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I'm looking into the possibility of using a non-causal time series filter for some data. The goal is filtering (for the purpose of anomaly detection). However, this is not particularly relevant.

I'm wondering how simulation of such a time series would work? Say for example your time series model were:

$$y_t = y_{t-1}+y_{t+1}+\eta_t$$

where $\eta$ corresponds to some standard random noise ($N(0,1)$ or something). My initial thoughts are to just simulate the random noise $\eta_t^*$ and then solve a linear system to get the $y_t$. However, this also sounds slow.. and I can't help but feel someone has probably already thought about this problem.

So, what is the standard practice here?

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  • $\begingroup$ you cant simulate $y_{t+1}$ you can only predict it. I just simulated for an earlier post without $y_{t+1}$ see here stats.stackexchange.com/questions/107551/… $\endgroup$ – forecaster Jul 12 '14 at 3:38
  • $\begingroup$ Sorry I don't understand what you mean - I mean to create a simulation for all t in [0,...,T]. I see that the end points will be a problem here though.. so that could remain fixed? $\endgroup$ – thebigdog Jul 12 '14 at 3:59
  • $\begingroup$ Ok got it , then yes. $\endgroup$ – forecaster Jul 12 '14 at 4:01
  • $\begingroup$ So you mean yes do it as a linear system, holding the ends fixed? I guess in my example it would be a fairly simple linear system.. $\endgroup$ – thebigdog Jul 12 '14 at 7:55
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Here's an easy way to accomplish this in this particular situation. Rewrite your first equation as: $\Delta y_t = -y_{t-1}+\epsilon_t$ where $\epsilon_t=-\eta_t$. Simulate $\epsilon_t$ as random $N(0,1)$, and start with an initial $y_0$ and $y_1$. Then create $y_2$ as $y_1-y_0+\epsilon_1$. You'll find it obeys the same relationship. Simulate a bunch and then burn some if you're worried about the starting point. I'm not sure if this helps for a more general version of your problem.

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  • $\begingroup$ Ah.. good point. I guess 'ignoring end points' all no causal time series can be rewritten as a casual version in this way. At least when the relationship is linear. $\endgroup$ – thebigdog Jul 15 '14 at 9:47

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