0
$\begingroup$

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:

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

  • 0
    I have now defined what I mean by uncorrelated with itself. Thanks for pointing that out.2011-11-14

2 Answers 2

2

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.

  • 0
    @Iconoclast, Could you explain: "The power in white Gaussian noise is infinite, becuase its PSD is a non-zero constant over (−∞,∞)."? Namely, could you build Gaussian White Noise with $ \mathbb{E} \left[ x \left( t \right) x \left( t + \tau \right) \right] = {\sigma}^{2} $ (Namely finite)?2018-06-16
1

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 $0$.

  • 0
    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).2018-06-16