# How to know the specific ARIMA model through its yt equation?

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

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

• 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. Apr 30, 2020 at 8:08
• thank you so much! :D Apr 30, 2020 at 8:31

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 <- 5
> for ( ii in 2:nn ) yy[ii] <- 5+0.3*yy[ii-1]+epsilon[ii]+0.9*epsilon[ii-1]
> mean(yy)
 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.

• Stephen very nice work on this .. I had a miscue here ... Apr 30, 2020 at 19:39

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.