I have a pretty basic question: If a Gaussian random process is uncorrelated with itself, does it imply that it is stationary? I think not, but I wanted to confirm my assertion.

Edit: Uncorrelated with itself means that any two samples of the random process (taken at different times) have zero covariance. In other words:

$$\text{Cov}_X(t_1, t_2)=\text{Cov}(X(t_1), X(t_2))=0~\text{for all}~t_1 \neq t_2$$

  • $\begingroup$ What precisely do you mean by uncorrelated with itself? $\endgroup$ – cardinal Nov 14 '11 at 10:08
  • $\begingroup$ I have now defined what I mean by uncorrelated with itself. Thanks for pointing that out. $\endgroup$ – Iconoclast Nov 14 '11 at 13:20

No, it does not. Uncorrelatedness simply means that the covariance matrix of the process is diagonal for any selection of base points. The variance, however, can still depend on the selection of base points. Only if it wouldn't, and the mean is constant (thanks Dilip), the process would be stationary.

Edit: I have to agree with Dilip that it would only be stationary if the mean is also the same for each location.

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    $\begingroup$ This answer worries me a bit. Suppose that $\text{var}(X(t)) = \sigma^2$ for all $t$ where $\sigma^2$ is a finite positive real number. Is it possible to have a Gaussian process for which $$\text{cov}(X(t_1), X(t_2)) = \begin{cases}0, & t_1\neq t_2,\\ \sigma^2, & t_1 = t_2,\end{cases}??$$ or is white Gaussian noise (for which we cannot say that $\text{var}(X(t))$ is finite) the only possibility? $\endgroup$ – Dilip Sarwate Nov 14 '11 at 12:58
  • $\begingroup$ @Dilip Yes, this is Gaussian white noise, and the variance is finite for every $t$. What makes you think otherwise? $\endgroup$ – user940 Nov 14 '11 at 13:27
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    $\begingroup$ @ByronSchmuland I am used to thinking of a (continuous-time) white Gaussian noise process as a zero-mean process having autocorrelation function $\frac{N_0}{2}\delta(t)$ where $\delta(t)$ is the Dirac delta or impulse and power spectral density $\frac{N_0}{2}, -\infty<f<\infty$ (as per Iconoclast's comment). For this process, the variance $E[(X(t))^2]$ cannot be said to be finite. For a discrete-time process, there is no issue. $\endgroup$ – Dilip Sarwate Nov 14 '11 at 13:59
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    $\begingroup$ @Dilip: I agree. There are even more pernicious (lack of) measurability issues at play here. See, e.g., the first few pages of Kallianpur's Stochastic Filtering Theory text. $\endgroup$ – cardinal Nov 14 '11 at 14:13
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    $\begingroup$ @DidierPiau Yes, this is what I call a white noise process. What I hadn't realized is that some, in particular Kallianpur, explicitly reject this as an appropriate model of white noise when $T$ is uncountable. $\endgroup$ – user940 Nov 15 '11 at 13:22

I wish to disagree with the accepted answer to a slight extent.

  • A discrete-time Gaussian process $\{X[n] \colon n \in \mathbb{Z}\}$ with the property that $$\text{cov}(X[m],X[n]) = \begin{cases}0, & n \neq m,\\\sigma^2, & n = m \end{cases} $$ has the property that the covariance matrix of $X[n_1], X[n_2], \ldots X[n_m]$ (where $n_1, n_2, \ldots n_m$ are distinct integers) is $\sigma^2I_m$. But the process is not necessarily stationary unless we also require that $E[X[n]]$ is the same for all $n$.

  • Suppose that there is a zero-mean continuous-time Gaussian process $\{X(t) \colon t \in \mathcal{R}\}$ with the property that $$\text{cov}(X(t_1),X(t_2)) = \begin{cases}0, & t_1 \neq t_2,\\\sigma^2, & t_1 = t_2.\end{cases}$$
    Then the distribution of $X(t_1), X(t_2), \ldots X(t_m)$ (where $t_1, t_2, \ldots t_m$ are distinct real numbers) is the same as the distribution of $X(t_1+\tau), X(t_2+\tau), \ldots X(t_m+\tau)$ as is needed for stationarity. However, the autocorrelation function of this process is $$R_X(\tau) = E[X(t)X(t+\tau)] =\begin{cases} \sigma^2, & \tau=0,\\0, & \tau \neq 0,\end{cases}$$ which is discontinuous at $\tau=0$ and has Fourier transform $0$, that is, the power spectral density is $0$.

  • $\begingroup$ Could you derive why for White Noise we have the Delta Term for the Auto Correlation Function? From your answer one motivation is to have PSD which is not zero (As you stated above). $\endgroup$ – Royi Jun 16 '18 at 19:46

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