On Distribution of Rice–Middleton Model

Abstract

The probability density function in Rice–Middleton model, which describes the behavior of the single sinusoidal random signal combined with Gaussian noise is expressed in three mutually independent ways: firstly, with the aid of an integral representation of the modified Bessel function of the first kind of integer order; secondly, by a hyperbolic cosine differential operator and thirdly, applying the Grünwald–Letnikov fractional derivative. The cumulative distribution functions are also described in all these cases, and also using the Nuttall Q–function. An associated, seemingly new, probability distribution is introduced which cumulative distribution function and the raw moments of general real order are obtained whilst the characteristic function’s power series form is inferred. The exposition ends with a discussion in which by–product summations are given for the considered Neumann series of the second type built by modified Bessel functions of the second kind having integer order.

1 Introduction and Motivation

In 1944 Rice [24, Sect. 3.10] studied the probability distribution of the signal which consists of the Gaussian noise plus sine wave. According to Rice, if we have a steady sinusoidal current

$$\begin{aligned} v_a = v_a(t) = A\cos (\omega _a t-\phi _a) \end{aligned}$$

(1)

where t is time, $A, \omega _a \in (0, \pi )$ are the amplitude and frequency, while $\phi _a$ stands for the phase angle, and if we pick times $t_1,\,t_2,\,\dots$ at random in (1) it is the same, statistically, as holding t constant and picking the phase angles $\phi _a$ at random from the interval $(0,2\pi )$. Rice also stated that if $v_a$ is regarded as a random variable defined by a random variable $\phi _a$, its characteristic function is $\varphi (z) = J_0(Az)$, where $J_0$ is the Bessel function of the first kind of zeroth order. Then, supposing that we have a noise current $v_N$ plus a sine wave we have a representation

$$\begin{aligned} v(t) = v = v_a+v_N = A\cos (\omega _a t-\psi _a)+\sum _{n=1}^M c_n\cos (\omega _n t-\psi _n), \end{aligned}$$

where $c_n^2 = 2 \omega (f_n) \Delta f$, $\omega _n = 2\pi f_n$, $f_n=n\Delta f$, $f_{k-n}=(k-n)\Delta f=f_k-f_n$ and $\phi _1,\,\phi _2,\dots ,\,\phi _M$ are random phase angles (see [24, pp. 125-126]). Further, if we note v at the random times $t_1,\,t_2,\dots$, since $v_p$ and $v_N$ may be regarded as independent random variables and since the characteristic functions for the sum of two such variables is the product of their characteristic functions, the characteristic function of v is

$$\begin{aligned} \varphi _v(z) = J_0(Az) \exp \left\{ \frac{-\sigma ^2z^2}{2}\right\} , \end{aligned}$$

having in mind that $\sigma ^2$ and $\exp \left\{ -\sigma ^2z^2/2\right\}$ are the variance and the characteristic function of the Gaussian noise (also consult Ho et al. [9]).

In 1960 Middleton [15] also studied such random variable and derived the appropriate probability density function in the form [9, p. 73, Eq. (4)], [15, p 422, Eq. (9.67a)]

$$\begin{aligned} p(x) = \dfrac{1}{\sqrt{2\pi }\sigma }\sum _{k\ge 0}\dfrac{(-x^2/2\sigma ^2)^k}{k!}{}_1F_1 \left[ \begin{array}{c} k+\frac{1}{2} \\ 1 \end{array} \Big | -\frac{A^2}{2\sigma ^2}\right] . \end{aligned}$$

Quite recently, Trifonov [29, 30] studied the distribution of the time dependent random variable (rv)

$$\begin{aligned} v(t) = A \cos (\omega t + \phi ) + u(t), \end{aligned}$$

(2)

where t is time, $A, \omega \in (0, \pi )$, $\phi \in (0, 2\pi )$, are the amplitude, frequency, and phase of the real–valued sinusoid, respectively, and u(t) is the measurement noise that follows a Gaussian distribution with zero mean and variance $\sigma ^2$. The probability density functions (PDF) of u(t) and $s(t) = A \cos (\omega t + \phi )$ are given by

$$\begin{aligned} p_u(x)=\dfrac{1}{\sigma \sqrt{2\pi }}{\textrm{e}}^{-\frac{x^2}{2\sigma ^2}},\qquad p_s(x)=\dfrac{1}{\pi \sqrt{A^2-x^2}}, \end{aligned}$$

respectively. The PDF p(x) of the signal v(t) in (2) is the convolution $p_s *p_u$ of $p_s$ and $p_u$, and equals [29, p. 984, Eq. (1)]

$$\begin{aligned} p(x) = \dfrac{1}{\sigma {\sqrt{2 \pi }}} \textrm{e}^{-\frac{x^2}{2\sigma ^2}} \sum _{n \ge 0} (-1)^n\beta _n I_{2n}\left( \sqrt{2\psi }\frac{x}{\sigma }\right) , \qquad x\in \mathbb {R}. \end{aligned}$$

(3)

Here, and in what follows, $\psi = \tfrac{1}{2} \, A^2 \sigma ^{-2}$ and it presents the signal–to–noise ratio, whilst

$$\begin{aligned} \beta _n=k_n\textrm{e}^{-\frac{\psi }{2}}\,I_n\left( \frac{\psi }{2}\right) >0,\qquad \sum _{n \ge 0}\beta _n=1, \end{aligned}$$

more precisely, a one–parameter representation (3) stands for the PDF of the instantaneous value of a single sinusoidal signal additively combined with Gaussian noise [29, p. 984]. Also, $I_\mu$ stands for the modified Bessel function of the first kind, of order $\mu$, which has the power series definition [3, p. 13]

$$\begin{aligned} I_\mu (z) = \sum _{n \ge 0} \dfrac{1}{\Gamma (\mu +n+1)\, n!}\, \left( \dfrac{z}{2}\right) ^{2n+\mu }, \qquad \Re (\mu )>-1,\, z \in \mathbb {C}, \end{aligned}$$

and $k_n$ is the coefficient defined by

$$\begin{aligned} k_n = {\left\{ \begin{array}{ll} 1, &{} n = 0 \\ 2, &{} n \ge 1 \end{array}\right. }. \end{aligned}$$

Considering these coefficients the PDF (3) can be rewritten in the form

$$\begin{aligned} p(x)&= \dfrac{\textrm{e}^{-\frac{x^2}{2\sigma ^2}-\frac{\psi }{2}}}{\sigma {\sqrt{2\pi }}}\, \left\{ I_0\left( \dfrac{\psi }{2}\right) I_0\left( \sqrt{2\psi }\frac{x}{\sigma }\right) \right. \nonumber \\&\qquad \left. + 2\sum _{n \ge 1}(-1)^nI_{n}\left( \dfrac{\psi }{2}\right) I_{2n}\left( \sqrt{2\psi }\frac{x}{\sigma }\right) \right\} . \end{aligned}$$

(4)

The analysis of the amplitude distribution of a single real sinusoid in noise has a long history, but an appropriate theoretical amplitude distribution has been given in the literature to the case of additive white Gaussian noise; an analytical expression for the appropriate PDF was considered by Rice [25], for the first time, on the other hand Middleton [15] and Ho et al. [9] derived a new formula for such PDF, but in terms of the confluent hypergeometric function ${}_1F_1$ and the Hermite polynomials (see Discussion section ${\textsf{A}}$, (29) for the latter).

Trifonov presented a new representation for that PDF in terms of a series of the exponentially scaled modified Bessel functions of the first kind $I_\mu$, of integer order $\mu$, given above in (3), pointing out that the advantage of such representation formula is that it converges much faster than the series consisting of Hermite polynomials or the confluent hypergeometric functions.

The problem of summing up the Neumann series of modified Bessel functions of the first kind

$$\begin{aligned} {\mathfrak {N}}_{\nu }(z) = \sum _{n \ge 0} \alpha _n\, I_{\nu +n}(z), \qquad z\in \mathbb {C}, \end{aligned}$$

where $\nu ,\alpha _n$ are constants and $I_\nu$ stands for the modified Bessel function of the first kind of order $\nu$, was considered in a number of publications in the mathematical literature; see e.g. [1, 6, 21]. Also, the Nist project [19], Wolfram formula collection, Hansen’s classical sums collection [8] and the classical book by Prudnikov et al. [21] contain an exhaustive list of summations for alike series. As a natural extension there is the second type Neumann series [2, 3, Chap. 2, pp. 27 et seq.]

$$\begin{aligned} {\mathscr {N}}_{a, b}^{\rho , \mu }(x, y) := \sum _{n \ge 0} \alpha _n I_{\rho n+a}(x)I_{\mu n+b}(y), \qquad a, b, \rho , \mu \in \mathbb {R}\, , \end{aligned}$$

(5)

whose terms consist from a product of (at least) two Bessel or modified Bessel functions. The second type Neumann series of products of different kind Bessel and/or modified Bessel functions terms are also thoroughly considered in the recent monograph [3] in which methods are developed for their summation. On the other hand, in the case when the arguments of modified Bessel functions in (5) are not the same, which is the case in (4), the published list of the summation formulae is very short, see [10, 11] and the adequate references therein.

Here, in (4) we are faced with the subclass of second type Neumann series built by modified Bessel functions of the first kind of specific form. In the recent investigation we develop several new methods for summing up the auxiliary Neumann series appearing in (4); these are (i) the integral representation of the modified Bessel functions in the summands, and (ii) the approach using the Grünwald–Letnikov fractional derivative. The results consist of integral forms of the initial PDF p(x) and the related cumulative distribution function (CDF) P(x) in both cases. Next, a seemingly new probability distribution function $p_\xi (s)$, related to a random variable $\xi$ is introduced in the Sect. 3. The explicit expression of the raw moment of general real order and the power series expansion of the characteristic function for the newly considered random variable $\xi$ are given there. The Sect. 4 presents another form of the initial input cumulative distribution function P(x) in terms of the Nuttall Q–function. Few by–products of the inferred results are also listed in the Discussion section, drawing the interested reader’s attention e.g. to the summation of Neumann series of Hermite polynomials (29).

2 New Integral Representations of the PDF (3)

Let us recall the integral definitions of the error function $\textrm{erf}(z)$ and the complementary error function $\textrm{erfc}(z)$ read [16, p. 3, Eqs. 1, 2]

$$\begin{aligned} \textrm{erf}(z) = \frac{2}{\sqrt{\pi }} \int _0^z \textrm{e}^{-x^2}\, \textrm{d}x, \qquad \textrm{erfc}(z) = \frac{2}{\sqrt{\pi }} \int _z^\infty \textrm{e}^{-x^2}\, \textrm{d}x, \qquad z \in {\mathbb {R}}, \end{aligned}$$

respectively, where $\textrm{erf}(x)$ is odd function. Their inter–connection formula is

$$\begin{aligned} \textrm{erf}(z) + \textrm{erfc}(z) = 1. \end{aligned}$$

Theorem 1

For all $x\in \mathbb {R}$ and $\sigma , \psi >0$ there holds

$$\begin{aligned} p(x) = \dfrac{\textrm{e}^{-\frac{x^2}{2\sigma ^2}-\frac{\psi }{2}}}{\sigma (2\pi )^{3/2}}\, \int _{-\pi }^\pi \textrm{e}^{-\sqrt{2\psi }\, \frac{x}{\sigma } \sin t+\frac{\psi }{2} \cos (2 t)}\,\textrm{d}t. \end{aligned}$$

Moreover, the related cumulative distribution function reads

$$\begin{aligned} P(x) = \frac{1}{2} + \dfrac{1}{4 \pi }\, \int _{-\pi }^\pi {\textrm{erf}}\left( \dfrac{x}{\sigma \sqrt{2}} + \sqrt{\psi }\sin t\right) \,\textrm{d}t\,. \end{aligned}$$

(6)

Proof

Consider the Neumann series in (4). Applying the unnecessarily complicated formula [27, p. 1713, Eq. (17)] in the simplified form:

$$\begin{aligned} I_n(\alpha ) = \dfrac{1}{2\pi } \int _{-\pi }^\pi \textrm{e}^{-\alpha \sin t}\,\cos \left( \left( t+\frac{\pi }{2}\right) n\right) \,\textrm{d}t, \end{aligned}$$

it follows that

$$\begin{aligned} I_{2n} \left( \sqrt{2\psi }\frac{x}{\sigma }\right) = \dfrac{(-1)^n}{2\pi } \int _{-\pi }^{\pi } \textrm{e}^{-\sqrt{2\psi }\frac{x}{\sigma }\sin t}\, \cos (2n t)\,\textrm{d}t, \end{aligned}$$

(7)

which yields at $\alpha = \psi /2$ the following form for the Neumann series in (4):

$$\begin{aligned} {\mathscr {N}}_{\sigma ,\psi }(x)&= \sum _{n \ge 1}(-1)^n I_n\left( \frac{\psi }{2}\right) I_{2n} \left( \sqrt{2\psi }\frac{x}{\sigma }\right) \nonumber \\&= \dfrac{1}{2\pi } \int _{-\pi }^\pi \textrm{e}^{-\sqrt{2\psi } \frac{x}{\sigma }\sin t} \sum _{n \ge 1} \cos (2n t) I_n\left( \frac{\psi }{2}\right) \textrm{d}t \end{aligned}$$

(8)

$$\begin{aligned}&= \dfrac{1}{4\pi } \int _{-\pi }^\pi \textrm{e}^{-\sqrt{2\psi }\frac{x}{\sigma }\sin t} \left[ \textrm{e}^{\frac{\psi }{2} \cos (2 t)} - I_0\left( \frac{\psi }{2}\right) \right] \,\textrm{d} t \nonumber \\&= \dfrac{1}{4\pi } \int _{-\pi }^\pi \textrm{e}^{-\sqrt{2\psi }\frac{x}{\sigma }\sin t + \frac{\psi }{2} \cos (2 t)}\,\textrm{d} t - \dfrac{1}{2} I_0\left( \frac{\psi }{2}\right) \,I_0\left( \sqrt{2\psi }\frac{x}{\sigma }\right) , \end{aligned}$$

(9)

where in (8) we use for $a =2 t$ the summation [21, p. 695, Eq. 5.8.5.3.]

$$\begin{aligned} \textrm{e}^{z\cos a}= I_0(z) + 2 \sum _{k \ge 1} \cos (a k)\, I_k(z), \end{aligned}$$

whilst (9) we recognize as (7) at $n = 0$. This proves the assertion.

The cumulative distribution function, P(x) say, we obtain with the help of [16, p. 18, Eq. 5 (A2)]

$$\begin{aligned} \int {\textrm{e}}^{-a^2s^2+b s}\,{\textrm{d}}s = \dfrac{\sqrt{\pi }}{2a}\,\textrm{e}^{\frac{b^2}{4a^2}} \, {\textrm{erf}}\left( a s-\dfrac{b}{2a}\right) , \end{aligned}$$

where in our setting $a = (\sigma \sqrt{2})^{-1}$, $b = -\tfrac{\sqrt{2\psi }\sin t}{\sigma }$. Hence,

$$\begin{aligned} \int _{-\infty }^x \textrm{e}^{-\frac{s^2}{2\sigma ^2}- \sqrt{2\psi }\,\frac{s}{\sigma }\, \sin t} \,\textrm{d}s&= \sigma \sqrt{\frac{\pi }{2}}\, \textrm{e}^{\psi \sin ^2t}\,{\textrm{erf}}\left( \dfrac{s}{\sigma \sqrt{2}} + \sqrt{\psi }\sin t\right) \Big |_{-\infty }^x\\&= \sigma \sqrt{\frac{\pi }{2}}\, \textrm{e}^{\psi \sin ^2t}\, \left\{ 1+{\textrm{erf}}\left( \dfrac{x}{\sigma \sqrt{2}} + \sqrt{\psi }\sin t\right) \right\} , \end{aligned}$$

being $\lim _{z \rightarrow \infty }{\textrm{erf}}(z) = 1$ by the error function’s oddity. In turn, this yields

$$\begin{aligned} P(x)&= \int _{-\infty }^x p(s) \textrm{d}s = \dfrac{\textrm{e}^{-\frac{\psi }{2}}}{\sigma (2\pi )^{3/2}} \int _{-\pi }^\pi \textrm{e}^{\frac{\psi }{2} \cos (2 t)} \left\{ \int _{-\infty }^x \textrm{e}^{-\frac{s^2}{2\sigma ^2} - \sqrt{2\psi } \frac{s}{\sigma }\, \sin t}\textrm{d}s\right\} \textrm{d}t \\&= \dfrac{\textrm{e}^{-\frac{\psi }{2}}}{4 \pi }\, \int _{-\pi }^\pi \textrm{e}^{\frac{\psi }{2} \cos (2 t)+\psi \sin ^2t} \, \left\{ 1 +{\textrm{erf}}\left( \dfrac{x}{\sigma \sqrt{2}}+\sqrt{\psi }\sin t\right) \right\} \,\textrm{d}t \\&= \dfrac{1}{4 \pi }\, \int _{-\pi }^\pi \left\{ 1+{\textrm{erf}}\left( \dfrac{x}{\sigma \sqrt{2}} + \sqrt{\psi }\sin t\right) \right\} \,\textrm{d}t, \end{aligned}$$

which completes the proof of (6). $\square$

The next result concerns another fashion expression for the PDF p(x). First, we introduce a hyperbolic cosine–type differential operator via the operator power series

$$\begin{aligned} \cosh \{2\, \sqrt{A}\,\}[f(z)] = \sum _{n \ge 0} \dfrac{A^n}{\left( \frac{1}{2}\right) _n\,n!}\,[f(z)], \end{aligned}$$

(10)

where A stands for the input differential operator which acts on a suitable function $f \in \textrm{C}^\infty ({\mathbb {R}})$ with respect to the variable z.

Theorem 2

For all $x \in \mathbb {R}$ and $\psi , \sigma >0$ there holds

$$\begin{aligned} p(x) = \dfrac{\textrm{e}^{-\frac{x^2}{2\sigma ^2}-\frac{\psi }{2}}}{\sigma \sqrt{2\pi }} \, \cosh \left\{ \frac{\sqrt{\psi }\,x}{\sigma }\,\left( 1-2\frac{\textrm{d}}{\textrm{d} \psi }\right) ^{\frac{1}{2}}\right\} \, \left[ I_0\left( \frac{\psi }{2}\right) \right] . \end{aligned}$$

Proof

In [30, p. 4, Theorem 2.2.] Trifonov proved that for any real variable $u\in \mathbb {R}$ and for any nonnegative parameter $\psi$ and coefficients $\beta _n$ defined above, the following integral representation holds:

$$\begin{aligned} \sum _{n \ge 0} (-1)^n \beta _n I_{2n}\left( 2 u \sqrt{\psi }\,\right) = \frac{2}{\sqrt{\pi }}\,\textrm{e}^{-\psi } \int _0^\infty \textrm{e}^{-y^2}I_0(2\sqrt{\psi }\sqrt{u^2+y^2})\,\textrm{d}y\,. \end{aligned}$$

(11)

By using the series definition of the modified Bessel I–function and the binomial formula rewrite the integral (11). Thus,

$$\begin{aligned} \int _0^\infty&\textrm{e}^{-y^2}I_0\left( 2\sqrt{\psi } \sqrt{u^2+y^2}\right) \,\textrm{d}y = \sum _{n \ge 0}\dfrac{\psi ^n}{n!} \sum _{k=0}^n \dfrac{u^{2k}}{(n-k)!\,k!}\int _0^\infty {\textrm{e}}^{-y^2}y^{2(n-k)}\,\textrm{d}y \nonumber \\&= \dfrac{\sqrt{\pi }}{2} \sum _{k \ge 0} \dfrac{(\psi u^2)^k}{(1)_k\,k!} \sum _{n \ge 0}\dfrac{\left( \frac{1}{2}\right) _n\,\psi ^n}{(1+k)_n\,n!} = \dfrac{\sqrt{\pi }}{2} \sum _{k \ge 0} \dfrac{(\psi u^2)^k}{(k!)^2}\, {}_1F_1\left( \tfrac{1}{2}; 1+k; \psi \right) \nonumber \\&= \textrm{e}^{\frac{\psi }{2}} \dfrac{\sqrt{\pi }}{2} \sum _{k \ge 0}\dfrac{(\psi u^2)^k}{2^k \left( \frac{1}{2}\right) _k k!}\, \sum _{p=0}^k \left( {\begin{array}{c}k\\ p\end{array}}\right) \left( -\frac{1}{2}\right) ^p \sum _{k=0}^p \left( {\begin{array}{c}p\\ k\end{array}}\right) I_{p-2k} \left( \dfrac{\psi }{2}\right) ; \end{aligned}$$

(12)

where we apply the special case of [31] for $n=0$ and $m=1+k$:

$$\begin{aligned} {}_1F_1\left( \tfrac{1}{2}; 1+k; \right) = \frac{k!\,\textrm{e}^{\frac{z}{2}}}{2^k \,\left( \frac{1}{2}\right) _k}\, \sum _{p=0}^k \frac{(-1)^p}{2^p} \left( {\begin{array}{c}k\\ p\end{array}}\right) \sum _{j=0}^p \left( {\begin{array}{c}p\\ j\end{array}}\right) \, I_{p-2j}\left( \frac{z}{2}\right) , \end{aligned}$$

and the Bailey–transform in summing up double infinite series

$$\begin{aligned} \sum _{n,k \ge 0}a_{k,n} = \sum _{n \ge 0} \sum _{k=0}^n a_{k,n-k}. \end{aligned}$$

Substituting $u = x\,(\sigma \sqrt{2}\,)^{-1}$ we proceed to the following transformation of (3) via (12):

$$\begin{aligned} p(x) = \dfrac{\textrm{e}^{-\frac{x^2}{2\sigma ^2}-\frac{\psi }{2}}}{\sigma \sqrt{2\pi }} \sum _{k \ge 0}\dfrac{\left( \frac{\psi x^2}{4\sigma ^2}\right) ^k}{\left( \frac{1}{2}\right) _k\, k!} \sum _{p=0}^k \left( {\begin{array}{c}k\\ p\end{array}}\right) \frac{(-1)^p}{2^p} \sum _{j=0}^p \left( {\begin{array}{c}p\\ j\end{array}}\right) I_{2j-p} \left( \dfrac{\psi }{2}\right) . \end{aligned}$$

Further on, the p–sum is in fact $I_0^{(p)}(\cdot )$, thus

$$\begin{aligned} \left( \frac{\textrm{d}}{\textrm{d} \psi }\right) ^p I_0\left( \frac{\psi }{2}\right) = 4^{-p} \sum _{j=0}^p \left( {\begin{array}{c}p\\ j\end{array}}\right) I_{2j-p}\left( \frac{\psi }{2}\right) . \end{aligned}$$

Hence, we get the operator description:

$$\begin{aligned} p(x)&= \dfrac{\textrm{e}^{-\frac{x^2}{2\sigma ^2}-\frac{\psi }{2}}}{\sigma \sqrt{2\pi }} \sum _{k \ge 0}\dfrac{\left( \frac{\psi x^2}{4\sigma ^2}\right) ^k}{\left( \frac{1}{2}\right) _k k!} \sum _{p=0}^k \left( {\begin{array}{c}k\\ p\end{array}}\right) (-1)^p 2^p \left( \frac{\textrm{d}}{\textrm{d}\psi }\right) ^p I_0\left( \frac{\psi }{2}\right) \\&= \dfrac{\textrm{e}^{-\frac{x^2}{2\sigma ^2}-\frac{\psi }{2}}}{\sigma \sqrt{2\pi }} \sum _{k \ge 0}\dfrac{\left( \frac{\psi x^2}{4\sigma ^2}\right) ^k}{\left( \frac{1}{2}\right) _k k!} \left( 1-2\frac{\textrm{d}}{\textrm{d} \psi }\right) ^k I_0\left( \frac{\psi }{2}\right) \,, \end{aligned}$$

which implies

$$\begin{aligned} p(x)&= \dfrac{\textrm{e}^{-\frac{x^2}{2\sigma ^2}-\frac{\psi }{2}}}{\sigma \sqrt{2\pi }} \sum _{k \ge 0}\dfrac{1}{\left( \frac{1}{2}\right) _k k!} \left[ \frac{\psi x^2}{4\sigma ^2}\,\left( 1-2\frac{\textrm{d}}{\textrm{d} \psi }\right) \right] ^k I_0\left( \frac{\psi }{2}\right) \\&= \dfrac{\textrm{e}^{-\frac{x^2}{2\sigma ^2}-\frac{\psi }{2}}}{\sigma \sqrt{2\pi }} \, \cosh \left\{ \frac{\sqrt{\psi }\,x}{\sigma }\,\left( 1-2\frac{\textrm{d}}{\textrm{d} \psi }\right) ^{\frac{1}{2}}\right\} \, \left[ I_0\left( \frac{\psi }{2}\right) \right] \,. \end{aligned}$$

Here the notation (10) is used for the input differential operator

$$\begin{aligned} A = \frac{\psi \,x^2}{4 \sigma ^2} \,\left( 1-2\frac{\textrm{d}}{\textrm{d} \psi }\right) , \end{aligned}$$

applied to the function $f(\psi ) = I_0\left( \tfrac{\psi }{2}\right) \in \textrm{C}^\infty ({\mathbb {R}})$. $\square$

Now, we present another fashion closed form expression for the PDF (3), this time applying the Grünwald-Letnikov (GL) fractional derivative ${\mathbb {D}}_x^\nu [f]$ of the order $\nu$ with respect to x of a suitable function f. The definition of this operator reads [26]

$$\begin{aligned} {\mathbb {D}}_x^\nu [f] = \lim _{h \downarrow 0} \dfrac{1}{h^\nu } \sum _{m=0}^\infty (-1)^m \left( {\begin{array}{c}\nu \\ m\end{array}}\right) f\left( x+(\nu -m)h\right) , \end{aligned}$$

and $h \downarrow 0$ means that in approaching zero h remains positive. The Grünwald-Letnikov fractional derivative of order $\nu$ of the exponential function is [20]

$$\begin{aligned} {\mathbb {D}}_x^\nu \left[ \textrm{e}^{\alpha x}\right] = \alpha ^\nu \textrm{e}^{\alpha x}. \end{aligned}$$

(13)

Theorem 3

For all $x \in {\mathbb {R}}$ and for all $\sigma ,\psi >0$ we have

$$\begin{aligned} p(x)&= \dfrac{\textrm{e}^{-\psi }}{4 \pi \sigma \sqrt{2}}\, {\mathbb {D}}_\psi ^{-\frac{1}{2}} \left[ \frac{\textrm{e}^\psi }{\sqrt{\psi }} \left\{ \textrm{erf} \left( \frac{\sigma \sqrt{2\psi }-x}{\sigma \sqrt{2}}\right) + \textrm{erf} \left( \frac{\sigma \sqrt{2\psi }+x}{\sigma \sqrt{2}}\right) \right\} \right] \nonumber \\&\qquad - \frac{\textrm{e}^{-\frac{x^2}{2\sigma ^2}-\frac{\psi }{2}}}{\sigma \sqrt{2 \pi }}\, I_0\left( \frac{\psi }{2}\right) \, I_0\left( \sqrt{2\psi } \dfrac{x}{\sigma }\right) \,. \end{aligned}$$

(14)

Moreover, the cumulative distribution function

$$\begin{aligned} P(x)&= \dfrac{\textrm{e}^{-\psi }}{4 \pi \sigma \sqrt{2}}\, {\mathbb {D}}_\psi ^{-\frac{1}{2}} \left[ \frac{\textrm{e}^\psi }{\sqrt{\psi }} \int _{-\infty }^x \left\{ \textrm{erf} \left( \frac{\sigma \sqrt{2\psi }-t}{\sigma \sqrt{2}}\right) + \textrm{erf} \left( \frac{\sigma \sqrt{2\psi }+t}{\sigma \sqrt{2}}\right) \right\} \,\textrm{d}t \right] \nonumber \\&\qquad - I_0^2\left( \frac{\psi }{2}\right) \, \left( \frac{1}{2} + \kappa (x)\right) \,, \qquad x \in {\mathbb {R}}\,, \end{aligned}$$

(15)

where

$$\begin{aligned} \kappa (x) = \frac{\textrm{e}^{-\frac{\psi }{2}}}{\sigma \sqrt{2 \pi }\,I_0(\tfrac{\psi }{2})}\, \int _0^x \textrm{e}^{-\frac{t^2}{2\sigma ^2}}\,I_0\left( \sqrt{2\psi } \dfrac{t}{\sigma }\right) \,\textrm{d}t. \end{aligned}$$

Proof

We begin with the formula [7, p. 193, Eq. 512.6a)]

$$\begin{aligned} I_{2n}(z)=(-1)^n\dfrac{2}{\pi }\int _0^{\pi /2}\cosh (z\sin x) \cos (2nx)\,{\textrm{d}}x, \qquad n\in \mathbb {Z} \end{aligned}$$

by which we treat again the Neumann series of the second type appearing in (4), that is in (8):

$$\begin{aligned} {\mathscr {N}}_{\sigma ,\psi }(x){} & {} = \dfrac{2}{\pi } \int _0^{\pi /2}\cosh \left( \sqrt{2\psi }\frac{x}{\sigma }\sin t\right) \sum _{n \ge 1} \cos (2nt) I_n\left( \frac{\psi }{2}\right) \,\textrm{d}t \\{} & {} = \dfrac{1}{\pi } \int _0^{\pi /2}\textrm{e}^{\frac{\psi }{2} \cos (2t)} \cosh \left( \sqrt{2\psi }\frac{x}{\sigma }\sin t\right) \,\textrm{d}t\\{} & {} \qquad - \frac{1}{\pi }I_0\left( \frac{\psi }{2}\right) \int _0^{\pi /2} \cosh \left( \sqrt{2\psi } \frac{x}{\sigma }\sin t\right) \,\textrm{d}t \\{} & {} = \dfrac{\textrm{e}^{\frac{\psi }{2}}}{\pi }\int _0^{\pi /2}\textrm{e}^{-\psi \sin ^2t} \cosh \left( \sqrt{2\psi }\frac{x}{\sigma }\sin t\right) \,\textrm{d}t \\{} & {} \qquad - \frac{1}{\pi }I_0\left( \frac{\psi }{2}\right) \int _0^{\pi /2} \cosh \left( \sqrt{2\psi } \frac{x}{\sigma }\sin t\right) \,\textrm{d}t \\{} & {} = \dfrac{\textrm{e}^{\frac{\psi }{2}}}{2\pi } \int _0^1 \frac{\textrm{e}^{-\psi u^2} \cosh \left( \sqrt{2\psi }\dfrac{x}{\sigma }u \right) }{\sqrt{1-u^2}} \,\textrm{d}u\\{} & {} \qquad - \frac{1}{\pi }I_0\left( \frac{\psi }{2}\right) \int _0^{\pi /2} \cosh \left( \sqrt{2\psi } \frac{x}{\sigma }\sin t\right) \,\textrm{d}t \\{} & {} := \dfrac{\textrm{e}^{\frac{\psi }{2}}}{2\pi } \cdot U_1 - \frac{1}{\pi }I_0\left( \frac{\psi }{2}\right) \cdot U_2\,. \end{aligned}$$

The evaluation of the integral $U_1$ we realize by the GL fractional derivative. First, the explicit expression for the following integral reads

$$\begin{aligned} I(a) = \int _0^1\dfrac{\textrm{e}^{-a x^2+ bx}}{\sqrt{1-x^2}} \textrm{d}x = \frac{\sqrt{\pi }}{2\, \textrm{e}^a} {\mathbb {D}}_a^{-\frac{1}{2}} \left[ \frac{1}{\sqrt{a}} \textrm{e}^{\frac{b^2}{4a} + a} \left\{ \textrm{erf} \left( \frac{2a-b}{2 \sqrt{a}}\right) + \textrm{erf} \left( \frac{b}{2 \sqrt{a}}\right) \right\} \right] . \end{aligned}$$

(16)

Indeed, having in mind the linearity of the GL derivative and (13) we have:

$$\begin{aligned} I(a)&= \textrm{e}^{-a} \int _0^1 \dfrac{\textrm{e}^{a (1-x^2) + b x}}{\sqrt{1-x^2}}\, \textrm{d}x \\&= \textrm{e}^{-a} \int _0^1 \textrm{e}^{ b x}\, {\mathbb {D}}_a^{-\frac{1}{2}} \big [ \textrm{e}^{a (1-x^2)}\big ]\, \textrm{d}x \\&= \textrm{e}^{-a}\, {\mathbb {D}}_a^{-\frac{1}{2}} \left[ \int _0^1 \textrm{e}^{a (1-x^2) + b x}\, \textrm{d}x \right] \\&= \frac{\sqrt{\pi }}{2}\,\textrm{e}^{-a}\, {\mathbb {D}}_a^{-\frac{1}{2}} \left[ \frac{1}{\sqrt{a}}\, \textrm{e}^{\frac{b^2}{4a} + a}\, \left\{ \textrm{erf} \left( \frac{2a-b}{2 \sqrt{a}}\right) + \textrm{erf} \left( \frac{b}{2 \sqrt{a}}\right) \right\} \right] \,, \end{aligned}$$

which completes the proof of (16). This integral enables to conclude that

$$\begin{aligned} J(a,b)&= \int _0^1\dfrac{\textrm{e}^{-a x^2} \cosh (b x)}{\sqrt{1-x^2}}\, \textrm{d}x \nonumber \\&= \frac{\sqrt{\pi }}{4 \textrm{e}^a} {\mathbb {D}}_a^{-\frac{1}{2}} \left[ \frac{\textrm{e}^{\frac{b^2}{4a} + a}}{\sqrt{a}} \left\{ \textrm{erf} \left( \frac{2a-b}{2 \sqrt{a}}\right) + \textrm{erf} \left( \frac{2a+b}{2 \sqrt{a}}\right) \right\} \right] , \end{aligned}$$

(17)

which implies the solution

$$\begin{aligned} U_1 = J\left( \psi , \sqrt{2\psi }\, \frac{x}{\sigma }\right) . \end{aligned}$$

Now, by virtue of the integral form of the modified Bessel function of the first kind [19, p. 252, Eq. 10.32.1] we get

$$\begin{aligned} U_2&= \int _0^{\frac{\pi }{2}}\cosh \left( \sqrt{2\psi }\frac{x}{\sigma }\sin t\right) \,{\textrm{d}}t = \frac{1}{2}\int _0^\pi \cosh \left( \sqrt{2\psi }\frac{x}{\sigma }\cos u\right) \,{\textrm{d}}u \\&= \frac{\pi }{2} I_0\left( \sqrt{2\psi } \dfrac{x}{\sigma }\right) \,. \end{aligned}$$

Collecting the values of $U_1$ and $U_2$, we get the sum of the Neumann series

$$\begin{aligned} {\mathscr {N}}_{\sigma ,\psi }(x)&= \dfrac{\textrm{e}^{\frac{x^2}{2\sigma ^2} -\frac{\psi }{2}}}{8 \sqrt{\pi }} \, {\mathbb {D}}_\psi ^{-\frac{1}{2}} \left[ \frac{\textrm{e}^\psi }{\sqrt{\psi }} \left\{ \textrm{erf} \left( \frac{\sigma \sqrt{2\psi }-x}{\sigma \sqrt{2}}\right) + \textrm{erf} \left( \frac{\sigma \sqrt{2\psi }+x}{\sigma \sqrt{2}}\right) \right\} \right] \\&\qquad - \frac{1}{2} I_0\left( \frac{\psi }{2}\right) I_0\left( \sqrt{2\psi } \dfrac{x}{\sigma }\right) \,. \end{aligned}$$

Since

$$\begin{aligned} p(x) = \frac{2}{\sigma \sqrt{2 \pi }}\, {\textrm{e}^{-\frac{x^2}{2\sigma ^2}-\frac{\psi }{2}}} \cdot {\mathscr {N}}_{\sigma ,\psi }(x), \end{aligned}$$

the way to (14) is open.

The CDF P(x) we obtain by integrating the PDF p(x):

$$\begin{aligned} P(x)&= \dfrac{\textrm{e}^{-\psi }}{4 \pi \sigma \sqrt{2}}\, {\mathbb {D}}_\psi ^{-\frac{1}{2}} \left[ \frac{\textrm{e}^\psi }{\sqrt{\psi }} \int _{-\infty }^x \left\{ \textrm{erf} \left( \frac{\sigma \sqrt{2\psi }-t}{\sigma \sqrt{2}}\right) + \textrm{erf} \left( \frac{\sigma \sqrt{2\psi }+t}{\sigma \sqrt{2}}\right) \right\} \,\textrm{d}t \right] \\&\qquad - \frac{\textrm{e}^{-\frac{\psi }{2}}}{\sigma \sqrt{2 \pi }}\, I_0\left( \frac{\psi }{2}\right) \, \int _{-\infty }^x \textrm{e}^{-\frac{t^2}{2\sigma ^2}}\, I_0\left( \sqrt{2\psi } \dfrac{t}{\sigma }\right) \,\textrm{d}t\,. \end{aligned}$$

Assuming $x>0$, splitting the integration interval in a suitable way and having in mind that the integrand is even, we have

$$\begin{aligned} \left( \int _{-\infty }^0 + \int _0^x\right) \textrm{e}^{-\frac{t^2}{2\sigma ^2}}\,I_0\left( \sqrt{2\psi } \dfrac{t}{\sigma }\right) \,\textrm{d}t =: V_1+V_2, \end{aligned}$$

where by the Laplace transform formula [23, p. 318, Eq. 10]

$$\begin{aligned} \int _0^\infty \textrm{e}^{-p\,x}\,x^{-\tfrac{1}{2}}\, I_\nu (a \sqrt{x}\,)\,\textrm{d}x = \sqrt{\frac{\pi }{p}}\, \exp \left\{ \frac{a^2}{8 p}\right\} \, I_{\frac{\nu }{2}}\left( \frac{a^2}{8 p}\right) \,, \end{aligned}$$

(18)

valid for all $\Re (\nu )>-1,\,|\textrm{arg}(a)|<\pi$, we conclude

$$\begin{aligned} V_1&= \frac{\sigma }{\sqrt{2\psi }} \int _0^\infty \textrm{e}^{-\frac{x^2}{4\psi }} I_0(x) \textrm{d}x\\&= \frac{\sigma }{2 \sqrt{2\psi }} \int _0^\infty \textrm{e}^{-\frac{x}{4\psi }} I_0(\sqrt{x}\,) \frac{\textrm{d}x}{\sqrt{x}} = \frac{\sigma \sqrt{\pi }}{\sqrt{2}} \textrm{e}^{\frac{\psi }{2}}\,I_0\left( \frac{\psi }{2}\right) . \end{aligned}$$

On the other hand, it is

$$\begin{aligned} V_2 = \int _0^x \textrm{e}^{-\frac{t^2}{2\sigma ^2}}\, I_0\left( \sqrt{2\psi } \dfrac{t}{\sigma }\right) \,\textrm{d}t = I_0^2\left( \frac{\psi }{2}\right) \, \kappa (x), \end{aligned}$$

mentioning that $\kappa (x)$ is odd function. It remains to calculate

$$\begin{aligned} \lim _{z \rightarrow \infty } \kappa (z)&= \frac{\textrm{e}^{-\frac{\psi }{2}}}{\sigma \sqrt{2 \pi }\,I_0\left( \tfrac{\psi }{2}\right) }\, \int _0^\infty \textrm{e}^{-\frac{t^2}{2\sigma ^2}}\,I_0\left( \sqrt{2\psi }\frac{t}{\sigma }\right) \,\textrm{d}t\\&= \frac{\textrm{e}^{-\frac{\psi }{2}} \sigma \sqrt{\pi }\,\textrm{e}^{\frac{\psi }{2}}}{\sigma \sqrt{2 \pi }\, I_0\left( \tfrac{\psi }{2}\right) \,\sqrt{2}}\,I_0\left( \frac{\psi }{2}\right) =\frac{1}{2}\,. \end{aligned}$$

However, this confirms the final form of the assertion (15). $\square$

3 A PDF Generated by $\kappa (x)$

The auxiliary function $\kappa (x)$ which appears in Theorem 3, possesses certain similarity with the CDF of the standard normal distribution ${\mathscr {N}}(0,1)$, that is, with the Laplace function

$$\begin{aligned} \Phi (x) = \frac{1}{\sqrt{2 \pi }} \int _0^{x} \textrm{e}^{-\frac{t^2}{2}}\, \textrm{d}t. \end{aligned}$$

Now, we define a random variable $\xi$, say, which bell–shaped PDF is built considering the integrand of $\kappa (x)$ introducing mutually the scaling parameter $\sigma >0$. After re–normalization the related density takes the following form:

$$\begin{aligned} p_\xi (x) = \frac{\textrm{e}^{-\frac{\psi }{2}}\,I_0\left( \sqrt{2\psi } \dfrac{x}{\sigma }\right) }{\sigma \sqrt{2 \pi }\, I_0\left( \frac{\psi }{2}\right) }\, \,\textrm{e}^{-\tfrac{x^2}{2\sigma ^2}}\,,\qquad \psi , \sigma >0;\, x \in {\mathbb {R}}\,, \end{aligned}$$

(19)

where the parameters originate back to the mathematical model presented in the Introduction. The shape of $p_\xi (x)$ dictates several properties of $\xi$, for instance, the existence of the raw moments $\mu _s = {\textsf{E}} \xi ^s$ of the order s; obviously the odd order raw moments $\mu _{2n-1} = {\textsf{E}} \xi ^{2n-1},\, n \in {\mathbb {N}}$ vanish, being the density an even function of x. Several graphs of the above presented PDF are illustrated in Fig. 1.

Fig. 1

The graph of the PDF (19) for $\sigma =10$ and $\psi \in \{0.1,\,0.5,\,0.7,\,1,\,5,\,10,\,15\}$

Theorem 4

For all $s>-1$ the raw moments of the order s of the rv $\xi$ read

$$\begin{aligned} \mu _s = \frac{2^{\frac{s}{2}-1}\,\sigma ^s\,\left( 1 + \textrm{e}^{\textrm{i}\pi s}\right) }{\sqrt{\pi }\, \textrm{e}^{\frac{\psi }{2}}\,I_0\left( \frac{\psi }{2}\right) }\,\Gamma \left( \frac{s+1}{2}\right) \, {}_1F_1\left( \frac{s+1}{2}; 1; \psi \right) . \end{aligned}$$

Proof

By definition

$$\begin{aligned} \mu _s&= {\textsf{E}} \xi ^s = \int _{{\mathbb {R}}} x^s\, p_\xi (x)\, \textrm{d}x \\&= \frac{\textrm{e}^{-\frac{\psi }{2}}}{\sigma \,\sqrt{2 \pi }\, I_0\left( \frac{\psi }{2}\right) } \int _{{\mathbb {R}}} x^s\, \textrm{e}^{-\tfrac{x^2}{2\sigma ^2}}\, I_0\left( \sqrt{2\psi } \dfrac{x}{\sigma }\right) \, \textrm{d}x \\&= \frac{\textrm{e}^{-\frac{\psi }{2}} \left( 1 + \textrm{e}^{\textrm{i}\pi s}\right) \,\sigma ^s}{2^{\frac{s}{2}+2}\,\psi ^{\frac{s+1}{2}}\, \sqrt{\pi }\, I_0\left( \frac{\psi }{2}\right) }\, \int _0^\infty \textrm{e}^{-\tfrac{x}{4 \psi }}\, x^{\frac{s-1}{2}}\, I_0\left( \sqrt{x} \right) \, \textrm{d}x\,, \end{aligned}$$

which is the Laplace transform (up to multiplicative constant) of the input function $x^{\frac{s-1}{2}}\, I_0\left( \sqrt{x} \right)$. According to the formula [23, p. 318, Eq. 5]

$$\begin{aligned} \int _0^\infty \textrm{e}^{-p x} x^\mu \, I_\nu \left( a \sqrt{x}\right) \, \textrm{d}x = \frac{\Gamma (\mu +\frac{\nu }{2}+1)}{\Gamma (\nu +1)\, p^{\mu +\frac{\nu }{2}+1}}\, \left( \frac{a}{2}\right) ^\nu \, {}_1F_1 \left( \mu +\frac{\nu }{2}+1;\nu +1;\frac{a^2}{4p}\right) , \end{aligned}$$

which holds for all $\Re (2\mu +\nu )>-2$ and $|\arg (a)|<\pi$, our integral reduces to

$$\begin{aligned} \Gamma \left( \tfrac{s+1}{2}\right) \,(4 \psi )^{\frac{s+1}{2}}\, {}_1F_1\left( \tfrac{s+1}{2}; 1; \psi \right) , \end{aligned}$$

for $s>-1$ and $a = 1$. Now, routine steps give the asserted expression.

$\square$

The next our result concern the characteristic function (CHF) $\varphi _\xi (t) = {\textsf{E}}\, \textrm{e}^{\textrm{i} t \xi }$. Since the parity of the PDF $p_\xi (x)$ the related characteristic function $\varphi _\xi (t)$ equals to the Fourier cosine transform of $p_\xi (x)$. We present here the power series representation of the CHF.

Theorem 5

For all $\psi , \sigma >0$ and for all $t \in {\mathbb {R}}$ we have

$$\begin{aligned} \varphi _\xi (t) = \frac{\textrm{e}^{-\frac{\psi }{2}}}{I_0\left( \frac{\psi }{2}\right) } \sum _{n \ge 0} \frac{(-1)^n}{n!}\, {}_1F_1\left( n+\tfrac{1}{2}; 1; \psi \right) \, \left( \frac{\sigma t}{32}\right) ^{2n}. \end{aligned}$$

(20)

Proof

Bearing in mind the property

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}\psi ^{-1}} = -\psi ^2 \frac{\textrm{d}}{\textrm{d}\psi }, \end{aligned}$$

(21)

we conclude by direct calculation:

$$\begin{aligned} \varphi _\xi (t)&= \frac{2\, \textrm{e}^{-\frac{\psi }{2}}}{\sigma \,\sqrt{2 \pi }\, I_0\left( \frac{\psi }{2}\right) }\, \int _0^\infty \cos (t x)\, \textrm{e}^{-\tfrac{x^2}{2\sigma ^2}}\,I_0\left( \sqrt{2\psi } \dfrac{x}{\sigma }\right) \, \textrm{d}x \nonumber \\&= \frac{2\, \textrm{e}^{-\frac{\psi }{2}}}{\sigma \,\sqrt{2 \pi }\, I_0\left( \frac{\psi }{2}\right) }\, \sum _{n \ge 0} \frac{(-1)^n t^{2n}}{(2n)!} \int _0^\infty x^{2n}\, \textrm{e}^{-\tfrac{x^2}{2\sigma ^2}}\,I_0\left( \sqrt{2\psi } \dfrac{x}{\sigma }\right) \, \textrm{d}x \nonumber \\&= \frac{2\, \textrm{e}^{-\frac{\psi }{2}}}{\sigma \,\sqrt{2 \pi }\, I_0\left( \frac{\psi }{2}\right) }\, \sum _{n \ge 0} \frac{t^{2n}}{(2n)!} \left( \frac{\sigma }{\sqrt{2 \psi }}\right) ^{2n+1} \left( \frac{\textrm{d}}{\textrm{d} \psi ^{-1}}\right) ^n \int _0^\infty \textrm{e}^{-\frac{x^2}{4 \psi }}\,I_0(x)\,\textrm{d}x \nonumber \\&= \frac{\textrm{e}^{-\frac{\psi }{2}}}{\sqrt{\pi \psi }\,I_0\left( \frac{\psi }{2}\right) } \sum _{n \ge 0} \frac{(-1)^n}{(2n)!} \left( \frac{\sigma ^2 t^2 \psi }{8} \right) ^n \left( \frac{\textrm{d}}{\textrm{d} \psi }\right) ^n \int _0^\infty \textrm{e}^{-\frac{x}{4 \psi }}\,I_0(\sqrt{x}\,)\,\frac{\textrm{d}x}{\sqrt{x}} \end{aligned}$$

(22)

$$\begin{aligned}&= \frac{\textrm{e}^{-\frac{\psi }{2}}}{\sqrt{\pi \psi }\,I_0\left( \frac{\psi }{2}\right) } \sum _{n \ge 0} \frac{(-1)^n}{(2n)!} \left( \frac{\sigma ^2 t^2 \psi }{8} \right) ^n \left( \frac{\textrm{d}}{\textrm{d} \psi }\right) ^n \left[ \sqrt{\psi }\, \textrm{e}^{\frac{\psi }{2}}\, I_0\left( \frac{\psi }{2}\right) \right] \nonumber \\&= \frac{\sqrt{2}\, \textrm{e}^{-\frac{\psi }{2}}}{\sqrt{\psi }\,I_0\left( \frac{\psi }{2}\right) } \sum _{n \ge 0} \frac{(-1)^n}{(2n)!} \left( \frac{\sigma ^2 t^2 \psi }{16} \right) ^n \left( \frac{\textrm{d}}{\textrm{d} x}\right) ^n \big [ \sqrt{x}\, \textrm{e}^x\, I_0(x)\big ]_{x = \frac{\psi }{2}}\,, \end{aligned}$$

(23)

whilst the integral (22) is solved using (18). Substitution $x^{-1} \mapsto x$ in the derivative [4, p. 33, Eq. 20.]

$$\begin{aligned} \left( \frac{\textrm{d}}{\textrm{d}x}\right) ^n \left[ x^{-\frac{1}{2}}\, \textrm{e}^{\frac{a}{x}}\, I_\nu \left( \frac{a}{x}\right) \right]&= \frac{(-1)^n (2a)^\nu }{x^{\nu +n+\frac{1}{2}}}\,\frac{\Gamma (\nu +n+\frac{1}{2})}{\sqrt{\pi }\,\Gamma (2\nu +1)}\\&\qquad \cdot {}_1F_1\left( \nu +n+\frac{1}{2}; 2\nu +1; \frac{2a}{x} \right) \,, \end{aligned}$$

for $a=1$ and $\nu =0$ results in

$$\begin{aligned} \left( \frac{\textrm{d}}{\textrm{d} x}\right) ^n \big [ \sqrt{x}\, \textrm{e}^x\, I_0(x)\big ] = x^{\frac{1}{2}-n}\, \left( \tfrac{1}{2}\right) _n\, {}_1F_1 \left( n+\tfrac{1}{2}; 1; 2x\right) , \end{aligned}$$

by virtue of the formula (21). Accordingly, (23) becomes

$$\begin{aligned} \varphi _\xi (t)&= \frac{\textrm{e}^{-\frac{\psi }{2}}}{I_0\left( \frac{\psi }{2}\right) } \sum _{n \ge 0} \frac{(-1)^n\,\left( \frac{1}{2}\right) _n}{(2n)!} \left( \frac{\sigma ^2\,t^2}{8} \right) ^n \, {}_1F_1 \left( n+\frac{1}{2}; 1; \psi \right) \,, \end{aligned}$$

which completes the proof, since the rest is obvious. $\square$

One of the remaining important questions is the modality of the rv $\xi$. The Fig. 1 suggests that the distribution is unimodal for $0< \psi \le 1$ and bimodal when $\psi >1$. The analytical proof of that hypothesis follows. Obviously, the mode of this distribution equals 0.

Theorem 6

The PDF $p_\xi (x)$ defines unimodal distribution for $\psi \in (0,1]$, whilst for $\psi >1$ the distribution is bimodal.

Proof

We consider

$$\begin{aligned} p_\xi (x) = C(\psi , \sigma )\,\textrm{e}^{-\tfrac{x^2}{2\sigma ^2}}\, I_0\left( \sqrt{2\psi } \frac{x}{\sigma }\right) , \qquad \psi , \sigma >0,\, x \in {\mathbb {R}}, \end{aligned}$$

where

$$\begin{aligned} C(\psi ,\sigma ) = \frac{\textrm{e}^{-\frac{\psi }{2}}}{\sigma \sqrt{2 \pi }\, I_0\left( \frac{\psi }{2}\right) }. \end{aligned}$$

The assertion holds if we prove that $(0, p_\xi (0))$ is the unique maximum of the PDF. Hence,

$$\begin{aligned} p_\xi '(x) = \frac{1}{\sigma }\,C(\psi , \sigma )\,\textrm{e}^{-\tfrac{x^2}{2\sigma ^2}}\,I_0\left( \sqrt{2\psi } \frac{x}{\sigma }\right) \, \left[ \sqrt{2\psi }\,\frac{I_1\left( \sqrt{2\psi } \frac{x}{\sigma }\right) }{I_0\left( \sqrt{2\psi } \frac{x}{\sigma }\right) } - \frac{x}{\sigma }\right] \,. \end{aligned}$$

(24)

Applying the inequality by Joshi and Bissu [12, p. 255]

$$\begin{aligned} \frac{I_{\nu +1}(z)}{I_\nu (z)} < \frac{z}{2(\nu +1)}, \qquad \nu>-1;\, z>0, \end{aligned}$$

for $\nu =0$ and $z = \sigma ^{-1} \sqrt{2\psi }\, x$, we get the estimate

$$\begin{aligned} p_\xi '(x) < \frac{C(\psi , \sigma )}{\sigma ^2}\,(\psi -1)\,x \,\textrm{e}^{-\tfrac{x^2}{2 \sigma ^2}}\, I_0\left( \sqrt{2\psi } \frac{x}{\sigma }\right) \le 0, \end{aligned}$$

which reduces to $\psi \le 1$ for all positive x. The remaining obvious steps and the parity of the PDF finish the proof of the first assertion.

Consider another inequality for the quotient of modified Bessel functions of the first kind by Joshi and Bissu [12, p. 255, Eq. (2.22)]:

$$\begin{aligned} \frac{z}{2(\nu +1)} - \frac{z^3}{8(\nu +1)^2(\nu +2)} < \frac{I_{\nu +1}(z)}{I_\nu (z)}, \qquad \nu>-1;\, z>0. \end{aligned}$$

The same specification as above $\nu =0$ and $z = \sigma ^{-1} \sqrt{2\psi }\, x$ gives

$$\begin{aligned} \sqrt{2\psi }\,\frac{I_1\left( \sqrt{2\psi } \frac{x}{\sigma }\right) }{I_0\left( \sqrt{2\psi } \frac{x}{\sigma }\right) } > \frac{\psi \,x}{\sigma }- \frac{\psi ^2}{4\,\sigma ^3} x^3, \end{aligned}$$

which implies mutatis mutandis by (24) that

$$\begin{aligned} p_\xi '(x) > \frac{C(\psi ,\sigma )}{\sigma ^2}\,x \,\textrm{e}^{-\tfrac{x^2}{2 \sigma ^2}}\, I_0\left( \sqrt{2\psi }\,\frac{x}{\sigma }\right) \, \left( \psi -1 - \frac{\psi ^2}{4 \sigma ^2}\,x^2\right) \ge 0. \end{aligned}$$

Hence, the PDF increases from $p_\xi (0) = C(\psi ,\sigma )$ (at least) on the interval $(0, 2\tfrac{\sigma }{\psi }\sqrt{\psi -1}\,)$ for $\psi >1$, and decreases outside of this interval to 0, as $p_\xi (x)$ vanishes for growing x. Being the PDF even the proof is complete. $\square$

We finish the exposition in this section with deriving the parameter estimators observing the d–dimensional vector of realizations (also called sample) $\varvec{x} = (x_1, \cdots , x_d)$ of the iid replicae $\xi _j,\,j= 1, \cdots , d$ of the input rv $\xi \sim p_\xi (x)$. Denote, as usual, the pth power sample mean

$$\begin{aligned} \overline{X_d^p} = \frac{1}{d} \sum _{j=1}^d x_j^p, \qquad p, d \in {\mathbb {N}}. \end{aligned}$$

Firstly, we present the maximum likelihood estimator (MLE) for the considered parameters.

From the log–likelihood function

$$\begin{aligned} \ell _{\psi , \sigma }({\varvec{x}})&= \log \prod _{j=1}^d p_\xi (x_j) = -d \left[ \log \sqrt{2 \pi } + \frac{\psi }{2} + \log \sigma + \log I_0\left( \frac{\psi }{2}\right) \right] \\&\qquad - \frac{1}{2 \sigma ^2} \sum _{j=1}^d x_j^2 + \sum _{j=1}^d \log I_0\left( \sqrt{2 \psi } \frac{x_j}{\sigma }\right) \,, \end{aligned}$$

we get the system

$$\begin{aligned} \frac{\partial }{\partial \sigma } \ell _{\psi ,\sigma }({\varvec{x}})&= -\frac{d}{\sigma }+ \frac{1}{\sigma ^3} \sum _{j=1}^d x_j^2 - \frac{\sqrt{2 \psi }}{\sigma ^2} \sum _{j=1}^d x_j\,\frac{I_1\left( \sqrt{2 \psi } \frac{x_j}{\sigma }\right) }{I_0\left( \sqrt{2 \psi } \frac{x_j}{\sigma }\right) } = 0 \nonumber \\ \frac{\partial }{\partial \psi }\ell _{\psi ,\sigma }({\varvec{x}})&=-\frac{d}{2} \left( 1+\frac{I_1\left( \frac{\psi }{2}\right) }{I_0\left( \frac{\psi }{2}\right) }\right) + \frac{1}{\sigma \sqrt{2 \psi }} \sum _{j=1}^d x_j\,\frac{I_1(\sqrt{2 \psi } \frac{x_j}{\sigma })}{I_0(\sqrt{2 \psi } \frac{x_j}{\sigma })} = 0\,, \end{aligned}$$

(25)

which imply the relation

$$\begin{aligned} \sigma ^2 = \frac{\overline{X_d^2}\, I_0\left( \frac{\psi }{2}\right) }{\psi \left[ I_0\left( \frac{\psi }{2}\right) +I_1\left( \frac{\psi }{2}\right) \right] }\,. \end{aligned}$$

(26)

Inserting $\sigma$ into the equation (25) of the above system, we conclude the transcendental equation in $\psi$:

$$\begin{aligned} \sqrt{1 + \frac{I_1(\tfrac{\psi }{2})\bigg )}{I_0(\tfrac{\psi }{2})}} = \frac{1}{d\,\sqrt{\overline{X_d^2}}} \sum _{j=1}^d x_j \dfrac{I_1\left( \frac{\sqrt{2}\,\psi x_j}{\sqrt{\overline{X_d^2}\,I_0\left( \frac{\psi }{2}\right) }} \sqrt{I_0\left( \frac{\psi }{2}\right) + I_1\left( \frac{\psi }{2}\right) }\,\right) }{I_0\left( \frac{\sqrt{2}\,\psi x_j}{\sqrt{\overline{X_d^2}\,I_0\left( \frac{\psi }{2}\right) }}\, \sqrt{I_0\left( \frac{\psi }{2}\right) + I_1\left( \frac{\psi }{2}\right) }\,\right) }. \end{aligned}$$

The solution is ${\widehat{\psi }}$, which in conjunction with (26) gives the estimator ${\widehat{\sigma }}$.

Our final task in this section is to establish the moment method estimators (MME) for the parameters $\psi , \sigma$ by means of the d–dimensional sample ${\varvec{x}}$. The MMEs ${\widetilde{\psi }}, {\widetilde{\sigma }}$ of the parameters $\psi , \sigma$ are the solutions of the system

$$\begin{aligned}&\frac{\overline{X_d^2}}{\sigma ^2} = 1+\psi + \psi \,\frac{I_1\left( \frac{\psi }{2}\right) }{I_0\left( \frac{\psi }{2}\right) }, \quad \frac{\overline{X_d^4}}{\sigma ^4} = 2 \psi ^2 + 6 \psi +3+2 \psi (\psi +2)\,\frac{I_1\left( \frac{\psi }{2}\right) }{I_0\left( \frac{\psi }{2}\right) }\,.\nonumber \\ \end{aligned}$$

(27)

As to the mathematical background of the MME method we note that the mean $\mu _1 = 0$ of the rv $\xi$, so $\mu _2 = \textrm{Var}(\xi )$, moreover

$$\begin{aligned}{} & {} \mu _2 = \sigma ^2\, \left( 1+\psi + \psi \,\frac{I_1\left( \frac{\psi }{2}\right) }{I_0\left( \frac{\psi }{2}\right) }\right) , \quad \\{} & {} \mu _4 = \sigma ^4\, \left( 2 \psi ^2 + 6 \psi +3+2 \psi (\psi +2)\,\frac{I_1\left( \frac{\psi }{2}\right) }{I_0\left( \frac{\psi }{2}\right) }\right) . \end{aligned}$$

Replacing the second and the fourth order moments $\mu _2, \mu _4$ with their counterpart sample means $\overline{X_d^2}$ and $\overline{X_d^4}$ we arrive at the inter–connection formula

$$\begin{aligned} \psi = \dfrac{\sigma ^4+\overline{X_d^4}}{2 \overline{X_d^2}\,\sigma ^2} - 2. \end{aligned}$$

Combining this relation with one of the equations in (27), we solve this system. The solutions in $\psi$ and $\sigma$ are the MMEs ${\widetilde{\psi }}, {\widetilde{\sigma }}$.

4 Another Form of the CDF P(x)

In order to derive an appropriate cumulative distribution function (CDF) for (3), let us first recall the Nuttall function $Q_{\mu ,\nu }(a,b)$, introduced by A. H. Nuttall in [17] as

$$\begin{aligned} Q_{\mu ,\nu }(a,b) =\int _b^\infty t^\mu \textrm{e}^{-\frac{t^2+a^2}{2}}\, I_{\nu }(at)\,\textrm{d}t, \end{aligned}$$

defined for all real values of the parameters a, b, $\mu$ and $\nu$ with the possible extension by analytic continuation for all finite complex a and b (for more details see [14, p. 523]); according to some authors the condition $b>0$ must be satisfied too [5, p. 34, Eq. (1)]. The Nuttall function became very popular and useful since its first appearance: it is often used in the context of communication theory [14]; precisely, applications include the digital communication performance evaluation, the error probability performance of non–coherent digital communication, the outage probability of wireless communication systems, the performance analysis and capacity statistics of uncoded MIMO (multiple–input multiple–output) systems, the extraction of the required loglikelihood ratio in decoding differential phase–shift keying (DPSK) signals using turbo codes for instance, see [5, 13, 18, 27, 28] and the relevant references therein.

Finally, we present a new formula for the CDF P(x) of a rv which presents a single sinusoidal signal plus Gaussian noise.

Theorem 7

For all $x \in \mathbb {R}$ and $\sigma , \psi >0$ the CDF reads

$$\begin{aligned} P(x) = \textrm{e} \sum _{n \ge 0} (-1)^n \beta _n I_n(1) - \dfrac{\textrm{e}^\psi }{\sqrt{2\pi }} \sum _{n \ge 0} (-1)^n \beta _n Q_{0,2n} \left( \sqrt{2\psi },\frac{x}{\sigma }\right) \,. \end{aligned}$$

Proof

By definition of CDF

$$\begin{aligned}&P(x) = \int _{-\infty }^x p(t)\,\textrm{d}t = \dfrac{1}{\sigma \sqrt{2\pi }} \sum _{n \ge 0} (-1)^n\beta _n \int _{-\infty }^x \textrm{e}^{-\frac{t^2}{2\sigma ^2}} I_{2n}\left( \sqrt{2\psi }\frac{t}{\sigma }\right) \,\textrm{d}t.\nonumber \\ \end{aligned}$$

(28)

The integral can be rewritten in the following form

$$\begin{aligned} \int _{-\infty }^x&\textrm{e}^{-\frac{t^2}{2\sigma ^2}} I_{2n}\left( \sqrt{2\psi }\frac{t}{\sigma }\right) \,\textrm{d}t = \sigma \textrm{e}^\psi \int _{-\infty }^{x/\sigma } \textrm{e}^{-\frac{u^2+2\psi }{2}}I_{2n}\left( \sqrt{2\psi }u\right) \,{\textrm{d}}u\\&= \sigma \textrm{e}^\psi \left( \int _{-\infty }^\infty \textrm{e}^{-\frac{u^2+2\psi }{2}}I_{2n}\left( \sqrt{2\psi }u\right) \,{\textrm{d}}u - \int _{x/\sigma }^\infty \textrm{e}^{-\frac{u^2+2\psi }{2}}I_{2n}\left( \sqrt{2\psi }u\right) \,{\textrm{d}}u \right) \\&= \sigma {\textrm{e}}^\psi \left( \sqrt{2\pi }{\textrm{e}}^{1-\psi }I_n(1)-Q_{0,2n}\left( \sqrt{2\psi },\frac{x}{\sigma }\right) \right) \,, \end{aligned}$$

which yields the stated result substituting this expression into (28). $\square$

5 Discussion, Final Remarks

${\textsf{A}}.$ An equivalent representation for the PDF p(x) can be obtained in terms of the probabilist’s writing of the Hermite polynomials $He_{2n}$ which Rodrigues formula reads

$$\begin{aligned} He_{2n}(x) = (-1)^n {\textrm{e}}^{\frac{x^2}{2}} \left( \dfrac{\textrm{d}}{\textrm{d}x} \right) ^{2 n} \textrm{e}^{-\frac{x^2}{2}}. \end{aligned}$$

According to Trifonov it is given as [15, § 9.3], [9, p. 74, Eq. (5)] and [30]

$$\begin{aligned} p(x) = \dfrac{\textrm{e}^{-\frac{x^2}{2\sigma ^2}}}{\sigma \sqrt{2\pi }} \sum _{n \ge 0} \dfrac{\psi ^n}{2^n (n!)^2}\, He_{2n}\left( \frac{x}{\sigma }\right) \,. \end{aligned}$$

(29)

However, the summation formula of this Neumann series built by Hermite polynomials is a simple consequence of Theorems 1 and 3. For instance, Theorem 3 implies the following closed form summation.

Corollary 7.1

For all $x \in {\mathbb {R}}$ and $\sigma , \psi >0$ we have

$$\begin{aligned} \sum _{n \ge 0} \dfrac{\psi ^n}{2^n (n!)^2} He_{2n}\left( \frac{x}{\sigma }\right)&= \dfrac{\textrm{e}^{\frac{x^2}{2\sigma ^2}-\psi }}{4 \sqrt{\pi }}\, {\mathbb {D}}_\psi ^{-\frac{1}{2}} \left[ \frac{\textrm{e}^\psi }{\sqrt{\psi }} \left\{ \textrm{erf} \left( \frac{\sigma \sqrt{2\psi }-x}{\sigma \sqrt{2}}\right) \right. \right. \\&\qquad \left. \left. + \textrm{erf} \left( \frac{\sigma \sqrt{2\psi }+x}{\sigma \sqrt{2}}\right) \right\} \right] - \textrm{e}^{-\frac{\psi }{2}}\, I_0\left( \frac{\psi }{2}\right) \, I_0\left( \sqrt{2\psi } \dfrac{x}{\sigma }\right) \,. \end{aligned}$$

${\textsf{B}}.$ We expressed the CDF P(x) in Theorem 1 in a condensed elegant form. However, the integral (6) can still be transformed, for instance, in the following way:

$$\begin{aligned} \int _{-\pi }^\pi \textrm{erf}(a\,\sin t&+ b)\, \textrm{d}t = 2 \int _{-1}^1 \frac{\textrm{erf}(a x+ b)}{\sqrt{1-x^2}}\, \textrm{d}x\nonumber \\&= 2 \sum _{n \ge 0} \frac{\left( \frac{1}{2}\right) _n}{n!} \int _1^1 \textrm{erf}(a x+b)\, x^{2n}\, \textrm{d}x \nonumber \\&= \frac{2}{a} \sum _{n \ge 0} \frac{\left( \frac{1}{2}\right) _n}{n!} \left( \frac{b}{a}\right) ^{2n} \sum _{k = 0}^n \frac{(-1)^k}{b^k} \left( {\begin{array}{c}2n\\ k\end{array}}\right) \, \int _{b-a}^{b+a} \textrm{erf}(y)\, y^k\, \textrm{d}y\,, \end{aligned}$$

(30)

where the last integral, $W_k$, say, reads

$$\begin{aligned} W_k = \int _{b-a}^{b+a} \textrm{erf}(y)\, y^k\, \textrm{d}y&= \frac{(b+a)^{k+1}}{k+1} \left\{ \left[ \textrm{erf}(a+b) + \frac{a+b}{\sqrt{\pi }}\, E_{-\frac{k}{2}}\left( (a+b)^2\right) \right] \right. \\&\left. \,\, - \left( \frac{b-a}{b+a}\right) ^{k+1} \left[ \textrm{erf}(b-a) + \frac{b-a}{\sqrt{\pi }} E_{-\frac{k}{2}}\left( (b-a)^2\right) \right] \right\} . \end{aligned}$$

Here $E_\mu$ stands for the exponential integral. Unfortunately, this transformation formula cannot help in summing up the series considered in (30).

${\textsf{C}}.$ The straightforward counterpart formula of (17) reads

$$\begin{aligned} \int _0^1\dfrac{\textrm{e}^{-a x^2} \sinh (b x)}{\sqrt{1-x^2}} \textrm{d}x&= \frac{\sqrt{\pi }}{4 \textrm{e}^a} \, {\mathbb {D}}_a^{-\frac{1}{2}} \left[ \frac{\textrm{e}^{\frac{b^2}{4a} + a}}{\sqrt{a}} \left\{ \textrm{erf} \left( \frac{2a-b}{2 \sqrt{a}}\right) \right. \right. \\&\qquad \left. \left. + 2\,\textrm{erf} \left( \frac{b}{2 \sqrt{a}}\right) - \textrm{erf} \left( \frac{2a+b}{2 \sqrt{a}}\right) \right\} \right] \,. \end{aligned}$$

The proof is based on the fact that the error function $\textrm{erf}$ is odd.

${\textsf{D}}.$ Concerning the result of Theorem 4, we point out that the sth order raw moment of the rv $\xi$ can be expressed in terms of the Laguerre function $L_\nu (\cdot )$ as

$$\begin{aligned} \mu _s = \frac{2^{\frac{s}{2}}\,\sigma ^s\,\Gamma \left( \frac{s+1}{2}\right) }{\sqrt{\pi }\, \textrm{e}^{\frac{\psi }{2}}\,I_0\left( \frac{\psi }{2}\right) }\,L_{-\frac{s+1}{2}}(\psi ). \end{aligned}$$

${\textsf{E}}.$ The ’control’ property $\varphi _\xi (0) = 1$ of the CHF is satisfied since [22, p. 579, Eq. 7.11.2.10]

$$\begin{aligned} {}_1F_1 \left( \frac{1}{2}; 1; \psi \right) = \textrm{e}^{\frac{\psi }{2}}\,I_0\left( \frac{\psi }{2}\right) , \end{aligned}$$

which is the only non–zero term in (20) at $t=0$.

${\textsf{F}}.$ The inter–connection between the auxiliary function $\kappa (x)$ which occurs in Theorem 3 and Nuttall Q function can be established by the following relation:

$$\begin{aligned} \kappa (x) = \frac{1}{2} \left\{ 1-\frac{\sqrt{2}\,\textrm{e}^{\frac{\psi }{2}}}{\sqrt{\pi }\,I_0\left( \frac{\psi }{2}\right) } Q_{0,0}\left( \sqrt{2 \psi }, \frac{x}{\sigma }\right) \right\} . \end{aligned}$$

To show this relation we start

$$\begin{aligned} \kappa (x)&= \frac{\textrm{e}^{-\frac{\psi }{2}}}{\sigma \sqrt{2 \pi } I_0\left( \frac{\psi }{2}\right) } \left\{ \int _0^\infty \textrm{e}^{-\frac{t^2}{2\sigma ^2}} I_0\left( \sqrt{2\psi } \dfrac{t}{\sigma }\right) \textrm{d}t - \int _x^\infty \textrm{e}^{-\frac{t^2}{2\sigma ^2}}\,I_0\left( \sqrt{2\psi } \dfrac{t}{\sigma }\right) \textrm{d}t\right\} \\&= \dfrac{\textrm{e}^{-\frac{\psi }{2}}}{\sigma \sqrt{2\pi }\, I_0\left( \frac{\psi }{2}\right) } \left\{ \frac{\sigma \sqrt{\pi }}{\sqrt{2}}\,\textrm{e}^{\frac{\psi }{2}}\,I_0\left( \frac{\psi }{2}\right) - \sigma \int _{\frac{x}{\sigma }}^\infty \textrm{e}^{-\frac{1}{2}s^2}\,I_0\left( \sqrt{2\psi }\, s\right) \,\textrm{d}s\right\} \\&= \dfrac{1}{\sqrt{2\pi }\,I_0\left( \frac{\psi }{2}\right) }\, \left\{ \sqrt{\frac{\pi }{2}}\,I_0\left( \frac{\psi }{2}\right) - \textrm{e}^{\frac{\psi }{2}}\,Q_{0,0}\left( \sqrt{2\psi },\frac{x}{\sigma }\right) \right\} \,, \end{aligned}$$

which is equivalent to the statement.

A related result is the different shape expression for the sth order raw moment of the rv $\xi$ having PDF $p_\xi (x)$ given by (19). Indeed,

$$\begin{aligned} \mu _s&= \frac{\textrm{e}^{-\frac{\psi }{2}}}{\sigma \,\sqrt{2 \pi }\, I_0\left( \frac{\psi }{2}\right) } \int _{{\mathbb {R}}} x^s\, \textrm{e}^{-\tfrac{x^2}{2\sigma ^2}}\, I_0\left( \sqrt{2\psi } \dfrac{x}{\sigma }\right) \, \textrm{d}x \\&= \frac{\textrm{e}^{-\frac{\psi }{2}} \left( 1 + \textrm{e}^{\textrm{i}\pi s}\right) }{\sigma \,\sqrt{2 \pi }\, I_0\left( \frac{\psi }{2}\right) }\, \int _0^\infty x^s\, \textrm{e}^{-\tfrac{x^2}{2\sigma ^2}}\, I_0\left( \sqrt{2\psi } \dfrac{x}{\sigma }\right) \, \textrm{d}x \\&= \frac{\textrm{e}^{-\frac{\psi }{2}} \left( 1 + \textrm{e}^{\textrm{i}\pi s}\right) }{\sigma \,\sqrt{2 \pi }\, I_0\left( \frac{\psi }{2}\right) }\, \,\sigma ^{s+1}\, \textrm{e}^{\frac{\psi }{2}}\, Q_{s,0}\left( \sqrt{2 \psi },0\right) \\&= \frac{\sigma ^s\,\left( 1 + \textrm{e}^{\textrm{i}\pi s}\right) }{\sqrt{2 \pi }\, I_0\left( \frac{\psi }{2}\right) }\, Q_{s,0}\left( \sqrt{2 \psi },0\right) \,, \qquad s>-1\,, \end{aligned}$$

where the general fractional order moment is expressed in terms of the Nuttall Q–function.