自相关函数是否完全描述了随机过程？

随机过程是否由其自相关函数完全描述？

如果不是，则需要哪些其他属性？

随机过程的完整描述是什么意思？好吧，在数学上，一个随机过程是随机变量的集合，在索引集，每个瞬时都有一个，其中通常是是整条实线或正实线，完整的描述意味着对于每个整数和时刻，我们知道的（联合）分布所述随机变量，，$\{X(t) : t \in {\mathbb T}\}$$t$ $\mathbb T$$\mathbb T$$n \geq 1$$n$$t_1, t_2, \ldots, t_n \in \mathbb T$$n$$X(t_1)$$X(t_2)$$\ldots, X(t_n)$。这是大量信息：我们需要知道每个时刻的的CDF，所有时刻选择的和的（二维）联合CDF。和，，和等三维）CDF 等，等等。$X(t)$$t$$X(t_1)$$X(t_2)$$t_1$$t_2$$X(t_1)$$X(t_2)$$X(t_3)$

因此，人们自然会寻找更简单的描述和更严格的模型。当过程对于时间原点的变化不变时，就会发生一种简化。这意味着

• 该过程中的所有随机变量具有相同的CDF：对于所有。$F_{X(t_1)}(x) = F_{X(t_2)}(x)$$t_1, t_2$
• 任何两个随机变量相隔一定的时间指定量具有相同的关节CDF作为任何其他对随机变量的分离由相同量的时间。例如，随机变量和相隔秒，随机变量和也因此$X(t_1)$$X(t_1 + \tau)$$\tau$$X(t_2)$$X(t_2 + \tau)$$F_{X(t_1), X(t_1 + \tau)}(x,y) = F_{X(t_2), X(t_2 + \tau)}(x,y)$
• 隔开和任意三个随机变量，，具有与，，也将和隔开，$X(t_1)$$X(t_1 + \tau_1)$$X(t_1 + \tau_1 + \tau_2)$$\tau_1$$\tau_2$$X(t_2)$$X(t_2 + \tau_1)$$X(t_2 + \tau_1 + \tau_2)$$\tau_1$$\tau_2$
• 对于所有多维CDF ，等等。有关多维案例的详细信息，请参见例如Peter K.的答案。

实际上，随机过程的概率描述不依赖于我们选择在时间轴上称呼原点的方式：移动所有时刻$t_1, t_2, \ldots, t_n$一定的固定量$\tau$到$t_1 + \tau, t_2 + \tau, \ldots, t_n + \tau$对随机变量给出相同的概率描述。此属性称为严格意义上的平稳性 享有此属性的随机过程称为严格平稳的随机过程，或更简单地说，称为平稳的随机过程。

请注意，严格的平稳性本身并不需要任何特定形式的CDF。例如，它没有说所有变量都是高斯变量。

形容词严格建议可以定义较宽松的平稳形式。如果$N^{\text{th}}$的阶联合CDF $X(t_1), X(t_2), \ldots, X(t_N)$是一样的$N^{\text{th}}$阶的关节CDF $X(t_1+\tau), X(t_2+\tau), \ldots, X(t_N +\tau)$为所有的选择$t_1,t_2, \ldots, t_N$和$\tau$，则该随机过程被认为是静止的，以顺序$N$和作为被称为$N^{\text{th}}$阶平稳随机过程。请注意，一个 $N^{\text{th}}$阶平稳随机过程也静止顺序$n$每个正$n < N$。（这是因为$n^{\text{th}}$阶关节CDF是极限$N^{\text{th}}$阶CDF作为$N-n$的参数接近$\infty$：的一般化 $F_X(x) = \lim_{y\to\infty}F_{X,Y}(x,y)$）。那么严格平稳的随机过程是对所有阶数$N$都静止的随机过程。

如果一个随机过程至少在$1$阶上是平稳的，则所有$X(t)$都具有相同的分布，因此，假设存在均值，则对于所有t，$E[X(t)] = \mu$都是相同的。类似地， E [ （X （t ））2 ]对于所有t都是相同的，称为过程的幂。所有物理过程都具有有限的能力，因此通常假定 E [$t$$E[(X(t))^2]$$t$$E[(X(t))^2] < \infty$在这种情况下，尤其是在较早的工程文献中，该过程称为二阶过程。名称的选择是不幸的，因为它邀请的困惑与二阶 平稳（参见我的上stats.SE这个答案），所以在这里我们将调用一个过程，这$E[(X(t))^2]$是有限的对于所有$t$（是否为$E[(X(t))^2]$是一个常数），为有限功率处理和避免这种混淆。但请再次注意

一阶平稳过程不必是有限幂过程。

考虑一个随机过程，该过程的阶数为$2$。现在，由于的联合分布$X(t_1)$和$X(t_1 + \tau)$是一样的联合分布函数$X(t_2)$和$X(t_2 + \tau)$，$E[X(t_1)X(t_1 + \tau)] = E[X(t_2)X(t_2 + \tau)]$，该值仅取决于$\tau$。这些期望对于有限功率过程是有限的，它们的值称为过程的自相关函数：$R_X(\tau) = E[X(t)X(t+\tau)]$是$\tau$的函数，时间间隔随机变量$X(t)$和$X(t+\tau)$，并且完全不依赖于$t$。另请注意，

$$E[X(t)X(t+\tau)] = E[X(t+\tau)X(t)] = E[X(t+\tau)X(t + \tau - \tau)] = R_X(-\tau),$$ 因此自相关函数是其参数的偶函数。

有限幂二阶平稳随机过程具有以下性质

1. 它的平均值$E[X(t)]$是一个常数
2. 它的自相关函数$R_X(\tau) = E[X(t)X(t+\tau)]$是$\tau$的函数，是随机变量$X(t)$和$X(t+\tau)$的时间间隔，而不是完全取决于$t$。

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平稳性的假设在某种程度上简化了随机过程的描述，但是，对于对从实验数据中构建模型感兴趣的工程师和统计学家而言，估算所有这些CDF并非易事，特别是当只有一部分样本路径时（或实现）$x(t)$在其上的测量可以进行。相对容易进行的两个测量（因为工程师已经在其工作台上配备了必要的仪器（或软件库中的MATLAB / Python / Octave / C ++中的程序））的直流值 $\frac 1T\int_0^T x(t)\,\mathrm dt$ of $x(t)$ and the autocorrelation function $R_x(\tau) = \frac 1T\int_0^T x(t)x(t+\tau)\,\mathrm dt$ (or its Fourier transform, the power spectrum of $x(t)$). Taking these measurements as estimates of the mean and the autocorrelation function of a finite-power process leads to a very useful model that we discuss next.

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A finite-power random process is called a wide-sense-stationary (WSS) process (also weakly stationary random process which fortunately also has the same initialism WSS) if it has a constant mean and its autocorrelation function $R_X(t_1, t_2) = E[X(t_1)X(t_2)]$ depends only on the time difference $t_1 - t_2$ (or $t_2 - t_1$).

Note that the definition says nothing about the CDFs of the random variables comprising the process; it is entirely a constraint on the first-order and second-order moments of the random variables. Of course, a finite-power second-order stationary (or $N^{\text{th}}$-order stationary (for $N>2$) or strictly stationary) random process is a WSS process, but the converse need not be true.

A WSS process need not be stationary to any order.

Consider, for example, the random process $\{X(t)\colon X(t)= \cos (t + \Theta), -\infty < t < \infty\}$ where $\Theta$ takes on four equally likely values $0, \pi/2, \pi$ and $3\pi/2$. (Do not be scared: the four possible sample paths of this random process are just the four signal waveforms of a QPSK signal). Note that each $X(t)$ is a discrete random variable that, in general, takes on four equally likely values $\cos(t), \cos(t+\pi/2)=-\sin(t), \cos(t+\pi) = -\cos(t)$ and $\cos(t+3\pi/2)=\sin(t)$, It is easy to see that in general $X(t)$ and $X(s)$ have different distributions, and so the process is not even first-order stationary. On the other hand,

$$E[X(t)] = \frac 14\cos(t)+ \frac 14(-\sin(t)) + \frac 14(-\cos(t))+\frac 14 \sin(t) = 0$$ for every $t$ while $$\begin{align} E[X(t)X(s)]&= \left.\left.\frac 14\right[\cos(t)\cos(s) + (-\cos(t))(-\cos(s)) + \sin(t)\sin(s) + (-\sin(t))(-\sin(s))\right]\\ &= \left.\left.\frac 12\right[\cos(t)\cos(s) + \sin(t)\sin(s)\right]\\ &= \frac 12 \cos(t-s). \end{align}$$ In short, the process has zero mean and its autocorrelation function depends only on the time difference $t-s$, and so the process is wide sense stationary. But it is not first-order stationary and so cannot be stationary to higher orders either.

Even for WSS processes that are second-order stationary (or strictly stationary) random processes, little can be said about the specific forms of the distributions of the random variables. In short,

A WSS process is not necessarily stationary (to any order), and the mean and autocorrelation function of a WSS process is not enough to give a complete statistical description of the process.

Finally, suppose that a stochastic process is assumed to be a Gaussian process ("proving" this with any reasonable degree of confidence is not a trivial task). This means that for each $t$, $X(t)$ is a Gaussian random variable and for all positive integers $n \geq 2$ and choices of $n$ time instants $t_1$, $t_2$, $\ldots, t_n$, the $N$ random variables $X(t_1)$, $X(t_2)$, $\ldots, X(t_n)$ are jointly Gaussian random variables. Now a joint Gaussian density function is completely determined by the means, variances, and covariances of the random variables, and in this case, knowing the mean function $\mu_X(t) = E[X(t)]$ (it need not be a constant as is required for wide-sense-stationarity) and the autocorrelation function $R_X(t_1, t_2) = E[X(t_1)X(t_2)]$ for all $t_1, t_2$ (it need not depend only on $t_1-t_2$ as is required for wide-sense-stationarity) is sufficient to determine the statistics of the process completely.

If the Gaussian process is a WSS process, then it is also a strictly stationary Gaussian process. Fortunately for engineers and signal processors, many physical noise processes can be well-modeled as WSS Gaussian processes (and therefore strictly stationary processes), so that experimental observation of the autocorrelation function readily provides all the joint distributions. Furthermore since Gaussian processes retain their Gaussian character as they pass through linear systems, and the output autocorrelation function is related to th input autocorrelation function as

$$R_y = h*\tilde{h}*R_X$$ so that the output statistics can also be easily determined, WSS process in general and WSS Gaussian processes in particular are of great importance in engineering applications.