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Why does it mean that the process/signal is not stationary when its variance varied with time? that is,

$VAR[X(t)]= \alpha \times t$,$t$ is time,and $\alpha$ is a constant,then $X(t)$ is not the WSS process

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    $\begingroup$ ... because that's the very definition of stationarity. Things are stationary that don't move. $\endgroup$ – Marcus Müller Jun 19 '18 at 11:05
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From the Wiki page - A stationary random process is a stochastic process whose unconditional joint probability distribution does not change when shifted in time. Consequently, parameters such as mean and variance, if they are present, also do not change over time.

Additional References are: Link1 and Link2.

The implication from the definition is that the mean and variance of random process do not vary over time. Since by your definition of the variance - it varies over time, then your process is non-stationary.

A couple of notes:

  1. Just because the mean and variance don't vary over time does not imply that the process is stationary. The definition of a stationary random process has more implications.
  2. A random process where the mean and auto-covariance are invariant to time shifts is often called a "Wide Sense Stationary" process. There are no conditions on the higher order moments (skewness, kurtosis, etc). This is a less stringent version of stationarity that is often more applicable for real-world signals.
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    $\begingroup$ Another note that the fact that a process is called WSS implies its two first moment do exist, while strictly stationary process may not. $\endgroup$ – AlexTP Jun 19 '18 at 12:40
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    $\begingroup$ Your 2. is incorrect. Wide sense stationarity requires more than just the mean and variance being constant; it is also necessary to consider the entire autocorrelation function of the process. $\endgroup$ – Dilip Sarwate Jun 19 '18 at 15:06
  • $\begingroup$ @DilipSarwate Yes, you're correct. I was trying to emphasize the point that the stationarity of the mean and autocovariance doesn't imply strict stationary rather than provide a strict definition of WSS. I modified my answer to reflect this. $\endgroup$ – David Jun 19 '18 at 18:16
  • $\begingroup$ David: Your explanation was nice but I think wide sense is actually strong than just stationary because it refers to the distribution rather than just the first two moments. But, Dilipe will hopefully correct me if I'm wrong. That's happened before. LOL !!!! $\endgroup$ – mark leeds Jun 19 '18 at 21:58

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