0
$\begingroup$

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.
Improve this question
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

Improve this answer
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

Improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.