One-Dimensional Quaternion Fourier Transform with Application to Probability Theory

Abstract

The Fourier transform occupies a central place in applied mathematics, statistics, computer sciences, and engineering. In this work, we introduce the one-dimensional quaternion Fourier transform, which is a generalization of the Fourier transform. We derive the conjugate symmetry of the one-dimensional quaternion Fourier transform for a real signal. We also collect other properties, such as the derivative and Parseval’s formula. We finally study the application of this transformation in probability theory.

1. Introduction

The classical Fourier transform is an indispensable tool in signal and image processing (see, e.g., [1,2]). On the one side, the quaternion Fourier transform (QFT) (for e.g., [3,4,5,6,7,8,9]) as a direct extension of the classical Fourier transform (FT) in the quaternion setting is also an indispensable tool for image processing for quaternion signals. A number of essential properties of the QFT have been demonstrated such as convolution, correlation, energy conservation, and inequalities. These properties are modifications of the corresponding properties of the FT. In [10], a general form of the QFT definition was discussed and the main properties and an application of the proposed transformation were also presented in detail. Moreover, many the mathematical problems are formulated in the language of quaternion algebras, such as the quaternion linear system [11] and linear quaternion differential equations [12].

It is well-nown that the FT plays crucial roles in probability theory. It is related to the characteristic function of any real-valued random variable used to compute the moment and the distribution function. Although the one-dimensional quaternion Fourier transform has been reported in [13,14,15] and the authors of [16] utilized it to construct the one-dimensional quaternion linear canonical transform, we have not yet come across its use in probability theory. Therefore, in this study, we first introduce the one-dimensional quaternion Fourier transform (1DQFT) and state some of its main properties. We develop its application in probability theory. In particular, we define the characteristic function and expected value in the quaternion setting. Distinct from the characteristic function and expected value in the real case, the quaternion characteristic function and expected value have four components. This fact causes several properties of the classical characteristic function to be modified in quaternion domains. We also study the relationship between the quaternion characteristic function and the 1DQFT. We apply this relation to obtain the moments and variance in the framework of a quaternion algebra. The results obtained play an important role in the development of probability theory in the context of a quaternion algebra [17,18].

The remainder of this paper is organized as follows. In Section 2, we briefly review the basic knowledge of quaternion used in the next section. In Section 3, we introduce some useful properties of the 1DQFT such as its convolution and correlation. An application of the properties of the 1DQFT is presented to obtain the inequality in that section. In Section 4, we present the application of the 1DQFT in probability theory. Some conclusions are drawn in Section 5.

2. Notations

Quaternions are a direct extension of complex numbers, which are associative but noncommutative over real numbers $\mathbb{R}$. Every element of a quaternion $\mathbb{H}$ can be written in the following form [19]:

$$\mathbb{H}=\{r=r^a+\mathbf{i}{r}^b+\mathbf{j}{r}^c+\mathbf{k}{r}^d:r^a,r^b,r^c,r^d\in \mathbb{R}\},$$

(1)

which obeys the following algebraic rules:

$$\begin{array}{cc}\hfill \mathbf{ij}& =-\mathbf{ji}=\mathbf{k},\mathbf{jk}=-\mathbf{kj}=\mathbf{i},\mathbf{ki}=-\mathbf{ik}=\mathbf{j},\hfill \\ \hfill {\mathbf{i}}^2& ={\mathbf{j}}^2={\mathbf{k}}^2=\mathbf{ijk}=-1.\hfill \end{array}$$

(2)

For any quaternion $r=r^a+\mathbf{i}{r}^b+\mathbf{j}{r}^c+\mathbf{k}{r}^d\in \mathbb{H}$, we call $r^a$ the scalar part of r. It is denoted by

$$\begin{array}{c}\hfill S_c\left(r\right)=r^a,\end{array}$$

(3)

and its vector part is

$$\begin{array}{c}\hfill V\left(r\right)=\mathbf{i}{r}^b+\mathbf{j}{r}^c+\mathbf{k}{r}^d=\mathbf{r}.\end{array}$$

From (2) we obtain the quaternion multiplications $rz$ in the form

$$rz=r^a{z}^a-\mathbf{r}·\mathbf{z}+r^a\mathbf{z}+z^a\mathbf{r}+\mathbf{r}\times \mathbf{z},$$

(4)

where

$$\begin{array}{cc}\hfill \mathit{r}·\mathit{z}& =r^b{z}^b+r^c{z}^c+r^d{z}^d\hfill \\ \hfill \mathit{r}\times \mathit{z}& =\mathbf{i}(r^c{z}^d-r^d{z}^c)+\mathbf{j}(r^d{z}^b-r^b{z}^d)+\mathbf{k}(r^b{z}^c-r^c{z}^d).\hfill \end{array}$$

According to (4), one can verify that the scalar part (3) satisfies a cyclic multiplication symmetry, i.e.,

$$S_c\left(rpq\right)=S_c\left(prq\right)=S_c\left(qpr\right),\forall r,p,q\in \mathbb{H}.$$

(5)

Analogous to the complex case, the quaternion conjugate of any quaternion r is defined by

$$\overline{r}=r^a-\mathbf{i}{r}^b-\mathbf{j}{r}^c-\mathbf{k}{r}^d,$$

(6)

which satisfies

$$\overline{rz}=\overline{z}\overline{r},\forall z,r\in \mathbb{H}.$$

(7)

It is easily seen from (7) that the quaternion conjugate changes the order of the multiplication. From (6), we obtain the norm or modulus of $r\in \mathbb{H}$ defined as

$$\left|r\right|=\sqrt{r\overline{r}}=\sqrt{{\left(r^a\right)}^2+{\left(r^b\right)}^2+{\left(r^c\right)}^2+{\left(r^d\right)}^2}.$$

(8)

It is routine to check that

$$|r^2{|=|r|}^2,\left|rz\right|=\left|r\right|\left|z\right|,\mathrm{and}|r+z|\le \left|r\right|+\left|z\right|,\forall z,r\in \mathbb{H}.$$

(9)

From the quaternion conjugate (6) and the modulus of r, we obtain the inverse of a nonzero quaternion $r\in \mathbb{H}$ by

$$r^{-1}=\frac{\overline{r}}{{\left|r\right|}^2}.$$

(10)

This shows that $\mathbb{H}$ is a normed division algebra. For $\left|r\right|=1,r$ is a unit quaternion and $r^a=0,r$ is called a pure quaternion.

Similar to a complex number, we may define the inner product for two functions $f,g:\mathbb{R}⟶\mathbb{H}$ as

$${(f,g)}_{L^2(\mathbb{R};\mathbb{H})}={\int }_{\mathbb{R}}f\left(x\right)\overline{g\left(x\right)}dx.$$

(11)

For $f=g$, we get the $L^2(\mathbb{R};\mathbb{H})$-norm as

$${\parallel f\parallel }_{L^2(\mathbb{R};\mathbb{H})}={\left({\int }_{\mathbb{R}}{\left|f\left(x\right)\right|}^{2}dx\right)}^{\frac{1}{2}}.$$

(12)

3. One-Dimensional Quaternion Fourier Transform with Properties

We start by introducing the definition of the one-dimensional quaternion Fourier transform (1DQFT). We recall its essential properties such as Parseval’s formula and the convolution theorem. We propose an application of the properties which is the main result of this section. More details regarding the 1DQFT are discussed in [3].

Definition 1.

The one-dimensional quaternion Fourier transform is defined by

$${\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right)={\int }_{\mathbb{R}}f\left(x\right)e^{\mathbf{i}\xi x}dx,$$

(13)

for$x,\xi \in \mathbb{R}$ and $f\in L^1(\mathbb{R};\mathbb{H})\cap L^2(\mathbb{R};\mathbb{H})$.

Below, we present the reconstruction formula for the 1DQFT.

Definition 2.

Let$f\in L^1(\mathbb{R};\mathbb{H})$and${\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\in L^1(\mathbb{R};\mathbb{H})$. The inverse transform of the 1DQFT is computed by

$$\begin{array}{c}\hfill {\mathcal{F}}_{\mathbf{i}}^{-1}\left[{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\right]\left(x\right)=f\left(x\right)=\frac{1}{2\pi }{\int }_{\mathbb{R}}{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right)e^{-\mathbf{i}\xi x}d\xi .\end{array}$$

(14)

The following results related to the 1DQFT are useful.

Theorem 1.

Let$f\in L^2(\mathbb{R};\mathbb{H})$, we have

$$\begin{array}{c}\hfill {\int }_{\mathbb{R}}{\left|f\left(x\right)\right|}^{2}dx=\frac{1}{2\pi }{\int }_{\mathbb{R}}{\left|{\mathcal{F}}_{\mathbf{i}}\left(\xi \right)\right|}^{2}d\xi ,\end{array}$$

(15)

which is known as Plancherel’s formula for the 1DQFT.

Theorem 2.

Let$f\in L^1(\mathbb{R};\mathbb{H})\cap L^2(\mathbb{R};\mathbb{H})$, one gets

$$\begin{array}{c}\hfill \overline{{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right)}={\mathcal{F}}_{\mathbf{i}}\left\{f^a\right\}(-\xi )-\mathbf{i}{\mathcal{F}}_{\mathbf{i}}\left\{f^b\right\}(-\xi )-\mathbf{j}{\mathcal{F}}_{\mathbf{i}}\left\{f^c\right\}\left(\xi \right)-\mathbf{k}{\mathcal{F}}_{\mathbf{i}}\left\{f^d\right\}\left(\xi \right).\end{array}$$

(16)

In particular, when$f\left(x\right)$is a real-valued function, (16) changes to

$$\begin{array}{c}\hfill \overline{{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right)}={\mathcal{F}}_{\mathbf{i}}\left\{f\right\}(-\xi ),\end{array}$$

(17)

which is known as the conjugate symmetry for the 1DQFT.

Proof. 

It follows from (13) that

$$\begin{array}{cc}\hfill & \overline{{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right)}\hfill \\ \hfill =& {\int }_{\mathbb{R}}{e}^{-\mathbf{i}\xi x}\overline{f\left(x\right)}dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}{e}^{-\mathbf{i}\xi x}\left(f^a\left(x\right)-\mathbf{i}{f}^b\left(x\right)-\mathbf{j}{f}^c\left(x\right)-\mathbf{k}{f}^d\left(x\right)\right)dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}{f}^a\left(x\right)e^{-\mathbf{i}\xi x}dx-\mathbf{i}{\int }_{\mathbb{R}}{f}^b\left(x\right)e^{-\mathbf{i}\xi x}dx-\mathbf{j}{\int }_{\mathbb{R}}{f}^c\left(x\right)e^{\mathbf{i}\xi x}dx-\mathbf{k}{\int }_{\mathbb{R}}{f}^d\left(x\right)e^{\mathbf{i}\xi x}dx\hfill \\ \hfill =& {\mathcal{F}}_{\mathbf{i}}\left\{f^a\right\}(-\xi )-\mathbf{i}{\mathcal{F}}_{\mathbf{i}}\left\{f^b\right\}(-\xi )-\mathbf{j}{\mathcal{F}}_{\mathbf{i}}\left\{f^c\right\}\left(\xi \right)-\mathbf{k}{\mathcal{F}}_{\mathbf{i}}\left\{f^d\right\}\left(\xi \right).\hfill \end{array}$$

This is the required result. □

It is known that one useful tool related to the 1DQFT is the convolution operator. Below, we briefly recall the convolution definition and its convolution theorem.

Definition 3.

The convolution of two quaternion function$f,g\in L^1(\mathbb{R};\mathbb{H})$is defined by

$$(f\ast g)\left(x\right)={\int }_{\mathbb{R}}f\left(y\right)g(x-y)dy.$$

(18)

With Definition 3 above, we get the following [13].

Theorem 3.

Suppose that the quaternion functions$f\in L^1(\mathbb{R};\mathbb{H})$and$g\in L^1(\mathbb{R};\mathbb{H})$. Then, the following holds:

$$\begin{array}{cc}\hfill {\mathcal{F}}_{\mathbf{i}}\{f\ast g\}\left(\xi \right)=& {\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right){\mathcal{F}}_{\mathbf{i}}\left\{g^a\right\}\left(\xi \right)+\mathbf{i}{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right){\mathcal{F}}_{\mathbf{i}}\left\{g^b\right\}\left(\xi \right)\hfill \\ \hfill & +\mathbf{j}{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right){\mathcal{F}}_{\mathbf{i}}\left\{g^c\right\}\left(\xi \right)+\mathbf{k}{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right){\mathcal{F}}_{\mathbf{i}}\left\{g^d\right\}\left(\xi \right).\hfill \end{array}$$

(19)

Moreover,

$$\begin{array}{cc}\hfill (f\ast g)\left(x\right)=& {\mathcal{F}}_{\mu }^{-1}[{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right){\mathcal{F}}_{\mathbf{i}}\left\{g^a\right\}\left(\xi \right)+\mathbf{i}{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right){\mathcal{F}}_{\mathbf{i}}\left\{g^b\right\}\left(\xi \right)\hfill \\ \hfill & +\mathbf{j}{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right){\mathcal{F}}_{\mathbf{i}}\left\{g^c\right\}\left(\xi \right)+\mathbf{k}{\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\left(\xi \right){\mathcal{F}}_{\mathbf{i}}\left\{g^d\right\}\left(\xi \right)]\left(x\right).\hfill \end{array}$$

Definition 4.

The correlation for the 1DQFT of two quaternion functions$f,g\in L^1(\mathbb{R};\mathbb{H})$ is given by the integral

$$(f\circ g)\left(x\right)={\int }_{\mathbb{R}}f(x+y)\overline{g\left(y\right)}dy.$$

(20)

The next result will be very useful for deriving the main results of this work.

Theorem 4.

Suppose$f\in L^1(\mathbb{R};\mathbb{H})$such that f(x) is continuous n-time differentiable; then, for${lim}_{x⟶\pm \infty }f\left(x\right)=0$, the following holds

$${\mathcal{F}}_{\mathbf{i}}\left\{\frac{d^n}{d{x}^n}f\right\}\left(\xi \right)={\mathcal{F}}_{\mathbf{i}}\left\{f\right\}{\xi }^n{(-\mathbf{i})}^n,n\in \mathbb{N}.$$

(21)

Proof. 

Consider first the case $n=1$. Direct computations yield

$$\begin{array}{cc}\hfill {\mathcal{F}}_{\mathbf{i}}\left\{\frac{d}{dx}f\right\}\left(\xi \right)=& {\int }_{\mathbb{R}}\left(\frac{d}{dx}f\left(x\right)\right)e^{\mathbf{i}\xi x}dx\hfill \\ \hfill =& e^{\mathbf{i}\xi x}{f\left(x\right)|}_{-\infty }^{\infty }-{\int }_{\mathbb{R}}f\left(x\right)\frac{d}{dx}{e}^{\mathbf{i}\xi x}dx\hfill \\ \hfill =& e^{\mathbf{i}\xi x}{f\left(x\right)|}_{-\infty }^{\infty }-{\int }_{\mathbb{R}}f\left(x\right)\left(\mathbf{i}\xi e^{\mathbf{i}\xi x}\right)dx\hfill \\ \hfill =& -{\int }_{\mathbb{R}}f\left(x\right)e^{\mathbf{i}\xi x}dx\xi \left(\mathbf{i}\right)\hfill \\ \hfill =& {\int }_{\mathbb{R}}f\left(x\right)e^{\mathbf{i}\xi x}dx\xi (-\mathbf{i})\hfill \\ \hfill =& {\mathcal{F}}_{\mathbf{i}}\left\{f\right\}\xi (-\mathbf{i}).\hfill \end{array}$$

(22)

Using the mathematical induction, we can finish the proof. □

Let us now present the use of the properties of the 1DQFT to explore an inequality (compare to [20]), which is the main result in this section.

Theorem 5

(Szokefalvi-Nagy’s inequality). Suppose that

$${\int }_{-\infty }^{\infty }{\left|f\left(x\right)\right|}^{2}dx=E,\mathrm{and}{\int }_{-\infty }^{\infty }{\left|\frac{df}{dx}\right|}^{2}dx=E_1;$$

(23)

then,

$$\left|f\left(x\right)\right|\le \sqrt[4]{E{E}_1}.$$

(24)

The equality holds for$x=x_0$ if and only if

$$f\left(x\right)=\sqrt[4]{E{E}_1}{e}^{-\alpha |x-x_0|},\alpha =\sqrt{\frac{E_1}{E}}.$$

(25)

Proof. 

For every $x_0$ and $\alpha >0$, we obtain from the inverse transform of the 1DQFT (14) that

$$\begin{array}{c}\hfill |f\left(x_0\right){|}^2=\frac{1}{4{\pi }^2}{\left|{\int }_{-\infty }^{\infty }{\mathcal{F}}_i\left(\xi \right)e^{-\mathbf{i}\xi x_0}d\xi \right|}^2.\end{array}$$

Applying the Cauchy-Schwartz inequality, we see that

$$\begin{array}{cc}\hfill |f\left(x_0\right){|}^2& =\frac{1}{4{\pi }^2}{\left|{\int }_{-\infty }^{\infty }\sqrt{{\alpha }^2+{\xi }^2}{\mathcal{F}}_{\mathbf{i}}\left(\xi \right)\frac{e^{-\mathbf{i}\xi x_0}}{\sqrt{{\alpha }^2+{\xi }^2}}d\xi \right|}^2\hfill \\ \hfill & \le \frac{1}{4{\pi }^2}{\int }_{-\infty }^{\infty }{\left|\sqrt{{\alpha }^2+{\xi }^2}{\mathcal{F}}_{\mathbf{i}}\left(\xi \right)\frac{e^{-\mathbf{i}\xi x_0}}{\sqrt{{\alpha }^2+{\xi }^2}}\right|}^{2}d\xi \hfill \\ \hfill & =\frac{1}{4{\pi }^2}{\int }_{-\infty }^{\infty }\left({\alpha }^2+{\xi }^2\right){\left|{\mathcal{F}}_{\mathbf{i}}\left(\xi \right)\right|}^{2}d\xi {\int }_{-\infty }^{\infty }\frac{d\xi }{{\alpha }^2+{\xi }^2}\hfill \\ \hfill & =\frac{1}{4{\pi }^2}\left({\int }_{-\infty }^{\infty }{\alpha }^2|{\mathcal{F}}_{\mathbf{i}}{\left(\xi \right)|}^{2}d\xi +{\int }_{-\infty }^{\infty }{\xi }^2{\left|{\mathcal{F}}_{\mathbf{i}}\left(\xi \right)\right|}^{2}d\xi \right)\frac{\pi }{\alpha }.\hfill \end{array}$$

By (15) and (21), we obtain

$$\begin{array}{cc}\hfill |f\left(x_0\right){|}^2& \le \frac{1}{4{\pi }^2}\left({\alpha }^2{\int }_{-\infty }^{\infty }|{\mathcal{F}}_{\mathbf{i}}{\left(\xi \right)|}^{2}d\xi +{\int }_{-\infty }^{\infty }{\xi }^2{\left|{\mathcal{F}}_{\mathbf{i}}\left(\xi \right)\right|}^{2}d\xi \right)\frac{\pi }{\alpha }\hfill \\ \hfill & =\frac{1}{4{\pi }^2}\left({\alpha }^{2}2\pi {\int }_{-\infty }^{\infty }{\left|f\left(x\right)\right|}^{2}dx+2\pi {\int }_{-\infty }^{\infty }\left|\frac{df}{dx}\right|dx\right)\frac{\pi }{\alpha }.\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill |f\left(x_0\right){|}^2& \le \left({\alpha }^{2}E+E_1\right)\frac{\pi }{\alpha }\hfill \\ \hfill & =\frac{\alpha E}{2}+\frac{E_1}{2\alpha }.\hfill \end{array}$$

(26)

Since $\alpha =\sqrt{\frac{E_1}{E}}$, we get

$$\begin{array}{c}\hfill |f\left(x_0\right){|}^4\le E{E}_1.\end{array}$$

The equality holds only if

$$\sqrt{{\alpha }^2+{\xi }^2}{\mathcal{F}}_{\mathbf{i}}\left(\xi \right)=\frac{k{e}^{-\mathbf{i}\xi x_0}}{\sqrt{{\alpha }^2+{\xi }^2}}.$$

(27)

Applying the inverse transform of the 1DQFT (14), relation (27) above leads to

$$f\left(x\right)=\frac{k}{\alpha }{e}^{-\alpha |x-x_0|}.$$

(28)

Substituting (28) into relation (23), we obtain

$$E=\frac{k^2}{4{\alpha }^3},E_1=\frac{k^2}{4\alpha }.$$

(29)

From Equation (29), it is easily checked that

$$\alpha =\sqrt{\frac{E_1}{E}}.$$

(30)

Inserting Equation (30) into Equation (29) gives

$$\begin{array}{c}\hfill k^2=4E{\alpha }^3=4{\alpha }^{2}E\alpha =4{\alpha }^{2}E\sqrt{\frac{E_1}{E}}.\end{array}$$

(31)

This means that

$$\begin{array}{c}\hfill k=2\alpha \sqrt[4]{E{E}_1}.\end{array}$$

(32)

This ends the proof of the theorem. □

4. One-Dimensional Quaternion Fourier Transform in Probability Theory

In this part, we present the utility of the one-dimensional quaternion Fourier transform in probability theory. To begin with, we introduce the following definition:

Definition 5.

[21] Let X be a real random variable. A quaternion-valued function$f_{\mathbf{X}}\left(x\right)=f_X^a\left(x\right)+\mathbf{i}{f}_X^b\left(x\right)+\mathbf{j}{f}_X^c\left(x\right)+\mathbf{k}{f}_X^d\left(x\right)$is called the quaternion probability density function of X if

$$\begin{array}{c}\hfill {\int }_{\mathbb{R}}{f}_X^i\left(x\right)dx=1,f_X^i\left(x\right)\ge 0,\forall x\in \mathbb{R},i=a,b,c,d.\end{array}$$

Here,$f_X^i$is a real probability density function. The quaternion cumulative distribution function is expressed as

$$f_X\left(x\right)=\frac{d}{dx}{F}_X\left(x\right),$$

(33)

where the probability P is related to $F_X$given by

$$F_X\left(x\right)=P(X\le x).$$

(34)

Definition 6

(Expected value). Let X be a real random variable with the quaternion probability density function $f_X\left(x\right)$. The expected value $m=\mathrm{E}\left[X\right]$ is defined through

$$\begin{array}{cc}\hfill m& =\mathrm{E}\left[X\right]\hfill \\ \hfill & ={\int }_{\mathbb{R}}x{f}_X\left(x\right)dx\hfill \\ \hfill & ={\int }_{\mathbb{R}}x(f_X^a\left(x\right)+\mathbf{i}{f}_X^b\left(x\right)+\mathbf{j}{f}_X^c\left(x\right)+\mathbf{k}{f}_X^d\left(x\right))dx\hfill \\ \hfill & ={\int }_{\mathbb{R}}x{f}_X^a\left(x\right)dx+\mathbf{i}{\int }_{\mathbb{R}}x{f}_X^b\left(x\right)dx+\mathbf{j}{\int }_{\mathbb{R}}x{f}_X^c\left(x\right)dx+\mathbf{k}{\int }_{\mathbb{R}}x{f}_X^d\left(x\right)dx\hfill \\ \hfill & =\mathrm{E}\left[X_a\right]+\mathbf{i}\mathrm{E}\left[X_b\right]+\mathbf{j}\mathrm{E}\left[X_c\right]+\mathbf{k}\mathrm{E}\left[X_d\right],\hfill \end{array}$$

(35)

Here,

$$\mathrm{E}\left[X_i\right]={\int }_{\mathbb{R}}x{f}_X^i\left(x\right)dx.$$

The expected value of the above definition is often called the mean in the quaternion setting. It is easily seen that

$$\begin{array}{c}\hfill \overline{\mathrm{E}\left[X\right]}=\mathrm{E}\left[X_a\right]-\mathbf{i}\mathrm{E}\left[X_b\right]-\mathbf{j}\mathrm{E}\left[X_c\right]-\mathbf{k}\mathrm{E}\left[X_d\right],\end{array}$$

(36)

and

$$\begin{array}{c}\hfill {\left|m\right|}^2={\left|\mathrm{E}\left[X\right]\right|}^2=\mathrm{E}\left[X\right]\overline{\mathrm{E}\left[X\right]}={\mathrm{E}}^2\left[X_a\right]+{\mathrm{E}}^2\left[X_b\right]+{\mathrm{E}}^2\left[X_c\right]+{\mathrm{E}}^2\left[X_d\right].\end{array}$$

(37)

Definition 7.

Let X be a real random variable with the quaternion probability density function$f_X\left(x\right)$. The characteristic function of X,${\varphi }_X:\mathbb{R}⟶\mathbb{H}$, is defined by the formula

$$\begin{array}{cc}\hfill {\varphi }_X\left(t\right)& =\mathrm{E}\left[e^{\mathbf{i}tX}\right]\hfill \\ \hfill & ={\int }_{\mathbb{R}}{f}_X\left(x\right)e^{\mathbf{i}tx}dx.\hfill \end{array}$$

(38)

Relation (38) above may be expressed in the form

$$\begin{array}{cc}\hfill {\varphi }_X\left(t\right)=& {\int }_{\mathbb{R}}(f_X^a\left(x\right)+\mathbf{i}{f}_X^b\left(x\right)+\mathbf{j}{f}_X^c\left(x\right)+\mathbf{k}{f}_X^d\left(x\right))e^{\mathbf{i}tx}dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}{f}_X^a\left(x\right)e^{\mathbf{i}tx}dx+\mathbf{i}{\int }_{\mathbb{R}}{f}_X^b\left(x\right)e^{\mathbf{i}tx}dx+\mathbf{j}{\int }_{\mathbb{R}}{f}_X^c\left(x\right)e^{\mathbf{i}tx}dx\hfill \\ \hfill & +\mathbf{k}{\int }_{\mathbb{R}}{f}_X^d\left(x\right)e^{itx}dx\hfill \\ \hfill =& {\varphi }_X^a\left(t\right)+\mathbf{i}{\varphi }_X^b\left(t\right)+\mathbf{j}{\varphi }_X^c\left(t\right)+\mathbf{k}{\varphi }_X^d\left(t\right),\hfill \end{array}$$

where

$${\varphi }_X^i\left(t\right)={\int }_{\mathbb{R}}x{f}_X^i\left(x\right)e^{\mathbf{i}tx}dx,i=a,b,c,d.$$

(39)

Definition 8.

Let X be a real random variable with the quaternion probability density function$f_X\left(x\right)$. If${\int }_{\mathbb{R}}\left|{\varphi }_X\left(t\right)\right|dt<\infty$, the function$f_X\left(x\right)$can be constructed using the formula

$$\begin{array}{c}\hfill f_X\left(x\right)=\frac{1}{2\pi }{\int }_{\mathbb{R}}{\varphi }_X\left(t\right)e^{-\mathbf{i}tx}dt.\end{array}$$

(40)

Some basic properties of the quaternion characteristic function are studied in the following results.

Lemma 1.

Let X be a real random variable. Then,

$$\begin{array}{c}\hfill \overline{{\varphi }_X\left(t\right)}={\varphi }_X^a(-t)-\mathbf{i}{\varphi }_X^b(-t)-\mathbf{j}{\varphi }_X^c\left(t\right)-\mathbf{k}{\varphi }_X^d\left(t\right).\end{array}$$

(41)

Proof. 

In fact, we have

$$\begin{array}{cc}\hfill \overline{{\varphi }_X\left(t\right)}=& {\int }_{\mathbb{R}}\overline{f_X\left(x\right)e^{\mathbf{i}tx}}dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}{e}^{-\mathbf{i}tx}\overline{f_X\left(x\right)}dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}{e}^{-\mathbf{i}tx}\left(f_X^a\left(x\right)-\mathbf{i}{f}_X^b\left(x\right)-\mathbf{j}{f}_X^c\left(x\right)-\mathbf{k}{f}_X^d\left(x\right)\right)dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}{f}_X^a\left(x\right)e^{-\mathbf{i}tx}dx-\mathbf{i}{\int }_{\mathbb{R}}{f}_X^b\left(x\right)e^{-\mathbf{i}tx}dx-\mathbf{j}{\int }_{\mathbb{R}}{f}_X^c\left(x\right)e^{\mathbf{i}tx}dx-\mathbf{k}{\int }_{\mathbb{R}}{f}_X^d\left(x\right)e^{\mathbf{i}tx}dx\hfill \\ \hfill =& {\varphi }_X^a(-t)-\mathbf{i}{\varphi }_X^b(-t)-\mathbf{j}{\varphi }_X^c\left(t\right)-\mathbf{k}{\varphi }_X^d\left(t\right).\hfill \end{array}$$

This is the desired result. □

Lemma 2.

Let X be a real random variable. Then,

$$\begin{array}{c}\hfill {\varphi }_{\overline{aX+b}}\left(t\right)={\varphi }_X(-at)e^{-\mathbf{i}tb},\end{array}$$

where a and b are real constants.

Proof. 

Simple computations yield

$$\begin{array}{cc}\hfill {\varphi }_{\overline{aX+b}}\left(t\right)=& {\int }_{\mathbb{R}}{f}_X\left(x\right)\overline{e^{\mathbf{i}t(ax+b)}}dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}{f}_X\left(x\right)e^{-\mathbf{i}tax}dx{e}^{-\mathbf{i}tb}\hfill \\ \hfill =& {\varphi }_X(-at)e^{-\mathbf{i}tb},\hfill \end{array}$$

and the proof is complete. □

It is easily seen that

$$\begin{array}{c}\hfill |{\varphi }_X\left(t\right)|=|{\int }_{\mathbb{R}}{f}_X\left(x\right)e^{itx}dx|\le {\int }_{\mathbb{R}}\left|f_X\left(x\right)\right|dx={\int }_{\mathbb{R}}{f}_X\left(x\right)dx=1.\end{array}$$

(42)

Lemma 3.

For any real random variable X, one has

$$\begin{array}{c}\hfill \overline{{\varphi }_{\overline{aX+b}}\left(t\right)}={\varphi }_X^a\left(at\right)e^{\mathbf{i}tb}-\mathbf{i}{\varphi }_X^b\left(at\right)e^{\mathbf{i}tb}-\mathbf{j}{\varphi }_X^c(-at)e^{-\mathbf{i}tb}-\mathbf{k}{\varphi }_X^d(-at)e^{-\mathbf{i}tb}.\end{array}$$

(43)

Proof. 

It follows from relation (38) that

$$\begin{array}{cc}\hfill \overline{{\varphi }_{\overline{aX+b}}\left(t\right)}=& {\int }_{\mathbb{R}}\overline{f_X\left(x\right)\overline{e^{\mathbf{i}t(ax+b)}}}dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}{e}^{\mathbf{i}t(ax+b)}\overline{f_X\left(x\right)}dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}\left(f_X^a\left(x\right)-\mathbf{i}{f}_X^b\left(x\right)\right)e^{\mathbf{i}tax}dx{e}^{\mathbf{i}tb}-{\int }_{\mathbb{R}}\left(\mathbf{j}{f}_X^c\left(x\right)-\mathbf{k}{f}_X^d\left(x\right)\right)e^{-\mathbf{i}tax}dx{e}^{-\mathbf{i}tb}\hfill \\ \hfill =& \left({\varphi }_X^a\left(at\right)-\mathbf{i}{\varphi }_X^b\left(at\right)\right)e^{\mathbf{i}tb}-\left(\mathbf{j}{\varphi }_X^c(-at)+\mathbf{k}{\varphi }_X^d(-at)\right)e^{-\mathbf{i}tb}.\hfill \end{array}$$

(44)

This is the desired result. □

With Definition 7, we obtain the following important result.

Theorem 6.

If the quaternion characteristic functions${\varphi }_X$and${\psi }_X$of the random variable X are defined by

$${\varphi }_X\left(t\right)={\int }_{\mathbb{R}}{f}_X\left(x\right)e^{\mathbf{i}tx}dx,{\psi }_X\left(x\right)={\int }_{\mathbb{R}}{g}_X\left(t\right)e^{\mathbf{i}tx}dt,$$

(45)

then the following holds:

$$\begin{array}{cc}\hfill {\int }_{\mathbb{R}}{g}_X\left(t\right){\varphi }_X\left(t\right)e^{-\mathbf{i}ty}dt=& {\int }_{\mathbb{R}}{f}_X\left(x\right){\psi }_X^a(x-y)dx+\mathbf{i}{\int }_{\mathbb{R}}{f}_X\left(x\right){\psi }_X^b(x-y)dx\hfill \\ \hfill & +\mathbf{j}{\int }_{\mathbb{R}}{f}_X\left(x\right){\psi }_X^c(y-x)dx+\mathbf{k}{\int }_{\mathbb{R}}{f}_X\left(x\right){\psi }_X^d(y-x))dx.\hfill \end{array}$$

(46)

Proof. 

By virtue of the characteristic function (40), we obtain

$$\begin{array}{cc}\hfill {\varphi }_X\left(t\right)e^{-\mathbf{i}ty}& ={\int }_{\mathbb{R}}{f}_X\left(x\right)e^{\mathbf{i}tx}dx{e}^{-ity}\hfill \\ \hfill & ={\int }_{\mathbb{R}}{f}_X\left(x\right)e^{\mathbf{i}t(x-y)}dx.\hfill \end{array}$$

(47)

Multiplying both sides of the above identity by $g_X\left(t\right)$ and then integrating with respect to $dt$, we see that

$$\begin{array}{cc}\hfill {\int }_{\mathbb{R}}{g}_X\left(t\right){\varphi }_X\left(t\right)e^{-\mathbf{i}ty}dt=& {\int }_{\mathbb{R}}{g}_X\left(t\right)\left({\int }_{\mathbb{R}}{f}_X\left(x\right)e^{\mathbf{i}t(x-y)}dx\right)dt\hfill \\ \hfill =& {\int }_{\mathbb{R}}{\int }_{\mathbb{R}}\left(g_X^a\left(t\right)+\mathbf{i}{g}_X^b\left(t\right)+\mathbf{j}{g}_X^c\left(t\right)+\mathbf{k}{g}_X^d\left(t\right)\right)f_X\left(x\right)e^{\mathbf{i}t(x-y)}dxdt\hfill \\ \hfill =& {\int }_{\mathbb{R}}{\int }_{\mathbb{R}}\left(g_X^a\left(t\right)+\mathbf{i}{g}_X^b\left(t\right)+\mathbf{j}{g}_X^c\left(t\right)+\mathbf{k}{g}_X^d\left(t\right)\right)f_X\left(x\right)e^{\mathbf{i}t(x-y)}dxdt.\hfill \end{array}$$

Fubini’s theorem allows us to obtain

$$\begin{array}{cc}\hfill & {\int }_{\mathbb{R}}{g}_X\left(t\right){\varphi }_X\left(t\right)e^{-\mathbf{i}ty}dt\hfill \\ \hfill =& {\int }_{\mathbb{R}}\left({\int }_{\mathbb{R}}{f}_X\left(x\right)g_X^a\left(t\right)e^{\mathbf{i}t(x-y)}dt+\mathbf{i}{f}_X\left(x\right){\int }_{\mathbb{R}}{g}_X^b\left(t\right)e^{\mathbf{i}t(x-y)}dt\right)dx\hfill \\ \hfill & +{\int }_{\mathbb{R}}\left(\mathbf{j}{f}_X\left(x\right){\int }_{\mathbb{R}}{g}_X^c\left(t\right)e^{\mathbf{i}t(y-x)}dt+\mathbf{k}{f}_X\left(x\right){\int }_{\mathbb{R}}{g}_X^d\left(t\right)e^{\mathbf{i}t(y-x)}dt\right)dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}\left(f_X\left(x\right){\psi }_X^a(x-y)+\mathbf{i}{f}_X\left(x\right){\psi }_X^b(x-y)\right)dx\hfill \\ \hfill & +{\int }_{\mathbb{R}}\left(\mathbf{j}{f}_X\left(x\right){\psi }_X^c(y-x)+\mathbf{k}{f}_X\left(x\right){\psi }_X^d(y-x)\right)dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}{f}_X\left(x\right){\psi }_X^a(x-y)dx+\mathbf{i}{\int }_{\mathbb{R}}{f}_X\left(x\right){\psi }_X^b(x-y)dx\hfill \\ \hfill & +\mathbf{j}{\int }_{\mathbb{R}}{f}_X\left(x\right){\psi }_X^c(y-x)dx+\mathbf{k}{\int }_{\mathbb{R}}{f}_X\left(x\right){\psi }_X^d(y-x))dx.\hfill \end{array}$$

This is the desired result. □

It follows from (33) that

$${\mathcal{F}}_i\left\{\frac{d}{dx}{F}_X\right\}\left(t\right)={\mathcal{F}}_i\left\{f_X\right\}\left(t\right)={\varphi }_X\left(t\right),$$

(48)

where $F_X\left(x\right)$ is the quaternion distribution function of random variable X. Furthermore, the application of (21) leads to

$${\varphi }_X\left(t\right)={\mathcal{F}}_{\mathbf{i}}\left\{F_X\right\}\left(t\right)t(-\mathbf{i}),$$

(49)

and thus

$${\mathcal{F}}_{\mathbf{i}}\left\{F_X\right\}\left(t\right)=\frac{\mathbf{i}}{t}{\varphi }_X\left(t\right),t\ne 0.$$

(50)

Based on (35), we define the nth moment of a real random variable X as

$$m_n=\mathrm{E}\left[X^n\right]={\int }_{\mathbb{R}}{x}^n{f}_X\left(x\right)dx,n=1,2,3,\cdots ,$$

(51)

provided the integral exists. It is obvious that for $n=1$ in (51), we obtain the first moment $m_1$, which is called the expectation of X.

Theorem 7.

If X is a real random variable, then there exists nth continuous derivatives for the quaternion characteristic function${\varphi }_X\left(t\right)$which is given by the formula

$$\frac{d^k}{d{t}^k}{\varphi }_X\left(t\right)={\int }_{\mathbb{R}}{f}_X\left(x\right)e^{\mathbf{i}tx}{x}^{k}dx{\mathbf{i}}^k.$$

(52)

Moreover,

$$m_k=\mathrm{E}\left[X^k\right]=\frac{d^k}{d{t}^k}{\varphi }_X\left(0\right){(-\mathbf{i})}^k,k=1,2,3,....n.$$

(53)

Proof. 

For $k=1$, direct computations reveal that

$$\begin{array}{cc}\hfill \frac{d}{dt}{\varphi }_X\left(t\right)=& \frac{d}{dt}{\int }_{\mathbb{R}}{f}_X\left(x\right)e^{\mathbf{i}tx}dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}{f}_X\left(x\right)\left(\frac{d}{dt}{e}^{\mathbf{i}tx}\right)dx\hfill \\ \hfill =& {\int }_{\mathbb{R}}{f}_X\left(x\right)e^{\mathbf{i}tx}xdx\mathbf{i}.\hfill \end{array}$$

(54)

In view of relation (54), we further get

$$\begin{array}{cc}\hfill \frac{d^2}{d{t}^2}{\varphi }_X\left(t\right)=& \frac{d}{dt}\left({\int }_{\mathbb{R}}{f}_X\left(x\right)e^{\mathbf{i}tx}xdx\mathbf{i}\right)\hfill \\ \hfill =& {\int }_{\mathbb{R}}{f}_X\left(x\right)e^{\mathbf{i}tx}{x}^{2}dx{\mathbf{i}}^2.\hfill \end{array}$$

(55)

This means that

$$\begin{array}{cc}\hfill \frac{d^k}{d{t}^k}{\varphi }_X\left(t\right)=& {\int }_{\mathbb{R}}{f}_X\left(x\right)e^{\mathbf{i}tx}{x}^{k}dx{\mathbf{i}}^k,\hfill \end{array}$$

(56)

and

$$\begin{array}{c}\hfill \frac{d^k}{d{t}^k}{\varphi }_X\left(0\right)={\int }_{\mathbb{R}}{f}_X\left(x\right)x^{k}dx{\mathbf{i}}^k.\end{array}$$

(57)

Hence,

$$\begin{array}{c}\hfill \frac{d^k}{d{t}^k}{\varphi }_X\left(0\right){\left({\mathbf{i}}^k\right)}^{-1}={\int }_{\mathbb{R}}{f}_X\left(x\right)x^{k}dx.\end{array}$$

Or, equivalently,

$$\begin{array}{cc}\hfill \frac{d^k}{d{t}^k}{\varphi }_X\left(0\right){(-\mathbf{i})}^k=& {\int }_{\mathbb{R}}{x}^k{f}_X\left(x\right)dx\hfill \\ \hfill =& \mathrm{E}\left[X^k\right].\hfill \end{array}$$

The assertion is proved. □

Definition 9.

Let X be a any real random variable. The variance of X in the quaternion setting is defined by

$$\begin{array}{cc}\hfill {\sigma }^2& =\mathrm{E}\left[(X-\mathrm{E}\left[X\right])\left(\overline{X-\mathrm{E}\left[X\right]}\right)\right]\hfill \\ \hfill & =\mathrm{E}\left[(X-\mathrm{E}\left[X\right])(X-\overline{\mathrm{E}\left[X\right]})\right]\hfill \\ \hfill & =\mathrm{E}\left[(X^2-X\overline{\mathrm{E}\left[X\right]}-X\mathrm{E}\left[X\right]+|\mathrm{E}\left[X\right]{|}^2\right]\hfill \\ \hfill & =\mathrm{E}\left[X^2\right]-\mathrm{E}\left[X\right]\overline{\mathrm{E}\left[X\right]}-\mathrm{E}\left[X\right]\mathrm{E}\left[X\right]+{\left|\mathrm{E}\left[X\right]\right|}^2]\hfill \\ \hfill & =\mathrm{E}\left[X^2\right]-{\left(\mathrm{E}\left[X\right]\right)}^2.\hfill \end{array}$$

(58)

The variance ${\sigma }^2$ of a real random variable in terms of the quaternion characteristic function can be obtained as

$$\begin{array}{c}\hfill {\sigma }^2=\frac{d^2}{d{t}^2}{\varphi }_X\left(0\right){(-\mathbf{i})}^2-{\left(\frac{d^2}{d{t}^2}{\varphi }_X\left(0\right)(-\mathbf{i})\right)}^2.\end{array}$$

(59)

The following example illustrates the use of the results mentioned above.

Example 1.

The random variable X has the probability density function

$$f\left(x\right)=\frac{1}{\left|\sigma \right|\sqrt{2\pi }}{e}^{-\frac{{(x-|m\left|\right)}^2}{{2\left|\sigma \right|}^2}}.$$

(60)

We find the first and second moments of X.

It follows from (40) that

$${\varphi }_X\left(t\right)={\int }_{\mathbb{R}}\frac{1}{\left|\sigma \right|\sqrt{2\pi }}{e}^{-\frac{{(x-|m\left|\right)}^2}{{2\left|\sigma \right|}^2}}{e}^{\mathbf{i}tx}dx.$$

Performing a change of variable $x-\left|m\right|=y$, it is easily seen that

$$\begin{array}{cc}\hfill {\varphi }_X\left(t\right)& ={\int }_{\mathbb{R}}\frac{1}{\left|\sigma \right|\sqrt{2\pi }}{e}^{-\frac{y^2}{{2\left|\sigma \right|}^2}}{e}^{\mathbf{i}t\left(\right|m|+y)}dy\hfill \\ \hfill & =\frac{e^{\mathbf{i}t\left|m\right|}}{\left|\sigma \right|\sqrt{2\pi }}{\int }_{\mathbb{R}}{e}^{-\frac{y^2}{{2\left|\sigma \right|}^2}+\mathbf{i}ty}dy.\hfill \end{array}$$

(61)

We further obtain

$$\begin{array}{cc}\hfill {\varphi }_X\left(t\right)& =\frac{e^{\mathbf{i}t\left|m\right|}}{\sigma \sqrt{2\pi }}{\int }_{\mathbb{R}}{e}^{-\frac{1}{{2\left|\sigma \right|}^2}\left(y^2-2{\sigma }^2\mathbf{i}ty\right)}dy\hfill \\ \hfill & =\frac{e^{\mathbf{i}t\left|m\right|}}{\left|\sigma \right|\sqrt{2\pi }}{\int }_{\mathbb{R}}{e}^{-\frac{1}{{2\left|\sigma \right|}^2}\left({\left(y-{\left|\sigma \right|}^2\mathbf{i}t\right)}^2-{\left({\left|\sigma \right|}^2\mathbf{i}t\right)}^2\right)}dy\hfill \\ \hfill & =\frac{e^{\mathbf{i}t\left|m\right|}}{\left|\sigma \right|\sqrt{2\pi }}{\int }_{\mathbb{R}}{e}^{\frac{{\left({\left|\sigma \right|}^2\mathbf{i}t\right)}^2}{{2\left|\sigma \right|}^2}}{e}^{-\frac{1}{2{\sigma }^2}{\left(y-{\left|\sigma \right|}^2\mathbf{i}t\right)}^2}dy\hfill \\ \hfill & =\frac{e^{\mathbf{i}t\left|m\right|}}{\left|\sigma \right|\sqrt{2\pi }}{e}^{\frac{{\left({\left|\sigma \right|}^2\mathbf{i}t\right)}^2}{{2\left|\sigma \right|}^2}}{\int }_{\mathbb{R}}{e}^{-\frac{1}{{2\left|\sigma \right|}^2}{\left(y-{\left|\sigma \right|}^2\mathbf{i}t\right)}^2}dy\hfill \\ \hfill & =\frac{e^{\mathbf{i}t\left|m\right|}}{\left|\sigma \right|\sqrt{2\pi }}{e}^{\frac{{\left|\sigma \right|}^2{\mathbf{i}}^2{t}^2}{2}}{\int }_{\mathbb{R}}{e}^{-a{x}^2}dx\hfill \\ \hfill & =\frac{e^{\mathbf{i}t\left|m\right|}}{\left|\sigma \right|\sqrt{2\pi }}{e}^{\frac{-{\left|\sigma \right|}^2{t}^2}{2}}\sqrt{{2\pi \left|\sigma \right|}^2}\hfill \\ \hfill & =e^{\mathbf{i}t\left|m\right|-\frac{{\left|\sigma \right|}^2{t}^2}{2}}.\hfill \end{array}$$

(62)

Therefore,

$$\begin{array}{cc}\frac{d}{dt}{\varphi }_X\left(t\right)\hfill & =\left(\mathbf{i}\left|m\right|-{t\left|\sigma \right|}^2\right)e^{\mathbf{i}t\left|m\right|-\frac{{\left|\sigma \right|}^2{t}^2}{2}}\hfill \\ \frac{d^2}{d{t}^2}{\varphi }_X\left(t\right)\hfill & =\left(\mathbf{i}\left|m\right|-{\left({t\left|\sigma \right|}^2\right)}^2-{\left|\sigma \right|}^2\right)e^{\mathbf{i}t\left|m\right|-\frac{{\left|\sigma \right|}^2{t}^2}{2}}\hfill \\ \frac{d^3}{d{t}^3}{\varphi }_X\left(t\right)\hfill & =\left(\mathbf{i}\left|m\right|-{\left({t\left|\sigma \right|}^2\right)}^3-{3\left|\sigma \right|}^2\left(\mathbf{i}\right|m|-{t\left|\sigma \right|}^2)\right)e^{\mathbf{i}t\left|m\right|-\frac{{\left|\sigma \right|}^2{t}^2}{2}}\hfill \\ \frac{d^4}{d{t}^4}{\varphi }_X\left(t\right)\hfill & =\left(\mathbf{i}\left|m\right|-{\left({t\left|\sigma \right|}^2\right)}^4-{6\left|\sigma \right|}^2\left(\mathbf{i}\right|m|-{t\left|\sigma \right|}^2{)+3|\sigma |}^4\right)e^{\mathbf{i}t\left|m\right|-\frac{{\left|\sigma \right|}^2{t}^2}{2}}.\hfill \end{array}$$

(63)

Combining (53) and (63) yields

$$\begin{array}{c}\hfill m_1=\mathrm{E}\left[X\right]=\frac{d}{dt}{\varphi }_X\left(0\right)(-\mathbf{i})=\left(\mathbf{i}\right|m\left|\right)(-\mathbf{i})=\left(-{\mathbf{i}}^2\right)\left|m\right|=\left|m\right|.\end{array}$$

(64)

Similarly,

$$\begin{array}{cc}{m}_2\hfill & =\mathrm{E}\left[X^2\right]\hfill \\ \hfill & =\frac{d^2}{d{t}^2}{\varphi }_X\left(0\right){(-\mathbf{i})}^2\hfill \\ \hfill & =\left(-{\sigma }^2+\mathbf{i}\left|m\right|\right){(-\mathbf{i})}^2\hfill \\ \hfill & =\left({\sigma }^2-{\mathbf{i}}^2\left|m\right|\right).\hfill \end{array}$$

(65)

5. Conclusions

In this paper, we introduced the one-dimensional quaternion Fourier transform (1DQFT) and utilized its properties for deriving the inequality related to this transformation. We demonstrated its use in probability theory. The characteristic function, expected value, and variance in the quaternion setting were studied in detail. These results play an important role in the development of probability theory in the context of quaternion algebra. In the future, uncertainty principles relating the quaternion probability density function and its characteristic function will be investigated.