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Can anyone help me out with this question I found on my practice paper?

The series $\{y_t\}$ is described by

$$ y_t = 5+0.3y_{t-1}+\epsilon_t+0.9\epsilon_{t-1}. $$

Identify the specific ARIMA model. What is the mean of the series $\{y_t\}$?

I think it is a ARMA (1,1) process? is that correct? And how do I calculate the mean of the ARMA model? Is it the same as AR model? If it is then

$$ E(y_t) = 5/ (1-0.30) = 7.14?$$

enter image description here

The sample ACF and PACF plot of a series are given above. Appraise the patterns and judge which ARIMA model fit the series. Justify your conclusion.

I identified as the AR(4). Is that correct? Since ACF has an exponential decay and PACF has a cut off? Differencing is not needed as the ACF decreases sufficiently to 0.

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  • $\begingroup$ Thank you for reposting! I took the liberty of tweaking the formatting a bit more. Next time, you may want to post two separate questions in a case like this, since there are two different things and can be treated independently. Also, please next time add the self-study tag. $\endgroup$
    – Stephan Kolassa
    Commented Apr 30, 2020 at 8:08
  • $\begingroup$ thank you so much! :D $\endgroup$
    – Shane Lum
    Commented Apr 30, 2020 at 8:31

2 Answers 2

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Your series is indeed ARIMA(1,0,1), or ARMA(1,1). Your calculation of the mean is also correct, since the innovations have an expectation of zero, so

$$ E(y_t) = 5+0.3\underbrace{E(y_{t-1})}_{=E(y_t)}+\underbrace{E(\epsilon_t)}_{=0}+0.9\underbrace{E(\epsilon_{t-1})}_{=0} $$

yields

$$ E(y_t) = 5+0.3E(y_t) \Longrightarrow E(y_t)=\frac{5}{1-0.3}.$$

We can also verify this by a quick simulation in R:

> nn <- 1000
> epsilon <- rnorm(nn,0,1)
> yy <- ts(rep(NA,nn))
> yy[1] <- 5
> for ( ii in 2:nn ) yy[ii] <- 5+0.3*yy[ii-1]+epsilon[ii]+0.9*epsilon[ii-1]
> mean(yy)
[1] 7.179631

For your ACF/PACF plots, I won't give you the answer straight away. But it's not an AR(4) series. In your particular situation, I would recommend that you take a look at the ACF/PACF plots of the series yy we just simulated:

acf(yy)
pacf(yy)

Notice anything? Best to also simulate an AR(4) series and take a look at its ACF/PACF plots, to see the difference.

Note that in general, it is not easy to deduce a model from the ACF/PACF plot alone. See here for more information.

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  • $\begingroup$ Stephen very nice work on this .. I had a miscue here ... $\endgroup$
    – IrishStat
    Commented Apr 30, 2020 at 19:39
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For part 2, we note that the ACF shows exponential decay that cuts off at lag 2.

On the other hand, the PACF shows a damped sinusoid that decays after lag 4. Since the ACF decays more exponentially than the PACF, we can conclude that the series is an MA(q) model, where q=2.

Hence, I believe the series fits an MA (2) model best.

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