Example in Section III.8

$\displaystyle E[\log Z]$ $\displaystyle=$
$\displaystyle\log\frac{t}{s}+\psi\left(\frac{s}{2}\right)-\psi\left(\frac{t}{2}\right),$
(96)

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