i'm pretty new to the topic and I'm trying to understand how to determine the stability of a process. I'm giving this discrete-time stochastic system:

$$ \cases{ s_t = 2s_{t-2} + 3w_{t-2} \\ y_t = s_t + v_t \\ w_t = wn(0, q) \\ v_t = wn(0, r) \\ w_t \perp v_t } $$ where $w_t = wn(0, q)$ means that the process $w_t$ is white noise with 0 mean and variance $\sigma^2 = q$ and $w_t \perp v_t$ means that the processes are uncorrelated.

I'm asked if the process $y_t$ is stationary and I'm finding some difficulties in deriving the answer.

I (think I)'ve written $s_t$ in terms of his transfer function (interpreting $z$ as the lag operator):

$$ s_t = \frac{3z^{-2}}{1 - 2z^{-2}} w_t $$

This transfer function has poles in $\pm \sqrt{2}$, outside the unit circle, hence I'd say that this process is not stable and that $y_t$ cannot be stable, either.

Regarding the stationarity of $y_t$: do I have to solve the discrete SDE and evaluate mean and variance of $y_t$ or is a better and smarter way to answer the question?

Can I give a state space representation of $s_t$ and work it out in that domain?

Thank you for your help!

Edit: @mark-leeds thank you very much, you have been very clear. I was unsure about this interpretation of an AR model but now I'm OK with it.

One last question regarding the state space representation: is it possible to express the process $s_t$ in a state space model? Something like

$$ x(t+1) = A(t)x(t) + B(t)e(t) \\ s(t) = C(t)x(t) + D(t)e(t) $$

Thank you very much again!

  • $\begingroup$ Hi: In time series lingo, you could say that $s_t$ is a non-stationary AR(1) ( which is the same as unstable ) because the absolute value of the AR(1) coefficient is greater than 1.0. This then makes the $y_t$ process not stationary also. $\endgroup$ – mark leeds Sep 15 '19 at 14:35
  • $\begingroup$ I've been introduced to AR models as $y(t) + a_1 y(t-1) + \dots a_n y(t-n) = e(t)$ where $e(t) = wn(0, \sigma_e^2)$ . Do you find the two writings consistent? $\endgroup$ – PGN Sep 15 '19 at 14:41
  • $\begingroup$ Hi: With time series, one always has to worry about notational differences-inconsistencies. But yes, if you replace $𝑠$ with $y$, use 2 lags instead of one and multiply $\epsilon_{t}$ by a constant, you still have an AR(1). Or you could call it an AR(2) where the first AR coefficient is zero. You are correct though that a more consistent way to write it would be $y_t = 2 𝑦_{t−1}+\epsilon_{t−2}$. In fact, the latter is still not standard because 𝜖, the error term, is usually time aligned with the left hand side. But that's okay. $\endgroup$ – mark leeds Sep 16 '19 at 6:44
  • $\begingroup$ One more thing since the notation can definitely be confusing. Note that the bottom line is that a lagged variable ( relative to the response ) has a coefficient whose absolute value is greater than 1.0. That makes it not-stable. $\endgroup$ – mark leeds Sep 16 '19 at 6:45

@w00zie: Hi: I'm putting this here just because I find it easier to write in here. This is not an adequate answer from a DSP point of view (and I would like to see one also) but let me just it explain more clearly since I babbled above and it's still not that clear.

Take the more standard AR(1):

$y_{t} = a y_{t-1} + \epsilon_{t}$. ( denote this as first equation ) This can be re-written as

$y_{t}(1 - a L) = \epsilon_t$ where L is the backshift operator which is kind of the analogue to $z^{-1}$ in DSP notation. This expression can be re-written as

$y_{t} = \frac{\epsilon_{t}}{(1 - a L)}$ which can be re-written as

$y_{t} = \sum_{i=0}^{\infty} a^{i} \epsilon_{t-i}$.

But the only way that the infinite sum on the RHS can converge is if $|a| < 1$. This is why that restriction arises for stability. I hope this helps even though it's still a time-series viewpoint rather than a DSP viewpoint.

Oh, and you asked why it looks so different from the standard AR(1) notationally.

You have: $s_t = 2 s_{t-2} + 3 w_{t-2}$.

If you let $y_t = s_t$ and $\epsilon_t = w_{t-2}$, then that gives

$y_t = 2 * y_{t-2} + 3 \epsilon_{t}$. ( denote this as last equation ).

The equivalence of "first equation" and l"last equation" can be argued in two different ways:

1) you can think of last equation as an AR(2) with a zero $y_{t-1}$ coefficient. But the restriction on $a$ for stability doesn't change.

2) you can think of last equation as an AR(1) where the time step is doubled.

Note that the difference in the lags of the error terms, $\epsilon_{t}$ and $w_{t-2}$ is okay because $w_t$ is iid and uncorrelated with all other terms. You are correct that, if $w_t$ was not iid and uncorrelated with everything else, then you couldn't use 1) or 2) as arguments for equivalence.

Note that I really shouldn't have used $y_{t}$ here since you use $y_{t}$ in a different context but my $y_t$ and your $y_{t}$ are not related.

Finally, your $y_{t}$ is not stationary (i.e: not stable ) because it's just $s_t$ + noise and $s_t$ is not stationary.

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  • $\begingroup$ @w00zie: I just saw your question about state space system. What you wrote is the innovations representation of the state space system. I played around and couldn't obtain that rep but I have a feeling it's possible. The two different noise terms are confusing me because innovations rep uses one. I'm sorry that I can't be of help here. Check out Kalaith's papers for where the innovations representation comes from. It looks like it's pulled out of thin air but there's an interesting argument behind it. $\endgroup$ – mark leeds Oct 16 '19 at 11:33

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