# Deformation of power law in the double Pareto distribution using uniformly
distributed observation time

Ken Yamamoto  Faculty of Science, University of the Ryukyus, Nishihara,
Okinawa 903-0213, Japan  Takashi Bando, Hirokazu Yanagawa  Production
Engineering Department, SCM Division, Measurement Business Group, ANRITSU
CORPORATION, Atsugi, Kanagawa 243-8555, Japan  Yoshihiro Yamazaki  School of
Advanced Science and Engineering, Waseda University, Shinjuku, Tokyo 169-8555,
Japan

###### Abstract

The double Pareto distribution is a heavy-tailed distribution with a power-law
tail, that is generated via geometric Brownian motion with an exponentially
distributed observation time. In this study, we examine a modified model
wherein the exponential distribution of the observation time is replaced with
a continuous uniform distribution. The probability density, complementary
cumulative distribution, and moments of this model are exactly calculated.
Furthermore, the validity of the analytical calculations is discussed in
comparison with numerical simulations of stochastic processes.

##  I Introduction

Stochastic models have substantially contributed to the analysis of
fluctuating or noisy systems [1, 2]. Furthermore, simple stochastic processes
have been proved to adequately describe real phenomena, and the connection
between stochastic processes and resultant probability distributions have been
established theoretically. A probability distribution having a subexponential
tail is referred to as a heavy-tailed distribution [3], which provides
theoretical support for statistical physics such as critical phenomena [4],
anomalous diffusion [5], and long-range memories [6]. Additionally, heavy-
tailed distributions are crucial for complex systems such as social [7, 8] and
biological [9] systems. Lognormal and power-law distributions are the heavy-
tailed distributions focused upon in this study.

A random variable $X$ is said to follow the lognormal distribution if the
logarithm of $X$ is normally distributed [10]. The probability density
function (PDF) of the lognormal distribution is expressed as

$f_{\text{LN}}(x;\mu,\sigma^{2})=\frac{1}{\sqrt{2\pi}\sigma
x}\exp\left(-\frac{(\ln x-\mu)^{2}}{2\sigma^{2}}\right),$

where $\mu$ and $\sigma$ are the mean and standard deviation of $\ln X$,
respectively. Its complementary cumulative distribution function (CCDF) is
expressed as

$F_{\text{LN}}(x;\mu,\sigma^{2})=\int_{x}^{\infty}f_{\text{LN}}(y;\mu,\sigma^{2})dy=\frac{1}{2}\operatorname{erfc}\left(\frac{\ln
x-\mu}{\sqrt{2}\sigma}\right),$

where

$\operatorname{erfc}(z)\coloneqq\frac{2}{\sqrt{\pi}}\int_{z}^{\infty}e^{-u^{2}}du$
(1)

is the complementary error function [11]. Furthermore, the $k$th moment of the
lognormal random variable $X$ is computed as

$E[X^{k}]=\exp\left(k\mu+\frac{k^{2}}{2}\sigma^{2}\right).$ (2)

Specifically, the mean and variance are

$E[X]=\exp\left(\mu+\frac{\sigma^{2}}{2}\right),\quad
V[X]=e^{2\mu+\sigma^{2}}(e^{\sigma^{2}}-1),$

respectively.

The lognormal distribution occurs in the multiplicative stochastic process,
which is given by

$X_{n}=M_{n}X_{n-1},$ (3)

where the initial value $X_{0}$ is a positive constant, and
$M_{1},M_{2},\ldots$ are random variables distributed independently and
identically. The distribution of $X_{n}$ for sufficiently large $n$ can be
approximated via a lognormal distribution, owing to the central limit theorem.
The process (3) has been used as a simplified model for the X-ray burst [12]
and the growth of organisms [13, 14]; this model was originally analyzed by
Kolmogorov [15].

As Eq. (3) has a simple form, various additional effects have been applied. By
modifying Eq. (3), the lognormal distribution can change qualitatively to
other distributions. For example, power-law distributions are obtained by
introducing additive noise [16], reset event [17], random stopping [18], and
temporal cumulative sum [19, 20]. The introduction of a lower bound yields the
power-law distribution [21], and a related model introducing the sample-
dependent lower bound yields a heavy-tailed but not power-law distribution
[22].

Geometric Brownian motion [23] can yield a lognormal distribution, and is
expressed as the stochastic differential equation

$dS_{t}=\mu S_{t}dt+\sigma S_{t}dB_{t},$ (4)

where $B_{t}$ indicates the Brownian motion, $\mu$ is a real constant, and
$\sigma$ is a positive constant. For simplicity, we assume that the initial
value $S_{0}$ is constant. By using Itô’s lemma [23], the value of $S_{t}$ at
a given time $t=T$ is lognormally distributed, whose PDF is
$f_{\text{LN}}(x;\ln S_{0}+\tilde{\mu}T,\sigma^{2}T)$, where
$\tilde{\mu}=\mu-\sigma^{2}/2$. The geometric Brownian motion is applied to
the Black–Scholes equation in mathematical finance [24].

When the observation time $T$ of the geometric Brownian motion (4) is changed
to an exponential random variable, $S_{T}$ does not follow the lognormal
distribution. Instead, the double Pareto (DP) distribution is observed [25,
26]. The PDF of $S_{T}$ can be expressed as

$f_{\text{DP}}(x)=\int_{0}^{\infty}\lambda e^{-\lambda t}f_{\text{LN}}(x;\ln
S_{0}+\tilde{\mu}t,\sigma^{2}t)dt=\int_{0}^{\infty}\frac{\lambda e^{-\lambda
t}}{\sqrt{2\pi t}\sigma x}\exp\left(-\frac{(\ln x-\ln
S_{0}-\tilde{\mu}t)^{2}}{2\sigma^{2}t}\right)dt,$

where $\lambda e^{-\lambda t}$ is the PDF of the exponential distribution with
mean $1/\lambda$. This integral represents the mixture of the time-dependent
lognormal size distribution with the exponential distribution as a weight.

To calculate this integral, $u=\sqrt{t}$ is introduced and the formula [11]

$\int_{0}^{\infty}\exp\left(-a^{2}u^{2}-\frac{b^{2}}{u^{2}}\right)du=\frac{\sqrt{\pi}}{2a}e^{-2ab}$
(5)

can be used. Finally, we obtain

$f_{\text{DP}}(x)=\begin{dcases}\frac{\lambda}{S_{0}\sqrt{\tilde{\mu}^{2}+2\sigma^{2}\lambda}}\left(\frac{x}{S_{0}}\right)^{-\alpha-1}&x\geq
S_{0},\\\
\frac{\lambda}{S_{0}\sqrt{\tilde{\mu}^{2}+2\sigma^{2}\lambda}}\left(\frac{x}{S_{0}}\right)^{\beta-1}&x<S_{0},\end{dcases}$
(6)

where

$\alpha=\frac{-\tilde{\mu}+\sqrt{\tilde{\mu}^{2}+2\sigma^{2}\lambda}}{\sigma^{2}},\quad\beta=\frac{\tilde{\mu}+\sqrt{\tilde{\mu}^{2}+2\sigma^{2}\lambda}}{\sigma^{2}}.$

The term “double Pareto” originates from the property that $f_{\text{DP}}(x)$
has two different power-law exponents depending on whether $x\geq S_{0}$ or
$x<S_{0}$. The DP distribution has been observed in various phenomena, such as
income [27] and microblog posting interval [28].

A natural generalization of the DP distribution involves replacing the
exponential distribution of $T$ by other distributions. Upon replacing
$\lambda e^{-\lambda t}$ with a general PDF $g(t)$, the PDF of $S_{T}$ is
formally expressed as

$f_{g}(x)=\int_{0}^{\infty}g(t)f_{\text{LN}}(x;\ln
S_{0}+\tilde{\mu}t,\sigma^{2}t)dt.$

However, the calculation of this integral cannot be performed for general
$g(t)$. This study focuses on the case wherein $T$ follows a uniform
distribution as a simple case. In this case, the PDF and CCDF of $S_{T}$ can
be exactly calculated although they have complicated forms. We further
calculate the moments of $S_{T}$ and compare the CCDF to the discrete-time
process (3).

##  II Probability density

In this section, we derive the PDF of the observed value $S_{T}$ of geometric
Brownian motion wherein the observation time $T$ follows the uniform
distribution on the interval $[0,T_{\text{max}}]$:

$g_{\text{uni}}(t)=\begin{dcases}\frac{1}{T_{\text{max}}}&0\leq t\leq
T_{\text{max}}\\\ 0&t>T_{\text{max}}.\end{dcases}$

The PDF of $S_{T}$ in this case is written as $f_{\text{uni}}(x)$ and is
expressed as

$\displaystyle f_{\text{uni}}(x)$
$\displaystyle=\int_{0}^{\infty}g_{\text{uni}}(t)f_{\text{LN}}(x;\ln
S_{0}+\tilde{\mu}t,\sigma^{2}t)dt$
$\displaystyle=\frac{1}{T_{\text{max}}}\int_{0}^{T_{\text{max}}}f_{\text{LN}}(x;\ln
S_{0}+\tilde{\mu}t,\sigma^{2}t)dt$
$\displaystyle=\frac{1}{T_{\text{max}}}\int_{0}^{T_{\text{max}}}\frac{1}{\sqrt{2\pi
t}\sigma x}\exp\left(-\frac{(\ln x-\ln
S_{0}-\tilde{\mu}t)^{2}}{2\sigma^{2}t}\right)dt.$

After changing the integral variable to $u=\sqrt{t/T_{\text{max}}}$ and some
manipulation, we obtain

$f_{\text{uni}}(x)=\frac{2}{\sqrt{2\pi T_{\text{max}}}\sigma
x}\int_{0}^{1}\exp\left(-\frac{(\ln x-\ln
S_{0}-T_{\text{max}}\tilde{\mu}u^{2})^{2}}{2T_{\text{max}}\sigma^{2}u^{2}}\right)du.$

By introducing the scaling transformations $\mu_{\ast}=T_{\text{max}}\mu$,
$\sigma_{\ast}=\sqrt{T_{\text{max}}}\sigma$, and
$\tilde{\mu}_{\ast}=\mu_{\ast}-\sigma_{\ast}^{2}/2=\tilde{\mu}T_{\text{max}}$,
the parameter $T_{\text{max}}$ can be eliminated:

$\displaystyle f_{\text{uni}}(x)$
$\displaystyle=\frac{2}{\sqrt{2\pi}\sigma_{\ast}x}\int_{0}^{1}\exp\left(-\frac{(\ln
x-\ln S_{0}-\tilde{\mu}_{\ast}u^{2})^{2}}{2\sigma_{\ast}^{2}u^{2}}\right)du$
$\displaystyle=\frac{2}{\sqrt{2\pi}\sigma_{\ast}x}\left(\frac{x}{S_{0}}\right)^{\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}\int_{0}^{1}\exp\left(-\frac{\tilde{\mu}_{\ast}^{2}}{2\sigma_{\ast}^{2}}u^{2}-\frac{(\ln(x/S_{0}))^{2}}{2\sigma_{\ast}^{2}u^{2}}\right)du.$
(7)

Therefore, although $f_{\text{uni}}$ originally involves three parameters
$\tilde{\mu}$, $\sigma$, and $T_{\text{max}}$, it can be essentially reduced
to two ($\tilde{\mu}_{\ast}$ and $\sigma_{\ast}$).

To calculate the integral in Eq. (7), we employed the following formula [11]:

$\int_{z}^{\infty}\exp\left(-a^{2}u^{2}-\frac{b^{2}}{u^{2}}\right)du=\frac{\sqrt{\pi}}{4a}\left[e^{2ab}\operatorname{erfc}\left(az+\frac{b}{z}\right)+e^{-2ab}\operatorname{erfc}\left(az-\frac{b}{z}\right)\right],$
(8)

where $a,b>0$. Equation (8) can be considered as a generalization (or
indefinite integral) of Eq. (5). In fact, Eq. (5) is obtained by considering
the $z\to 0+$ limit and using the limit values
$\operatorname{erfc}(+\infty)=0$ and $\operatorname{erfc}(-\infty)=2$. Using
this relation, we obtain

$\displaystyle\int_{0}^{1}\exp\left(-a^{2}u^{2}-\frac{b^{2}}{u^{2}}\right)du$
$\displaystyle=\left(\int_{0}^{\infty}-\int_{1}^{\infty}\right)\exp\left(-a^{2}u^{2}-\frac{b^{2}}{u^{2}}\right)du=\frac{\sqrt{\pi}}{4a}\left[e^{-2ab}(2-\operatorname{erfc}(a-b))-e^{2ab}\operatorname{erfc}(a+b)\right]$
$\displaystyle=\frac{\sqrt{\pi}}{4a}\left[e^{-2ab}\operatorname{erfc}(b-a)-e^{2ab}\operatorname{erfc}(a+b)\right],$
(9)

where $\operatorname{erfc}(-z)=2-\operatorname{erfc}(z)$ (see Ref. [11]) is
used in the final equality.

The integral in Eq. (7) for $\tilde{\mu}_{\ast}\neq 0$ can be calculated by
inserting $a=|\tilde{\mu}_{\ast}|/(\sqrt{2}\sigma_{\ast})$ and
$b=|\ln(x/S_{0})|/(\sqrt{2}\sigma_{\ast})$ in Eq. (9). The factor $e^{\pm
2ab}$ is reduced to

$e^{\pm
2ab}=\exp\left(\pm\frac{|\tilde{\mu}_{\ast}|}{\sigma_{\ast}^{2}}\left|\ln\frac{x}{S_{0}}\right|\right)=\left(\frac{x}{S_{0}}\right)^{\pm\operatorname{sgn}(x-S_{0})|\tilde{\mu}_{\ast}|/\sigma_{\ast}^{2}},$

where $\operatorname{sgn}$ is the signum function defined by

$\operatorname{sgn}(x)=\begin{cases}+1&x>0,\\\ 0&x=0,\\\ -1&x<0.\end{cases}$

Therefore,

$\displaystyle f_{\text{uni}}(x)$
$\displaystyle=\frac{1}{2|\tilde{\mu}_{\ast}|x}\left[\left(\frac{x}{S_{0}}\right)^{(1-\operatorname{sgn}(x-S_{0}))|\tilde{\mu}_{\ast}|/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{|\ln(x/S_{0})|-|\tilde{\mu}_{\ast}|}{\sqrt{2}\sigma_{\ast}}\right)\right.$
$\displaystyle\qquad\left.-\left(\frac{x}{S_{0}}\right)^{(1+\operatorname{sgn}(x-S_{0}))|\tilde{\mu}_{\ast}|/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{|\ln(x/S_{0})|+|\tilde{\mu}_{\ast}|}{\sqrt{2}\sigma_{\ast}}\right)\right].$

When $\tilde{\mu}_{\ast}>0$, the absolute value can be simply removed:
$|\tilde{\mu}_{\ast}|=\tilde{\mu}_{\ast}$. When $\tilde{\mu}_{\ast}<0$,
$|\tilde{\mu}_{\ast}|=-\tilde{\mu}_{\ast}$, and it can be confirmed in a
straightforward manner that the expression of $f_{\text{uni}}(x)$ becomes the
same as in $\tilde{\mu}_{\ast}>0$. Thus, we finally obtain

$\displaystyle f_{\text{uni}}(x)$
$\displaystyle=\frac{1}{2\tilde{\mu}_{\ast}x}\left[\left(\frac{x}{S_{0}}\right)^{(1-\operatorname{sgn}(x-S_{0}))\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{|\ln(x/S_{0})|-\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\right.$
$\displaystyle\qquad\left.-\left(\frac{x}{S_{0}}\right)^{(1+\operatorname{sgn}(x-S_{0}))\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{|\ln(x/S_{0})|+\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\right].$
(10)

The $f_{\text{uni}}(x)$ for $\tilde{\mu}_{\ast}=0$ is written as

$f_{\text{uni}}(x)=\frac{2}{\sqrt{2\pi}\sigma_{\ast}x}\int_{0}^{1}\exp\left(-\frac{(\ln(x/S_{0}))^{2}}{2\sigma_{\ast}^{2}u^{2}}\right)du=\frac{2}{\sqrt{2\pi}\sigma_{\ast}x}\int_{0}^{1}\exp\left(-\frac{b^{2}}{u^{2}}\right)du,$

where $b=\frac{|\ln(x/S_{0})|}{\sqrt{2}\sigma_{\ast}}$, as above. By
introducing a new integration variable $v=|b|/u$ and integrating by parts, we
obtain

$\displaystyle f_{\text{uni}}(x)$
$\displaystyle=\frac{2}{\sqrt{2\pi}\sigma_{\ast}x}\left[-\frac{|b|}{v}e^{-v^{2}}\right]_{|b|}^{\infty}-\frac{2}{\sqrt{2\pi}\sigma_{\ast}x}\int_{|b|}^{\infty}2|b|e^{-v^{2}}dv=\frac{2e^{-b^{2}}}{\sqrt{2\pi}\sigma_{\ast}x}-\frac{\sqrt{2}|b|}{\sigma_{\ast}x}\operatorname{erfc}(|b|)$
$\displaystyle=\frac{2}{\sqrt{2\pi}\sigma_{\ast}x}\exp\left(-\frac{(\ln(x/S_{0}))^{2}}{2\sigma_{\ast}^{2}}\right)-\frac{|\ln(x/S_{0})|}{\sigma_{\ast}^{2}x}\operatorname{erfc}\left(\frac{|\ln(x/S_{0})|}{\sqrt{2}\sigma_{\ast}}\right).$
(11)

The derived PDFs in Eqs. (10) and (11) are complicated compared to the DP
distribution (6).

We derive the asymptotic form of $f_{\text{uni}}(x)$ in the $x\to\infty$ and
$x\to 0+$ limits. Using the asymptotic expansion of the $\operatorname{erfc}$
function [11]

$\operatorname{erfc}(z)\sim\frac{1}{\sqrt{\pi}z}e^{-z^{2}}\quad(z\to\infty)$
(12)

and the relation

$\left(\frac{x}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}}\exp\left(-\frac{(\ln(x/S_{0})+\tilde{\mu}_{\ast})^{2}}{2\sigma_{\ast}^{2}}\right)=\exp\left(-\frac{(\ln(x/S_{0})-\tilde{\mu}_{\ast})^{2}}{2\sigma_{\ast}^{2}}\right),$

we obtain

$f_{\text{uni}}(x)\sim\sqrt{\frac{2}{\pi}}\frac{\sigma_{\ast}}{x(\ln
x)^{2}}\exp\left(-\frac{(\ln(x/S_{0})-\tilde{\mu}_{\ast})^{2}}{2\sigma_{\ast}^{2}}\right)=\frac{2\sigma_{\ast}^{2}}{(\ln
x)^{2}}f_{\text{LN}}(x;\ln
S_{0}+\tilde{\mu}_{\ast},\sigma_{\ast}^{2})\quad(x\to\infty,0+).$

Notably, both limits $x\to\infty$ and $x\to 0+$ have the same asymptotic form.
Owing to the $(\ln x)^{-2}$ factor, $f_{\text{uni}}(x)$ decays slightly faster
than the lognormal PDF.

###  II.1 Shape of $f_{\text{uni}}(x)$ graph

(a) (b)
(c) (d)

Figure 1:  Graphs of $f_{\text{uni}}(x)$. (a) $\tilde{\mu}_{\ast}=1$ and
$\sigma_{\ast}=0.5,1,2$, and $4$. (b) Log-log graph of (a). (c)
$\sigma_{\ast}=1$ and $\tilde{\mu}_{\ast}=-1,0,1$, and $2$. (d) Log-log graph
of (c). The vertical dashed line indicates $x=S_{0}=1$.

The graph of $f_{\text{uni}}(x)$ is shown in Fig. 1. For any
$\tilde{\mu}_{\ast}$ and $\sigma_{\ast}$, $f_{\text{uni}}(x)$ is continuous
for all $x>0$ and is not differentiable at $x=S_{0}$, which corresponds to the
discontinuity point of $\operatorname{sgn}(x-S_{0})$. Panel (a) shows the
graph of fixed $\tilde{\mu}_{\ast}=1$ and varying $\sigma_{\ast}$, and (b)
shows the log-log graph of (a). Panel (c) shows the graph of fixed
$\sigma_{\ast}=1$ and varying $\tilde{\mu}_{\ast}$, and (d) shows the log-log
graph of (c). The decay of $f_{\text{uni}}(x)$ as $x\to\infty$ is slower for
larger $\sigma_{\ast}$ and $\tilde{\mu}_{\ast}$. The graph for
$\tilde{\mu}_{\ast}=1$ and $\sigma_{\ast}=4$ in Fig. 1(a) appears to increase
monotonically as $x\to 0+$; however, this is not true. As $\lim_{x\to
0+}f_{\text{uni}}(x)=0$ for any $\tilde{\mu}_{\ast}$ and $\sigma_{\ast}$, this
graph attains a maximum value for a very small (but positive) $x$.

In Fig. 1, the graph of $f_{\text{uni}}(x)$ has a peak at $x=S_{0}$
for$(\tilde{\mu}_{\ast},\sigma_{\ast})=(1,0.5),(1,1),(0,1)$, and $(2,1)$; the
peak for the remaining graphs are not at $x=S_{0}$. We investigate the reason
for this qualitative change, by deriving the condition for
$(\tilde{\mu}_{\ast},\sigma_{\ast})$ such that $f_{\text{uni}}(x)$ has a peak
at $x=S_{0}$. For sufficiently small $\delta>0$,

$f_{\text{uni}}(S_{0}\pm\delta)\simeq f_{\text{uni}}(S_{0})\pm
f_{\text{uni}}^{\prime}(S_{0}^{\mathord{\pm}})\delta,$

where

$f_{\text{uni}}^{\prime}(S_{0}^{\mathord{+}})=\lim_{h\to
0+}\frac{f_{\text{uni}}(S_{0}+h)-f_{\text{uni}}(S_{0})}{h},\quad
f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})=\lim_{h\to
0+}\frac{f_{\text{uni}}(S_{0})-f_{\text{uni}}(S_{0}-h)}{h}$

are the right and left derivatives of $f_{\text{uni}}(x)$ at $x=S_{0}$,
respectively. We have to consider one-sided derivatives because the function
$f_{\text{uni}}(x)$ is not differentiable at $x=S_{0}$. The condition such
that $f_{\text{uni}}(x)$ has a peak at $x=S_{0}$ is

$f_{\text{uni}}^{\prime}(S_{0}^{\mathord{+}})<0,\quad
f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})>0.$

The differentiation of Eqs. (10) and (11) yields

$f_{\text{uni}}^{\prime}(S_{0}^{\mathord{+}})=\begin{dcases}\frac{1}{\tilde{\mu}_{\ast}S_{0}^{2}}\left[\left(1-\frac{\tilde{\mu}_{\ast}}{\sigma_{\ast}^{2}}\right)\operatorname{erfc}\left(\frac{\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)-1\right]&\tilde{\mu}_{\ast}\neq
0,\\\
-\frac{2}{\sqrt{2\pi}\sigma_{\ast}S_{0}^{2}}&\tilde{\mu}_{\ast}=0,\end{dcases}$

and

$f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})=\begin{dcases}\frac{1}{\tilde{\mu}_{\ast}S_{0}^{2}}\left[\frac{2\tilde{\mu}_{\ast}}{\sigma_{\ast}^{2}}-1+\left(1-\frac{\tilde{\mu}_{\ast}}{\sigma_{\ast}^{2}}\right)\operatorname{erfc}\left(\frac{\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\right]&\tilde{\mu}_{\ast}\neq
0,\\\
\frac{1}{\sigma_{\ast}^{2}S_{0}^{2}}-\frac{2}{\sqrt{2\pi}\sigma_{\ast}S_{0}^{2}}&\tilde{\mu}_{\ast}=0.\end{dcases}$
(13)

First, we prove that $f_{\text{uni}}^{\prime}(S_{0}^{\mathord{+}})<0$ always
holds. This is trivial for $\tilde{\mu}_{\ast}=0$; thus, we examine the
$\tilde{\mu}_{\ast}\neq 0$ case:

$f_{\text{uni}}^{\prime}(S_{0}^{\mathord{+}})=-\frac{1}{S_{0}^{2}\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)-\frac{1}{\tilde{\mu}_{\ast}S_{0}^{2}}\left[1-\operatorname{erfc}\left(\frac{\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\right].$

The first term on the right-hand side is always negative because the
$\operatorname{erfc}$ function always takes a positive value. The factor
$1-\operatorname{erfc}(\tilde{\mu}_{\ast}/(\sqrt{2}\sigma_{\ast}))$ in the
second term becomes positive for positive $\tilde{\mu}_{\ast}$ and becomes
negative for negative $\tilde{\mu}_{\ast}$. Thus,
$1-\operatorname{erfc}(\tilde{\mu}_{\ast}/(\sqrt{2}\sigma_{\ast}))$ always has
the same sign as $\tilde{\mu}_{\ast}$, and the second term is always negative.

In contrast, whether $f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})>0$ depends
on $\tilde{\mu}_{\ast}$ and $\sigma_{\ast}$. Figure 2 shows the
$f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})$ values with $S_{0}=1$ for
$-1\leq\tilde{\mu}_{\ast}\leq 3$ and $0<\sigma_{\ast}\leq 2$. To ensure clear
visualization, we restrict the color bar within the interval $[-6,6]$,
although certain $(\tilde{\mu}_{\ast},\sigma_{\ast})$ yield values outside
this range [e.g., $f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})\approx 49$ for
$(\tilde{\mu}_{\ast},\sigma_{\ast})=(1,0.2)$]. The dashed curve in the figure
represents the contour of $f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})=0$.
Thus, $f_{\text{uni}}(x)$ exhibits a peak at $x=S_{0}$ when
$(\tilde{\mu}_{\ast},\sigma_{\ast})$ is on the right side of this curve. By
using the limit value $\operatorname{erfc}(+\infty)=0$, it can be proven using
Eq. (13) that the curve for $f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})=0$
asymptotically draws the parabola $\tilde{\mu}_{\ast}=\sigma_{\ast}^{2}/2$ for
$\sigma_{\ast}\gg 1$. When $\tilde{\mu}_{\ast}=0$,
$f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})>0$ can be exactly solved to
attain $\sigma_{\ast}<\sqrt{\pi/2}\approx 1.25$. However, the solution for the
$\tilde{\mu}_{\ast}\neq 0$ case will not be obtained exactly, owing to the
$\operatorname{erfc}$ function.

Figure 2:  Diagram of $f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})$ on the
$(\tilde{\mu}_{\ast},\sigma_{\ast})$ plane. The dashed curve represents the
contour of $f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})=0$. The graph of
$f_{\text{uni}}(x)$ yields a peak at $x=S_{0}$ if the point
$(\tilde{\mu}_{\ast},\sigma_{\ast})$ lies on the right side of this curve.

Further analysis shows that $f_{\text{uni}}(x)$ is unimodal for any
$\tilde{\mu}_{\ast}$ and $\sigma_{\ast}$ (further detailes have been presented
in Appendix A). Depending on the $(\tilde{\mu}_{\ast},\sigma_{\ast})$ values,
the peak position of $f_{\text{uni}}(x)$ is either at $x=S_{0}$ (corresponding
to $f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})\geq 0$) or $x<S_{0}$
($f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})<0$). The peak position $x$ for
the latter case is characterized by the solution of the equation
$f_{\text{uni}}^{\prime}(x)=0$. The explicit form of this equation is
presented in Eq. (16) in Appendix A, which is a transcendental equation
involving the $\operatorname{erfc}$ function. Rather than providing the exact
solution of the peak position, we present a numerical result for the peak
position in Fig. 3. The peak tends to be located at small $x$ with decreasing
$\tilde{\mu}_{\ast}$ and increasing $\sigma_{\ast}$.

Figure 3:  Numerical result for the peak position depicted in the
$(\tilde{\mu}_{\ast},\sigma_{\ast})$ plane. The color bar represents the peak
position $x$ divided by $S_{0}$. The dashed curve shows the contour of
$f_{\text{uni}}^{\prime}(S_{0}^{\mathord{-}})=0$, which is shown in Fig. 2. To
the right of this curve, $f_{\text{uni}}(x)$ shows a peak at $x=S_{0}$.

##  III Complementary cumulative distribution

In principle, the CCDF
$F_{\text{uni}}(x)=\int_{x}^{\infty}f_{\text{uni}}(y)dy$ is derived by
integrating the PDF $f_{\text{uni}}(x)$ in Eqs. (10) and (11). However, as the
absolute value in the $\operatorname{erfc}$ function cannot be handled in a
straightforward manner, we first remove the absolute value symbols.

When $x<S_{0}$ and $\tilde{\mu}_{\ast}\neq 0$,

$f_{\text{uni}}(x)=\frac{1}{2\tilde{\mu}_{\ast}x}\left[\operatorname{erfc}\left(\frac{\ln(x/S_{0})-\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)-\left(\frac{x}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{\ln(x/S_{0})+\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)+2\left(\left(\frac{x}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}-1\right)\right],$

where the relation $\operatorname{erfc}(-z)=2-\operatorname{erfc}(z)$ is used.
The difference from the case $x>S_{0}$ is the last term. Therefore, for any
$x>0$,

$f_{\text{uni}}(x)=\frac{1}{2\tilde{\mu}_{\ast}x}\left[\operatorname{erfc}\left(\frac{\ln(x/S_{0})-\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)-\left(\frac{x}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{\ln(x/S_{0})+\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)+(1-\operatorname{sgn}(x-S_{0}))\left(\left(\frac{x}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}-1\right)\right]$

is valid.

For the derivation of $F_{\text{uni}}(x)$, we calculate the following two
integrals as a preliminary step:

$I_{1}=\int_{x}^{\infty}\operatorname{erfc}\left(\frac{\ln(y/S_{0})-\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\frac{dy}{y},\quad
I_{2}=\int_{x}^{\infty}\left(\frac{y}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{\ln(y/S_{0})+\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\frac{dy}{y}.$

By setting $z=(\ln(y/S_{0})-\tilde{\mu}_{\ast})/(\sqrt{2}\sigma_{\ast})$ and
$z_{\mathord{-}}=(\ln(x/S_{0})-\tilde{\mu}_{\ast})/(\sqrt{2}\sigma_{\ast})$,

$\displaystyle I_{1}$
$\displaystyle=\sqrt{2}\sigma_{\ast}\int_{z_{\mathord{-}}}^{\infty}\operatorname{erfc}(z)dz=\left[\sqrt{2}\sigma_{\ast}z\operatorname{erfc}(z)\right]_{z_{\mathord{-}}}^{\infty}+2\sqrt{\frac{2}{\pi}}\sigma_{\ast}\int_{z_{\mathord{-}}}^{\infty}ze^{-z^{2}}dz$
$\displaystyle=-\left(\ln\frac{x}{S_{0}}-\tilde{\mu}_{\ast}\right)\operatorname{erfc}\left(\frac{\ln(x/S_{0})-\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)+\sqrt{\frac{2}{\pi}}\sigma_{\ast}\exp\left(-\frac{(\ln(x/S_{0})-\tilde{\mu}_{\ast})^{2}}{2\sigma_{\ast}^{2}}\right).$

We use integration by parts in the second equality. Similarly, by setting
$z=(\ln(y/S_{0})+\tilde{\mu}_{\ast})/(\sqrt{2}\sigma_{\ast})$ and
$z_{\mathord{+}}=(\ln(x/S_{0})+\tilde{\mu}_{\ast})/(\sqrt{2}\sigma_{\ast})$,

$\displaystyle I_{2}$
$\displaystyle=\sqrt{2}\sigma_{\ast}e^{-2\tilde{\mu}_{\ast}^{2}/\sigma_{\ast}^{2}}\int_{z_{\mathord{+}}}^{\infty}\exp\left(\frac{2\sqrt{2}\tilde{\mu}_{\ast}}{\sigma_{\ast}}z\right)\operatorname{erfc}(z)dz$
$\displaystyle=\left[\frac{\sigma_{\ast}^{2}}{2\tilde{\mu}_{\ast}}\exp\left(-\frac{2\tilde{\mu}_{\ast}^{2}}{\sigma_{\ast}^{2}}+\frac{2\sqrt{2}\tilde{\mu}_{\ast}}{\sigma_{\ast}}z\right)\operatorname{erfc}(z)\right]_{z_{\mathord{+}}}^{\infty}+\frac{\sigma_{\ast}^{2}}{\tilde{\mu}_{\ast}\sqrt{\pi}}e^{-2\tilde{\mu}_{\ast}^{2}/\sigma_{\ast}^{2}}\int_{z_{\mathord{+}}}^{\infty}\exp\left(-z^{2}+\frac{2\sqrt{2}\tilde{\mu}_{\ast}}{\sigma_{\ast}}z\right)dz$
$\displaystyle=-\frac{\sigma_{\ast}^{2}}{2\tilde{\mu}_{\ast}}\left(\frac{x}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{\ln(x/S_{0})+\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)+\frac{\sigma_{\ast}^{2}}{\sqrt{\pi}\tilde{\mu}_{\ast}}\int_{z_{\mathord{+}}-\sqrt{2}\tilde{\mu}_{\ast}/\sigma_{\ast}}^{\infty}e^{-z^{2}}dz$
$\displaystyle=-\frac{\sigma_{\ast}^{2}}{2\tilde{\mu}_{\ast}}\left(\frac{x}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{\ln(x/S_{0})+\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)+\frac{\sigma_{\ast}^{2}}{2\tilde{\mu}_{\ast}}\operatorname{erfc}\left(\frac{\ln(x/S_{0})-\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right).$

The CCDF for $x\geq S_{0}$ becomes

$\displaystyle F_{\text{uni}}(x)$
$\displaystyle=\frac{1}{2\tilde{\mu}_{\ast}}\int_{x}^{\infty}\left[\operatorname{erfc}\left(\frac{\ln(y/S_{0})-\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)-\left(\frac{y}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{\ln(y/S_{0})+\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\right]\frac{dy}{y}$
$\displaystyle=\frac{I_{1}-I_{2}}{2\tilde{\mu}_{\ast}}$
$\displaystyle=\frac{1}{2\tilde{\mu}_{\ast}}\left[\sqrt{\frac{2}{\pi}}\sigma_{\ast}\exp\left(-\frac{(\ln(x/S_{0})-\tilde{\mu}_{\ast})^{2}}{2\sigma_{\ast}^{2}}\right)+\frac{\sigma_{\ast}^{2}}{2\tilde{\mu}_{\ast}}\left(\frac{x}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{\ln(x/S_{0})+\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\right.$
$\displaystyle\qquad\left.-\left(\ln\frac{x}{S_{0}}-\tilde{\mu}_{\ast}+\frac{\sigma_{\ast}^{2}}{2\tilde{\mu}_{\ast}}\right)\operatorname{erfc}\left(\frac{\ln(x/S_{0})-\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\right],$

and for $x<S_{0}$,

$F_{\text{uni}}(x)=\frac{I_{1}-I_{2}}{2\tilde{\mu}_{\ast}}+\frac{1}{\tilde{\mu}_{\ast}}\int_{x}^{S_{0}}\left(\left(\frac{y}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}-1\right)\frac{dy}{y}.$

The remaining integral can be calculated as

$\frac{1}{\tilde{\mu}_{\ast}}\int_{x}^{S_{0}}\left(\left(\frac{y}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}-1\right)\frac{dy}{y}=\frac{1}{\tilde{\mu}_{\ast}}\left(\frac{\sigma_{\ast}^{2}}{2\tilde{\mu}_{\ast}}-\frac{\sigma_{\ast}^{2}}{2\tilde{\mu}_{\ast}}\left(\frac{x}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}+\ln\frac{x}{S_{0}}\right).$

Consequently, we obtain

$\displaystyle F_{\text{uni}}(x)$
$\displaystyle=\frac{1}{2\tilde{\mu}_{\ast}}\left[\sqrt{\frac{2}{\pi}}\sigma_{\ast}\exp\left(-\frac{(\ln(x/S_{0})-\tilde{\mu}_{\ast})^{2}}{2\sigma_{\ast}^{2}}\right)+\frac{\sigma_{\ast}^{2}}{2\tilde{\mu}_{\ast}}\left(\frac{x}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\frac{\ln(x/S_{0})+\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\right.$
$\displaystyle\qquad\left.-\left(\ln\frac{x}{S_{0}}-\tilde{\mu}_{\ast}+\frac{\sigma_{\ast}^{2}}{2\tilde{\mu}_{\ast}}\right)\operatorname{erfc}\left(\frac{\ln(x/S_{0})-\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\right]$
$\displaystyle\quad+\frac{1-\operatorname{sgn}(x-S_{0})}{2\tilde{\mu}_{\ast}}\left(\frac{\sigma_{\ast}^{2}}{2\tilde{\mu}_{\ast}}-\frac{\sigma_{\ast}^{2}}{2\tilde{\mu}_{\ast}}\left(\frac{x}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}+\ln\frac{x}{S_{0}}\right),$
(14)

which is valid for all $x>0$. This result is extremely complex, but it is
exact.

By integrating Eq. (11) or taking the $\tilde{\mu}_{\ast}\to 0$ limit in Eq.
(14), $F_{\text{uni}}(x)$ for $\tilde{\mu}_{\ast}=0$ is obtained as

$F_{\text{uni}}(x)=\frac{1}{2}\operatorname{erfc}\left(\frac{\ln(x/S_{0})}{\sqrt{2}\sigma_{\ast}}\right)-\frac{1}{2\sqrt{2\pi}\sigma_{\ast}}\exp\left(-\frac{(\ln(x/S_{0}))^{2}}{2\sigma_{\ast}^{2}}\right)+\frac{1}{2\sigma_{\ast}^{2}}\left(\ln\frac{x}{S_{0}}\right)^{2}\operatorname{erfc}\left(\frac{\ln(x/S_{0})}{\sqrt{2}\sigma_{\ast}}\right).$

(a) (b)

Figure 4:  Log-log graphs of $F_{\text{uni}}(x)$ for $\tilde{\mu}_{\ast}=1$
and $\sigma_{\ast}=0.5,1,2$, and $4$ (a), and for $\sigma_{\ast}=1$ and
$\tilde{\mu}_{\ast}=-1,0,1$, and $2$ (b). The vertical dashed line indicates
$x=S_{0}=1$.

Figure 4 shows the graph of $F_{\text{uni}}(x)$ on a log-log scale. The graph
decays slower for larger $\sigma_{\ast}$ [in (a)] and larger
$\tilde{\mu}_{\ast}$ [in (b)]. The graphs do not exhibit significant variety
in shape as $f_{\text{uni}}(x)$ in Fig. 2, and the overall shapes are similar
for different $\tilde{\mu}_{\ast}$ and $\sigma_{\ast}$,

##  IV Calculation of moments

The $k$th moment of $S_{T}$ can be calculated as

$\displaystyle E[S_{T}^{k}]$
$\displaystyle=\int_{0}^{\infty}x^{k}f_{\text{uni}}(x)dx$
$\displaystyle=\int_{0}^{\infty}x^{k}\frac{1}{T_{\text{max}}}\int_{0}^{T_{\text{max}}}f_{\text{LN}}(x;\ln
S_{0}+\tilde{\mu}t,\sigma^{2}t)dtdx$
$\displaystyle=\frac{1}{T_{\text{max}}}\int_{0}^{T_{\text{max}}}\int_{0}^{\infty}x^{k}f_{\text{LN}}(x;\ln
S_{0}+\tilde{\mu}t,\sigma^{2}t)dxdt$
$\displaystyle=\frac{S_{0}^{k}}{T_{\text{max}}}\int_{0}^{T_{\text{max}}}\exp\left(k\tilde{\mu}t+\frac{k^{2}}{2}\sigma^{2}t\right)dt$
$\displaystyle=\frac{2}{k^{2}\sigma_{\ast}^{2}+2k\tilde{\mu}_{\ast}}\left[\exp\left(\frac{k^{2}\sigma_{\ast}^{2}}{2}+k\tilde{\mu}_{\ast}\right)-1\right]S_{0}^{k}.$

The moment of the lognormal distribution [Eq. (2)] is used in the fourth
equality. In contrast to the DP distribution, $E[S_{T}^{k}]$ is not divergent
for any $k$.

Specifically, the mean and variance of $S_{T}$ for $\mu_{\ast}\neq 0$ and
$\mu_{\ast}\neq-\sigma_{\ast}^{2}$ becomes

$E[S_{T}]=\frac{e^{\mu_{\ast}}-1}{\mu_{\ast}}S_{0},\quad
V[S_{T}]=\frac{e^{\sigma_{\ast}^{2}+\mu_{\ast}}-1}{\sigma_{\ast}^{2}+\mu_{\ast}}S_{0}^{2}-\frac{(e^{\mu_{\ast}}-1)^{2}}{\mu_{\ast}^{2}}S_{0}^{2}.$

We obtain $E[S_{T}]=S_{0}$ when $\mu_{\ast}=0$ and

$V[S_{T}]=S_{0}^{2}-\frac{(e^{\mu_{\ast}}-1)^{2}}{\mu_{\ast}^{2}}S_{0}^{2}$

when $\mu_{\ast}=-\sigma_{\ast}^{2}$.

Figure 5 shows the semi-log graphs of $E[S_{T}^{k}]$ for $S_{0}=1$ with
different $\tilde{\mu}_{\ast}$ and $\sigma_{\ast}$. $E[S_{T}^{k}]$ becomes
large for large $\tilde{\mu}_{\ast}$ and $\sigma_{\ast}$. The graph for
$\tilde{\mu}_{\ast}=-1$ and $\sigma_{\ast}=1$ in Fig. 5(b) is not
monotonically increasing; $E[S_{T}^{1}]\approx 0.787<1=E[S_{T}^{0}]$. We can
easily prove that

$\frac{2}{k^{2}\sigma_{\ast}^{2}+2k\tilde{\mu}_{\ast}}\left[\exp\left(\frac{k^{2}\sigma_{\ast}^{2}}{2}+k\tilde{\mu}_{\ast}\right)-1\right]<1$

when

$\frac{k^{2}\sigma_{\ast}^{2}}{2}+k\tilde{\mu}_{\ast}<0.$

Therefore, this phenomenon can occur for the $\tilde{\mu}_{\ast}<0$ case.

(a) (b)

Figure 5:  The $k$th moment of $S_{T}$ ($S_{0}=1$) up to $k=5$ for
$\tilde{\mu}_{\ast}=1$ and $\sigma_{\ast}=0.5,1,2$, and $4$ (a), and for
$\sigma_{\ast}=1$ and $\tilde{\mu}_{\ast}=-1,0,1$, and $2$ (b).

##  V Comparison to discrete-time multiplicative process

We investigate whether the above results are valid for the discrete-time
multiplicative stochastic process (3).

If $X_{0}$ is a constant value, $\ln X_{n}=\ln M_{1}+\cdots+\ln M_{n}+\ln
X_{0}$ approximately follows the normal distribution with mean $\mu n$ and
variance $\sigma^{2}n$, where $\mu=E[\ln M_{i}]$ and $\sigma^{2}=V[\ln
M_{i}]$. Therefore, $X_{n}$ approximately follows the lognormal distribution
with $f_{\text{LN}}(x;\ln X_{0}+\mu n,\sigma^{2}n)$. This lognormality is
similar to $S_{T}$ for the geometric Brownian motion having
$f_{\text{LN}}(x;\ln S_{0}+\tilde{\mu}T,\sigma^{2}T)$.

If the observation time $n$ is drawn from the discrete uniform distribution on
$\\{1,2,\ldots,N\\}$, the PDF and CCDF of $X_{n}$ are expected to become
approximately similar to $f_{\text{uni}}(x)$ and $F_{\text{uni}}(x)$ derived
in Sections II and III, respectively. We compare the numerically generated
CCDF and $F_{\text{uni}}(x)$ using two examples for the distribution of
$M_{i}$.

The first example is the case wherein $M_{i}$ is uniformly distributed over
the interval $[a,b]$ ($a>0$). The mean $\mu$ and variance $\sigma^{2}$ of $\ln
M_{i}$ become

$\mu=\int_{a}^{b}\frac{1}{b-a}\ln ydy=\frac{b\ln b-a\ln
a}{b-a}-1,\quad\sigma^{2}=\int_{a}^{b}\frac{1}{b-a}(\ln
y-\mu)^{2}dy=1-ab\left(\frac{\ln b-\ln a}{b-a}\right)^{2}.$

In the numerical calculation, we set $a=1/2$ and $b=3/2$ so that
$\mu\approx-0.0452$ and $\sigma^{2}\approx 0.0948$. The CCDFs for $N=2,5,10$,
and $20$ are shown in Fig. 6(a) with points, calculated from $10^{4}$
independent samples each. The solid curves represent $F_{\text{uni}}(x)$ in
Eq. (14) by using $\tilde{\mu}_{\ast}=\mu N$ and
$\sigma_{\ast}^{2}=\sigma^{2}N$.

The second example is where $M_{i}$ follows the power-law distribution. The
mean $\mu$ and variance $\sigma^{2}$ of $\ln M_{i}$ can be computed to be

$\mu=\int_{m_{0}}^{\infty}\frac{\nu-1}{m_{0}}\left(\frac{y}{m_{0}}\right)^{-\nu}\ln
ydy=\frac{1}{\nu-1}+\ln
m_{0},\quad\sigma^{2}=\int_{m_{0}}^{\infty}\frac{\nu-1}{m_{0}}\left(\frac{y}{m_{0}}\right)^{-\nu}(\ln
y-\mu)^{2}dy=\frac{1}{(\nu-1)^{2}},$

respectively, where the power-law PDF with exponent $\nu(>1)$ and lower bound
$m_{0}(>0)$ is expressed as

$\frac{\nu-1}{m_{0}}\left(\frac{y}{m_{0}}\right)^{-\nu}.$

In the numerical calculation, we used $\nu=3$ and $m_{0}=1$, so that $\mu=1/2$
and $\sigma^{2}=1/4$. Figure 6(b) shows the numerical result for $N=2,5,10$,
and $20$ with points and $F_{\text{uni}}(x)$ with curves.

(a) (b)

Figure 6:  Comparison between numerically calculated CCDF of $X_{n}$ for
$N=2,5,10$, and $20$ (points) and the corresponding $F_{\text{uni}}(x)$
(curves). The random variable $M_{i}$ is drawn from the uniform distribution
on the interval$[1/2,3/2]$ in (a) and the power-law distribution with exponent
$\nu=3$ and lower bound $m_{0}=1$ in (b).

A common property in these two examples is that the CCDF of $X_{n}$ gradually
approaches $F_{\text{uni}}(x)$ with increase in $N$. The deviation from
$F_{\text{uni}}(x)$ in small $N$ is attributed to the fact that the lognormal
approximation based on the central limit theorem fails. In other words, the
distribution of $X_{n}$ for small $N$ is dependent on the individuality of the
distribution of $M_{i}$. For example, the numerical CCDF of $X_{n}$ for the
uniform $M_{i}$ decays faster than $F_{\text{uni}}(x)$, whereas that for the
power-law $M_{i}$ decays slower than $F_{\text{uni}}(x)$. This observation is
consistent with the property that the lognormal distribution has a tail
heavier than the uniform distribution and the power-law distribution has an
even further heavier tail.

##  VI Discussion

This study examined the geometric Brownian motion $S_{T}$ under uniformly
distributed observation time $T$. This model is a deformation of the DP
distribution in which the observation time follows an exponential
distribution. The PDF $f_{\text{uni}}(x)$ and CCDF $F_{\text{uni}}(x)$ are
exactly calculated, although they have nontrivial complicated forms compared
to those of the DP distribution. Consequently, the power-law form of the DP
distribution is realized owing to the balance of the geometric Brownian motion
and the exponential distribution of $T$.

By changing the observation time distribution, the DP distribution can be
generalized. However, it remains unclear whether the exact expressions of the
PDF and CCDF can be obtained in such a generalized case. In particular, the
difficulty will arise in the computation of the (indefinite) integral, which
involves $\exp(-a^{2}u^{2}-b^{2}/u^{2})$ factor. Therefore, observation time
distributions that provide exact PDF and CCDF of $S_{T}$ will be highly
limited.

In this study, we assume that the initial value of the geometric Brownian
motion, $S_{0}$, is a constant, which is common to the DP distribution [26]. A
possible extension of this study involves changing $S_{0}$ to a random
variable. When $S_{0}$ is distributed lognormally and the observation time $T$
is exponentially distributed, the distribution of $S_{T}$ is exactly
calculated and is referred to as the double Pareto-lognormal distribution [25,
29]. In the context of the double Pareto-lognormal distribution, an
interesting challenge associated with this study is the derivation of the
distribution of $S_{T}$ with uniform $T$ when the constant $S_{0}$ in this
study is replaced with a lognormal random variable.

In Section V, the applicability of $F_{\text{uni}}(x)$ to discrete-time
stochastic process (3) is examined. $F_{\text{uni}}(x)$ is expected to provide
a reasonable estimate when the maximum number $N$ of steps is large, whereas
the distribution of $X_{n}$ is strongly dependent on the property of the
random variable $M_{i}$ for small $N$. A further systematic and quantitative
study is required to establish the connection between continuous- and
discrete-time processes.

From a practical viewpoint, the exploration of empirical datasets that exhibit
the proposed distribution is an important future research direction. Because
the double Pareto distribution has been observed in various fields related to
human activities and social phenomena [27, 28], we believe that the proposed
distribution, which can be regarded as a modification of the DP distribution,
is useful in the analysis of empirical data. In future studies, the practical
importance of this study will be tested by its application to data analysis.

##  VII Conclusion

The double Pareto (DP) distribution, a double-sided power-law distribution,
can be obtained by geometric Brownian motion with a constant initial value and
exponentially distributed observation time. This study investigates a
deformation of the DP distribution by replacing the exponential distribution
for the observation time with a continuous uniform distribution.

The exact forms of the probability density function (PDF) and complementary
cumulative distribution function (CCDF) are derived using the error function
[see Eqs. (10) and (14)]. Furthermore, the detailed shape of the PDF, e.g.,
the asymptotic form, unimodality, and peak position, is analyzed. The main
finding of this study is the establishment of a parametric distribution that
is an extension of the lognormal and power-law distributions.

##  Appendix A Unimodality of $f_{\text{uni}}(x)$

Here we show that the PDF $f_{\text{uni}}(x)$ given in Eqs. (10) and (11) is
unimodal. That is, $f_{\text{uni}}(x)$ has exactly one peak, at $x=S_{0}$ or
$0<x<S_{0}$.

Before the main proof of the unimodality, we prove that the function

$Q(w)=\ln[e^{w^{2}}\operatorname{erfc}(w)]$

is strictly convex. The second derivative of $Q(w)$ becomes

$Q^{\prime\prime}(w)=\frac{2e^{-w^{2}}}{\operatorname{erfc}(w)^{2}}\left[e^{w^{2}}\operatorname{erfc}(w)^{2}+\frac{2}{\sqrt{\pi}}w\operatorname{erfc}(w)-\frac{2}{\pi}e^{-w^{2}}\right]\eqqcolon\frac{2e^{-w^{2}}}{\operatorname{erfc}(w)^{2}}q(w).$

Notably, the key to a simple proof is to separate $e^{-w^{2}}$ in the
numerator. We analyze the function $q(w)$. The derivative of $q(w)$ becomes

$q^{\prime}(w)=2e^{w^{2}}\operatorname{erfc}(w)\left[w\operatorname{erfc}(w)-\frac{e^{-w^{2}}}{\sqrt{\pi}}\right]=-2e^{w^{2}}\operatorname{erfc}(w)\int_{w}^{\infty}\operatorname{erfc}(u)du,$

where

$\int_{w}^{\infty}\operatorname{erfc}(u)du=-w\operatorname{erfc}(w)+\frac{e^{-w^{2}}}{\sqrt{\pi}}$

can be derived by integration by parts. The $\operatorname{erfc}$ function
always takes a positive value, and its integral always becomes positive.
Therefore, $q^{\prime}(w)<0$, i.e., $q(w)$ is a decreasing function. Using the
asymptotic expansion (12) of the $\operatorname{erfc}$ function, we obtain

$\lim_{w\to\infty}q(w)=0.$

Consequently, $q(w)$ takes a positive value for any $w$. Thus,
$Q^{\prime\prime}(w)>0$, i.e., the strict convexity of $Q(w)$, is proven. For
reference, Fig. 7 shows graphs of $Q(w)$, $Q^{\prime}(w)$, and
$Q^{\prime\prime}(w)$.

(a) (b) (c)

Figure 7:  Graphs of $Q(w)$, $Q^{\prime}(w)$, and $Q^{\prime\prime}(w)$ in
$-5\leq w\leq 5$. (a) The function $Q(w)$ is strictly convex. (b) The
derivative $Q^{\prime}(w)$ is increasing and negative-valued. (c) The second
derivative $Q^{\prime\prime}(w)$ is positive-valued.

If $f_{\text{uni}}(x)$ has a peak at $x\neq S_{0}$, this peak position is
characterized by the equation $f_{\text{uni}}^{\prime}(x)=0$. First, we prove
the unimodality for the $\tilde{\mu}_{\ast}=0$ case. By differentiating Eq.
(11), the equation $f_{\text{uni}}^{\prime}(x)=0$ becomes

$\left(1-\ln\frac{x}{S_{0}}\right)\operatorname{erfc}\left(\pm\frac{\ln(x/S_{0})}{\sqrt{2}\sigma_{\ast}}\right)=\mp\sqrt{\frac{2}{\pi}}\sigma_{\ast}\exp\left(-\frac{(\ln(x/S_{0}))^{2}}{2\sigma_{\ast}^{2}}\right),$

where the upper and lower signs refer to the cases of $x>S_{0}$ and $x<S_{0}$,
respectively. Taking the logarithm of this equation for $x<S_{0}$ and
introducing $w=-\ln(x/S_{0})/(\sqrt{2}\sigma_{\ast})$, we obtain

$Q^{\prime}(w)=-\frac{\sqrt{2}}{\sigma_{\ast}},$ (15)

where

$Q^{\prime}(w)=-\frac{2e^{-w^{2}}}{\sqrt{\pi}\operatorname{erfc}(w)}+2w.$

Owing to the strict convexity of $Q(w)$, the derivative $Q^{\prime}(w)$ is an
increasing function. Moreover, $Q^{\prime}(w)$ satisfies

$\lim_{w\to-\infty}Q^{\prime}(w)=-\infty,\quad\lim_{w\to\infty}Q^{\prime}(w)=0,$

where the asymptotic expansion (12) is required to compute
$Q^{\prime}(\infty)$. Therefore, Eq. (15) has a unique solution $w$ for any
$\sigma_{\ast}>0$. If this solution is $w>0$, the peak position of
$f_{\text{uni}}(x)$ is $x=S_{0}\exp(-\sqrt{2}\sigma_{\ast}w)<S_{0}$, and
otherwise, $f_{\text{uni}}(x)$ is monotonically increasing in $0<x<S_{0}$ and
the peak is at $x=S_{0}$. The solution $w$ is positive if and only if

$-\frac{\sqrt{2}}{\sigma_{\ast}}<Q^{\prime}(0)=-\frac{2}{\sqrt{\pi}},$

which is equivalent to

$\sigma_{\ast}>\sqrt{\frac{\pi}{2}}.$

This threshold value $\sqrt{\pi/2}$ can be obtained as the solution of
$f_{\text{uni}}^{\prime}(S_{0}^{-})=0$ stated in Section II.1.

The equation for $x>S_{0}$ becomes

$Q^{\prime}(w)=\frac{\sqrt{2}}{\sigma_{\ast}},$

with $w=\ln(x/S_{0})/(\sqrt{2}\sigma_{\ast})$. This equation does not have
solutions because $Q^{\prime}(w)$ is always negative, as stated above;
$f_{\text{uni}}(x)$ does not have a peak in $x>S_{0}$. Thus, the unimodality
of $f_{\text{uni}}(x)$ for $\tilde{\mu}_{\ast}=0$ is proven.

Next, we show the unimodality of $f_{\text{uni}}(x)$ for
$\tilde{\mu}_{\ast}\neq 0$. The equation $f_{\text{uni}}^{\prime}(x)=0$
becomes

$\operatorname{erfc}\left(\pm\frac{\ln(x/S_{0})-\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)=\left(1-\frac{2\tilde{\mu}_{\ast}}{\sigma_{\ast}^{2}}\right)\left(\frac{x}{S_{0}}\right)^{2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}}\operatorname{erfc}\left(\pm\frac{\ln(x/S_{0})+\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right),$
(16)

where $\pm$ indicates “$+$” for $x>S_{0}$ and “$-$” for $x<S_{0}$, as in the
$\tilde{\mu}_{\ast}=0$ case above. Since the $\operatorname{erfc}$ function is
positive-valued, the existence of a solution $x$ requires
$1-2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}>0$. Otherwise, if
$1-2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}\leq 0$, Eq. (16) does not have
solutions, which means that the peak of $f_{\text{uni}}(x)$ is $x=S_{0}$.
Taking the logarithm of Eq. (16) and introducing
$w=\pm\ln(x/S_{0})/(\sqrt{2}\sigma_{\ast})$, we obtain

$Q\left(w+\frac{\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)-Q\left(w-\frac{\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)=\begin{dcases}-\ln\left(1-\frac{2\tilde{\mu}_{\ast}}{\sigma_{\ast}^{2}}\right)&x>S_{0},\\\
\ln\left(1-\frac{2\tilde{\mu}_{\ast}}{\sigma_{\ast}^{2}}\right)&x<S_{0}.\end{dcases}$
(17)

The condition $1-2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2}>0$ ensures that the
term $\ln(1-2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2})$ takes a real value.

Let us focus on the $\tilde{\mu}_{\ast}>0$ case. According to the theory of
convex functions [30], for any strictly convex function $\psi$ and constant
$a>0$, the function $\psi(w+a)-\psi(w-a)$ is an increasing function of $w$.
Hence, the left-hand side of Eq. (17) is an increasing function of $w$, and
satisfies

$\lim_{w\to-\infty}\left[Q\left(w+\frac{\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)-Q\left(w-\frac{\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\right]=-\infty,\quad\lim_{w\to\infty}\left[Q\left(w+\frac{\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)-Q\left(w-\frac{\tilde{\mu}_{\ast}}{\sqrt{2}\sigma_{\ast}}\right)\right]=0.$

That is, the left-hand side of Eq. (17) is always negative and can take any
negative value by tuning $w$. Noting that
$\ln(1-2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2})<0$ for $\tilde{\mu}_{\ast}>0$,
we conclude that Eq. (17) for $x>S_{0}$ does not have solution and for
$x<S_{0}$ has a unique solution. As stated in the $\tilde{\mu}_{\ast}=0$ case,
the solution $w<0$ corresponds to $x>S_{0}$, and the peak of
$f_{\text{uni}}(x)$ is at $x=S_{0}$.

For the $\tilde{\mu}_{\ast}<0$ case, the left-hand side of Eq. (17) becomes
positive-valued decreasing function and
$\ln(1-2\tilde{\mu}_{\ast}/\sigma_{\ast}^{2})>0$. Therefore, Eq. (17) for
$x>S_{0}$ does not have solution and for $x<S_{0}$ has a unique solution,
which indicates that the $f_{\text{uni}}(x)$ is unimodal.

## Acknowledgments

The authors are grateful to referees for providing details regarding the
double Pareto-lognormal distribution. One of the authors (K.Y.) was supported
by a Grant-in-Aid for Scientific Research (C) 19K03656 and 23K03264 from Japan
Society for the Promotion of Science.

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