A função de autocorrelação descreve completamente um processo estocástico?

Um processo estocástico é completamente descrito por sua função de autocorrelação?

Caso contrário, quais propriedades adicionais seriam necessárias?

O que se entende por uma descrição completa de um processo estocástico? Bem, matematicamente, um processo estocástico é uma coleção $\{X(t) : t \in {\mathbb T}\}$ de variáveis ​​aleatórias, uma para cada instante $t$ em um conjunto de índices $\mathbb T$ , onde geralmente $\mathbb T$ é a linha real inteira ou a linha real positiva, e uma descrição completa significa que para cada número inteiro $n \geq 1$ e $n$ instantes de tempo $t_1, t_2, \ldots, t_n \in \mathbb T$ , conhecemos as distribuições (conjuntas) das$n$ variáveis ​​aleatórias$X(t_1)$ ,$X(t_2)$ ,$\ldots, X(t_n)$ . É umaquantidadeenormede informações: precisamos conhecer o CDF de$X(t)$ para cada instante$t$ , o CDF ( bidimensional) conjunto de$X(t_1)$ e$X(t_2)$ para todas as opções de tempo instantes$t_1$ e$t_2$ , os CDFs ( tridimensionais) de$X(t_1)$ ,$X(t_2)$ e$X(t_3)$ , etc. etc. etc.

Então, naturalmente, as pessoas procuravam descrições mais simples e modelos mais restritivos. Uma simplificação ocorre quando o processo é invariável a uma mudança na origem do tempo. O que isso significa é que

• Todas as variáveis ​​aleatórias no processo têm CDFs idênticos: $F_{X(t_1)}(x) = F_{X(t_2)}(x)$ para todos os$t_1, t_2$ .
• Quaisquer duas variáveis ​​aleatórias separadas por uma quantidade de tempo especificada têm o mesmo CDF conjunto que qualquer outro par de variáveis ​​aleatórias separadas pela mesma quantidade de tempo. Por exemplo, as variáveis ​​aleatórias $X(t_1)$ e $X(t_1 + \tau)$ são separadas por $\tau$ segundos, assim como as variáveis ​​aleatórias $X(t_2)$ e $X(t_2 + \tau)$ e, portanto, $F_{X(t_1), X(t_1 + \tau)}(x,y) = F_{X(t_2), X(t_2 + \tau)}(x,y)$
• Quaisquer três variáveis ​​aleatórias $X(t_1)$ , $X(t_1 + \tau_1)$ , $X(t_1 + \tau_1 + \tau_2)$ espaçadas $\tau_1$ e $\tau_2$ afastadas têm a mesma CDF conjunta que $X(t_2)$ , $X(t_2 + \tau_1)$ , $X(t_2 + \tau_1 + \tau_2)$ que, como também espaçados$\tau_1$ e$\tau_2$ ,
• e assim por diante para todos os CDFs multidimensionais. Veja, por exemplo, a resposta de Peter K. para obter detalhes do caso multidimensional.

Efetivamente, as descrições probabilísticas do processo aleatório não dependem do que escolhemos chamar de origem no eixo do tempo: deslocando todos os instantes de tempo $t_1, t_2, \ldots, t_n$ por alguma quantidade fixa $\tau$ para $t_1 + \tau, t_2 + \tau, \ldots, t_n + \tau$ fornece a mesma descrição probabilística das variáveis ​​aleatórias. Essa propriedade é chamada de estacionariedade de sentido estrito e um processo aleatório que goza dessa propriedade é chamado de processo aleatório estritamente estacionário ou, mais simplesmente, processo aleatório estacionário de s.

Observe que a estacionariedade estrita por si só não requer nenhuma forma específica de CDF. Por exemplo, não diz que todas as variáveis ​​são gaussianas.

O adjetivo sugere estritamente que é possível definir uma forma mais flexível de estacionariedade. Se o $N^{\text{th}}$ -order CDF conjunta de $X(t_1), X(t_2), \ldots, X(t_N)$ é o mesmo que o $N^{\text{th}}$ -order CDF conjunta de $X(t_1+\tau), X(t_2+\tau), \ldots, X(t_N +\tau)$ para todas as opções de$t_1,t_2, \ldots, t_N$ e$\tau$ , em seguida, o processo aleatório é dito ser estacionário a ordem$N$ e é referido como um$N^{\text{th}}$ processo aleatório -order estacionário. Note-se que uma $N^{\text{th}}$ processo aleatório -order estacionário também é estacionário a ordem$n$ para cada positivo$n < N$ . (Isto acontece porque o$n^{\text{th}}$ -order CDF comum é o limite da$N^{\text{th}}$ -order CDF como$N-n$ of the arguments approach $\infty$: a generalization of $F_X(x) = \lim_{y\to\infty}F_{X,Y}(x,y)$). A strictly stationary random process then is a random process that is stationary to all orders $N$.

If a random process is stationary to (at least) order $1$, then all the $X(t)$'s have the same distribution and so, assuming the mean exists, $E[X(t)] = \mu$ is the same for all $t$. Similarly, $E[(X(t))^2]$ is the same for all $t$, and is referred to as the power of the process. All physical processes have finite power and so it is common to assume that $E[(X(t))^2] < \infty$ in which case, and especially in the older engineering literature, the process is called a second-order process. The choice of name is unfortunate because it invites confusion with second-order stationarity (cf. this answer of mine on stats.SE), and so here we will call a process for which $E[(X(t))^2]$ is finite for all $t$ (whether or not $E[(X(t))^2]$ is a constant) as a finite-power process and avoid this confusion. But note again that

a first-order stationary process need not be a finite-power process.

Consider a random process that is stationary to order $2$. Now, since the joint distribution of $X(t_1)$ and $X(t_1 + \tau)$ is the same as the joint distribution function of $X(t_2)$ and $X(t_2 + \tau)$, $E[X(t_1)X(t_1 + \tau)] = E[X(t_2)X(t_2 + \tau)]$ and the value depends only on $\tau$. These expectations are finite for a finite-power process and their value is called the autocorrelation function of the process: $R_X(\tau) = E[X(t)X(t+\tau)]$ is a function of $\tau$, the time separation of the random variables $X(t)$ and $X(t+\tau)$, and does not depend on $t$ at all. Note also that

$$E[X(t)X(t+\tau)] = E[X(t+\tau)X(t)] = E[X(t+\tau)X(t + \tau - \tau)] = R_X(-\tau),$$ and so the autocorrelation function is an even function of its argument.

A finite-power second-order stationary random process has the properties that

1. Its mean $E[X(t)]$ is a constant
2. Its autocorrelation function $R_X(\tau) = E[X(t)X(t+\tau)]$ is a function of $\tau$, the time separation of the random variables $X(t)$ and $X(t+\tau)$, and does not depend on $t$ at all.

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The assumption of stationarity simplifies the description of a random process to some extent but, for engineers and statisticians interested in building models from experimental data, estimating all those CDFs is a nontrivial task, particularly when there is only a segment of one sample path (or realization) $x(t)$ on which measurements can be made. Two measurements that are relatively easy to make (because the engineer already has the necessary instruments on his workbench (or programs in MATLAB/Python/Octave/C++ in his software library) are the DC value $\frac 1T\int_0^T x(t)\,\mathrm dt$ 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.