# wide sense stationary of dynamic process

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.
• 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. – mark leeds Aug 17 '20 at 4:49
• Please, don't use unrelated tags. dsp-core has literally nothing to do with this. – Marcus Müller Aug 17 '20 at 9:41

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.

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.

• 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, $cor(X_1, X_{m+1}) = \phi^m = cor(X_2, X_{m+2})$. – mark leeds Aug 17 '20 at 14:00
• @markleeds yep, it's the constituent equation of an AR mechanism, something must be wrong. – Marcus Müller Aug 17 '20 at 15:12

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.

• Thanks for explaining covariance part. But WSS implies mean to be constant. What is the mean here? @mark leeds – jrvinayak Aug 18 '20 at 2:15
• 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. – mark leeds Aug 18 '20 at 5:58