# 数学代写|傅里叶分析代写Fourier analysis代考|Weakly Stationary Stochastic Processes

## 数学代写|傅里叶分析代写Fourier analysis代考|Weakly Stationary Stochastic Processes

Throughout this section, $T=\mathbb{R}, \mathbb{Z}$ or $\mathbb{N}$ and $(\Omega, \mathcal{E}, P)$ is a probability space. $X(t, \omega): T \times \Omega \rightarrow \mathbb{C}$ is a stochastic process. In the case of $T=\mathbb{Z}$ or $\mathbb{N}$, we sometimes write $X_t(\omega)$ instead of $X(t, \omega)$.

Definition 8.3 A stochastic process $X(t, \omega)$ is strongly stationary if the distribution of $\left(X\left(t_1+t, \omega\right), X\left(t_2+t, \omega\right), \cdots, X\left(t_n+t, \omega\right)\right)$ is independent of $t$ for any $\mathbf{t}=\left(t_1, t_2, \cdots, t_n\right) \in \mathcal{T}$
We define $\Phi_{t_1, t_2, \cdots, t_n}(E)$ for each $E \in \mathcal{B}\left(\mathbb{C}^n\right)$ by
$$\Phi_{t_1, t_2, \cdots, t_n}(E)=P\left{\omega \in \Omega \mid\left(X\left(t_1, \omega\right), \cdots, X\left(t_n, \omega\right)\right) \in E\right}$$
Then $X(t, \omega)$ is strongly stationary if and only if
$$\Phi_{t_1, t_2, \cdots, t_n}=\Phi_{t_1+t, t_2+t, \cdots, t_n+t}, \quad n \in \mathbb{N}, \quad t_1, t_2, \cdots, t_n \in T$$
Definition 8.4 $X(t, \omega)$ is said to be weakly stationary if the following conditions are satisfied:
(i) The absolute moment of the second order is finite:
$$\mathbb{E}|X(t, \omega)|^2<\infty \quad \text { for each } \quad t \in T$$
(ii) The expectation is constant throughout time:
$$\mathbb{E} X(t, \omega)=m(t)=m \quad \text { constant for all } \quad t \in T$$
(iii) The covariance depends only upon the difference $u=s-t$ of times:
$$\rho(s, t)=\rho(s+h, t+h) \quad \text { for any } \quad s, t, h \in T$$
In this case, the convariance $\rho(s, t)$ is a function of $s-t$. Hence we write it as $\rho(u), u=s-t$ instead of $\rho(s, t) .^6$

## 数学代写|傅里叶分析代写Fourier analysis代考|Periodicity of Weakly Stationary Stochastic Process

Theorem 8.6′ (Spectral representation of $\rho: T=\mathbb{R}$ ) If $X: \mathbb{R} \times \Omega \rightarrow \mathbb{C}$ is a measurable and weakly stationary process, its covariance function $\rho(u)$ can be expressed as the Fourier transform of certain positive Radon measure $v$ on $\mathbb{R}$ :
$$\rho(u)=\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} e^{-i u t} d v(t) .$$
Such a measure $v$ is determined uniquely.
Since $X(t, \omega)$ is measurable, $\rho(u)$ is continuous by Theorem 8.5. We have already confirmed the positive semi-definiteness of $\rho(u)$. Hence Theorem 8.6′ follows from Bochner’s Theorem 6.11 (p. 150).
$\nu$ is not necessarily a probability measure because $\rho(0)=1 / \sqrt{2 \pi}$ is not necessarily satisfied.

The Radon measure $v$ appearing Theorem 8.6 or Theorem $8.6^{\prime}$ is called the spectral measure of $X(t, \omega)$.
The function $F: \mathbb{R} \rightarrow \mathbb{R}$ defined by
$$F(\alpha)=v((-\infty, \alpha)), \quad \alpha \in \mathbb{R}$$
is called the spectral distribution function of $X(t, \omega)$. If $v$ is absolutely continuous with respect to the Lebesgue measure $d t$, the Radon-Nikodým derivative $p(t) \in$ $\rho^1(\mathbb{R}, \mathbb{R}), p(t) \geqq 0$ is called the spectral density function of $X(t, \omega)$; i.e.
$$v(E)=\int_E p(t) d t$$
The spectral density function, if it exists, is unique. We will discuss later on the condition which assures the existence of a spectral density function of a stochastic process.

# 傅里叶分析代写

## 数学代写|傅里叶分析代写Fourier analysis代考|Weakly Stationary Stochastic Processes

$$\Phi_{t_1, t_2, \cdots, t_n}=\Phi_{t_1+t, t_2+t, \cdots, t_n+t}, \quad n \in \mathbb{N}, \quad t_1, t_2, \cdots, t_n \in T$$

(i) 二阶的绝对矩是有限的:
$$\mathbb{E}|X(t, \omega)|^2<\infty \quad \text { for each } \quad t \in T$$
(ii) 期望在整个时间是不变的:
$$\mathbb{E} X(t, \omega)=m(t)=m \quad \text { constant for all } \quad t \in T$$
(iii) 协方差仅取决于差异 $u=s-t$ 次数:
$$\rho(s, t)=\rho(s+h, t+h) \quad \text { for any } \quad s, t, h \in T$$

## 数学代写|傅里叶分析代写Fourier analysis代考|Periodicity of Weakly Stationary Stochastic Process

$$\rho(u)=\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} e^{-i u t} d v(t) .$$

$$F(\alpha)=v((-\infty, \alpha)), \quad \alpha \in \mathbb{R}$$

$$v(E)=\int_E p(t) d t$$

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