Questions about the stability (and stationarity) of a system and state space representations

Question

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!

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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!

Answer 1

@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.

Answer 2

You are given $$ \begin{align} s_t & = 2s_{t-2} + 3w_{t-2} \\ y_t & = s_t + v_t \\ w_t & = wn(0, q) \\ v_t & = wn(0, r) \\ w_t & \perp v_t \end{align}. \tag 1 $$

You can turn this into a state-space model almost by inspection. Let $\mathbf x$ be your state vector. Note that $s_t$ is affected by inputs and states no more than two steps in the past, and the line defining it is the only line in (1) that depends on prior values of anything. This means that you can let $\mathbf x_t = \begin{bmatrix}s_t & s_{t-1}\end{bmatrix}^T$. By the size of $\mathbf x$, system is second-order. Then (1) can be rewritten as $$ \begin{align} \mathbf x_{t+1} &= \begin{bmatrix}0 & 2 \\ 1 & 0\end{bmatrix} \mathbf x_t + \begin{bmatrix}0 \\ \frac{3}{2}\end{bmatrix} w_t \\ y_t &= \begin{bmatrix}1 & 0\end{bmatrix}\mathbf x_t + v_t \end{align}. \tag 2 $$

You can show that the process isn't stationary by counter-example. Assume that $\mathbf x \in \mathcal N\left(\mu_x, \mathbf P \right)$. Let $$\mathbf A = \begin{bmatrix}0 & 2 \\ 1 & 0\end{bmatrix}$$ and $$\mathbf B = \begin{bmatrix}0 \\ \frac{3}{2}\end{bmatrix}$$ For the process to be stationary, the variance of $\mathbf x$ must remain constant over time; thus there must be a $\mathbf P $ where $$\mathbf P = \mathbf A \mathbf P \mathbf A^T + \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}q \tag 1$$

I cannot give you a rigorous proof of this, but in general if $\mathbf A$ has eigenvalues on the stability boundary then (2) cannot be solved, and $P_k$ increases linearly; if $\mathbf A$ has eigenvalues outside of the stability boundary then (2) cannot be solved and $P_k$ increases exponentially.