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I am trying to understand the definition of wide sense stationary on my own and probably have some silly questions. Wikipedia says, wide sense stationary is a process with constant mean and autocorrelation function over time. Now, lets assume a process,

$$X(n) = 0.9X(n-1) + V(n) + U(n),$$ where $V$ is white noise and $U$ is an external input.

  1. Lets assume $U(n) = 0\quad \forall n$, i.e. $X_1(n) = 0.9X_1(n-1)+V(n)$.
    In this case can $X_1$ be called WSS? What does mean being constant actually implies? Can I assume $0.9X(n-1)$ is mean since it is deterministic and say mean is decreasing over time, and this is not WSS?
  2. Lets assume $U(n) = C \text{ const.}$, i.e. $X_2(n) = 0.9X_2(n-1) + V(n) + C$
    If $X_1$ was WSS, I believe this should be WSS as well. Please confirm.
  3. If $U(n)$ is an external signal changing over time, i.e. $X_3=X$
    then would still be WSS, if $X_p= 0.9X_p(n-1)+V(n)$ were originally? In this case $X$ is depended on a signal which is not part of the original process.
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  • $\begingroup$ Hi: I don't know how the external input $u$ affects things but, without it, you have an stationary AR(1) with mean zero and auto-covariance ( at the various lags ) not changing so it's wide sense stationary. $\endgroup$ – mark leeds Aug 17 '20 at 4:49
  • $\begingroup$ Please, don't use unrelated tags. dsp-core has literally nothing to do with this. $\endgroup$ – Marcus Müller Aug 17 '20 at 9:41
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Ok, let's rule out 3. first. Whether that is WSS depends on $U$ and you don't know that. For example, if $U$ has unbounded variance, then so does $X_3$, and bounded variance is a necessary condition for WSS.

For the others:

The aspect of the definition you're looking for is that you can define an autocorrelation function that only depends on the time difference:

$$\phi_{X_kX_k}(n,m)= E\left(X_k(n)X_k^*(m)\right) \overset != E\left(X_k(n)X_k^*(n-h)\right)=:\phi_{X_kX_k}(h)$$

for all times $n$ (and $m$).

Now, let's apply this definition of the autocorrelation function to your $X_1$:

\begin{align} \phi_{X_1X_1}(n,m) &= E\left(X_1(n)X_1^*(m)\right) \\ &=E\left[\left(0.9X_1(n-1)+V(n)\right)\left(0.9X_1(m-1)+V(m)\right)^*\right]\\ &=E\left[0.81 X_1(n-1)X^*_1(m-1)\right]+E\left[X_1(n-1)V^*(m)\right]+E\left[X^*_1(m-1)V(n)\right]+E\left[V(n)V^*(m)\right]\\ &\text{due to noise being }\color{blue}{\text{independent}}\text{ from signal and } \color{gray}{\text{white}}\\ &=0.81E\left[X_1(n-1)X^*_1(m-1)\right]+\color{blue}{E\left[X_1(n-1)\right]E\left[V^*(m)\right]}+\color{blue}{E\left[X^*_1(m-1)\right]E\left[V(n)\right]}+\color{grey}{0}\\ &\text{white noise must have a }\color{green}{\text{zero mean}}\\ &=0.81E\left[X_1(n-1)X^*_1(m-1)\right]+E\left[X_1(n-1)\right]\color{green}{0}+E\left[X^*_1(m-1)\right]\color{green}{0}\\ &=0.81E\left[X_1(n-1)X^*_1(m-1)\right]\tag Ä\label{fail} \end{align}

But as seen in \eqref{fail}, $\phi_{X_1X_1}$ depends on the actual points in time $n$ and $m$, and hence, $X_1$ can't be WSS.

Let's address

If $X_1$ was WSS, I believe this should be WSS as well. Please confirm.

Neither $X_1$ nor $X_2$ are WSS, same derivation as above, just that the expectations of noise aren't zero, but still time independent whereas the $X_2$-product isn't.

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  • $\begingroup$ Marcus Muller: 1) and 2) are definitely wide sense stationary because it's an AR(1). For an AR(1), the auotcorrelation of X ( easier to use X ) at any lag only depends on the lag number. I can't prove it right now ( no time nor space ) but, for an AR(1), the autcorrelation of $X$ at lag $m$ say is $\phi^m$ and only depends on the lag difference. So, $cor(X_1, X_{m+1}) = \phi^m = cor(X_2, X_{m+2})$. $\endgroup$ – mark leeds Aug 17 '20 at 14:00
  • $\begingroup$ @markleeds yep, it's the constituent equation of an AR mechanism, something must be wrong. $\endgroup$ – Marcus Müller Aug 17 '20 at 15:12
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I'll change the notation slightly and leave out $u_t$. ( Not clear on how that effects things ).

$X_t = \phi X_{t-1} + \epsilon_t = \sum_{i=0}^{\infty} \phi ^{i} \epsilon_{t-i}$

Also,

$X_{t+m} = \phi X_{t+ m-1} + \epsilon_{t+m} $

$ = \sum_{j=0}^{\infty} \phi^{j} \epsilon_{t+m-j}$

So, if one takes the covariance of these two expressions with the sums, the only places where you get non-zero terms is when the $\epsilon$ subscripts are the same. Well, when are they the same ?. This happens when $abs(i-j) = m$ which is the same as saying that the $\epsilon$ terms are $m$ periods apart. In this case, it's easy to show :) that $cov(X_{i}, X_{j})$ = $\phi^m \sigma^2$. So, the covariance only depends on $m$ which is the lag distance. Therefore, the AR(1) is covariance stationary.

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  • $\begingroup$ Thanks for explaining covariance part. But WSS implies mean to be constant. What is the mean here? @mark leeds $\endgroup$ – jrvinayak Aug 18 '20 at 2:15
  • $\begingroup$ Hi: The mean for this AR(1) is zero. Also, any initial condition, say $X_{0}$, dies out as long as $\phi < 1 $. This is the stationarity condition for an AR(1). Note though that an AR(1) with non-zero mean such as $X_t = \mu + \phi X_{t-1} + \epsilon_t$ is still WSS because the mean is then $\mu$ which is constant. $\endgroup$ – mark leeds Aug 18 '20 at 5:58

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